TITLE: Good overview of singularity theory
QUESTION [6 upvotes]: Can anyone recommend a good overview of singularity theory? In particular, quotient singularities...

REPLY [3 votes]: There is the classic Singular Points of Complex Hypersurfaces by J. W. Milnor. Try also Topics in real and complex singularities by A. Dimca.<|endoftext|>
TITLE: What is soliton
QUESTION [9 upvotes]: I am new to this word.. This is not research level problem and it is soft question in nature.  Just for curiosity, i am asking.. 
In literature, i am finding following words:(Wikipedia+ others).
Soliton is a self-reinforcing solitary wave
Solition is a phenomenon.
Solition is a property
Solitonic solution
As wikipedia also says, single definition is difficult to find. Can somebody explain this term according to you... It will be better if you give the idea of Soliton more mathematically rather only intuitively.
More precisely my question is WHAT IS SOLITON.

REPLY [3 votes]: Leaving the Ricci solitons aside, there is a number of different (and not equivalent) definitions. It is mostly agreed that solitons in nonlinear systems are solitary waves which balance dispersion and nonlinearity and maintain their shape, even after elastic interactions like collisions, see e.g. the Scholarpedia article and the discussion at Physics.SX.<|endoftext|>
TITLE: Topological spaces made by identifying opposite faces of a cube?
QUESTION [20 upvotes]: My bashful, nameless, colleague asked me:
When you identify opposite faces of a square, then depending on where you twist or not, you get a torus, Klein bottle, or projective plane.

What spaces can you get when identifying opposite faces of a cube?

He was hoping for a reference.

REPLY [16 votes]: B. Everitt. 3-manifolds from platonic solids.  Topology and its applications, 2004. 

Covers everything you're asking for and more.<|endoftext|>
TITLE: Closed formula for colored Jones polynomial of the trefoil? (reference request)
QUESTION [6 upvotes]: (EDIT: Powers of $q$ in the formula corrected.)
I've been doing some computations with skein modules, and I found the following formula for the N-th colored Jones polynomial of the trefoil:
$\frac{1}{q^2-q^6}\sum_{i=1}^{N+1} (-1)^i q^{6(N^2-i^2)}(q^{10i-4}-q^{2i})$
I'm interested in comparing this to formulas that appear in the literature. I'm pretty sure I've seen similar formulas before, but I haven't been able to find them again.

Where can I find explicit formulas for colored Jones polynomials of the trefoil (or even better, for some family of "small" knots that includes the trefoil)?

(To be precise I'll explain my normalization. Identify the Kauffman bracket skein module of the tubular neighborhood of the (0-framed) trefoil with $\mathbb C[x]$, where $x^n$ is identified with $n$ parallel copies of the trefoil. When I say "N-th colored Jones polynomial," I mean the Chebyshev polynomial $S_N(x)$ viewed as an element of the skein module of $S^3$. In particular, the first Jones polynomial of the unknot is $-q^2-q^{-2}$.) (Again, to be precise, $S_0=1, S_1=x$, and $S_{n+1}(x) = xS_n(x) - S_n(x)$.)

REPLY [3 votes]: This appears to be done by K. Habiro in "On the colored jones polynomials of some simple links" (the trefoil is the last section of the paper).<|endoftext|>
TITLE: Can you name these orthogonal polynomials?
QUESTION [14 upvotes]: I have a collection of orthogonal polynomials in infinitely commuting variables $x_1, x_2, x_3, \ldots$.  I think they must be well known (perhaps Schur or Hermite polynomials or some variant thereof), but I haven't succeeded in finding them in the literature in a form that's recognizable to me.  If anyone can point me to an appropriate reference I would be grateful.  I suspect the answer to this must be very familiar to many people, but I'm not one of those people.
The polynomials are indexed by Young diagrams (partitions) of all sizes (i.e. [], [1], [2], [1,1], [3], [2,1], [1,1,1], [4], ...).  
The measure respect to which they are orthogonal is
$$
 \prod_{k=1}^\infty \frac{1}{\sqrt{2\pi k}}e^{-\frac{x_k^2}{2k}}dx_k
$$
In other words, a product of gaussian measures, with the width proportional to $\sqrt{k}$.  (From some points of view it is more natural to replace $x_k$ with $x_k-1$ when $k$ is even; i.e. shift the gaussian to be centered at 1 instead of 0 when $k$ is even.)
The multiplication rule for the polynomials is more complicated than Littlewood-Richardson.  Multiplying polynomials corresponding to Young diagrams of sizes $a$ and $b$ results in Young diagrams of sizes ranging from $|a-b|$ to $a+b$.  (The highest order part of the multiplication rule is Littlewood-Richardson.)  For example $[1] * [2,1] = [2] + [1,1] + [3,1] + [2,2] + [2,1,1]$.

ADDED:
Empirically, it seems to be true that if you sum the polynomials for all Young diagrams of size $n$, weighted by the dimension of the Young diagram, you get the $n$-th Hermite polynomial in the variable $x_1$.  (Hat-tip to Suvrit for suggesting that I look at Hermite polynomials.)

REPLY [6 votes]: The answer appears to be as follows.  ("Appears" because I haven't yet written out a detailed proof.)
Let $H^{[k]}_n(x)$ denote the variant of Hermite polynomials which are orthogonal with respect to the measure
$$ \frac{1}{\sqrt{2\pi k}}e^{-x^2/2k}dx . $$
Since the measure in the question is a product of the above measures (over all positive integers $k$), we have a family of orthogonal multivariable polynomials
$$ H^{[1]}_{n_1}(x_1) H^{[2]}_{n_2}(x_2) \cdots H^{[j]}_{n_j}(x_j)  ,$$
indexed by tuples $(n_1,\ldots,n_l)$.  The orthogonal polynomials of the questions are linear combinations of these.  More specifically, let $N = \sum_i i\cdot n_i$.  Think of $(n_1,\ldots,n_l)$ as encoding a conjugacy class in the symmetric group $S_N$, where $n_i$ is the number of $i$-cycles in a permutation.  We can use the character table of $S_N$ to change basis from conjugacy-class-bump-functions to characters-of-representations.  Applying an analogous change of basis to the above products of Hermite polynomials (separately for each $N$) yields the polynomials described in the question.  There are some normalization factors I have not mentioned, related to the fact that some of the bases mentioned above are orthogonal but not orthonormal.
Thanks again to Suvrit and John Wiltshire-Gordon for pointing me in the right direction.  I have not yet looked into Richard Borcherds' suggestion that these might also be specializations of Macdonald polynomials.<|endoftext|>
TITLE: formally smooth definition in SGA 1
QUESTION [19 upvotes]: SGA 1 introduces formally smooth in a very non-canonical way.  The way I usually saw it introduced was through the universal lifting property, i.e., for all $A$-algebra $C$ and all $J\subset C$ nilpotent, every homomorphism $B\to C/J$ lifts to a homomorphism $B\to C$.
Grothendieck defers this definition to section 2, however, and instead spends extensive time treating the definition of formally smooth given by:
Let $u: A\to B$ be a local homomorphism of local rings, and suppose the residue field of $B$ is finite over the residue field of $A$.  Then $u$ is formally smooth (or, Grothendieck states, $B$ is formally smooth over $A$) if there exists a locally finite $\hat{A}$-algebra $A'$ which is free over $\hat{A}$ such that the (and I hope I translated the French correctly here) localizations of the semi-local ring $\hat{B}\otimes_{\hat{A}}A'$ are $A'$-isomorphic to the formal series over $A'$.
I guess this definition is deferred to EGA for more intuition in the footnote, but I was wondering why this helps with our understanding of formally smooth, and how this relates to previous concepts Grothendieck introduced which would help with our understanding (e.g., does it generalize, in some sense, quasi-finite?)

REPLY [16 votes]: I guess you are referring to exposé III in SGA 1. The key point in infinitesimal properties in algebraic geometry is that infinitesimal isomorphism does not imply local isomorphism. In other words, an etale map can not be trivialized in small opens because Zariski opens are not small enough. Therefore something smooth that should be "locally an affine space" becomes locally an affine space followed by an etale morphism. This is the "working definition" that it is used in exposé 2, that I recommend you to get some intuition on the meaning of the definition.
That said, Grothendieck wanted a definition that did not involve a finiteness condition. Notice that in EGA IV smooth means "formally smooth + finite presentation". So, what is the nonfinite analog on of an affine space? A ring of formal power series over the completion at a certain point. Now the formal counterpart of "locally an affine space followed by an etale morphism" is at the completion of the point "a formal power series after replacing the completed local ring of the point by a flat cover". (Flat cover = finite locally free morphism). I guess you can figure out the analogy now.
The great idea of Grothendieck is that this condition can be expressed in a more compact way as a lifting property of infinitesimal morphisms. With this notion one is not plagued by noetherian hypothesis and it's unnecessary to appeal to the completion along the maximal ideal. In a sense, it is a very conceptual and useful definition and that is the approach taken in EGA IV.
The price one has to pay for this slick way of defining infinitesimal properties is that now the intuition behind smoothness is lost and beginners have to trace their way back to the simple cases to fully grasp the meaning of the definition.<|endoftext|>
TITLE: getting rid of existential quantifiers
QUESTION [14 upvotes]: It seems to me that for most of the twentieth century, axiomatic foundations for mathematical theories were constructed with the (mostly allied) goals of minimizing the number of primitive notions and minimizing the number of axioms.  But one could equally well be guided by the goal of minimizing the logical depth of the axioms, i.e., minimizing the use of quantifiers.
Consider group theory, for instance.  I have seen formalizations of group theory that say "There exists an element $h \in G$ such that for all $g \in G$, $gh=hg=g$" and then go on to prove that $h$ is unique and to name it $e$.  But I have also seen formalizations that say "A group is a set $G$ equipped with an element $e$ such that..." which eliminates the need for a quantifier in an axiom.  Likewise, if one includes the operation $g \mapsto g^{-1}$ as a primitive, one can avoid the axiom "For all $g$ there exists $h$ such that $gh=hg=e$" (or worse, "For all $g$ there exists $h$ such that for all $h'$, $ ghh'=hgh'=h'gh=h'hg"$, which I haven't seen, but which I can imagine a certain sorts of purist preferring!).
I know that in constructive mathematics, the status of existential quantifiers is suspect to begin with, so I imagine that there's already quite a bit of writing in foundations and philosophy of mathematics (and maybe computable mathematics as well) that addresses this issue.  Some pointers to relevant literature would be appreciated.

REPLY [2 votes]: Hilbert's epsilon-calculus  is  another method to completely get rid of all existential quantifiers as well as all universal quantifiers.  It is almost, but not quite, entirely unlike Skolemization.  
For any formula $A(x)$, the term $\epsilon x(A(x))$ is read as "any $x$ satisfying property $A$, if there is any -- and anything else otherwise".  Thus the formula $A( \,\epsilon x(A(x))$ says that $A$ holds for such an $x$, i.e., $\exists x\, A(x)$ holds.  The formula $\lnot A( \ \epsilon x(A(x) \ )$  then says that $A$ does not even hold for the $x$ which, if at all possible, satisfies $A$, or in other words:  $\lnot \exists x(A(x))$.   
The universal formula $\forall x (B(x))$ can therefore be rewritten as 
$B(\ \epsilon x (\lnot B(x))\ )$. ("Pick some $x$, any $x$ -- if there is one at all -- which fails property $B$.  Then $B$ will hold even of this $x$. So you failed to pick a counterexample. So $B$ holds for all $x$.") 
Theoretically this is wonderful -- not need for quantifiers, and you can use $\epsilon$ like a real term, you can plug terms into terms, etc.   In practice it did not catch on at all -- probably because formulas get very complicated very quickly, and also because the "meaning" of $\forall x (B(x))$ is so much easier to understand than the meaning of $B(\ \epsilon x (\lnot B(x))\ )$.<|endoftext|>
TITLE: Any progress on the Firoozbakht Conjecture?
QUESTION [10 upvotes]: Let $p_n$ be the n-th prime. The Firoozbakht Conjecture is a lesser known conjecture in the theory of primes but it has important consequences. It states that 
$$
p_n^{\frac{1}{n}} > p_{n+1}^{\frac{1}{n+1}}
$$
This truth of this immediately imply the Cramer's conjecture. In fact Firoozbakht conjecture is slightly better than the Cramer's conjecture in the sense that it would imply that 
$$p_{n+1} - p_n < \ln^2p_n - \ln p_n.$$
Notice that while Firoozbakht Conjecture will automatically imply the Cramer conjecture, it will also disprove the Cramer-Granville Conjecture. 
What has been the progress in this conjecture? Using computer calculation the conjecture has been verified, for all n upto 1.69x$10^{16}.$

REPLY [14 votes]: Significantly rewritten, yet the main message stays the same.
It is quite likely that this conjecture is false yet no counter example was found so far. 
The reason why this  seems likely is that there seems to be no supporting evidence for this conjecture beyond numerics. 
And, there are investigations base on quite natural random models of the primes that contradict it. What is commonly known as Cramér's conjecture, that is that maximal gaps between consecutive primes are of sizes at most $(\log p_n)^2$ (up to lower order terms) does not contradict this conjecture, and one might even think it supports it.
However, on the one hand it is not quite clear Cramér even conjectured this in precisely this form; he conjectured gaps are $O((\log p_n)^2)$ and somehow implied that about  $(\log p_n)^2$ might be true. On the other hand, and more importantly, Granville note that a finer investigations of Cramér's reasoning rather suggests maximal gaps of size $2 e^{-\gamma} (\log p_n)^2$ (up to lower order terms). And if this were true it would contradict the conjecture mentioned in OP (this is what is referred to in OP as Cramér-Granville conjecture). 
It should however be noted that Granville did not conjecture that the gaps are of this size, yet pointed out what taking an additional aspect into account would mean for Cramér's reasoning. For Granville on this matter on MO see Consequences of Legendre's conjecture
For details on Granville's arguments see for example 
http://www.dartmouth.edu/~chance/chance_news/for_chance_news/Riemann/cramer.pdf<|endoftext|>
TITLE: Careless packing
QUESTION [27 upvotes]: The sequence $\frac{1}{2}, \frac{1}{2\cdot 2}, \frac{1}{3\cdot 4}, \frac{1}{4\cdot 6}, \frac{1}{5\cdot 8}, \frac{1}{6\cdot 10},\ldots$ has a curious property, as follows:
a) the series with these terms sums to 1; 
b) no process of sequentially packing open intervals with these lengths into the unit interval $[0,1]$ can ever come to an impasse.
Many other sequences also enjoy this property.
Question: has this type of phenomena ever appeared in the literature?
In particular I wonder about possible decompositions (up to a set of measure zero)
of, say, the unit square (or unit sphere) into open sets which enjoy the corresponding property.

REPLY [14 votes]: This is a rigorous justification of Johan Wästlund's intuition. Namely, I will show that if we tile a round ball $B$ of area $\pi\zeta(\alpha)$ by round balls of area $\pi/n^\alpha$ for some $1<\alpha<1.1716$, then we never get stuck provided we have placed enough balls already. 
For later use note that the radius of $n$'th ball is $n^{-\alpha/2}$.
Suppose we have placed the first $N-1$ balls. Let $U$ be the union of them, and let $U'$ be the complement of $B$. We can place $N$'th ball iff the $N^{-\alpha/2}$-neighborhood of $U\cup U'$ does not contain all of $B$. We can bound the area of the neighborhood of $U$ by $$\sum_{n < N} \pi(n^{-\alpha/2}+N^{-\alpha/2})^2=\sum_{n < N} \pi(n^{-\alpha}+2n^{-\alpha/2}N^{-\alpha/2}+N^{-\alpha})=\pi(\Sigma_1+\Sigma_2+\Sigma_3).$$ We have $\Sigma_1\approx \zeta(\alpha)-\frac{N^{1-\alpha}}{\alpha-1}$, $\Sigma_2\approx 2N^{-\alpha/2} \frac{N^{1-\alpha/2}}{1-\alpha/2}=\frac{2}{1-\alpha/2}N^{1-\alpha}$ and $\Sigma_3\approx N^{1-\alpha}$. The area of the neighborhood of $U'$ is less than $2\pi\zeta(\alpha)^{1/2}N^{-\alpha/2}=o(N^{1-\alpha})$. The result follows since $$\frac{1}{\alpha-1}-\frac{2}{1-\alpha/2}-1$$ is positive for $\alpha<4-2\sqrt{2}=1.17157\ldots$.
Edit: Actually, the argument works for any centrally symmetric convex shapes. The only thing I used about balls is that the Minkowski sum of a ball and a ball is a ball of the correct size.
Edit 2: It is clear that if one wants a stronger conclusion that one never gets stuck, then one needs to make explicit errors in the asymptotic estimates above. Then one can either decrease $\alpha$ to subsume those errors, or to consider the balls of area $\pi m^{-\alpha},\pi(m+1)^{-\alpha},\dotsc$ in a ball of total area $\pi\sum_{n\geq m} n^{-\alpha}$ to reduce the errors. This mirrors the suggestion of John Shier in the write-up linked  above.<|endoftext|>
TITLE: Numerically equivalent effective divisors and semiampleness 
QUESTION [9 upvotes]: Recall that a divisor $M$ on a variety $X$ is said to be semiample if $kM$ is base point free for a certain $k > 0$.
Being semiample is not a numerical property (take for example torsion and a non-torsion divisor of degree 0 on a curve, or for more sophisticated examples just look at Lazarsfeld - Positivity in Algebraic Geometry II - Ex. 10.3.3), therefore I was wondering:
It is possible to find a smooth projective variety $X$ and two effective divisors $E,D$ on $X$ such that $E \equiv D$, but $E$ is semiample while $H^0(X,kD)=\mathbb{C}$ for every $k$?

REPLY [5 votes]: Does Example 2.4 from http://www.math.utah.edu/~hacon/commA.dvi answer your question?<|endoftext|>
TITLE: Given several beta distributions, what is the probability that one is the highest?
QUESTION [6 upvotes]: Given several random variables distributed according to different beta distributions, how can I calculate the probability that any one of those random variables is actually the highest?
The application for this is that I have several ads which people can view and possibly click.  Over time I collect more and more data about each ad (views and clicks, which I can use to determine a beta distribution), and I'm interested in picking the ad that people most like to click on.
I need to determine when it is safe to select the ad with the highest click-through rate, so I need to know when a particular ad exceeds a threshold probability of being the best.
ps. I'm a software engineer not a mathematician, I would appreciate it if answers could bear this in mind by avoiding any complex notation ;)

REPLY [4 votes]: For the case of two beta random variables, and the beta parameters alpha and beta both being integers, which I think they should be given that the use case is views versus clicks, there is a closed form analytical solution for the probability that one distribution is greater than the other, from Evan Miller's web article :http://www.evanmiller.org/bayesian-ab-testing.html#cite1.
Specifically eqn 6 of the article gives a closed form solution and there is also some pseudocode included.<|endoftext|>
TITLE: Intersection on Singular Varieties
QUESTION [7 upvotes]: Let $X$ be a variety over some algebraically closed field $k$. In order to define the intersection product of the Chow ring one usually demands $X$ to be smooth. This is for example well explained by Fulton in his book Intersection Theory.
I was wondering whether it is possible to weaken this assumption. I would like it to work on normal varieties but I guess that is too much to hope for. Generalizations that only work on small dimensional varieties might also be helpful for me.
In my specific situation I also have a birational map $f:X \to Y$ where $Y$ is a smooth variety. However, I'm unsure whether this helps at all for defining the product.

REPLY [14 votes]: One thing that works under pretty general conditions is intersecting curves with $\mathbb Q$-Cartier divisors. This covers for instance the example mentioned by David.
Just in case:
Definition Let $X$ be a normal variety and $D$ a Weil divisor on $X$. Then $D$ is called $\mathbb Q$-Cartier if there exists an $m\in\mathbb N$ such that $mD$ is Cartier. A normal variety $X$ is called $\mathbb Q$-factorial if all Weil divisors are $\mathbb Q$-Cartier. 
Now if $X$ is a normal variety, $D$ is a $\mathbb Q$-Cartier divisor on $X$ and $C\subset X$ is a proper curve, then one can define
$$
C\cdot D := \frac {C\cdot mD}m
$$
for any $m$ such that $mD$ is a Cartier divisor and of course $C\cdot mD$ is defined the usual way: If $L$ is a Cartier divisor and $\gamma:\widetilde C\to C$ is the normalization of $C$, then 
$$
C\cdot L:= \deg_{\widetilde C} \ \gamma^*\left(\mathscr O_X(L)|_C\right).
$$
In David's example a line through the singular point is not Cartier, but twice the line is, because it is linearly equivalent to the hyperplane section. The intersection of a hyperplane section and a line is clearly $1$, so the intersection of two lines through the singular point has to be $\dfrac 12$.
It turns out that $\mathbb Q$-factorial varieties exist in large numbers. For instance the minimal model program (this is currently in char $0$) produces $\mathbb Q$-factorial varieties and there is even a process of $\mathbb Q$-factorialization, which is a partial resolution and on not too bad singularities it is actually a small resolution, so the divisors are not effected. 
In the particular case of a normal surface this means that if it is $\mathbb Q$-factorial, then  you can intersect any two $1$-cycles, but of course the intersection numbers may be rational numbers and not integers even between integral curves. Although the intersection between a Cartier divisor and an integral curve will still always be an integer.
For a concrete example one can mention that rational surface singularities are always $\mathbb Q$-factorial. The example in David's answer falls in this category. (This is not true in higher dimensions).
As far as trying to use the resolution, the problem is that pulling back cycles does not respect linear equivalence. The only reasonable way to do it is that Cartier divisors can be pulled back as line bundles and then you can extend this numerically to $\mathbb Q$-Cartier divisors. However, then you're back to the same restriction as above.
Alternatively, one can try numerical pull-back as in Sakai's paper mentioned by Francesco. I am not sure how well that works in higher dimensions. I think Batyrev had some work on that and more recently Araujo. I don't remember off the top of my head.
For problems with trying to intersect with non-$\mathbb Q$-Cartier divisors see Tom G's answer.

REPLY [5 votes]: Normal is not enough to define an intersection product in higher dimensions.  In Fulton, I think you can find the example of a cone over a quadric surface on which it is impossible to define a well behaved intersection theory.  (Intersecting a plane over a line of one ruling with a line of the opposite ruling should give one point, but intersecting with a line of the same ruling should give zero.  Unfortunately, a line in one ruling is rationally equivalent to a line in the other ruling, since you can deform either to be a line through the vertex.)
On the other hand, if you have a morphism to a smooth variety Y, you can at least intersect any cycle on X with the pullback of a cycle on Y.  Perhaps this will be enough for you?<|endoftext|>
TITLE: Conjugation in GL(n) (p-adic setting) 
QUESTION [8 upvotes]: In $GL(n, \mathbb{Q}_p)$, what are the orbits under conjugation of $GL(n, \mathbb{Z}_p)$?

REPLY [15 votes]: The problem can be reduced to that of classifying $GL(n,\mathbf{Z}_p)$ conjugacy classes in $M(n,\mathbf{Z}_p)$. The situation for general $n$ is complicated, but for $n=2$ the problem is settled by the following.
Let $F\in M(2,\mathbf{Z}_p)$ be any matrix, let $f(x)$ be its characteristic polynomial, and
let
$n(F)=\sup_i(i\in\mathbf{Z}_{\ge 0},$ $F$ mod $p^i$ is multiplication by a scalar $)$.
Then the $GL(2,\mathbf{Z}_p)$ conjugacy class of $F$ is uniquely determined by $f(x)$ and $n(F)$.
If $n(F)$ is infinite, then $F$ is a scalar matrix and therefore central.
If $n(F)$ is finite then there exists a unique integer $\lambda\in\mathbf{Z}$, with $0\le\lambda\le p^{n(F)}-1$ such that $F$ is conjugate to
$\begin{pmatrix}
\lambda&0\\
0&\lambda
\end{pmatrix}+p^{n(F)}\begin{pmatrix}
0&-a_0\\
1&-a_1
\end{pmatrix}$.
Here $a_0$ and $a_1$ are the constant and linear term of the polynomial
$f_0(x):=p^{-2n(F)}f(p^{n(F)}x+\lambda)$, which has coefficients in $\mathbf{Z}_p$.
We have that $p^{n(F)}$ is the index of the ring $\mathbf{Z}_p[F]$ inside the ring $R_F:=\mathbf{Q}_p[F]\cap M(2, \mathbf{Z}_p)$.
All rings are viewed as subrings of $M(2,\mathbf{Q}_p)$. We will sometime think of $F$ and of elements of $R_F$ as endomorphisms of the standard lattice $\mathbf{Z}_p^2$ inside $\mathbf{Q}_p^2$.
Proof: if $n(F)$ is infinite then there is not much to prove, therefore we assume $n(F)$ finite, and $F$ not central. The ring $R_F$ as defined above contains $\mathbf{Z}_p[F]$ with finite index, since they both are finite free $\mathbf{Z}_p$-modules of rank two. This is clear for $\mathbf{Z}_p[F]$, since $f(x)$ is the minimal polynomial of $F$, since $F$ is not central. For $R_F$ it follows from the fact that $\mathbf{Q}_p[F]\cap M(2, \mathbf{Z}_p)$ is open, compact, and non-empty in $\mathbf{Q}_p[F]$, which has rank two over $\mathbf{Q}_p$, since $F$ is not central.
The ring $R_F$ has a $\mathbf{Z}_p$-basis of the form $(1, F')$ where $F'=(a+bF)/p^{h}$, for some $a,b\in\mathbf{Z}_p$ not both divisible by $p$, and where $p^h$, with $h\ge 0$, is the index of $\mathbf{Z}_p[F]$ in $R_F$. The $p$-adic integer $b$ is a unit, for otherwise $p^{h-1}F'-b'F=a/p$, with $b'=b/p\in\mathbf{Z}_p$, would belong to $R_F$, which is not possible since $a/p$ is not a $p$-adic integer. This shows that the natural action of $F$ on $\mathbf{Z}_p^2/(p^h)$ is multiplication by $-ab^{-1}$ mod $p^h$. Therefore $h\leq n(F)$.
On the other hand, if $\lambda$ is any integer such that $F-\lambda$ is zero mod $p^{n(F)}$, then $F'':=(F-\lambda)/p^{n(F)}$ is an element of $R_F$, since $F-\lambda$ commutes with $F$ and it is divisible by $p^{n(F)}$ in $M(2,\mathbf{Z}_p)$. Therefore $n(F)\leq h$. Thus $n(F)=p^h$ and $(1, F'')$ is a $\mathbf{Z}_p$-basis of $R_F$ (since it spans a lattice of the correct index).
Now, by the maximality of $n(F)$ we see that $F''$ does not act via scalar multiplication on $\mathbf{Z}_p^2/p$. This implies that there is a $\mathbf{Z}_p$-basis $(e_1, e_2)$ of $\mathbf{Z}_p^2$ such that
$F''(\mathbf{Z}_p\cdot e_1)\not\equiv \mathbf{Z}_p e_1$ mod $p$. It follows that $(e_1, F''(e_1))$ is also a $\mathbf{Z}_p$-basis of $\mathbf{Z}_p^2$.
With respect to this basis the action of $F''$ is given by a matrix of the form
$\begin{pmatrix}
0&-a_0\\
1&-a_1
\end{pmatrix}$, where $a_0$ and $a_1$ are the constant and the linear term of the characteristic polynomial of $F''$, which is $f_0(x):=p^{-2n(F)}f(p^{n(F)}x+\lambda)$ and has coefficients in $\mathbf{Z}_p$. By picking $\lambda$ in the range $0,\ldots,p^{n(F)}-1$, we see that the action of $F$ with respect to the basis $(e_1, F''(e_1))$ is that given by the statement.

Notice that this shows that $\mathbf{Z}_p^2$ is a free $R_F$-module of rank one, and classifying the action of $F$ on $\mathbf{Z}_p^2$ is roughly equivalent to finding $R_F$. I was interested exactly in this in the context of Tate modules of elliptic curves over finite fields ($F=$Frobenius). Probably there is a more conceptual/simpler proof. I would be interested to hear what you get in higher dimension. It won't be that easy, I expect. What makes this case simple is that orders of $\mathbf{Q}_p[F]$ containing $F$ are classified by the index with which $\mathbf{Z}_p[F]$ sits in them.<|endoftext|>
TITLE: Equivalence between $E_\infty$-spaces and connective spectra
QUESTION [11 upvotes]: It is well know that the $\infty$-category of group-like $E_\infty$-spaces and the $\infty$-category of connective spectra are equivalent, see e.g. 
May - "$E_\infty$-spaces, group completions and permutative categories" or
Lurie - "Higher Algebra", Remark 5.1.3.17
Now the category of $E_\infty$-spaces (here space means simplicial set) carries a model structure as well as the category of spectra. Is there a direct (left) Quillen functor 
$E_\infty$-space $\to$ Spectra
whose derived functor restricts to such an equivalence? I have been unable to find a discussion of this in the litertatur. The only thing I can find are indirect functors going through $\Gamma$-spaces or related categories. The Bar-construction which is usually used is not left Quillen (!?).

REPLY [6 votes]: Of course, as several people have noted, the answer depends on
the choice of details.  There is a variant of my original passage
from $E_{\infty}$ spaces to spectra that certainly works, as was
noted in ``Units of ring spectra and Thom spectra'' by Ando, 
Blumberg, Gepner, Hopkins, and Rezk (arXiv: 0810.4535v3). 
Take the Steiner $E_{\infty}$ operad for definiteness and 
denote the monad on based spaces associated to it by $\mathbf{C}$.
Take spectra to mean Lewis-May spectra since it is very convenient
to have the $(\Sigma^{\infty},\Omega^{\infty})$ adjunction for the
question at hand, and that is incompatible with symmetric monoidal
categories of spectra. Of course, that means I'm not using simplicial
sets, but I don't suffer from a prejudice in their favor: when I write
space I prefer to actually mean space.   
Then, as discussed in modern terms in my 
paper "What precisely are $E_{\infty}$ ring spaces and $E_{\infty}$ 
ring spectra?'' Geometry & Topology Monographs 16(2009), 215--282,
the spectrum associated to a $\mathbf{C}$-space $X$ is the two-sided
bar construction $B(\Sigma^{\infty},\mathbf{C},X)$.  For cofibrant 
$X$, this is equivalent to the ``tensor product''
$\Sigma^{\infty}\otimes_{\mathbf{C}}X$, which is defined
by an obvious coequalizer.  This functor from $\mathbf{C}$-spaces to
spectra is left adjoint to $\Omega^{\infty}$. Further details are as
one would expect.<|endoftext|>
TITLE: weak*-closed subspaces
QUESTION [7 upvotes]: Recall that a closed subspace $Y$ of a Banach space $X$ is weakly complemented if the set 
$$Y^{\bot}:= \{ f\in X^\*| f(y) = 0 \forall y\in Y\}$$ 
is a complemented subspace of $ X^*$. For example, $c_0$ is a weakly complemented subspace of $l_{\infty}$.
Question: Is there a Banach space $X$ such that there is a weak${}^*$-closed subspace $Y$ which is weakly complemented but not complemented in $X$.

REPLY [7 votes]: No. You get $Y^{**}=Y^{\perp\perp}$ complemented in $X^{**}$ and $Y$, being a dual space, is norm one complemented in $Y^{**}$.<|endoftext|>
TITLE: The `set' of all principal G bundles over `all' spaces
QUESTION [5 upvotes]: What is a good notation for the 'set' (or stack if you insist)
of all principal G bundles over 'all' spaces for given G?
BG is way over used.  How about Bun(G)?

REPLY [2 votes]: In algebraic geometry, $BG$ only ever refers to a stack, viewed as a fibered category over the category of schemes, or affine schemes, depending on your conventions.  The objects are principal $G$-bundles $P \to X$ of schemes (or if $G$ is a group sheaf, it is a map of sheaves).  The morphisms are $G$-equivariant fiber squares.
More to the point, there is no alternative object in the algebro-geometric universe that takes the name $BG$, so there is no source of confusion there.
The name $\operatorname{Bun}_G$ is usually reserved for the enriched Hom-stack construction $\underline{\operatorname{Hom}}(-, BG)$.  In other words, $\operatorname{Bun}_G(X)(T) = BG(X \times T)$.<|endoftext|>
TITLE: Rational forms of simple Lie algebras
QUESTION [12 upvotes]: I am more or less familiar with the classification of real forms of complex semisimple Lie algebras. But as soon as I wander off into the domain of very-non-algebraically closed fields, things seem to become considerably more complex.

Is there a classification of the rational forms of complex semisimple finite-dimensional Lie algebras?

Of course, there is such a classification, and one can in principle carry it out using standard ideas involving Galois cohomology &c. The question is, rather, has it been written down explicitly and with a nice parametrization?
In particular:

How many rational forms does the complex algebra of type $G_2$ have?

REPLY [3 votes]: (Disclaimer: I am not an expert and easily may overlook something).
Results about forms of exceptional types are technically difficult and scattered over the literature. For example, the only known to me full description of forms of $D_4$ is buried (somewhat implicitly) inside the book: Knus, Merkurjev, Rost, Tignol, The Book of Involutions, AMS Colloq. Publ, Vol. 44., 1998, http://www.mathematik.uni-bielefeld.de/~rost/BoI.html . The works of Skip Garibaldi (using the language of algebraic groups) are also highly relevant.<|endoftext|>
TITLE: (Second) Chern class of projective space, blown up in a linear subvariety
QUESTION [5 upvotes]: I already asked the same question at stack exchange but got no response for quite a while, so I thought I'd ask here. I also know that this has a certain resemblance to this question, but I cannot really make much sense of the answer given there.
Write $\mathbb{P}^n:=\mathbb{P}^n_\Bbbk$ for projective space over some (algebraically closed) field $\Bbbk$ and assume that $X\subseteq\mathbb{P}^n$ is a linear subvariety, $\mathbb{P}^m\cong X$, say. I now consider the blow-up of $Y:=\mathbb{P}^n$ in $X$, yielding a blow-up diagram
$$\begin{matrix} \tilde{X} & \xrightarrow{\;j\;} & \tilde{Y} \\
\hphantom{\scriptstyle g}\downarrow {\scriptstyle g} &&
\hphantom{\scriptstyle f}\downarrow {\scriptstyle f} \\
X &\xrightarrow{\;i\;} & Y
 \end{matrix}$$
My question is, what is the second chern class $c_2(\tilde Y):=c_2(\mathcal{T}_{\tilde{Y}})$ of the tangent sheaf of $\tilde{Y}$?
Remark: I am ultimately interested in the degree of $c_2(\tilde Y)c_1^{n-2}(\tilde Y)$. If I could understand the total chern class $c(\tilde Y)$, that would be even better.
My thoughts so far: The chern classes of $Y$ (and $X$) have well-known representation, and there is a formula for computing the chern classes of blown-up varieties in Fulton's book Intersection Theory, namely Theorem 15.4. For brevity, I will quote his Example 15.4.3, which gives a formula for $c_2$:
$$c_2(\tilde Y) = f^\ast(c_2(Y)) - j_\ast\left( (d-1) g^\ast(c_1(X)) + \tfrac{d(d-3)}{2} \zeta + (d-2) g^\ast(c_1(\mathcal{N})) \right)$$
Here, $\mathcal{N}=\mathcal{N}_{X/Y}$ is the normal bundle of $X$ in $Y$ and $\zeta$ denotes $c_1(\mathcal{O}_{\tilde{X}}(1))$.
From Fulton's Example 3.2.12, we know that $c_1(X)=(m+1-d)\cdot\xi$ and $c_1(\mathcal{N})=d\cdot\xi$ with $\xi = c_1(\mathcal{O}_X(1))$. Now, I am kinda stuck. I am not sure what the push-forwards and pullbacks really do - in particular, what is $g^\ast(\xi)$ in terms of $\zeta$? What is $f^\ast$, applied to the class of a hyperplane? What does $j_\ast$ do? More importantly, are these the right questions to ask? 
Ultimately, I thought (hoped) it would be possible to express $c_2(\tilde Y)$ or even $c(\tilde Y)$ as a sum of intersections of "obvious" cycles in $\tilde Y$, possibly involving only the class of the (strict) transform of a hyperplane and the exceptional divisor.

REPLY [2 votes]: Since you are blowing up $\mathbb{P}^n$ along a linear subvariety, which in turn is a smoooth complete intersection in $\mathbb{P}^n$, you can compute its Chern classes very easily via a formula of Aluffi (Lemma 1.3 in http://www.math.fsu.edu/~aluffi/archive/paper348.pdf). After denoting the first Chern class of a hyperplane in $\mathbb{P}^n$ by $H$, $f:\tilde{\mathbb{P}^n}\to \mathbb{P}^n$ the blowup of $\mathbb{P}^n$ along a linear subvariety say of codimension $k$ and the exceptional divisor by $E$, then
$$
c(\tilde{\mathbb{P}^n})=\frac{(1+E)(1+f^*H-E)^k}{(1+f^*H)^k}f^*c(\mathbb{P}^n).
$$ 
(To get the individual Chern classes, just replace $H$ by $t\cdot H$ and $E$ by $t\cdot E$ in the expression above, expand the expression as a powerseries in $t$, then the coefficient of $t^m$ will be $c_m$.) Then if you want to compute any Chern number, take the product of the corresponding Chern classes and push them forward to $\mathbb{P}^n$, which amounts to just knowing the pushforwards of self-intersections of the exceptional divisor via the projection formula, which can be easily computed via Lemma 2.1 in http://arxiv.org/pdf/1211.6077v1.pdf.<|endoftext|>
TITLE: Infinite closed partition of the real line with no closed infinite unions
QUESTION [6 upvotes]: Is there a partition of the real line into infinitely many closed subsets so that no infinite union of these subsets (except the whole space) is closed?
This question was asked also at math.stackexchange.com. While an interesting construction of a partition with no closed uncountable unions was given, there has not yet been a conclusive answer.
Two observations:

No infinite subcollection of the partition can be locally finite since the union of a locally finite collection of closed sets is closed.
The partition must be uncountable since the real line cannot be partitioned into countably many (and $\geq 2$) closed subsets by a theorem of Sierpiński.

REPLY [3 votes]: Let $\mathcal{F}$ be this partition. As noted above it must be uncountable. We may as well assume it lives on the interval $(0,1)$ and add the set $ \lbrace 0, 1\rbrace$ to each member to obtain a family of closed subsets of $[0,1]$. Now $\mathcal{F}$ is an uncountable subset of the space of closed subsets of $[0,1]$ endowed with the Hausdorff metric. The latter space is separable metric, so $\mathcal{F}$ contains a non-trivial sequence $\langle F_n:n\in\mathbb{N}\rangle$ that converges to a point $F$ of $\mathcal{F}$.
The union $F\cup\bigcup_n F_n$ is closed: if $x$ is outside the union, in particular outside $F$, let $\epsilon=\frac12 d(x,F)$. Then there is an $N$ such that $F_n\subseteq B(F, \epsilon)$ for $n\ge N$. This now easily implies that $x$ is not in the closure of the union.<|endoftext|>
TITLE: Decomposition of semisimple Lie group into almost simple factors
QUESTION [6 upvotes]: Can anyone suggest a reference that defines or explains that a semisimple real Lie group can be decomposed into a product of almost simple factors? In some papers that I read recently people keep talk about "semisimple Lie groups withoug compact factors" without explanation and it appears to be some standard notion. But the only reference I can find is Margulis' book "discrete subgroups of semisimple Lie groups" where this decomposition is claimed only for algebraic groups instead of for Lie groups.

REPLY [8 votes]: Concerning references, there exist many books which treat the structure and classification of semisimple Lie groups, usually with a wider agenda involving for example symmetric spaces, harmonic analysis, infinite dimensional representations.   Older and newer authors include Chevalley, Helgason, Knapp, Wallach, Onishchik-Vinberg, Bump, etc.    (Other books concentrate more heavily on compact groups.)
Though the coverage in such books varies a lot, the basic outline is usually similar: start with the notion of (real) Lie group and the associated Lie algebra (originally called the "infinitesimal group"), study the solvable radical, pass to semisimple groups and their Lie algebras, then complexify the situation in order to use more algebraic methods.   
Ultimately the structure of a complex semisimple group lifts from the structure of a complex semisimple Lie algebra.   Here relatively elementary methods, based on nondegeneracy of the Killing form, decompose the Lie algebra into a direct sum of simple ideals (which can be readily classified).  Then the nice correspondence between the groups and their Lie algebras allows most of this structure to be found in the group as well, though the "simple" groups may in fact just be "almost simple".    Decomposing the group directly into simple factors is not an attractive project, though it might be done indirectly using Chevalley's approach via linear algebraic groups over arbitrary algebraically closed fields. 
Only after such results are in hand for the complex Lie algebras and groups can one adapt the structure and classification to the real case.    I don't think it's practical to get a direct factorization of a semisimple (real) Lie group into its simple factors, but on the other hand the existence of unique compact real forms makes the less direct comparison of real and complex cases doable.
None of the standard Lie groups books can be viewed as easy reading, since by its nature Lie group theory merges ideas from analysis, topology, algebra in a
sophisticated way.   In any case, algebraic group ideas have their limits in the study of real Lie groups, since covering groups arise which are not algebraic.<|endoftext|>
TITLE: When do basepoints matter in homotopy theory?
QUESTION [20 upvotes]: [EDIT: This question as written contains a glaring error. It's sort of "not a real question"? But I am resisting the temptation to delete it, because I am going to make it into a real question.]
I tend to pay a lot of attention to when basepoints matter and when they don't. Currently something is bothering me:
Consider the "homotopy category" of unbased $1$-connected (i.e. simply-connected) spaces: that is, start with spaces that have one path component and trivial fundamental group, with unbased maps as the morphisms, and invert the weak homotopy equivalences. Or, if you prefer, start with $CW$ spaces of that kind and pass to homotopy classes. I suppose that this is NOT actually (equivalent to) the homotopy category of any model category, which is why I have used quotation marks above. 
[EDIT: As Fernando Muro points out in his answer, the previous sentence is obviously wrong. The model structure described in the following paragraph, with its artificial use of a basepoint, does the job!]
If I do the same thing with basepoints, there's no problem. For example, I can take the usual category of based spaces but declare the weak equivalences to be the maps inducing isomorphisms of $\pi_j$ for all $j\ge 2$ at the basepoint. Then every object is equivalent to a $1$-connected space (in fact, if you use the usual fibrations, then all cofibrant objects are $1$-connected), and the homotopy category is as intended. The same kind of thing works for $k$-connected based spaces for other values of $k$. 
When $k=0$ it is easy to see that the "homotopy category" of $k$-connected (i.e. path-connected) unbased spaces is not a homotopy category of a model category. In fact, it lacks coproducts. (Exercise: there is no universal example, up to homotopy, of a path-connected space equipped with two maps from the circle.) This is enough, because the coproduct (colimit of discrete diagram) of cofibrant objects in a model category is always a homotopy colimit.
When $k=-1$ (so that $k$-connected spaces means nonempty spaces) it is even easier to see: The homotopy category of nonempty spaces has no initial object.
[EDIT: I will leave this as it stands even though the answer is, come to think of it, obviously "no".]
QUESTION 1: Is it true that the homotopy category of (unbased) $1$-connected spaces cannot be the homotopy category of a model category?
This is not so obvious. You can't just say something like "a homotopy category must have all limits and colimits" because homotopy limits and colimits are not in fact limits and colimits in a homotopy category in general -- the homotopy category of spaces does not have pullbacks.
(I presume that most of what I said above more or less applies to other popular ways of axiomatizing homotopy theory, such as $\infty$-categories.) 
Apart from this technical question, I have a vague philosophical question:
QUESTION 2: If the answer to Question 1 is "yes", what is the right response?
I am thinking of rational homotopy theory, for example. I recall that when Quillen worked out an indirect equivalence between on the one hand $1$-connected spaces and rational equivalences and on the other hand $0$-connected commutative differential graded coalgebras over $\mathbb Q$ he used based spaces and coaugmented coalgebras. It's clear why. But Sullivan's equivalence between rational homotopy theory of simply connected spaces and commutative dgas works without basepoints, right? He doesn't do it in a model category framework and (this is really beside the point) he needs finite type conditions. It seems unnatural to insist on based spaces just for the sake of (model category or other) axioms. On the other hand they are such nice axioms ...

REPLY [2 votes]: I'll throw in a point (!) I have been making since 1967 that certainly for the 1-dimensional case it is convenient to use a set of base points chosen according to the geometry. Thus one finds that in van Kampen type situations one is calculating homotopy 1-types: groupoids are useful because they have structure in dimensions 0 and 1. 
Analogously,  in higher dimensions one needs algebraic objects with structure in a range of dimensions, especially as low dimensional identifications have high dimensional influence on homotopy types. It turns out that this can be accomplished to a useful extent if one works with filtered spaces, or $n$-cubes of spaces. With these structures, one can calculate precisely some homotopy $n$-types in gluing situations. 
The implications of having many base points are unclear in, for example, the theory of loop spaces, and their iterations, or for model categories. But it is clearly sensible to have a set of base points in discussing group actions, where the action  group $G$ acts also on the fundamental groupoid $\pi_1(X,A)$ if and only if $A$ is  union of orbits. Thus the natural theorem on orbit spaces involves  orbit groupoids.<|endoftext|>
TITLE: A property of unimodal sequences
QUESTION [12 upvotes]: It is well-known that $(-1)^j \sum_{i=0}^j (-1)^i\binom{n}{i} \geq 0$. This inequality can be used to prove Bonferroni's inequalities for example.  Recently I noticed that a similar inequality applies to non-negative unimodal sequences with alternating zero sum: if $a_0 \leq a_1 \leq \ldots \leq a_k \geq a_{k+1} \geq \dots \geq a_{n}$ with $\sum_{i=0}^n (-1)^i a_i = 0$ then $(-1)^j \sum_{i=0}^j (-1)^i a_i \geq 0$ for $j=0 \ldots n$. Of course the binomial sequence is just a special case.  My question is if anyone has any knowledge of this property of unimodal sequences having appeared or having been used anywhere in the literature?  Thanks.
Jose A Rodriguez

REPLY [6 votes]: Closely related ideas come up in discrete Morse theory too, in proving the Morse inequalities.  
You can prove your inequality combinatorially by a sign-reversing involution argument, i.e. a type of matching argument, as follows.  Make a graded partially ordered set with $a_i$ elements at rank $i$, i.e. in the $i$-th level of the poset, for each $i$.  Now include all possible edges between ranks $i$ and $i+1$ for each $i$.  I am assuming the $a_i$'s are integers, but you can do a weighted version otherwise.
Greedily start at rank $0$, taking each element of rank $i$ for $i < k$ which hasn't already been matched with an element of rank $i-1$ and matching it with an element of rank $i+1$. Likewise, start at rank $n$ and proceed downwards through the poset, greedily taking element of rank $n-i$ for $n-i > k$ which hasn't already been matched with an element of rank $n-i+1$ and matching it with an element of rank $n-i-1$.  When we get to rank $k+1$, we must only match with those elements of rank $k$ that weren't already matched with elements of rank $k-1$.  The fact that the alternating sum is 0 tells us we can get a complete matching this way.  
Now we can interpret $(-1)^j\sum_{i=0}^j (-1)^i a_i$ as the number of elements at rank $j$ that have been matched with elements at rank $j+1$, so in particular as a nonnegative integer.  To see this, assign weight $+1$ to each element at rank $j$ or at any other rank of the same parity as $j$; assign weight $-1$ to the elements whose rank differs from $j$ by an odd number.   The alternating sum you consider is now the sum of the weights for ranks $0$ to $j$.  Each matching edge completely contained in the lowest $j$ ranks contributes 0 to this alternating sum, whereas each element of rank $j$ that is matched with an element of rank $j+1$ contributes 1 to this alternating sum.  
I agree with Vladimir that this is a very nice observation you've made!<|endoftext|>
TITLE: Generalization of Rigid Body Motion to arbitrary (compact) Lie Groups
QUESTION [7 upvotes]: The classical dynamics of a rigid body in three dimensions may be described as the motion of a point on a configuration space given by the Lie group $SO(3)$, governed by Euler's equations for rigid body motion. It seems to me that there should be a straightforward generalization of this from $SO(3)$ to any (compact) Lie group. (Compact, presumably, because we want the Hamiltonian to be bounded below.) Would someone be so kind as to point me to some literature that discusses the configuration space, Lagrangian, canonical momenta, Hamiltonian and equations of motion in this more general case?  I would also like to gain some insight as to how the left and right actions of the Lie group on itself are related to the physical concepts of inertial and rotating frames.  My initial attempts took me immediately into some areas of integrable systems and algebraic geometry that, while interesting, assume that the simpler question(s) I am asking have already been understood by the reader.

REPLY [4 votes]: Here is the link that should satisfy you:
http://ncatlab.org/nlab/show/Hamiltonian+dynamics+on+Lie+groups .
It is also interesting to note that if you consider in this context infinite dimensional Lie group (eg. group of volume-preserving diffeomorphisms of some manifold), you'd recover hydrodynamics of the ideal fluid:
http://terrytao.wordpress.com/2010/06/07/the-euler-arnold-equation/ .
Except for above two references I'd recommend Arnold's "Mathematical methods of classical mechanics", "Topological methods in Hydrodynamics". Also, take a look at "Symplectic techniques in physics" by Sternberg and Guillemin.<|endoftext|>
TITLE: How "should" I define "absolutely continuous" functions on e.g. n-spheres?
QUESTION [6 upvotes]: (Am writing this post in a rush, out of office, so cannot give adequate links etc right now.)
There is a classical and well-understood definition of what it means for a continuous function $f:[a,b]\to{\mathbb R}$
 to be absolutely continuous: we should have
$$ \sup \{ \sum_{i=1}^n (t_i-t_{i-1}) \vert f(t_i)-f(t_{i-1}) \vert \} < \infty $$
where the supremum is over all partitions $a=t_0 < t_1 < \dots < t_n=b$.
From my limited reading on Wikipedia and various other online searches, it transpires that there are several different notions of "functions of bounded variation" defined on higher-dimensional "domains", with the quotes indicating that I am uncertain of the precise technical qualifiers. (The setting given in Wikipedia is for open subsets of ${\mathbb R}^n$; googling has also shown links to papers defining BV for Riemannian manifolds.)
It is not clear to me how these definitions in higher dimensions could be approached in a naive way via approximating sums, as in one variable. So my rather naive question is: can such a definition be made to work on, say, the $2$-sphere? What about higher spheres?
Also, since BV/AC are metric notions and not just topological ones, would I get definitions closer to the one-dimensional case if I worked with a suitable polyhedron or a triangulation of the $2$-sphere? (and if so, does such an approach obviously generalise to higher spheres?)
(The motivation for this question is slightly indirect, and I may try to give more context when I am next in the office with free time and working brain.)

REPLY [4 votes]: There is a difference  between  BV and AC.            AC functions on an interval are  integrable functions  whose  distributional derivatives are   representable by integrable functions.  BV function on an interval  are  integrable functions whose  distributional derivatives are distributions representable  by (signed) measures.
We can then  define  the space  AC functions on a Riemann manifold to be the Sobolev space $W^{1,1}$ (one derivative in $L^1$), i.e.,  integrable  functions with integrable gradients.   These functions are definitely not continuous  so they may not be as appealing to you.<|endoftext|>
TITLE: Least cardinality of a set of points in the plane
QUESTION [5 upvotes]: What is the least possible cardinality $K$, of a set S of points in the plane, such that there exists a point P in the plane and an open ball B centered at P, such that for all points X in B, not all the euclidean distances from X to points of S are rational?
Can we do better then $3\leq K \leq 2^{\aleph_0}$?
What can be said about K if all the points of S must be at integer distance from eachother, can it be proved to be finite?

REPLY [4 votes]: As Boris Bukh points out, three points suffice, but I'd like to point out that your question is related to this MO question.  
Here is a summary of the information in the previous question.  For the second part of your question, the author (me) conjectures that for any finite set $S$ with all rational distances, no such point $P$ exists.  As I noted in the comments, this is true when $|S|=3$, proven by Almering.
It is not known if there is a point with all rational distances to the unit square.  However, it is known that there are no points at rational distance from all vertices of a regular $n$-gon, except perhaps when $n=4,6,8,12,24$.  
Some more tangential remarks are that it is not known if there is a dense set of points in the plane with all distances rational, although it is conjectured that there is none.
Even more tangential, it is not known if every planar graph can be straight-line embedded in the plane with all edges having rational length, although it is conjectured to be true.<|endoftext|>
TITLE: Why do I need densities in order to integrate on a non-orientable manifold?
QUESTION [35 upvotes]: Integration on an orientable differentiable n-manifold is defined using a partition of unity and a global nowhere vanishing n-form called volume form. If the manifold is not orientable, no such form exists and the concept of a density is introduced, with which we can integrate both on orientable and non-orientable manifolds.
My question is: On a non-orientable n-manifold, every n-form vanishes somewhere, but shouldn't I be able to chose an n-form with say a countable number of zeros, which would then constitute a set of measure zero and thus allow me to use n-forms (with zeros) for global integration also on non-orientable n-manifolds?

REPLY [4 votes]: In fact, for the purpose of integrating functions on a non-orientable manifold, you don't directly need densities. Every (connected) manifold $M$ has an orientable double cover (connected if and only $M$ is not orientable), $\pi: \tilde M \to M$. Then, upon fixing an orientation and a volume form $\mathcal V$ on $\tilde M$, you can define the integral of a function $f$ on $M$ by
$$\frac{1}{2}\;\int_{\;\tilde M}\left(f\circ\pi\right) \mathcal V\,.$$
Of course, if $M$ is not orientable, the volume form $\mathcal V$ cannot be chosen so as to be invariant under the involution that swaps the two points in each fiber of $\pi$, otherwise it would be the pullback of a volume form on $M$. 
This has the disadvantage of not being very canonical in general. But, if $M$ is equipped with a Riemannian metric $g$, then there is a canonical volume form on $\tilde M$, namely that induced by $\pi^* g$. So, for instance, the volume of a Riemannian Moebius strip or of a Riemannian Klein bottle is well defined.<|endoftext|>
TITLE: Blow-up along a subscheme and along its associated reduced closed subscheme
QUESTION [13 upvotes]: Let $X$ be a noetherian scheme and let $Y$ be a closed subscheme of $X$. 
What relation is there between $\mathrm{Bl} _ {Y}(X)$ and $\mathrm{Bl} _{ Y _{\mathrm{red}}}(X)$ ?
Thanks.

REPLY [16 votes]: There is no map from one blow up to the other, and definitely not an isomorphism.  Please see my comments to J.C. Ottems answer.  
However, if you replace radical by integral closure, then everything is fine.  
Here's what I mean, if $I$ is an ideal and $J$ is its integral closure, then you always have an everywhere defined map
$$Bl_J X \to Bl_I X.$$
This need not be an isomorphism, indeed the integral closure of $(x^2, y^2)$ is $(x^2, xy, y^2)$.  The blow up of the latter ideal is the normalization of the blow up of the former.  
The other way you can get a map is if $J = \sqrt I$, and also if we can write $I = J \cdot \mathfrak{a}$ for some other ideal $\mathfrak{a}$.  Then the blow up of $I$ is always the blow up of $\mathfrak{a}$ pulled back to $Bl_J X$.
In general, you should expect no relation between the blow up of two ideals with the same radical unless there is some integral closure relation between them and/or one ideal is the product of the other (and something else).<|endoftext|>
TITLE: Power of two plus integer is not prime
QUESTION [6 upvotes]: Is it true that for every integer $k\neq 1$ there exists infinitely many natural numbers $n$ for which $2^{2^n}+k$ is a composite number?

REPLY [18 votes]: Yes it is true, here is a proof. Suppose that $2^s$ is the highest power of $2$ dividing $k-1$. Suppose also that all but finitely many numbers $2^{2^n}+k$ are primes. Take a very large such prime $p=2^{2^l}+k$, $l>s$. Then the maximal power of $2$ dividing $p-1$ is $2^s$ and $m=\frac{p-1}{2^s}$ is odd. Let $f=\phi(m)$ (the Euler function). Then $2^{ft}\equiv 1 \mod m$ for every $t\ge 1$. Then $2^{ft+s}\equiv 2^s \mod p-1$. Hence $2^{ft+l}=2^l \mod p-1$. Hence $2^{2^{ft+l}}\equiv 2^{2^l} \mod p$ and $2^{2^{ft+l}}+k\equiv 0 \mod p$, so it is never prime for $t\ge 1$. 
 Update 1. One can of course replace $2$ by any natural number $>1$ in the above proof. It is not clear what to do with numbers $m^{n^b}+k$ for fixed $m,n,k$ when $m\ne n$, but see a comment above.
 Update 2.  Actually the case $m\ne n$ is not very difficult either. Take a very large prime $p$ of the form $m^{n^l}+k$. We would like to have infinitely many $b$ such that  $n^b\equiv n^l \mod p-1$. The only thing needed for this is that $GCD(n^l,p-1)=GCD(n^c,p-1)$ for all big enough $c$. That seems to require very little work.<|endoftext|>
TITLE: Results in the Presentation of Finite Groups
QUESTION [7 upvotes]: I've been looking at combinatorial group theory, but all the results seem to be about infinite groups. Are there any important results about the presentations finite groups specifically (or are useful for finite groups?). About how the minimum number of relations implies something about the structure of the group? 
I'd prefer results that are applicable to all finite groups or to all finite simple or all simple groups.

REPLY [3 votes]: The result by Guralnick, Kantor, Kassabov and Lubotzky (which is mentioned already in Mark Sapir's answer) is one of the most striking results in finite group theory of the last decade, in my opinion. It shows that there is a uniform bound on the length of the presentation for (probably all) nonabelian finite simple groups. This is much stronger than what anyone before this result even conjectured, so it's absolutely amazing. More precisely, they show the following:

All nonabelian finite simple groups of rank $n$ over a field of size $q$, with the possible exception of the Ree groups $^2G_2$, have presentations with at most 80 relations and bit-length $O(\log n + \log q)$.

(By the way, the assumption "nonabelian" is essential, the theorem is false for cyclic groups of prime order!)
Their results appeared in a series of 3 papers:

Presentation of finite simple groups: a quantitative approach
Presentation of finite simple groups: cohomological and profinite approaches
Presentation of finite simple groups: a computational approach

I would recommend you to have a look at the first of those 3 papers, it contains a wealth of information about presentations of finite (simple) groups, and it is written in a very readable and enjoyable style.<|endoftext|>
TITLE: Is there a really big ring of differential operators in characteristic p?
QUESTION [8 upvotes]: $k$ is a field of characteristic $p$.
$k[t]$ has canonical first-order differential operator $\partial$
As an endomorphism of $k[t]$, $\partial^p=0$.
First way to fix it:
Use the divided power differential operator $\partial^{[p]}$.
Shortfall:
As an endomorphism of $k[t]$, ${\partial^{[p]}}^p=0$
Second way to fix it:
Use crystalline differential operators.
Shortfall:
No higher order operators on $k[t]$.
Question:
Is there a really big ring of differential operators which contains the divided powers $\partial^[n]$ for all $n$ and which also has a natural evaluation map to $End_k(k[t])$?

REPLY [3 votes]: Let me give some more details on Mariano's comment: The ring of differential operators a la EGA4 in this particular case will be a free  $k[t]$-algebra generated by the following operators: We write 
$$\partial_t^{(n)}$$ for the operator
which is defined by 
$$\partial_t^{(n)}(t^m)={m\choose n}t^{m-n}.$$
Because of this, sometimes the notation
$$\partial_t^{(n)}=\frac{1}{n!}\frac{\partial^n}{\partial t^n}$$
is used.
Actually, to generate the ring, the operators $\partial_t^{(p^n)}$ suffice. 
Now this ring is not noetherian, but it is an increasing union of noetherian subalgebras, lets denote them by $D^{(m)}$, which are the subalgebras generated by operators of degree $\leq p^m$.
Using partially divided powers, Berthelot abstractly defines rings $\mathcal{D}^{(m)}$ such that the full ring of differential operators $\mathcal{D}$ is the direct limit of the $\mathcal{D}^{(m)}$. The image of $\mathcal{D}^{(m)}$ in $\mathcal{D}$ is then precisely the $D^{(m)}$ that I defined ad-hoc above.  The crystalline operators that you defined in the question correspond to Berthelot's $\mathcal{D}^{(0)}$.<|endoftext|>
TITLE: On the least prime in arithmetic progressions
QUESTION [8 upvotes]: My question concerns the least prime (denoted $p(a, q)$) in the arithmetic progression $a \pmod q$ where $a$ and $q$ are coprime. Quite a time ago Linnik demonstrated that 
$$p(a, q) \ll q^L$$
for some absolute constant $L$. Wiki page for this theorem lists a number of papers that estimate $L$ with the most recent result by Xylouris who proved that $L \leq 5.2$.
It is also known that the Generalized Riemann Hypothesis implies
$$p(a, q) \ll (q\log q)^2 \text{,}$$
while in 1978, Heath-Brown conjectured even tighter bound:
$$p(a, q) \ll q(\log q)^2 \text{.}$$
I'm wondering whether this last bound, if true (it is still an open problem), implies something non-trivial about $L$-functions?

REPLY [4 votes]: I haven't looked at such types of questions, however my first thought is no: I don't see (or haven't seen yet) how the existence of one prime (or a small number of them) would force the Dirichlet L-functions (I guess these are which you meant) to look in a certain way.
As an example, take the explicit formula which counts the number of primes in arithmetic progressions by using zeros of some L-functions. 
Now, if you know that the "left side" in the particular equation (counting the primes) is 1 instead of 0, this does not seem to force anything noteworthy for the zeros which appear in the sum of the right side....Some small change in the imaginary parts of a couple of zeros (with very large imaginary part) might be enough to change the total value by 1, which then implies my answer for the special case if you only use the explicit formula. However, most arguments in analytic number theory (where this theorem on the least prime number comes from) tend to give similar behavior.
I hope, you understand the point I am trying to make besides my unclear presentation.<|endoftext|>
TITLE: Measure on the Boundary of a Hyperbolic Group
QUESTION [12 upvotes]: Let $\Gamma$ be a non-elementary Gromov's $\delta$-hyperbolic group. Let $B(1,n)$ be the set of elements at distance at most $n$ from the identity and let $\partial B(1,n)$ be the elements at distance $n$ (with the word metric). Consider the probability measures $\mu_n$ and $\nu_n$ defined as
$$
\mu_n := \frac{1}{|B(1,n)|}\sum_{\gamma\\, \in B(1,n)}{\delta_{\gamma}}
$$
and
$$
\nu_n := \frac{1}{|\partial B(1,n)|}\sum_{\gamma\\,\in\partial B(1,n)}{\delta_{\gamma}}.
$$ 
It is clear that these measures converge weakly (there is a convergent sub-sequence) on the compact space $\Gamma\cup\partial\Gamma$. Moreover, the limit measure is supported on $\partial\Gamma$.
My question are:

Does anybody have study the limit measures?
Are these related with the Hausdorff measure on the boundary? to the Patterson-Sullivan measure? to the harmonic measure?

REPLY [2 votes]: Letting $\mu_\infty$ be the Patterson-Sullivan measure on $\partial \Gamma$, any limit measure of the sequence $(\mu_n)$ is equivalent to $\mu_\infty$. Moreover, the Radon-Nikodym derivatives are bounded from above and below.
This is given by Lemma 2.13 in Gouëzel, Mathéus and Maucourant's paper Entropy and drift in word hyperbolic groups (Invent. math. (2018) 211:1201–1255). As suggested by Lee Mosher, the arguments follow from arguments of Coornaert's paper.
However, even if the Poisson boundary coincides with the Gromov boundary as a measured space, the harmonic measure on the boundary is not in general equivalent to the Patterson-Sullivan measure, see Theorem 1.5 in the same paper of Gouëzel, Mathéus and Maucourant.<|endoftext|>
TITLE: any software to compute multivariable resultant?
QUESTION [6 upvotes]: Are there any software to computer resultant for a system of equations (more than 2) with more than 2 variables?

REPLY [2 votes]: Singular and Macaulay2 do it according to the documentation, though I haven't personally tried using either of them for this purpose.  They are likely to be faster than Maple; I don't know about MARS.<|endoftext|>
TITLE: Is there a homotopy theory of unbased simply connected spaces?
QUESTION [10 upvotes]: If $\mathcal C$ and $\mathcal D$ are two categories each equipped with a class of morphisms called weak equivalences, then by an equivalence of homotopy theories I will mean functors $F:\mathcal C\to \mathcal D$ and $G:\mathcal D\to\mathcal C$ such that each of the compositions $G\circ F$ and $F\circ G$ is related to the relevant identity functor by a zigzag of natural weak equivalences. This is much stronger than just requiring $F$ and $G$ to induce inverse equivalences of homotopy categories: For an example, let $\mathcal C$ be the category of chain complexes over a field with quasi-isomorphisms as weak equivalences and $\mathcal D$ the subcategory of complexes with zero boundary map. ($F$ is the homology functor and $G$ is the inclusion.) You can see that this is not giving an equivalence of homotopy categories, because if it did then it would also give an equivalence of homotopy categories of $I$-diagrams for any $I$. Of course, if $\mathcal C$ and $\mathcal D$ are model categories then a Quillen equivalence gives an equivalence in the sense I mean (after restricting to categories of fibrant-cofibrant objects). But let's look at this simpler notion.
This is meant to be an improvement on my previous question When do basepoints matter in homotopy theory?. I was tempted to delete the latter, but decided to let it stand for its possible educational value and as a hairshirt for myself. I seriously suggest that nobody should upvote this question without downvoting that one. 
Consider the category of simply-connected spaces (a.k.a. $1$-connected spaces, i.e those having exactly one path component and trivial fundamental group) and continuous maps. We can universally invert the weak homotopy equivalences and get what might be called the homotopy category of $1$-connected spaces. This can be described as a full subcategory of the homotopy category associated to the Quillen model structure on $Top$. Alternatively, it can be described as the category whose objects are $1$-connected $CW$ complexes and whose morphisms are homotopy classes of maps. Note that I am using unbased spaces and unbased maps here. 
QUESTION 1: Is there an equivalence of homotopy theories in the sense defined above between $1$-connected spaces and (the fibrant-cofibrant objects of) some model category.
Another good example of a functor inducing an equivalence of homotopy categories but NOT an equivalence of homotopy theories is the forgetful functor from based $1$-connected spaces to unbased $1$-connected spaces. The positive statement just means that every unbased map between based $1$-connected spaces is homotopic to a based map, and that two such based maps are based homotopic if they are homotopic. The negative statement can be proved by using what I said above about categories of diagrams; for example, you can get a contradiction by picking an action of a group on a $1$-connected space such that the homotopy fixed point set is empty.
The category of $1$-connected based spaces is equivalent in this sense to a model category. You can colocalize the category of based spaces. The main idea is this: In the category of based spaces every object has a universal $1$-connected object over it. 
As some kind of evidence for a "no" answer to the question above, note that the statement in the last sentence is false for unbased $1$-connected spaces. More evidence: a "yes" answer would seem to lead to a reasonable theory of homotopy limits and colimits within unbased $1$-connected spaces. How could that go? In the based setting, hocolim is usual hocolim and holim is universal covering space of basepoint component of usual holim.
All of the above statements, questions, and suggestions about $1$-connected spaces have analogues in which weak homotopy equivalence is replaced by rational equivalence. Sullivan's version
of rational homotopy theory shows that the theory of unbased $1$-connected spaces of finite type is equivalent to that of the opposite of $1$-connected commutative DGAs (differential graded algebras) of finite type. I presume that this can be used to see that the theory of (unbased) $1$-connected spaces is equivalent to that of $1$-connected commutative DGCs (differential graded coalgebras). Quillen's version gives an alternative route, but with basepoints: a chain of Quillen equivalences between based $1$-connected spaces and $1$-connected based (i.e. coaugmented) commutative DGCs (differential graded coalgebras). 
QUESTION 2: Assuming the answer to Q1 is "no", do you have a good point of view to offer regarding this tradeoff between the artificiality of working with based objects and the disturbing lack of homotopy (co)limits in the unbased setting?

REPLY [9 votes]: It seems to me that the answer to (1) is "no", for essentially the reason you've given.
If $\mathcal{C}$ is a category with weak equivalences $W$, you can extract a simplicially
enriched category $\mathcal{C}[W^{-1}]$ using Dwyer-Kan localization, say. If this simplicially enriched category comes from a model category, then it has homotopy coproducts. That is, for every pair of objects $X$ and $Y$, there is another object $Z$ and a pair of maps
$X \rightarrow Z \leftarrow Y$ with the following property: for every object
$W$, the induced map Hom(Z,W) $\rightarrow$ Hom(X,W) $\times$ Hom(Y,W) is a weak homotopy equivalence of simplicial sets. The simplicially enriched category of simply connected CW complexes doesn't have these: for example, there is no homotopy coproduct of a point with itself.
As for $(2)$: you don't have arbitrary homotopy colimits, but you do have lots of them.
For example, you have all homotopy colimits indexed by diagrams with simply connected nerve.
This includes filtered colimits, geometric realizations of simplicial objects,
and the formation of pushouts. This is enough to allow for some useful techniques: for example, you can write every simply connected space as a geometric realization of a simplicial simply connected space, each term of which is a bouquet of $2$-spheres.<|endoftext|>
TITLE: Amenable groups of finite cohomological dimension
QUESTION [8 upvotes]: I seek an example of an amenable group of finite cohomological dimension that is not virtually polycyclic and has finite abelianization.
Remarks.

Elementary amenable groups of finite cohomological dimension are virtually solvable.
There are lots of virtually abelian groups of finite cohomological dimension that have finite abelianization. The simplest one is the fundamental group of a certain flat 3-manifold, the Hantsche-Wendt manifold.
Virtually polycyclic groups are finitely presented, so perhaps an example can be found among virtually solvable groups that are not finitely presentable, which do exist.
Elementary amenable groups of cohomological dimension two are classified here, see Theorem 3, and they have infinite abelianization (except for the trivial group). 
Another idea would be to search for an amenable group that is not elementary amenable.
There are very few known examples, and I do not know if any of the examples have finite cohomological dimension and finite abelianization.  See here for a list of torsion-free examples (of course, finite cohomological dimension implies torsion-free.

REPLY [5 votes]: Igor, what about doing the following. 
Let $A$ be a virtually polycyclic group of finite cohomological dimension whose abelianization is finite, but has has ${\mathbb Z}_n$, $n$ even, among its cyclic factors. Say, the fundamental group of a Hantsche-Wendt manifold would do the job. Next, take $B$, the semidirect product of $\mathbb Q$ and of ${\mathbb Z}$, where ${\mathbb Z}$ acts (through ${\mathbb Z}_2$) by $q\to -q$. Now, make $A$ act on $B$ where the action on $\mathbb Q$ is trivial and the action on ${\mathbb Z}$ is by $n\to -n$ (this is indeed an action as $\mathbb Z$ is abelian). Lastly, let $G$ the semidirect product of $B$ and $A$ using the above action. By construction $G$ has finite abelianization. Also $G$ has finite cohomological dimension because its semidirect factors have this property ($\mathbb Q$ is locally cyclic). Finally, $G$ is not virtually polycyclic as it contains $\mathbb Q$, and subgroups of virtually polycyclic groups are finitely generated. (This paragraph is edited by Igor Belegradek).
You can also imitate the construction of H-W groups, where ${\mathbb Z}_2^n$ would act 
on ${\mathbb Q}^n$ rather than ${\mathbb Z}^n$, but I did not think about the details.<|endoftext|>
TITLE: How to recognize a Hopf algebra?
QUESTION [30 upvotes]: Suppose we are handed an algebra $A$ over a field $k$.  What should we look at if we want to determine whether $A$ can or cannot be equipped with structure maps to make it a Hopf algebra?
I guess in order to narrow it down a bit, I'll phrase it like this: what are some necessary conditions on an algebra for it to be a Hopf algebra?
Thoughts so far:
The first obvious condition is that $A$ must be augmented, i.e. there must be a nontrivial character $\varepsilon : A \to k$.  Since this is generally not that hard to determine if we are given the algebra in some fairly concrete way, let's suppose that $A$ is given to us with an augmentation map.
If $A$ is finite-dimensional, then $A$ must be a Frobenius algebra.  But not every finite-dimensional Frobenius algebra is a Hopf algebra, e.g. $\Lambda^\bullet(k^2)$ is not a Hopf algebra if the characteristic of $k$ is not 2.  And generally I am more interested in the infinite-dimensional case.
All I can come up with is this: the category of finite-dimensional $A$-modules must be a (left) rigid monoidal category.  But I don't know if that is a helpful observation: given a category with a forgetful functor to finite-dimensional vector spaces over some field, how can one prove that it can't be given the structure of a braided rigid monoidal category?
And perhaps there are some homological invariants that one can look at?
To sum up, the question is:
Question
Given a $k$-algebra $A$ and a nonzero character $\varepsilon : A \to k$, are there invariants we can look at in order to show that $A$ cannot be given the structure of a Hopf algebra?

REPLY [7 votes]: I'm a bit late, but here's a simple observation. Consider a topological version of the question: given a topological space $X$, how can we recognize when $X$ can be given the structure of a topological group? A simple necessary condition is that that $X$ must be homogeneous, and in particular each point should have homeomorphic neighborhoods, because $X$ must act transitively on itself.
An analogous statement about Hopf algebras is the following. Let $H$ be a commutative Hopf algebra over a field $k$. Then $G = \text{Spec } H$ is an affine group scheme over $k$; moreover, the group $G(k)$ acts on $G$, and this action is transitive on $k$-points. In particular the dimension of the Zariski tangent space at each $k$-point of $H$ must be the same. So any commutative algebra without this property can't be the underlying commutative algebra of a Hopf algebra. Examples are given by the ring of functions on any singular variety, such as the cuspidal cubic $k[x, y]/(y^2 - x^3)$ ($k$ a field of characteristic other than $2$ or $3$); in this case the dimension of the tangent space at any point is $1$ except at $(0, 0)$ where it's $2$. 
I think this is essentially the point of David Speyer's answer, modulo some technicalities.
(A topological analogue of the observation about $\text{Ext}_A(k, k)$ being graded commutative is that if $X$ is a topological group then $\Omega X$ is an $E_2$-algebra, and in particular $\pi_1(X)$ is abelian by the Eckmann-Hilton argument.)<|endoftext|>
TITLE: What does the Lefschetz principle (in algebraic geometry) mean exactly?
QUESTION [36 upvotes]: This principle claims that every true statement about a variety over the complex number field $\mathbb{C}$ is true for a variety over any algebraic closed field of characteristic 0.
But what is it mean? Is there some "statement" not allowed in this principle?
Is there an analog in char p>0?
Is there reference about this topic? I tried to find some but in vain.
Thanks:)

REPLY [27 votes]: I'm not sure I should admit this in public, but although I am aware of the precise formulations using first order logic and beyond (mentioned in the above answers), 
I tend not to use them. Rather I view the Lefschetz principle as more of a philosphical principle of what ought to be possible in general, and do the necessary verifications as and when I need them (but perhaps  only implicitly). 
I suspect this attitude is pretty common among many algebraic geometers.
To give an example,
for many years the only known proofs* of the Kodaira vanishing theorem were analytic.
But since coherent cohomology behaves well under field extensions, Kodaira vanishing 
is valid over arbitrary (not necessarily algebraically closed) fields of characteristic $0$.
On the other hand, for certain kinds of arguments, one needs a big enough field to
carry out the argument. This typically happens when one is forced to remove a countable
union of exceptional sets.  Curiously, the Noether-Lefschetz theorem is one
such case. Here the Lefschetz principle in the most naive sense
won't work.
*(Added Footnote.) There is now an algebraic proof due to Deligne and Illusie, which involves
reduction to positive characteristic. This is yet another kind of transfer.<|endoftext|>
TITLE: Singular values of $X+iY$ where $X$ and $Y$ are Hermitian
QUESTION [7 upvotes]: Lets have two Hermitian $n\times n$ matrices $X$ and $Y$.
Are there any known properties of the singular values of
$$Z = X + i Y.$$
I am the most interested in bounding from above a few first singular values of $Z$ by the eigenvalues of $X$ and $Y$. And sth that is stronger than:
$$\sum_{i=1}^k \sigma_i^2(Z)\leq \sum_{i=1}^k \left( \lambda_i^2(X)+\lambda_i^2(Y)+\lambda_i(i[X,Y]) \right)$$
for $1\leq k \leq n$ and (singular/eigen)values sorted in the decreasing order.

REPLY [4 votes]: Some results in this direction that you might find useful are listed below.

Theorem (Bhatia and Kittaneh). Let $X$, $Y$, and $Z$ be as in the question above. Then,
  \begin{equation*}
\| (X^2+Y^2)^{1/2} \|_p \le \|Z\|_p \le 2^{1/2-1/p}\| (X^2+Y^2)^{1/2} \|_p,
\end{equation*}
  where $2 \le p \le \infty$, and $\|\cdot\|_p$ denotes the Schatten-$p$ norm. The inequality above gets reversed for $1\le p \le 2$. Also, these inequalities are sharp.

Even more directly relevant is the following theorem that discusses majorization of singular values of $X+Y$ by those of $Z$.

Theorem (Bhatia and Kittaneh). Let $X$, $Y$, and $Z$ be as in the question. Then
  \begin{equation*}
\sigma(X+Y)\quad \prec_w\quad \sqrt{2}\sigma(Z)
\end{equation*}
  If $X$ is psd, then the above weak majorization can be replaced by weak log-majorization, that is,
  \begin{equation*}
\prod_{j=1}^k \sigma_j(X+Y) \le \prod_{j=1}^k\sqrt{2}\sigma_j(Z).
\end{equation*}
  Finally, if both $X$ and $Y$ are psd, then we have even stronger inequalities:
  \begin{equation*}
  \sigma_j(X+Y) \le \sqrt{2}\sigma_j(Z)\quad 1 \le j \le n.
\end{equation*}

Bhatia and Kittaneh also discuss some applications of the above theorem to commutator inequalities.
References

R. Bhatia and F. Kittaneh. "The singular values of $A+B$ and $A+iB$." Linear Algebra and its Applications, 431(2009), pp. 1502-1508.
R. Bhatia and F. Kittaneh. "Cartesian decompositions and Schatten norms." Linear Algebra and its Applications, 318(2000), pp. 109--116.<|endoftext|>
TITLE: What has happened to Lang's Files and other political texts?
QUESTION [25 upvotes]: For some background on Lang and his files, one can read the first part of Lang's obituary in the AMS Notices at http://www.ams.org/notices/200605/fea-lang.pdf.
The book "Challenges" was published in 1997. I guess Lang went on to gather documentation regarding several cases dealt with in the book after its publication. Consequently, several related files must have contained much more documentation at the time of Lang's demise than at the time of "Challenges" publication. Besides, "Challenges" does not deal with all the cases Lang has been involved with and interested in.
In addition, the AMS obituary mentions that Lang "has some unpublished books of a political nature".

Are the full (i.e. the whole documentation that Lang had gathered and arranged) files regarding the "Challenges" cases (i.e. those dealt in that book) published somewhere? If not, are they available in some other form? What has happened to Lang's related personal papers?
What about other files not reproduced in "Challenges"? What are they, and where is the documentation?
Same question for Lang's "unpublished books of a political nature".

EDIT 1 (27 November 2012). For some bad reasons, it has taken me several months to decide to write to the Yale librarian. I can now provide a partial answer to my question, that is: these papers have not been deposited at Yale's library. I will follow the librarian's suggestion to write to the University of Texas. If someone there (or elsewhere) reads this post and wants to provide help to locate Lang's papers, please do not hesitate to contact me. I have copied below the message I have received from Yale's librarian following my query. I have hidden the name of the sender, although that may be unnecessary.
November 26, 2012
Dear Mr. Chiche:
I am writing in reply to your e-mail inquiry of October 22, 2012, regarding former Yale professor Serge Lang.
Our on-line catalog shows that the Yale University Library holds a number of Professor Lang’s published works but, if I am interpreting your query correctly, you wish to know about his unpublished writings.   The Manuscripts and Archives department does not hold a collection of Professor Lang’s papers.  They were offered to us following his death but we declined since the subject matter was outside our collecting areas.  We suggested the University of Texas, where a number of mathematical collections are held.  I searched OCLC’s World Cat to see if I could locate the papers but I did not find any results.  You may wish to contact the library directly to see if the Serge Lang Papers were ever received. Information and assistance is available at http://www.cah.utexas.edu/collections/math.php
If you have any additional questions, please do not hesitate to contact us.
Sincerely,
XXX
Archivist
Manuscripts and Archives

REPLY [12 votes]: I have a lengthy document entitled "The Fire Without, The Fire Within" written by Lang in 1969 about the Vietnam War, student unrest, why he left Columbia U, etc. I'd be glad to donate it to any interested institution or person. Please contact me for more info.<|endoftext|>
TITLE: When is a classifying space a topological manifold?
QUESTION [13 upvotes]: Let $G$ be a discrete group and $BG$ some model for the classifying space of $G$. So $BG$ is an aspherical path-conected topological space.
Under which conditions is $BG$ a topological manifold or only homotopy equivalent to a topological manifold?

REPLY [11 votes]: Here is a more detailed answer. 
Theorem. $K(G,1)$ is homotopy-equivalent to a (textbook) topological manifold if and only if $G$ is countable and has finite cohomological dimension (over ${\mathbb Z}$). 
Sketch of the proof. One direction is clear, so suppose that $G$ is countable and has finite cohomological dimension (say, $n$). Then, by Eilenberg-Ganea theorem (see Theorem 1 in their paper "On the Lusternik-Schnirelmann category of abstract groups", see also Brown's book "Cohomology of Groups", Theorem 7.1), there exists a countable CW complex $X$ of dimension $m\le n+1$ which is $K(G,1)$. This theorem is usually stated without countability assumption/conclusion, but the same proof works in the countable context. 
Now, by Whitehead's theorem (Theorem 13 from Whitehead's "Combinatorial Homotopy-I"), $X$ is homotopy-equivalent to an $m$-dimensional locally-finite CW complex $Y$. Without loss of generality, we can assume that $Y$ is simplicial. Then, by Whitney's embedding theorem (in the context of locally-finite simplicial complexes), there exists a PL embedding $Y\to {\mathbb R}^{2m+1}$. Next, take a suitable open regular neighborhood  $N$ of $Y$ in ${\mathbb R}^{2m+1}$. Then $N$ is homotopy-equivalent to $X$ and, hence, provides a manifold which is $K(G,1)$.<|endoftext|>
TITLE: Are there known non-real zeros of derivatives of Riemann zeta with 0 < Re(s) < 1/2?
QUESTION [8 upvotes]: According to New zero free regions for the derivatives of the Riemann zeta function
assuming the Riemann Hypothesis, $\zeta^{(k)}(s)$ has
at most a finite number of non-real zeros with $\operatorname{Re}(s) < \frac12$ , for $k \geq 1$.
For $k \leq 3$ there are no zeros $0 \leq \operatorname{Re}(s) < \frac12$ (assuming RH).

Are there known non-real zeros for $k>3$ and $\operatorname{Re}(s) < \frac12$ and even $0 \leq \operatorname{Re}(s) < \frac12$?

REPLY [6 votes]: Here are some more zeros of $\zeta^{(k)}(s)$ with $\operatorname{Re}(s) < \frac12$ found with sage:
k  zero
2, -0.35508433021047637343 + 3.5908393243989674267 i
3, -2.1101457926534055013 + 2.5842247720404084808 i
4, -3.2402527145532669326 + 1.6896108197133965315 i
4, -0.83754471075524375517 + 3.8476752430859881333 i
5, -4.2739256609673291219 + 0.66239093925605351714 i
5, -4.2739256609673291219 - 0.66239093925605351714 i
5, -2.1841011922856061145 + 3.0795001018135066045 i
5, 0.28760642074073336717 + 4.6944346849390517355 i
6, -3.1693828756887856638 + 2.289409321602478671 i
6, -1.2725578081548747958 + 4.0741784672743930742 i
7, -3.8750437984692690102 + 1.4917785168218934496 i
7, -2.3934461266307994251 + 3.4062662320064732901 i
7, -0.41331687971848942853 + 4.8452970349839958557 i
8, -4.5682125648555486581 + 0.81115476881064826173 i
8, -3.2523204085919068777 + 2.7169913803460029554 i
8, -1.6702731549827933708 + 4.2784414287490737296 i
8, 0.41829665603240684742 + 5.4752676150632522598 i
9, -3.945849046353689844 + 2.0451787413259943742 i
9, -2.6409589700825760293 + 3.6749136820039963426 i
9, -0.96722009647711882409 + 4.9985369336041978752 i
10, -4.5121442160084114922 + 1.3320691841545877923 i
10, -3.4228591767532529916 + 3.0609427521799546739 i
10, -2.0391435059208024397 + 4.4684432318278930578 i
10, -0.27483240240270603573 + 5.6133090845225833824 i
11, -5.0309783500584905982 + 0.76405382608151195607 i
11, -4.0769374324171281166 + 2.4384470640345412807 i
11, -2.9061882184131317446 + 3.9131827555009648537 i
11, -1.4413185933633706999 + 5.1493402952351168812 i
11, 0.41063992065310427805 + 6.1502251675161887368 i
12, -4.6217542104762414501 + 1.8307305267522831431 i
12, -3.6306820396013356257 + 3.3459003484685151686 i
12, -2.387362966467342712 + 4.6486063663821024758 i
12, -0.84520925190624343306 + 5.7472734659279260752 i
13, -5.1019135182813204449 + 1.1817501251741196135 i
13, -4.2445336021969936603 + 2.7739661902545951775 i
13, -3.1788394218248838054 + 4.1282947333700902908 i
13, -1.8653042754449209073 + 5.2971387830938879447 i
13, -0.24996831630193368259 + 6.2810810441530943019 i<|endoftext|>
TITLE: Are these abelian groups free?
QUESTION [10 upvotes]: Suppose we have a countable, torsion-free abelian group $A$ with the property that for each element $a\neq 0$ the set $D_a=\{x\in A|\exists n\in \mathbb{Z}:nx=a\}$ is finite. 
Is $A$ already a free abelian group?
If one drops the condition "countable" the infinite direct product of countably many copies of $\mathbb{Z}$ is a counterexample.

REPLY [14 votes]: No, there are non-free abelian groups of rank 2 (i.e., subgroups of $\mathbb Q^2$) in which every subgroup of rank 1 is free.  (I assume you intended $a\neq 0$ in the question; otherwise the only such group would be the zero group.)  In fact, such a rank-2 group can be so far from free that its quotient by any pure rank-1 subgroup is divisible.  That result is due to L. Fuchs and F. Loonstra in "On the cancellation of modules in direct sums over Dedekind domains" (Indag. Math. 33 (1971) 163-169).<|endoftext|>
TITLE: Are Chern connections on flat bundles flat?
QUESTION [8 upvotes]: More precisely formulated: We are given a hermitian holomorphic vector bundle $(E,\langle\cdot,\cdot\rangle,\bar{\partial})$ on a $\mathbb{C}$-manifold M such that as a topological bundle, it is flat (i.e. there is some connection, not necessarily compatible with $\langle\cdot,\cdot\rangle$ or $\bar{\partial}$, whose curvature vanishes).
Is it true that the curvature of the Chern connection necessarily vanishes?

REPLY [9 votes]: The answer is yes if $M$ is compact Kähler (EDIT: and you allow a conformal change in the metric) and $L$ is a line bundle and no in general, already in the case of line bundles. 
Take for example $M$ to be the standard Hopf surface $(\mathbb{C}^2\backslash \{(0,0)\})/[(z_1,z_2)\sim (2z_1,2z_2)]$. Then $M$ is diffeomorphic to $S^1\times S^3$ so every line bundle on $M$ has first Chern class zero, therefore it is topologically trivial and it admits a flat connection. 
Let us now see that the canonical bundle $K_M$ is not Chern flat (for any Hermitian connection). First of all, write down the explicit Hermitian metric on $M$
$$g_{i\overline{j}}=\frac{\delta_{ij}}{r^2}, r^2=|z_1|^2+|z_2|^2.$$
It induces a Hermitian metric on $K_M$ whose curvature is the first Chern form which can be easily calculated in local coordinates
$$\alpha=-i\partial\overline{\partial}\log\det (g_{i\overline{j}})=\frac{2}{r^2}\left(\delta_{kl}-\frac{\overline{z}_k z_l}{r^2}\right)i dz_k\wedge d\overline{z}_\ell,$$
and $\alpha$ is a nonnegative-definite Hermitian form, which is not identically zero. If you had another Hermitian connection on $K_M$ with zero first Chern form, by the transgression formula it would imply that $\alpha=i\partial\overline{\partial}u$ so $u$ would be a global plurisubharmonic function on $M$, which must be constant by the maximum principle, and you'd get that $\alpha=0$ which is false.
However, if $M$ is compact Kähler then the $\partial\overline{\partial}$-lemma shows that topologically trivial holomorphic line bundles admit flat Hermitian connections. Just start with any Hermitian connection, its Chern curvature form will be $\alpha$, say, which is cohomologous to zero, so $\alpha=i\partial\overline{\partial}u$ by the $\partial\overline{\partial}$-lemma. Then conformally scaling the Hermitian metric by $e^u$ gives you a new Hermitian connection with vanishing Chern curvature.
For a higher rank holomorphic bundle $E$ over a compact Kähler manifold $(M,\omega)$ you have that $E$ admits a flat Hermitian metric iff it has a Hermitian-Yang-Mills connection $H$ with $\omega^{n-1}\wedge F_H=0$ and its second Chern class vanishes, iff it is $[\omega]$-polystable and its first and second Chern classes vanish. This is essentially the Donaldson-Uhlenbeck-Yau Theorem. 
So if $E$ admits a flat connection its Chern classes indeed vanish, so it will admit a flat Hermitian metric iff it is $[\omega]$-polystable. I think there should be plenty of examples of such bundles which are not polystable.<|endoftext|>
TITLE: cumulant problem
QUESTION [6 upvotes]: A couple of days after I posted this to stackexchange, no one's answered:
I take the problem of cumulants to be this: given a sequence $(\kappa_1,\kappa_2,\kappa_3,\ldots)$ of real numbers, is it the sequence of cumulants of some probability distribution?  In one sense, this is trivially equivalent to the problem of moments: the $n$th moment is a polynomial in the first $n$ cumulants and vice-versa.  But cumulant sequences have a nice property that moment sequences don't have: the set of all such sequences is closed under addition.  So draw a ray out from the origin $(0,0,0,\ldots)$.  If the ray bumps into a cumulant sequence $(\kappa_1,\kappa_2,\kappa_3,\ldots)$, then $2(\kappa_1,\kappa_2,\kappa_3,\ldots),3(\kappa_1,\kappa_2,\kappa_3,\ldots),\ldots$ are also cumulant sequences.
For infinitely divisible distributions, for every real $t\ge 0$, the sequence $t(\kappa_1,\kappa_2,\kappa_3,\ldots)$ is a cumulant sequence.
Besides the nonnegative integers and the nonnegative reals, there are other sets of nonnegative reals closed under addition.
For which sets $T$ of nonnegative reals that are closed under addition is it the case that for some cumulant sequence $(\kappa_1,\kappa_2,\kappa_3,\ldots)$, for every real $t\ge 0$,
$$
t\in T \quad\text{iff}\quad t(\kappa_1,\kappa_2,\kappa_3,\ldots)\text{ is a cumulant sequence ?}
$$
I'm guessing only closed sets, but there should be more than that to say about it, I would think.  Even describing or classifying the (topologically) closed subsets of $\mathbb{R}^+$ that are closed under addition might be something of a chore, and I suspect many of them are not of this form.
That moderately complicated things might happen is at least hinted at by the case of $p\times p$ Wishart matrices, for which the number of degrees of freedom can be anywhere in the set $\lbrace 0,\dots,p-1\rbrace\cup (p-1,\infty)$.  But that's for matrix-valued, rather than real-valued, random variables, so it's at most a hint.

REPLY [5 votes]: This quote might explain why your question on MSE went unanswered:

There is one obstacle that stands in the way of dealing with
  cumulants. We do not know simple necessary and sufficient inequalities
  on cumulants that ensure that a given sequence of real numbers is the
  sequence of cumulants of a random variable. For characteristic
  functions, such a necessary and sufficient condition is Bochner's
  theorem. To be sure, such inequalities exist in abundance if one
  expresses moments in terms of cumulants. The problem is finding the
  simplest such inequalities. The explicit determination of fundamental
  inequalities satisfied by the cumulants of a random variable is surely
  the most important open problem in probability theory.

Gian-Carlo Rota, 1998: Twelve problems in probability no one likes to bring up - Problem eleven: cumulants.
A status update has been given by Di Nardo and Senato, 2009, unfortunately behind a pay wall.<|endoftext|>
TITLE: Length spectrum for Riemannian metrics in the projective plane
QUESTION [9 upvotes]: Are there (known) examples of non-isometric Riemannian metrics on the projective plane that have the same length spectrum?
This question is related to MO questions Length spectrum and Zoll surfaces of revolution and Length spectrum of spheres . There Zoll surfaces appear as counterexamples of the analogous question on $S^2$ (and spoil all the fun), so maybe one should concentrate on the projective plane where the only Zoll Riemannian metric is the the canonical metric. 
Remark. Notice that there are tons of very nice reversible Finsler Zoll metrics on the projective plane. Indeed, here is the Busemann recipe to cook up smooth reversible Finsler metrics on $RP^2$ such that all geodesics are projective lines:

Take a smooth strictly positive measure on the unit sphere in $\mathbb{R}^3$ that is invariant under the antipodal map.
If $x$ and $y$ are distinct, non-antipodal points, let $X$ and $Y$ denote the great circles obtained by intersecting the sphere with the subspaces orthogonal to $x$ and $y$.
The union of $X$ and $Y$ cuts the sphere into four connected components. 
Define the distance between $x$ and $y$ as the measure of the smallest of these components. 
Voilà, you have a metric on the sphere that being invariant under the antipodal map projects down to a metric on the projective plane. It is easy to see that projective lines are geodesics and not too hard to see that it is Finsler.

REPLY [5 votes]: The answer is positive; in fact any smooth manifold has two nonisometric metrics with conjugate geodesic flows. A construction is in C. Croke, B. Kleiner, Conjugacy and rigidity for manifolds with a parallel
vector field. J. Differential Geom. 39(1994), 659–680. 
The idea is quite simple: consider the followins two building blocks:   
the round one: it is the round  sphere without a small  ball around the north pole. 
the exotic  one: take  a Zoll metric of revolution 
 such that  it has  the standard metric near the north pole, and cut out a small  ball around the  north pole  (of  the same radius as in the round building block). 
Because the building blocks are isometric near boundary, if a manifold has a bump which is isometric to the round building block, then  one can replace it by the exotic building block: the result is a smooth manifold and one can show that its geodesic flow is conjugate to  the geodesic flow of the initial metric<|endoftext|>
TITLE: polynomials with minimal $L_\infty$ norm on multiple disjoint intervals
QUESTION [8 upvotes]: It is well-known that Chebyshev polynomials are the polynomials of minimal $L_\infty$ norm on [-1,1] with leading coefficient 1. But what if you want the minimal $L_\infty$ polynomial on two disjoint intervals $[-1,1]$ and $[a,b]$? 
An upper bound can be obtained by seeking the minimal $L_\infty$ norm polynomial over the "filled-in" interval $I = [\min(-1,a), \max(1,b)]$. This would be a translated and scaled Chebyshev polynomial. My question is, can you do significantly better than this? Intuitively, can you exploit the fact that there is empty space between $[-1,1]$ and $[a,b]$ where the polynomial is not required to have a small value? Or is this empty space essentially useless?
My first instinct was to try a product of scaled Chebyshevs which are small on $[-1,1]$ and $[a,b]$ respectively. However there is no growth control on one Chebyshev in the others' interval, so there is no guarantee that you are doing any better. Because of this I am pessimistic that one can do much better than the upper bound, but I would love to be proven wrong.
I am interested in this question because I am studying the convergence of Krylov subspace methods, where such approximations play an important role. I want to understand the convergence rate of conjugate gradients when there are multiple clusters of eigenvalues contained in different intervals, rather than just a single cluster.

REPLY [3 votes]: There exists a complete theory for minimal $L^\infty$ norm polynomials on two intervals.
It is due to N. I. Akhiezer. You can look at his books: Lectures on Approximation theory
or, another book, Elliptic functions.
There is also a nice survey paper on this problem for any number of intervals,
by Sodin and Yuditskii, Functions that deviate least from zero on closed subsets of the real axis. Algebra i Analiz 4 (1992), no. 2, 1--61.<|endoftext|>
TITLE: Can one prove vanishing of higher direct images fiber-wise?
QUESTION [6 upvotes]: Let $\pi:X\to Y$ be a  proper map of algebraic varieties (over $\mathbb C$) which is a bi-rational equivalence.
are the following statements equivalent?

The derived direct image of $O_X$ is $O_Y$.
For any $y \in Y$ we have $H^*(O_{\pi^{-1}(y)})= \mathbb C$

Remarks:

I do not assume that $X$ or $Y$ are smooth.
The fiber is considered scheme theoretically
By point one can mean a scheme theoretic point or a closed point or a geometric point. I think it is does not matter.
I also have 2 variations of the question:
a) What  happens  we consider  conditions  (1) and (2) only on the level of the zero cohomology?
b) What  happens  we consider conditions (1) and (2) only on the level of higher cohomologies?
Variation (b) make sense without the conditions on $\pi$. I need the answer only for the case I described (since I'm interested in rational singularities) but I'll be happy to know what happens in general

REPLY [3 votes]: Karl is correct. Here is an example which at least sheds some light, I hope. Embed the projective line with a large degree line bundle in $n$-space, so that the embedding is not linearly normal. Let $Y$ be the cone (polynomial ring in $n+1$ variables modulo the ideal of forms vanishing on the curve), so that $Y$ is not normal. Let $X$ be the blow up of the irrelevant maximal ideal. Then one can easily check that $X$ is smooth (a $\mathbb{P}^1$ bundle over $\mathbb{P}^1$), $\pi:X\to Y$ is birational with only one exceptional point, namely the vertex of $Y$ and the scheme theoretic inverse image of this point is just a $\mathbb{P}^1$. Thus condition b) is satisfied. But, $\pi_*O_X$ is the normalization of $O_Y$.<|endoftext|>
TITLE: On the joints problem in finite fields
QUESTION [6 upvotes]: The original version of the so-called "joints problem" consists of the following:
Let $L$ be a set of lines in $\mathbb{R}^{3}$. Determine the maximum number of "joints" determined by these lines, where by joint we understand a point of $\mathbb{R}^{3}$ lying on three lines from the set $L$ but which do not all lie in the same plane (i.e. they are non-coplanar).
The conjectured answer by Sharir was that this number of joints is $\leq C |L|^{\frac{3}{2}}$, for some positive constant $C$; this was proven by Guth and Katz using a rather simple polynomial mathod which easily generalies to $\mathbb{R}^{d}$ in which case the upper bound becomes $C |L|^{\frac{d}{d-1}}$. (references can be found very easily on google; for example, see http://www.dagstuhl.de/Materials/Files/09/09111/09111.SharirMicha.Other.pdf)
Now, it seems to me that this polynomial method does not generalize to finite fields; so, my question is, can we get some upper bound in this case? My thoughts for now are to use the graph $G$ having as vertices the lines and to connected them if the lines intersect. Then the number of joints is the number of triangles of $G$ minus $\frac{1}{2}\left(\binom{|L|}{2} - k\right)$, where $k$ is the number of distinct planes determined by the $|L|$ lines... but I feel that I'll be getting really weak bounds if I majorize this (using graph theoretic stuff about the number of triangles and Beck's theorem or related things for $k$).
So, any other ideas or knowledge about the finite field case in literature?
Thanks.

REPLY [3 votes]: I know I'm digging up an old thread, but I figured out how to extend Kaplan, Sharir and Shustin's proof (similar to Quilodran's proof) to finite fields last summer, then later realized that Dvir indicates the proof in a set of lecture notes.
Anyway, the trick is this: if $Q$ is a polynomial over $\mathbb{F}_q[x_1,...,x_n]$ whose gradient vanishes identically, then $Q$ is the $p$th power of some other polynomial $Q_1$ (where $q$ is a power of $p$). Since $Q_1$ is zero if and only if $Q_1^p$ is zero, if we're assuming that $Q$ is the minimum degree polynomial that vanishes on the set of lines forming our joints, we get a contradiction.
Dvir mentions how to do this in his lecture notes: http://www.cs.princeton.edu/~zdvir/teaching/incidence12/6.%20The%20polynomial%20method.pdf
I couldn't get Hasse derivatives to work, but it might be possible. I forgot what the trouble was with them.<|endoftext|>
TITLE: Are Banach Manifolds intrinsically interesting?
QUESTION [21 upvotes]: In the introduction to 'A convenient setting for Global Analysis', Michor & Kriegl make this claim: "The study of Banach manifolds per se is not very interesting, since they turnout to be open subsets of the modeling space for many modeling spaces."
But finite-dimensional manifolds are found to be interesting even though they can be embedded in some Euclidean space (of larger dimension). (Actually this seems to me, to make the above claim intuitively plausible, so that claim should be no more than we should expect).
But they do go on to say that "Banach manifolds are not suitable for many questions of Global Analysis, as ... a Banach Lie group acting effectively on a finite dimensional smooth manifold it must be finite dimensional itself.", which does seem a rather strong limitation.

REPLY [9 votes]: I am of two minds on this topic. It is much easier to work on  Banach manifolds because the implicit function theorem on such spaces    has a simple formulation.  On the other hand,   as the examples of gauge theory or    the theory of pseudo-holomorphic curves show, in these contexts one works   not with one Banach manifold, but with several, determined by    stronger and stronger Sobolev  norms. One  important  part of the game  is to   conclude that objects with a priori weaker Sobolev regularity are   in fact smooth.  This feels   very much like we are implicitly working on a    Frechet manifold.
One draw back of Banach spaces is that they do not have  many smooth functions on them, and the notion of  real analycity on such spaces is problematic.          Let's take the example of Seiberg-Witten equations.    These are quadratic equations  in its variables, so intuitively they ought to be real analytic, though I do not know how to formulate this  rigorously in a Sobolev context.
Why do I care  about real analycity? In the real analytic  context one can formulate an intersection theory involving not necessarily smooth objects. For example, the  point $0\in\mathbb{R}$ is a solution  of the quadratic  equation $x^2=0$. It is a degenerate zero, and from the point of view  of intersection theory it   has multiplicity $0$.  My hope  is  that this real analytic point of view would allow one to deal with mildly degenerate solutions of the the Seiberg-Witten equations,  and assign multiplicities   to such solutions.<|endoftext|>
TITLE: Proving the Gauss-Bonnet theorem for embedded surfaces using triangulations
QUESTION [11 upvotes]: I heard this really neat elementary proof of the "Gauss-Bonnet Theorem" :
Let $S$ be a surface embedded in $\mathbb{R}^3$. Now take a triangulation of that surface, and approximate the surface with a triangular mesh ("flatten" the triangles).
Call $\delta(v)$ the angle defect at a vertex $v$, that is $2\pi$ minus the sum of the angles around the vertex $v$. Now, the Euler characteristic is defined at $\chi(S) = V - E + F$. Also, if we assume that the surface is without boundary then every face has $3$ bounding edges and every edge bounds $2$ faces, which means $2E=3F$, so $F=2(E-F)$.
Let's compute the sum of all the angle defects at all the vertices of the mesh in two different ways. First, sum the angle defect at each vertex, that is $\sum_{v} \delta(v)$. Second, for each triangle the sum of the angles is exactly $\pi$ since they are euclidean triangles in $\mathbb{R}^3$, so we can sum of the angles of all the faces and get $F\pi$, and subtract that from what the total angle sum would be if the surface were flat to get $2 \pi V - F \pi$. Using the formula for the Euler characteristic, we get
$\sum_{v} \delta(v) = 2 \pi V - \pi F$
$\sum_{v} \delta(v) = 2\pi (V - E + F) = 2\pi\chi(S)$
Now, to make that into the "real" Gauss-Bonnet Theorem I heard the argument "take an infinitesimal triangulation, then the angle defect becomes the curvature". Is there any way to make this rigorous with a reasonable amount of not too advanced tools? If so, can I get a reference?
Thanks!

REPLY [2 votes]: You only need to take a geodesic traingulation, i.e., a triangulation such that the edges are geodesic arcs.  Then you can use  a  local version of the  Gauss-Bonnet theorem that says that the integral of the curvature  over such a triangle  equals the     angular defect, defined as   
$$ 2\pi - \mbox{sum  of the exterior angles at the vertices}. $$
Then you proceed following your outline. For more details, see my favorite source on   curves and surfaces

Dirk J. Struik, Lectures on Classical Differential Geometry, 2nd Edition, Dover, 1988

The proof that I have just outlined can be found in Sec. 4.8  in the above reference. On page 157 of the same book you will also find a description of the curvature  as  an "angular defect per unit of area".  This description follows by  considering a   sequence of geodesic triangles that concentrates around a point.<|endoftext|>
TITLE: The Rabinowitz Trick
QUESTION [25 upvotes]: The recent question about problems which are solved by generalizations got me thinking about the Rabinowitz trick, which is used to prove a statement of Hilbert's Nullstellensatz, specifically, the inclusion of the ideal generated by an affine variety $V(J)$ over an algebraically closed field into the radical of $J.$
Let $0\neq f\in J,$ as above. In the course of the proof, one extends the given polynomial ring by a single indeterminate  and writes its elements as, 
$$\sum_{i=1}^l h_ig_i + h(X_n\cdot f - 1),$$
where $h_i,h\in k[X_1,\dots,X_{n+1}]$ and $g_i\in k[X_1,\dots,X_n].$
One then applies the weak Nullstellensatz, to see that, indeed, every element of $k[X_1,\dots,X_{n+1}]$ can be written in the above form. Then, mapping back to the smaller polynomial ring, via $X_{n+1} \mapsto \frac{1}{f}$ yields the result, by simply clearing denominators. 
My question is this: While the trick uses some exceedingly clever algebra, does it have some sort of deeper geometric meaning? Why does it make sense to try this in the first place?

REPLY [46 votes]: Perhaps the "Rabinowitz trick" is more clear if one writes down the proof backwards in the following way:
Let $I \subseteq k[x_1,\dotsc,x_n]$ be an ideal and $f \in I(V(I))$, we want to prove $f \in \mathrm{rad}(I)$. In other words, we want to prove that $f$ is nilpotent in $k[x_1,\dotsc,x_n]/I$, or in other words, that the localization $(k[x_1,\dotsc,x_n]/I)_f$ vanishes. By general nonsense this algebra is isomorphic to $k[x_1,\dotsc,x_n,y]/(I,fy-1)$. But, clearly $V(I,fy-1)=\emptyset$ and therefore the Weak Nullstellensatz implies that $(I,fy-1)=(1)$, i.e. that the quotient vanishes.<|endoftext|>
TITLE: whether a kind of surgery can go on infinitely many steps?
QUESTION [5 upvotes]: let $M$ be a closed ortientable irreducible 3-mfd, let $T$ be a non-separating torus in $M$, we cut $M$ along $T$ and glue two solid tori along the two boundary tori, we get a new closed 3-mfd $M_1$ (we need $M_1$ also is irreducible). Now If there exists nonseperable torus $T_1$ in $M_1$, we go on the above process, we get a new closed 3-mfd $M_2$ ...
My question is, whether you can find a $M$, choose suitable $T_i$, glue solid tori suitablely, the process will go infinitely?
or can you prove that it is impossible to find such an example? (for example, from $M$---> $M_1$, some invariant of 3-mfds decrease strictly).

After Kevin's example, I added the condition "$M_1$ also is irreducible". This condition is natural
in the original field (for this question): 3-mfd with Anosov flow.

REPLY [8 votes]: The answer is "no", although it seems that a homological argument is not enough as Kevin's and Bin's examples show. I describe an argument which uses geometrization.
There is a quantity which decreases strictly at each operation. It is crucial to suppose that both $M$ and $M_1$ are irreducible. The quantity for an irreducible manifold $M$ is the triple $(\|M\|, n(M), s(M))$ of real numbers, where 

$\|M\|$ is Gromov's norm (i.e. the sum of the volumes of the hyperbolic pieces in the JSJ decomposition of $M$)
 $n(M)$ is the number of tori in the JSJ decomposition of $M$
 $s(M)$ is the sum of the $-\chi(S)$ for each Seifert piece of the decomposition with some base surface $S$.
Triples $(\|M\|, n(M), s(M))$ are ordered lexicographically. I prove below that the surgery you describe (cut along a non-separating torus and glue two solid tori) indeed strictly decreases this quantity, supposing that the resulting manifold $M_1$ is still irreducible (this hypothesis is important). The result then follows because Gromov norms of 3-manifolds form a well-ordered set.
Let $T$ be the torus you cut. If it is adjacent to at least one hyerbolic piece, the filled manifold $M_2$ has strictly smaller Gromov norm thanks to Thurston's Dehn filling theorem. It remains to consider the case $T$ is adjacent to two (possibly coinciding) Seifert pieces and Gromov norm does not decrease. If $T$ is a torus of the JSJ then $n(M)$ decreases. If $T$ is a torus inside a Seifert piece, then $s(M)$ decreases. You use here the following fact: the hypothesis that $M_2$ is irreducible ensures that the meridian of your Dehn filled tori are not fiber-parallel, and hence only add some (possibly non-singular) fiber at the adjacent Seifert pieces. Therefore the JSJ of the new manifold is easily controlled by the JSJ of the old manifold.
It is possible that this argument extends to the relaxed case where you only suppose $T$ to be incompressible (and $M$, $M_1$ are any manifolds).<|endoftext|>
TITLE: Whitehead lemmas in Lie algebra cohomology for non-algebraically closed fields
QUESTION [7 upvotes]: I read in Weibel's homological algebra that Whitehead's first and second lemmas are true for any characteristic 0 field. I mean the following: 
Whitehead Lemma(s): Let g be a semisimple Lie algebra over a characteristic 0 field, V any finite-dimensional representation of g, then H^1(g, V) = H^2(g, V) = 0. 
The proof uses
Theorem 7.8.9 : H^i(g, M) = 0 for any nontrivial irreducible representation M of g if g is semisimple over char 0 field.
The proof of this uses the Casimir operator. The Casimir operator is said to act by a scalar, but doesn't this assertion use Schur's lemma? Schur's lemma requires the field to be algebraically closed. 
Whitehead's first lemma implies the complete reducibility of finite-dimensional g-modules, but isn't complete reducibility only true for semisimple Lie algebras over an algebraically closed field of char 0?
Whitehead's second lemma implies the Levi decomposition. Does Levi decomposition require the field to be algebraically closed?
Does anyone know where I can find a proof of the Whitehead Lemma(s) (assuming it is true) when the field is not assumed to be algebraically closed?

REPLY [7 votes]: There is no problem about the Whitehead lemmas over an arbitrary field of characteristic 0  in the context of (1) complete reducibility of finite dimensional representations of a semisimple Lie algebra, (2) existence of a Levi decomposition of a finite dimensional Lie algebra.   Besides the modern treatment by Weibel, there are various older treatments such as Bourbaki, Chapter 1 of Groupes et algebres de Lie (1960) and Jacobson, Sections III.7, III.9 of Lie Algebras (1962) followed by III.10 on Lie algebra cohomology.    Along the way, one can make use of a Casimir element in the universal enveloping algebra: this is a commuting operator in the representations studied, which is what counts.   (Of course, when the field is large enough and the representation is irreducible, Schur's Lemma provides a scalar, but this is a corollary of the general Schur formulation.) 
For a treatment emphasizing highest weight representations, such as I gave in my short graduate text, it's not satisfactory to work over such an arbitrary field.   At the least, one needs a splitting field for a semisimple Lie algebra; my convention, not needed for some of the generalities, was that the field be algebraically closed.   So I may be partly responsible for confusing the issue of just how large the field needs to be at each step of the theory.
In particular, complete reducibility does not require that a Casimir element operate on a representation space by a scalar.   However, irreducible representations over arbitrary fields of characteristic 0 might be larger than those over a splitting field.   The Cartan-Weyl theory is best understood over a large enough field, where eigenvalues are available when needed. 
P.S. I've just tracked down my copy of Weibel.   Concerning his treatment of these topics, I've never gone through this in detail and find it difficult to sort out his proof of the Whitehead lemmas without going back into the earlier part of his book.   He does want the Casimir element to act by scalars when needed even though the field is arbitrary of characteristic 0, but this argument doesn't seem to enter into the other more self-contained treatments mentioned above.   Anyway, it's worth looking at Jacobson's Chapter III.<|endoftext|>
TITLE: How to compute homology of symmetric products of complexes?
QUESTION [11 upvotes]: First, I would appreciate references on the notion of derived symmetric powers of perfect modules over various kinds of derived commutative algebras (say cdgas in characteristic zero, simplicial commutative rings, or $E_{\infty}$-rings). 
Here is a very special case of the kind of computation one would like to do:
Let $R$ be a (discrete) commutative $k$-algebra over a field $k$ of characteristic zero, and let $M^{*}=M^{i} \rightarrow M^{i+1}$ be a two-term complex of free or projective $R$-modules. There is a notion of symmetric power, $Sym^{p}(M^{*})$, which people say is just the quotient of ${M^{*}}^{\otimes p}$ by the action of the symmetric group $\Sigma_{p}$. I take this to mean that we should take the quotient of the subcomplex generated by the images of $Id -\sigma$ for all $\sigma \in \Sigma_{p}$. (Is that right?) (Here, tensor products are taken over $R$, not $k$.)
QUESTION 1: How is the homology of $Sym^{p}M^{*}$ related to the homology of $M^{*}$?
QUESTION 2: The symmetric product of classical modules commutes with base change. Is that true for these derived versions?
I'm assuming characteristic zero, since otherwise I gather homology of the symmetric group causes interesting complications. Comments on such complications are also welcome.

REPLY [5 votes]: Building on the results established by Tom and Tyler, it seems to me that the structure of  $H_\ast(Sym^n M)$ can be further explored: 
Since the characteristic of $k$ is assumed to be zero, $H_\ast(Sym^n M) = H_\ast(M^{\otimes n})_{\Sigma_n}$ by Tom's and Tyler's answer and by Künneth formula 
$H_\ast(M^{\otimes n}) = H_\ast(M)^{\otimes n}$. Hence as an intermediate result
$$H_\ast(Sym^n M) = H_\ast(M)^{\otimes n}_{\Sigma_n}$$
Consider $H_\ast(M)$ as a graded $k$-vector space and choose a basis $\lbrace x_j \mid j \in J\rbrace$ where $J$ is assumed to be totally ordered (denote the order by $\le$). Then the elements $x_{j_1} \otimes \cdots \otimes x_{j_n}=: \underline{x}_j$, $\;j = (j_1,...,j_n) \in J^n$ form a basis of $H_\ast(M)^{\otimes n}$ and the action of $\Sigma_n$ is given by 
$$\sigma^{-1} \cdot x_{j_1} \otimes \cdots \otimes x_{j_n} = x_{i_1} \otimes \cdots \otimes x_{i_n}$$
where 
$$(i_1,...,i_n) = (\sigma(j_1),...,\sigma(j_n))=: \sigma^{-1} \cdot (j_1,...,j_n).$$
The latter defines an action of $\Sigma_n$ on $J^n$ as well. We can subdivide $J^n$ into those tuples having exactly $l$ equal elements $(l= 1,...,n)$. This leads to the following set of representatives $E = J^n/\Sigma_n = \coprod_{l=1}^n\; E_l$ where 
$$ E_1 = \lbrace (j_1,...,j_n) \mid j_1 < ... < j_n\rbrace $$ 
$$ E_l = \lbrace (j,...,j,j_1,...,j_{n-l}) \mid j_1 < ... < j_{n-l},\;\; j \neq j_i\; \forall i\rbrace\quad(2 \le l \le n)$$
The stabilizer of $j \in E_l$ is $\Sigma_l \le \Sigma_n$. Now by Corollary III.5.4 in Brown, Cohomology of Groups: 
$$H_\ast(M)^{\otimes n} = \bigoplus_{l=0}^n \;\bigoplus_{j \in E_l}\; \operatorname{Ind}^{\Sigma_n}_{Stab(j)}k\cdot  \underline{x}_j$$ 
$$\hspace{60pt} = \bigoplus_{l=0}^n \;\bigoplus_{j \in E_l}\; k[\Sigma_n] \otimes_{k[\Sigma_l]}\;k\cdot  \underline{x}_j$$
Hence by Brown, II (2.1): 
$\qquad H_\ast(M)^{\otimes n}_{\Sigma_n} = k \otimes_{k[\Sigma_n]} H_\ast(M)^{\otimes n}$ $= \bigoplus_{l=0}^n \;\bigoplus_{j \in E_l}\; k \otimes_{k[\Sigma_l]} \;k\cdot  \underline{x}_j$ 
$\hspace{60pt}= \bigoplus_{l=0}^n \;\bigoplus_{j \in E_l}\; k\cdot  \underline{x}_j$ 
So as final result we keep hold: 

$\qquad\qquad H_\ast(Sym^n M) = \bigoplus_{j \in E}\; k\cdot  \underline{x}_j\quad, \qquad E= J^n/\Sigma_n$

In particular, if $m := \dim_k H_\ast(M) < \infty$ then 

$\dim_k H_\ast(Sym^n M) = |E| = \binom{m}{n} + \sum_{l=0}^{n-2} \;\binom{m-1}{l}\;m$

For example, if $n > m$ then $\dim_k H_\ast(Sym^n M) = 2^{m-1}\;m$. 
Remark: The question requests that there is at most one $r$ such that $N_1 := H_r(M), N_2 := H_{r+1}(M) \neq 0$. Using a (more or less obvious) result of Dold: Homology of Symmetric Products and other Functors of Complexes (1958), (8.4), the structure of $H_\ast(M)^{\otimes n}_{\Sigma_n}$ can be refined as 
$$H_\ast(M)^{\otimes n}_{\Sigma_n} = \bigoplus_{p+q=n} (N_1)_{\Sigma_p} \otimes_k  (N_2) _{\Sigma_q}$$ 
Then each $(N_i)_{\Sigma_p}$ can be separately analyzed as before.<|endoftext|>
TITLE: Reorienting a ladder among $\mathbb{Z}^2$ poles
QUESTION [10 upvotes]: Imagine an object, which I'll call a ladder $\cal{L}$, a "racetrack" shape
composed of a rectangle of length $L$ capped at either end by
semicircles of radius $r$; so it is $L+2r$ tip-to-tip.
View every lattice point of $\mathbb{Z}^2$ as a point obstacle
(a pole).  The ladder is forbidden to include a pole in its
interior, but $\cal{L}$ may touch poles on its boundary.
$\cal{L}$ is initially sitting in the plane in a horizontal orientation.
          


My question is:

Q. 
  Under what conditions on $L \ge 0$ and $r \le \frac{1}{2}$can $\cal{L}$ be reoriented
  via continuous motions to a vertical orientation?

Throughout the motions, no pole may be interior to $\cal{L}$.
If $L=0$, $\cal{L}$ is a disk of radius $r$;
if $r=0$, $\cal{L}$ is a segment of length $L$.
I believe that, for $r=0$, there is no upper bound on $L$: an arbitrarily
long segment can be reoriented.
(I made this an exercise (7.3) in a
textbook.)
And certainly when $r=\frac{1}{2}$, and, say $L=1$, reorientation
is not possible.
I can see that, for $r=\frac{1}{2}$,
any $L \le \sqrt{2}-1$ can be reoriented
(because then $\cal{L}$ fits inside a circle of radius $\sqrt{2}/2$).
Almost everything else is unclear to me.
                    


An analogous question may be posed for an appropriately defined "ladder"
amidst $\mathbb{Z}^d$ obstacles.
Any insights, even for specific $(L,r)$ values, or corrections
to my beliefs above, are welcomed.  Thanks!

REPLY [10 votes]: At the EuroCG conference that ended as I post this, Sándor Fekete (Braunschweig)
solved this question modulo a few details, with useful input from Günter Rote (Berlin). 
Sándor's crucial observation is that, when the ladder
is in an extreme configuration, it is touching three lattice points, which form an empty lattice
triangle, which by Pick's Theorem, has area $\frac{1}{2}$, so the $L \times r$ rectangle
has area 1, and so $r \sim 1/L$.  There are many details to convert this to a formal proof,
but it seems this insight yields the right dependency between $r$ and $L$ to allow reorientability.
Curiously, at the conference a banquet bus was forced to execute a maneuver not dissimilar
from the 15-move back-&-forth path illustrated in my first figure! :-)<|endoftext|>
TITLE: Metric associated to a Connection on a Vector Bundle
QUESTION [6 upvotes]: General question: Given a vector bundle $E \rightarrow M$ on a complex manifold $M$, and a connection $\nabla$ on $E$, is it possible to find an Hermitian structure on $E$ such that $\nabla$ is the associated metric connection (i.e. the unique connection compatible with both the metric and the complex structure)?
Specific question: Given a line bundle $L \rightarrow X$ on a compact Riemann surface $X$ equipped with a flat connection $\nabla$, is it possible to find an Hermitian structure on $L$ such that $\nabla$ is the associated metric connection? 
Motivation: I´m trying to prove that a degree zero line bundle on a compact Riemann surface always admits an harmonic Hermitian metric. By a classical result of Weil and Atiyah every degree zero vector bundle admits a flat connection, moreover I think (though I still did not prove it) that the flatness condition on the connection could be translated (by computation on the vector fields $\partial z$ and $\partial \overline{z}$) into the harmonicity condition on the "associated" (in the sense of the question) Hermitian metric.
The question is clearly related to the well discussed question: When can a Connection Induce a Riemannian Metric for which it is the Levi-Civita Connection. But I´m not able to adjust the proof given in the mentioned question to an answer for my own. I´m asking also a general version of the question because I´m just curious about the answer.
Thank you for your time!
Edit: As pointed out by David Speyer the metric connection is constructed taking as input an Hermitian structure on the bundle and not an Hermitian metric on the manifold. I changed both the questions consequently.

REPLY [6 votes]: alvarezpaiva's answer shows that the answer to your question is no in general.
I will answer your motivation question instead, for which the answer is positive.
If $L$ is a holomorphic line bundle over a compact Kähler manifold $(M^n,\omega)$
with $\omega$-degree zero, then there is a Hermitian metric on the fibers of $L$ with curvature $F$ and $\omega$-trace of the curvature equal to zero (what you call a harmonic Hermitian metric)
$$\omega^{n-1}\wedge F=0.$$
This is just the Hermitian-Yang-Mills equation, and I am saying that it can always be solved in the case of line bundles.
This follows from simple Hodge theory. Start with any Hermitian metric on $L$, and call $F$ its curvature $(1,1)$-form. Since $L$ has $\omega$-degree zero, you have that
$$\int_M \omega^{n-1}\wedge F=\int_M \Lambda F \omega^n=0.$$
Now solve the Poisson equation 
$$\Delta h=\Lambda F,$$
which can be done because of the integral of the RHS being zero.
It follows that the new Hermitian metric on $L$ that you obtain by conformally rescaling the one you have by $e^h$, has curvature
$$F_h=F-i\partial\overline{\partial}h$$
and by construction
$$\omega^{n-1}\wedge F_h=0,$$
which is what you want.<|endoftext|>
TITLE: Is a retract of a free object free?
QUESTION [17 upvotes]: I wonder whether this is true in the categories of groups, monoids, commutative algebras, associative algebras, Lie algebras?

REPLY [12 votes]: A few months after the last activity on this question, Neena Gupta gave a proof that over a field $k$ of positive characteristic, a retract of a polynomial algebra need not be a polynomial algebra: http://arxiv.org/abs/1208.0483.
In fact, she gives a counterexample to the cancellation problem: there is an algebra $A$ such that $A[t]$ is isomorphic to $k[x_1,x_2,x_3,x_4]$ but $A$ is not isomorphic to $k[y_1,y_2,y_3]$. Composing the isomorphism $k[x_1,x_2,x_3,x_4]\to A[t]$ with the evaluation map $A[t]\to A$ at $t=0$ expresses $A$ as a retract of $k[x_1,x_2,x_3,x_4]$.<|endoftext|>
TITLE: Is exponent of discrete-analytic function also discrete-analytic?
QUESTION [11 upvotes]: Lets define a discrete analytic function such a function that is equal to its Newton series:
$$f(x) = \sum_{k=0}^\infty \binom{x}k \Delta^k f\left (0\right)$$
Is function $g(x)=e^{f(x)}$ also discrete-analytic?
This question arose from the following considerations.
As you know the difference equation 
$$\Delta y(x) = F(x)$$ 
has multiple solutions that differ only by an arbitrary 1-periodic function $C(x)$:
$$y(x)=y_1(x)+C(x)$$
At the same time there can be no more than one (up to a constant term) discrete-analytic solution which we can consider to be the natural solution of the equation.
But when considering multiplicative-difference equation $\frac{y(x+1)}{y(x)}=F(x)$ we come to a similar situation, this equation has multiple solutions which differ by an arbitrary 1-periodic factor:
$$y(x)=C(x)y_1(x)$$
Of these solutions, similarly, no more than one (up to a constant factor) is discrete-analytic which allows us to define the distinguished solution.
But on the other hand the following rule holds for indefinite product and sum:
$$\prod_x f(x)= e^{\sum_x \ln f(x)}$$
This means that we can obtain the solution to the equation $\frac{y(x+1)}{y(x)}=F(x)$ in the following form:
$$y(x)=e^{\sum_x \ln F(x)}$$
This allows us to select the distinguished solution by another method, that is taking the natural solution to the sum and taking exponent of it. The result will have a constant factor, but it is unevident whether it will be discrete-analytic or not, and as such, whether the both distinguished solutions coincide.
UPDATE
Due to the answer by David Speyer it is evident now that counter-examples exist among complex-valued functions and also there are instances when function $f(x)$ is discrete-analytic while the Newton series of its exponent does not converge.
So the question should be formulated more precisely: we assume that $f(x)$ is real-valued and Newton series for its exponent converges.
I started a bounty for this question
ADDENDUM
It would be even more great if somebody could prove a more general theorem about a composition of monotonous discrete-analytic functions. Whether the composition is also discrete-analytic and under what conditions.

REPLY [2 votes]: No. Take $f(x) = 2 \pi i x$.

There is also a more subtle way I can cheaply answer the question: It is easy to give functions $f(x)$ such that the Newton series of $f$ converges to $f$, but the Newton series of $e^f$ diverges. Take $f(x)=x^2$ or, more subtly, $f(x) = \cos (2 \pi x/8)$. I assume that the right formulation of the question is "If $f(x)$ is real, the Newton series of $f$ converges to $f$, and the Newton series of $e^f$ converges, does the Newton series of $e^f$ converge to $e^f$?<|endoftext|>
TITLE: Order of vanishing at the cusps for the modular theta function
QUESTION [8 upvotes]: I am trying to examine the behavior of the theta function $\theta(z)=\sum_{n\in\mathbb{Z}} e^{2\pi i n^2 z}$, which is modular for $\Gamma_0(4)$ of weight 1/2, at the cusps 0 and 1/2.  My calculations seem to show that it vanishes at least at one of these cusps.  I would like to calculate the order of vanishing.
Apostol discusses this theta function in some detail in his second number theory book.  He gives a transformation formula as $\theta(-1/z)=(-iz)^{1/2}\theta(z)$.  This transformational formula doesn't make clear what the function's behavior is at the cusp at $\infty$.
Koblitz talks some about the order's of vanishing at the 0 cusp, but doesn't mention the order of vanishing at other cusps.
Does anyone have a good reference for this problem?
I've also tried looking for the valence formula for $\Gamma_0(4)$ to help with the calculation, but couldn't find it written down anywhere.  A source for this would be helpful as well.

REPLY [15 votes]: [Edited for consistency with the normalization
$\theta(z) = \sum_{n=-\infty}^\infty e^{\pi i n^2 z}$ of the proposer
(and of my own article!);
expansions in powers of $e^{2\pi i z}$ are more common,
but for this modular form $q = e^{\pi i z}$ does seem to be the better choice.]
The function
$$
\theta(z) = \sum_{n=-\infty}^\infty q^{n^2}
$$
(where $q = e^{\pi i z}$) has a product formula converging in $|q|<1$,
and thus for all $z$ in the upper half-plane:
$$
\theta(z) = \frac{\eta(z)^5}{\bigl(\eta(z/2)\eta(2z)\bigr)^2}
= \frac{1+q}{1-q} \cdot
 \frac{1-q^2}{1+q^2} \cdot
 \frac{1+q^3}{1-q^3} \cdot
 \frac{1-q^4}{1+q^4} \cdot
 \frac{1+q^5}{1-q^5} \cdot
 \frac{1-q^6}{1+q^6} \cdots
$$
(where
  $\eta(z) = e^{\pi i z/12} \prod_{m=1}^\infty (1-e^{2\pi i m z})$
as usual).
It follows from this formula that $\theta$ cannot vanish except at a cusp:
as long as $|q|<1$, the factors converge quickly to $1$ and none of them
has a zero or pole.
But $\theta$ is a modular form of positive weight, so it must vanish
somewhere.  It is clear from the sum formula that $\theta \neq 0$
as $q \rightarrow 1$ and $q \rightarrow 0$, so $\theta$ does not
vanish at the cusps $z=0$ and $z=\infty$.  Hence it must vanish
at the remaining cusp of $\Gamma(2)$.  Indeed each factor
$(1 \pm q^n) / (1 \mp q^n)$ vanishes at $q = -1$
(and stays in $(0,1)$ for $-1 < q < 0$),
so $\theta$ must vanish at the corresponding cusp $z=1$.
The author of the question also asked for a source for "a good reference
for this problem".  It's not clear what level of exposition is most
appropriate here.  Perhaps 
this paper might be of use, since I needed to use the vanishing order
of $\theta$ in a context where I could not assume the reader had any
background in modular forms:

Elkies N.D.: A characterization of the ${\bf Z}^n$ lattice,
  Math. Research Letters 2 (1995), 321-326 (arxiv:math.NT/9906019).<|endoftext|>
TITLE: Applications of Zariski topology outside alg. geometry 
QUESTION [14 upvotes]: Are there applications of the Zariski topology in mathematics that are not within the scope of algebraic geometry (including schemes and algebraic groups) ? 
There is an older question with a similar title (What is the Zariski topology good/bad for? ) but the answers given there are mainly concerned with the geometry stuff.

REPLY [3 votes]: (In some sense, there is some overlap with Ralph's answer)
Gelfand Naimark theorem.
For a commutative $C^\star$ algebra $A$, the spectrum of $A$ is the set of primitive ideals (=kernel of functionals). With the Zariski topology ($C^\star$ algebraist prefer the notion Jacobson topology/hull-kernel topology), they become a topological space $X$ and we have $C_0(X) \cong A$. This yields an anti equivalence between locally compact Hausdorff spaces with commutative $C^\star$ algebras. This equivalence generalizes to so called to sober spaces, where the dual objects are complete Heyting algebras. So from this experience, it seems natural to topologize the dual of an algebra and see how much is encoded.
Pontryagin duality:
The Gelfand Naimark theorem can be enhanced to the Pontryagin duality of locally compact abelian groups.
Note that the Gelfand Naimark theorem was first, and probably inspired some of the constructions in algebraic geometry. Similar things are happening with spectral triples in Arakelov theory now, I guess.<|endoftext|>
TITLE: Is a semicontinuous real function  Borel measurable?
QUESTION [7 upvotes]: Let $f(x,u): [0,1]^2 \mapsto \mathbb{R}$ be a continuous
function.
[Q] Is $g(x) = \inf_{u\in [0,1]} f(x,u)$ always Borel measurable?
If not, can one find a counter-example?
Note that, for any $c$, 
we have
$$(x: g(x) < c) = \text{Proj}_x ((x,u): f(x,u) < c),$$
where $\text{Proj}_x$ is a projection operator to $x$-axis.
In the context of measurable selection theorem,
the projection of Borel set $((x,u): f(x,u) < c)$  of
$\mathbb{R}^2$ is
not necessarily a Borel set of $\mathbb{R}$.
But, I can not find a counter-example.
If there exists a proper counter-example, then it also
implies that a semicontinuous real function is not necessarily  Borel measurable.
Thanks.

REPLY [2 votes]: I think every (lower) semicontinuous function $f:X \to \mathbb{R}$ is Borel measurable, since you have the following characterization: for every $a \in \mathbb{R}$ the set
$$ f^{-1}((-\infty, a])$$
is closed in the topology that you are considering in $X$.
Since you only have to check the measurability property for a generating class of the Borelians in $\mathbb{R}$ you are done.<|endoftext|>
TITLE: example of special lagrangian submanifold
QUESTION [5 upvotes]: are there any examples of a real analytic riemannian manifold that cannot be isometrically embedded as a special lagrangian submanifold of a calabi-yau manifold ?
peter hara

REPLY [7 votes]: If the question is "Are there examples of compact real-analytic Riemannian manifolds that cannot be isometrically embedded as a special Lagrangian submanifold of a compact Calabi-Yau manifold?", then the answer is "yes".
If the question is "Are there known, explicit examples of compact real-analytic Riemannian manifolds that cannot be isometrically embedded as a special Lagrangian submanifold of a compact Calabi-Yau manifold?", then the answer is "probably".
If the question is "Are there known, explicit examples of compact real-analytic Riemannian manifolds for which a proof is known that they cannot be isometrically embedded as a special Lagrangian submanifold of a compact Calabi-Yau manifold?", then the answer is "no" (to my knowledge).

For the first question, just note that, already for dimension 2, the space of compact Calabi-Yau surfaces is a finite-dimensional space, and the metrics that can be realized on compact complex curves in such a Calabi-Yau fall into a countable union of finite dimensional families.  (Remember that special Lagrangian surfaces in a Calabi-Yau are complex curves in a different Calabi-Yau metric in the canonical $S^2$-family of Calabi-Yau metrics.) Thus, the set of such realizable metrics, even on the $2$-sphere, constitutes a countable union of finite dimensional families.  This could never account for all of the real-analytic metrics on the $2$-sphere.  Thus, some example exists, though we don't know one explicitly.
For the second question, consider the fact that it is highly unlikely that the induced metric on any complex curve in a Calabi-Yau surface has constant Gaussian curvature.  The 'reason' is that most (non-flat) Ricci-flat Kahler metrics contain no complex curves with constant Gaussian curvature. It would be remarkable indeed if one of the Ricci-flat Kahler metrics on a (non-flat) compact 4-manifold had such a curve.  In particular, I regard it as highly likely that the standard round metric on the $2$-sphere cannot be isometrically embedded as a complex curve in any compact Calabi-Yau surface.
My answer to the third question is just an affirmation of my ignorance.
A remark about the local story:  peter h asked about what I would call the 'local case', i.e., whether a real analytic Riemannian manifold can be isometrically embedded as a special Lagrangian submanifold in some Calabi-Yau, with no assumptions about completeness of the ambient manifold.  In particular, he raised the question for surfaces.
Now, in the case of a real-analytic metric on a Riemann surface, the answer would be 'yes', according to a paper in 2000 by D. Kaledin, "Hyperkaehler structures on total spaces of holomorphic cotangent bundles", which is available on the arXive (arXiv:alg-geom/9710026v1).  (It's 100 pages, and I don't claim that I have read it, I'm just pointing out that it is there.)  The main theorem of this paper is that, given any real-analytic Kahler manifold $M$, there exists a hyperKahler metric on a neighborhood of the $0$-section of the cotangent bundle $T^\ast M$ that is compatible with the natural complex and holomorphic structures on $T^\ast M$ and that induces the original metric on the $0$-section.  
When the (real) dimension of $M$ is $2$, this would apply to show that $M$ is isometrically imbedded as a complex curve in a Calabi-Yau (complex) surface, and then one can apply the 'rotation trick' to turn this into a special Lagrangian surface when the ambient $4$-manifold is regarded as a complex surface with respect to one of the orthogonal complex structures.  Thus, the case of surfaces would be covered by this theorem.
In fact, this would work in any even dimension when the given real-analytic metric is actually Kahler.  
There would remain the question (which I raised in my original paper) of whether every real-analytic metric on $S^4$ can be realized by an embedding as a special Lagrangian submanifold of a $4$-dimensional Calabi-Yau.<|endoftext|>
TITLE: Set theories without "junk" theorems?
QUESTION [97 upvotes]: Clearly I first need to formally define what I mean by "junk" theorem.  In the usual construction of natural numbers in set theory, a side-effect of that construction is that we get such theorems as $2\in 3$, $4\subset 33$, $5 \cap 17 = 5$ and $1\in (1,3)$ but $3\notin (1,3)$ (as ordered pairs, in the usual presentation).  
Formally: Given an axiomatic theory T, and a model of the theory M in set theory, a true sentence $S$ in the language of set theory is a junk theorem if it does not express a true sentence in T.
Would it be correct to say that structural set theory is an attempt to get rid of such junk theorems?
EDIT: as was pointed out $5 \cap 17 = 5$ could be correctly interpreted in lattice theory as not being a junk theorem.  The issue I have is that (from a computer science perspective) this is not modular: one is confusing the concrete implementation (in terms of sets) with the abstract signature of the ADT (of lattices).  Mathematics is otherwise highly modular (that's what Functors, for example, capture really well), why not set theory too?

REPLY [4 votes]: The problems you mention occur as a result of two related reasons:

Objects such as the set of real numbers, which do not intrinsically belong to set theory, are 'encoded' as a set, so we can ask meaningless questions and obtain junk answers.
The encoding is not natural or canonical, and different encodings of the same object give rise to different sets of junk theorems.

It appears that Homotopy Type Theory addresses these issues:

Firstly, you cannot ask meaningless questions which involve treating a term of one type as though it were another type (such as treating an ordered pair as a set, or as a real).
Secondly, the equality type $A = B$ of two types is defined as the space of isomorphisms between those types, so isomorphic objects are equal (in the sense that the equality type is inhabited). This means that different encodings of the same object (such as the set of real numbers) correspond to equal types.

The first of these bullet points applies equally well to ordinary Martin-Löf type theory; the second relies on Voevodsky's powerful univalence axiom.<|endoftext|>
TITLE: Maximal dimension of abelian ideals of a Lie algebra and extensions of the ground field
QUESTION [5 upvotes]: For a Lie algebra $L$ of dimension $n$ over a field ${\mathbb F}$ we denote by $\beta(L)$ the maximal dimension of abelian ideals of $L$. In general, $\beta(L)$ is not preserved under extensions of the ground field (see e.g. Example 2.7 in http://homepage.univie.ac.at/dietrich.burde/papers/burde_39_max_ab.pdf). 
Do you know any example in which $\beta(L)<\beta(L\otimes_{\mathbb F} \bar{{\mathbb F}})=n-1$, where $\bar{\mathbb F}$ is the algebraic closure of ${\mathbb F}$? 
(In other words, is it possible that $L\otimes_{\mathbb F} \bar{{\mathbb F}}$ contains an abelian ideal of codimension 1 and $L$ has no abelian ideal of codimension 1?)
I am mainly interested in the case where $L$ is a restricted Lie algebra over a field of characteristic $p>0$.

REPLY [4 votes]: There's no such example.
Since this is convenient, I denote by $L$ the Lie algebra over the algebraic closure. Let $A$ be a codimension 1 abelian ideal and let us show that some (possibly other) abelian ideal $A'$ is defined over the ground field, i.e. is a hyperplane that can be defined by a linear equation with coefficients in $K$.
Since $L/A$ is abelian, we have $[L,L]$ contained in $A$. In particular, $A$ is contained in the centralizer of $[L,L]$. In case $A$ is equal to the centralizer of $[L,L]$, this is defined over the ground field and thus we are done. So I now assume that the centralizer of $[L,L]$ is all of $L$ (so $L$ is nilpotent of step 2).
The case when $L$ is abelian is trivial. If the derived subalgebra of $L$ is 1-dimensional, then the Lie algebra law can be viewed as an alternate form. Since $A$ is a codimension 1 isotropic subspace for this form, it is easy to check that the kernel of this alternate form has codimension 2 (and is defined over the ground field) and contains $[L,L]$ because $L$ is nilpotent of step 2. Every hyperplane $A'$ containing this kernel is an abelian ideal; we can pick it to be defined over the ground field.
If the derived subalgebra of $L$ is at least 2-dimensional, there exist two linear forms $f_1,f_2$ on $L$ such that the alternate bilinear forms $(x,y)\mapsto b_i(x,y)=f_i([x,y])$, $i=1,2$ are not proportional. They can be chosen to be defined over the ground field. Let $K_i$ be the kernel of $b_i$. Then $K_i$ is contained in $A$ (otherwise $b_i$ would be zero). Besides, $K_1$ and $K_2$ have codimension 2 (because $A$ is an isotropic subspace for $b_i$) and are not equal, because otherwise $b_1$ and $b_2$ would be alternate forms on the plane $L/K_1$ and would thus be proportional as the set of antisymmetric matrices of size 2 is 1-dimensional. So the codimension of $K_1+K_2$ is at most 1. Since it's contained in $A$, we deduce that $A=K_1+K_2$. So $A$ is defined over the ground field and the proof is complete.<|endoftext|>
TITLE: Grothendieck on Topological Vector Spaces
QUESTION [17 upvotes]: In a short biography article on Alexander Grothendieck, it is mentioned that after Grothendieck submitted his first thesis on Topological Vector Spaces (TVS), apparently, he told Bernard Malgrange that "There is nothing more to do, the subject is dead."
Also, after nearly two decades, while listing 12 topics of his interest, Grothendieck gave the least priority to Topological Tensor Products and Nuclear Spaces.
Now, the questions I have are:

What led Grothendieck to make this pronouncement on TVS?
Could somebody indicate some significant problems or contributions in this area after Grothendieck? My interest is not in the applications or the impact of the subject on other areas of mathematics, but I am interested in knowing about the growth of TVS theory itself.

Thank you, in advance, for your answer.

REPLY [2 votes]: An important contribution in this general area which came after Grothendieck's work is the development of probability theory on such spaces like $S'$, $D'$, etc. This started in the fifties with the articles of Itô, Gel'fand and Minlos but only reached a mature stage in the thesis by Fernique [1] during the sixties.
[1] Xavier Fernique, Processus linéaires, processus généralisés, Annales de l'institut Fourier, Volume 17 (1967) no. 1 , p. 1-92, doi:10.5802/aif.249<|endoftext|>
TITLE: Which closed orientable $4$-dimensional manifolds cannot be embedded in $6$-space?
QUESTION [14 upvotes]: This question is a follow-up to  my previous question . The statement of the question is the title. 
Note that the $4$-dimensional real projective space is non-orientable and a characteristic class argument gives that it does not embed in $7$-space. Right now, I am more interested in orientable $4$-manifolds.

REPLY [14 votes]: $\mathbb{CP}^2$ does not even immerse in $\mathbb{R}^6$. Proof: If such an immersion exists, then the normal Euler class has the property, that its square is the normal Pontrjagin class, and that is $-3$ times the signature (when evaluated on the fundamental homology class). But in $H^2(\mathbb{CP}^2)$ there is no such a class $x$, for which $x^2$ evaluated on the fundamental class is $-3$. QED.
Moreover the following theorem is true (It is essentially due to Hughs)
Theorem: In the 4-dimensional oriented cobordism group $\Omega_4 \cong \mathbb{Z}$ precisely the even elements contain a manifold that admits an immersion into $\mathbb{R}^6$.
About embeddings: The conditions (the manifold must be spin and have zero signature) are clearly necessary: An embedded manifold in a Euclidean space has zero normal Euler class. Hence in the present case both $p_1$ and $w_2$ are zero. The opposite is non-trivial, it is the content of Ruberman's paper mentioned above.<|endoftext|>
TITLE: Wanted: differential coming from higher genus surface in Heegaard Floer homology
QUESTION [7 upvotes]: I am interested in studying moduli of complex surfaces which arise in computing the differential on the Heegaard Floer homology chain complex.  In particular, I am interested in the generic case, when the holomorphic discs in $\operatorname{Sym}^g\Sigma$ are as "bad" as possible.  I have searched online for some examples of "generic"/"complicated" Whitney disks in the symmetric product, but most papers I see deal with special cases (namely where one can see by inspection and the Riemann mapping theorem that certain homotopy classes of Whitney disks are uniquely representable by holomorphic disks).
To elaborate on what I mean by "complicated" holomorphic disks, let's recall a certain (now standard) perspective on the Whitney disks.  If we have a holomorphic disk $\phi:\mathbb D^2\to\operatorname{Sym}^g\Sigma$, then we can consider the fiber product:
$$\begin{matrix}
S&\xrightarrow{\tilde\phi}&\Sigma\times\operatorname{Sym}^{g-1}\Sigma\cr
\downarrow& &\downarrow\cr
\mathbb D^2&\xrightarrow\phi &\operatorname{Sym}^g\Sigma
\end{matrix}$$
Then $S\to\mathbb D^2$ is a $g$-fold ramified cover (and the fiber over a point $p\in\mathbb D^2$ is "the $g$ points in $\Sigma$ given by $\phi(p)$").  Thus another way of viewing holomorphic disks in $\operatorname{Sym}^g\Sigma$ is as $g$-fold ramified maps $S\to\mathbb D^2$ along with a map $S\to\Sigma$.  Thus, even though we consider only disks mapping to $\operatorname{Sym}^g\Sigma$, moduli spaces of more complicated Riemann surfaces naturally come in to play in Heegaard Floer homology as moduli of the ramified cover $S$.
I am interested in the following "complicated" behaviour of $S$ and of the map $\tilde\phi:S\to\Sigma$:

Can the map $S\to\Sigma$ fail to be an immersion? (I believe the answer is yes; in fact I think I know where to look for more information on this, I just haven't followed up on it yet).

In general, what do the "slits" or "cuts" along the $\alpha$ and $\beta$ curves look like in $S$, and how does $S$ degenerate as these slits vary in length?  Do the slits ever interact with each other (e.g. by colliding), or do they all give "independent" degenerations of $S$? (this is a bit vague, but if someone has an enlightening example, I'd really like to see it)

Can $S$ have positive genus?  This is the question I'm really most interested in.  I have on good authority that the answer is yes, so actually what I really want is an example where $S$ has positive genus and contributes to the differential (or at least has Maslov index one).

REPLY [10 votes]: Yes, $S \to \Sigma$ can fail to be an immersion.  The failure is a branch point of the map $S \to \Sigma$, since it is (close to) a holomorphic map. This is fairly common as soon as the multiplicity of some region in $\Sigma$ gets higher than $1$.  Of course, in the simple examples you can actually compute, this tends not to happen.

A "slit" along the $\alpha$ or $\beta$ curves looks like a perfectly ordinary piece of the boundary upstairs in $S$, including at the end of the slit.  It's best to think about the end of the slit as a boundary branch point, looking locally like the map $z \to z^2$ restricted to the upper half-plane in the domain.  Slits cannot collide, due to boundary monotonicity: above each point on $\partial \mathbb{D}^2$, the $g$ different points map to distinct $\alpha$-curves.  I don't know what you mean by "independent" degenerations, sorry....

Yes, higher genus images can happen, and it's not too hard to construct examples that are forced by gluing, although again in most cases where you're able to compute Heegaard Floer homology by directly counting curves it does not.  The index formula in Corollary 4.3 of Lipshitz's paper "A cylindrical reformulation of Heegaard Floer homology" will let you easily construct examples of high genus surface with index 1, and a little more playing around should let you see that some of these must actually have representatives.<|endoftext|>
TITLE: What would be some major consequences of the inconsistency of ZFC?
QUESTION [23 upvotes]: Update (21st April, 2019). Removed the reference / initial trigger behind my question (please see comment thread below for the reasons). Am retaining, of course, the actual question, noted both in the title as well as in the post below.
The key questions of this post are the following:
1. How "disastrous" would inconsistency of ZFC really be? 
2. A slightly more refined question is: what would be the major consequences of different types of alleged inconsistencies in ZFC?

Old material (the "no-longer relevant" part of the question).
I was happily surfing the arXiv, when I was jolted by the following paper:
Inconsistency of the Zermelo-Fraenkel set theory with the axiom of choice and its effects on the computational complexity by M. Kim, Mar. 2012.

Abstract. This paper exposes a contradiction in the Zermelo-Fraenkel set theory with the axiom of choice (ZFC). While Godel's incompleteness theorems state that a consistent system cannot prove its consistency, they do not eliminate proofs using a stronger system or methods that are outside the scope of the system. The paper shows that the cardinalities of infinite sets are uncontrollable and contradictory. The paper then states that Peano arithmetic, or first-order arithmetic, is inconsistent if all of the axioms and axiom schema assumed in the ZFC system are taken as being true, showing that ZFC is inconsistent. The paper then exposes some consequences that are in the scope of the computational complexity theory.

Now this seems to be a very major claim, and I lack the background to be able to judge if the claim is true, or there is some subtle or even obvious defect in the paper's arguments. But picking on this paper itself is not the purpose of my question.
If you feel that my questions might not admit "clearly right" answers, I will be happy to make this post CW.

REPLY [4 votes]: (This began as a comment to Henry Cohn’s answer, which which I mostly agree, but I’m over the character limit.) I think the specifics of the example in case 4 actually overlap with 1; an easily-remedied technical inconsistency in the Peano Axioms mightn’t call into question the whole axiomatic approach to mathematics, but perhaps only some details of it, such as using first-order rather than higher-order logic.  To reverse your mention of Frege, in the nineteenth century, it wouldn’t have been that surprising if Frege’s higher-order arithmetic had been the only workable approach, and the first-order axiomatization failed.  (Boolos made some good arguments in this context about the value of higher-order logic.)  Indeed, the existence of non-standard models of the Peano Axioms proves that they don’t capture our intuition about the integers, so it might not be the end of formal mathematics if those particular axioms failed in the other direction as well.
    Let me note that I’m just addressing the hypothetical reaction to the inconsistency of PA, not the actual plausibility of it, and certainly not the extended consequences given, for instance, the available consistency proofs.  See, for instance, Harvey Friedman’s rejoinder to Angus Macintyre’s Vienna lecture http://www.cs.nyu.edu/pipermail/fom/2011-June/015572.html.  Angus was my undergraduate supervisor, but I don’t necessarily disagree with Friedman’s points.  (Though I do disagree with his capitalization of Angus’s surname, a frequent error :-)
    My answer to the original question would be analogous.  I don’t see any reason to believe in the inconsistency of ZFC, but the details of the axiomatization don’t exactly capture the underlying intuition:  The iterative conception of set only justifies Replacement for formulæ which quantify over already-constructed levels of the cumulative hierarchy.  Randall Holmes has argued that the theory justified by the iterative conception is actually Zermelo Set Theory with Σ2 replacement. (http: //math.boisestate.edu/~holmes/holmes/sigma1slides.ps.  According to Professor Holmes, “this contain[s] an error, which Kanamori pointed out to me and which I know how to fix.”)  So the reaction to an inconsistency in ZFC might be more like the adjustments for the Russell Paradox to Frege’s arithmetic than to his set theory—Frege’s development of arithmetic holds up remarkably well, with a few adjustments to his formalism:  see, e.g., John Burgess’s Fixing Frege and Richard G. Heck’s Reading Frege's Grundgesetze.<|endoftext|>
TITLE: Hochschild (co)homology of differential operators
QUESTION [16 upvotes]: I googled the title on the internet, and arrived at the following result
$$HH_k(D)\cong H_{DR}^{2n-k}(M).$$
Here $M$ is a smooth manifold of dimension $n$, and $D$ is the ring of differential operators on $M$. My first question is
(A): is there a known chain map realizing the above isomorphism?
I think the above isomorphism remains true if we replace $D$ by the ring of differential operators on $M$ with coefficients in a vector bundle $E$. But I am not so sure. If this is indeed the case, my second question is
(B): question (A) with coefficients in $E$?

REPLY [2 votes]: It is not literally what you want, but very close: check Proposition 2.3 and related results in "A Riemann-Roch-Hirzebruch formula for traces of differential operators" by Markus Engeli and Giovanni Felder (http://arxiv.org/abs/math/0702461).<|endoftext|>
TITLE: residually finite groups with the same finite quotients
QUESTION [8 upvotes]: Let $G , H$ be two finitely generated residually finite groups such that $F(G)=F(H)$. Where $F(G)$ denotes the isomorphism classes of finite quotients of $G$. Can we say that $G\cong H$?

REPLY [5 votes]: There are infinitely many metabelian groups with the same finite quotients, see Pickel, P. F.
Metabelian groups with the same finite quotients. 
Bull. Austral. Math. Soc. 11 (1974), 115–120.
On the other hand, for many relatively free groups, including free metabelian groups, the genus (i.e. the number of groups with the same finite quotients) is finite, see Gupta; Noskov, G. A.
On the genus of certain metabelian groups. Algebra Colloq. 5 (1998), no. 1, 49–66. 
See also Grunewald, Fritz; Zalesskii, Pavel;
Genus for groups. J. Algebra 326 (2011), 130–168.<|endoftext|>
TITLE: Which comma categories are topoi?
QUESTION [17 upvotes]: It is well known that for a topos C and an object x of C, the
slice category C/x is also a topos ("topos" here as "elementary topos").
Now there is the more general concept of a comma category (F,G), with
functors F:A->C, G:B->C. I believe that there are reasonable generalisations
of the above fact, but I can't find anything in the literature.
So my question: Which criterions are known for functors F, G, so that
the comma category (F,G) is a topos?

REPLY [15 votes]: The Artin gluing of a functor $G\colon B \to C$ is the comma category $(1_C \Downarrow G)$ (or in the notation of the question, $(1_C, G)$).  If $B$ and $C$ are toposes and $G$ preserves pullbacks, then its Artin gluing is also a topos.  
I learned this from:

Aurelio Carboni, Peter Johnstone, Connected limits, familial representability and Artin glueing.  Mathematical Structures in Computer Science 5 (1995), 441-459.  Corrigenda: Mathematical Structures in Computer Science 14 (2004), 185-187.

They attribute the result to:

Gavin Wraith, Artin glueing.  Journal of Pure and Applied Algebra 4 (1974), 345-348.

According to Carboni and Johnstone, Wraith proved it under the hypothesis that $G$ preserves all finite limits, but they add that it was observed very soon afterwards (by whom, they don't say) that it's enough to assume that $G$ preserves pullbacks.
(Incidentally, I don't know what that extra "e" in "glueing" is doing there.  My dictionary says that's wrong.)
The strengthened version of Wraith's result includes the famous result on slices, since if $B = 1$ then a pullback-preserving functor $B \to C$ is just an object $c$ of $C$, and its Arting gluing is then $C/c$.<|endoftext|>
TITLE: Computational complexity of unconstrained convex optimisation
QUESTION [6 upvotes]: What is known about the relationship between unconstrained convex optimisation and computational complexity? For example, for which optimisation problems and which gradient descent algorithms is one guaranteed to get to the optima in a polynomial number of iterations? How about getting to within some $\epsilon>0$ of the optima?

REPLY [7 votes]: Since we are dealing with real number computation, we cannot use the traditional Turing machine for complexity analysis. There will always be some $\epsilon$s lurking in there.
That said, when analyzing optimization algorithms, several approaches exist:

Counting the number of floating point operations 
Information based complexity (so-called oracle model)
Asymptotic local analysis  (analyzing rate of convergence near an optimum)

A very popular, and in fact very useful model is approach 2: information based complexity. This, is probably the closest to what you have in mind, and it starts with the pioneering work of Nemirovksii and Yudin.  
The complexity depends on the structure of the convex function: Lipschitz continuous gradients help, strong convexity helps, a certain saddle point structure helps, and so on.
Even if your convex function is not differentiable, then depending on its structure, different results exist, and some of these you can chase by starting from Nesterov's "Smooth minimization of nonsmooth functions" cited by Brian in his answer.
But for the sake of clarity, I illustrate below the simplest upper-bound on the (information based) complexity of gradient descent. This complexity is worse than the lower bound (optimal methods that attain this do exist). In any case, the references cited will help you study this subject in much greater detail. 
Concrete example
In summary, in the informational complexity model, one assumes access to an oracle that given an input vector $x$, outputs the objective and gradient values $(f(x), \nabla f(x))$. Then, complexity of a convex optimization algorithm is measured in terms of how many calls to the oracle an algorithm makes to be able to obtain an $\epsilon$-accurate solution (e.g., in terms of objective value, norm of gradient, or distance to optimal).
Suppose $f$ is $C^1$ convex function defined over the reals, and we wish to solve the unconstrained optimization problem
\begin{equation*}
\min_x\quad f(x)
\end{equation*}
Suppose we solve the above problem using the gradient-descent iteration
\begin{equation*}
 x^{k+1} = x^k - \alpha_k\nabla f(x^k),\quad k=0,1,\ldots.
\end{equation*}
$\newcommand{\reals}{\mathbb{R}}$
Theorem. (Nesterov)  Let $f$ be as above, and let it have a Lipschitz continuous gradient with constant $L > 0$. Let $\alpha_k = 1/L$. let $x^*$ be an optimal solution to the above problem. Define the diameter $D:=\|x_0-x^*\|$. Then, the iterates $(x^k)$ produced by gradient-descent satisfy
$$
    f(x^k) - f(x^*) \le \frac{2LD^2}{k+4}.
$$
As a corollary, we see that gradient-descent requires $O(1/\epsilon)$ calls to the oracle to obtain an $\epsilon$-accuracy solution (i.e., to ensure that $f(x^k)-f(x^*) \le \epsilon$).
There also exist lower-bounds (e.g., $O(1/\sqrt{\epsilon})$ for the above class) on number of oracle calls, and methods such as Nesterov's optimal gradient method that actually achieve the lower-bound. 
More generally, several other specialized cases of information based complexity depending on different assumptions that one makes about the oracle (stochastic, deterministic, sparse), and the function (strongly convex, nonsmooth, etc.). Please have a look at the references cited below for more information.
References.

Nemirovsky, A. S., and Yudin, D. B. 1983. Problem complexity and method efficiency in optimization. Wiley-Interscience. Translated by: E.~R.~Dawson.
Nesterov, Yu. 2004. Introductory Lectures on Convex Optimization: A Basic Course
Kluwer Academic (now Springer).

REPLY [6 votes]: Some books to start with for background reading would include:
Y. Nesterov, Introductory Lectures on Convex Optimization: A Basic Course, Springer, 2003.
Y. Nesterov and A. Nemirovsky, Interior Point Polynomial Methods in Convex Programming : Theory and Algorithms, SIAM, 1993. (Really just about the computational complexity of conic optimization problems, particularly LP, SOCP, and SDP.)  
Stephen Vavasis, Nonlinear Optimization: Complexity Issues, Oxford University Press, 1991.
A seminal paper is 
Y. Nesterov.  Smooth Minimization of Non-Smooth Functions  Mathematical Programming 103:127-152. 2005.
In the analysis of subgradient descent methods for convex optimization problems, the normal approach is to analyze the number of iterations to get the objective function value to within $\epsilon$ of the optimal value (this is not the same as finding a solution within $\epsilon$ of the optimal solution!)  The key results are that for a smooth problem with Lipschitz continuous gradient with Lipschitz constant $L$, this can be done in $O(\sqrt{L/\epsilon})$ iterations (e.g. by using an algorithm published by Nesterov in a 1983 paper), and that for nonsmooth problems, $O(1/\epsilon^{2})$ iterations are required, and this is in general the best possible bound.  There are explicit counterexamples that show you can't do better.  
However, Nesterov shows how in many cases it is possible to smooth a nonsmooth problem and then apply an optimal method to the smoothed problem to get an $\epsilon$ optimal solution to the original nonsmooth problem in $O(1/\epsilon)$ time.   Doing this requires that the problem have some specific structure that can be exploited so as to produce a smooth problem with a Lipschitz continuous gradient and Lipschitz constant such that the optimal method for the smoothed problem takes $O(1/\epsilon)$ iterations.   Nesterov gives several examples of this process in the 2005 paper.     
This theoretical work hit at just the moment that interest in compressive sensing problems (which are nonsmooth convex optimization problems with special structure) was growing.  In particular, many people are interested in solving very large scale problems of the form
$ \min \| Ax - b \|^{2}\_{2} + \lambda \| x \|_{1} $
Nesterov's ideas are certainly applicable in this context, but practical methods for these problems don't all follow this approach.  
It isn't clear why you're asking about this, and there's quite a lot of research going on in this area.  Chances are that a more specific question would yield a much more informative answer.<|endoftext|>
TITLE: Sums of three non-zero squares
QUESTION [10 upvotes]: It is a well-known result of Legendre that a positive integer is sum of three squares unless it is of the form $4^a(8b+7)$.
In 

Grosswald, E.; Calloway, A.; Calloway, J. The representation of integers by three positive squares. Proc. Amer. Math. Soc. 10 1959 451–455. (MR0104623 (21 #3376)),

it is shown that there is a finite set $T$ such that any positive integer is a sum of three non-zero squares unless $n$ is of the form $4^a(8b+7)$ or of the form $4^am$ where $m\in T$.
The set $T$ is essentially identified, see 

Grosswald, Emil. Representations of integers as sums of squares. Springer-Verlag, New York, 1985. xi+251 pp. ISBN: 0-387-96126-7 (MR0803155 (87g:11002)):

Either
 $$ T=\{1,2,5,10,13,25,37,58,85,130\}, $$ 
or else the Riemann hypothesis fails, and $T$ consists of these 10 numbers, and at most another one, $k$, that must be larger than $5\cdot10^{10}$. The conjecture is that $|T|=10$, of course. 
I could not find any updates on the question of whether the conjecture has been settled, and would appreciate any information or pointers to the relevant literature.

REPLY [8 votes]: Googling on the title of the Grosswald paper produced a link to math.uab.edu/~simanyi/Goswick_et_al_final.pdf which (backing up the url to the ~simanyi) indicates it's a recent paper in JNT.  (Re-posted from comments at the OP's suggestion.)<|endoftext|>
TITLE: Under what conditions does $\mathcal{M} \vDash \mathsf{PA}$ and $\mathcal{K} \vDash \mathsf{PA}$ such that $\mathcal{M} \ncong \mathcal{K}$?
QUESTION [5 upvotes]: I'm currently learning some introductory model theory from Marker's "Model Theory: An Introduction", Kaye's "Models of Peano Arithmetic", and Hodges' "Model Theory", and I am confused by the Wikipedia article on Peano Arithmetic. I am interested in the question I posed in the title and this article is really confusing me.
It states there that,

A model of the Peano axioms is a triple $(\mathbb{N}, 0, S)$, where $\mathbb{N}$ an infinite set, $0 \in \mathbb{N}$ and $S : \mathbb{N} \rightarrow \mathbb{N}$ satisfies the axioms. Dedekind proved in his 1888 book, What are numbers and what should they do (German: Was sind und was sollen die Zahlen) that any two models of the Peano axioms (including the second-order induction axiom) are isomorphic. In particular, given two models $(\mathbb{N}_A, 0_A, S_A)$ and $(\mathbb{N}_B, 0_B, S_B)$ of the Peano axioms, there is a unique homomorphism $f : \mathbb{N}_A \rightarrow \mathbb{N}_B$ satisfying, $$f(0_A)=0_B$$ and $$f(S_{A}(n))=S_{B}(f(n))$$ and it is a bijection.

Doesn't Tennenbaums theorem, and the existence of non-standard models, show that not all models of peano arithmetic are isomorphic? If this is the case, then what result did Dedekind actually prove and does anyone know where I can find a reference to the theorem he proved or a proof of it?

REPLY [9 votes]: Your question is answered by the distinction between the first-order and second-order Peano axioms.
The categoricity result of Dedekind refers to the second-order Peano axioms rather the first-order axiomation PA that gives rise to the nonstandard models and other phenomenon you mention. 
The second order axiomatization includes the axiom that every subset of the model containing $0$ and closed under successor $S$ is equal to the entireity of the model. This axiom is second-order, because it refers to arbitrary subsets of the universe of the model. It is not difficult to see that any two models of the second order Peano axioms are isomorphic, since each initial segment of one maps uniquely to an initial segment of the other (proved by induction), and these maps union to an isomorphism. 
Meanwhile, the first-order axioms of PA are usually stated in a larger language, with symbols for additiona and multiplication, and one has the induction scheme only for subsets that are definable in this language. Meanwhile, the theorems of elementary model theory give rise to nonstandard models of this first-order version of PA. None of these nonstandard models is a model of the second-order axiomatization, since the standard cut of a nonstandard model is a subset containing $0$ and closed under successor, but is not the whole model. 
There is quite an interesting interplay between first-order arithmetic, second-order arithmetic and first-order set theory here, because the second-order logic involved in the second-order Peano axioms used by Dedekind can be treated naturally in first-order set theory, such as ZFC (for the subsets of the model of arithmetic are just first order objects, sets, in set theory). In short, one may formalize Dedekind's argument as a result in ZFC.  So ZFC proves that there is unique structure of arithmetic $\mathbb{N}$. But meanwhile, we know that different models of ZFC can have different non-isomorphic versions of this unique structure $\mathbb{N}$. So the situation is that there are many different models $M$ of ZFC, each insisting that its own version of arithmetic $\mathbb{N}^M$ is the one-and-only absolute concept of arithmetic, the unique structure of the second-order Peano axioms, but externally, we can see that these different $\mathbb{N}^M$'s are not all isomorphic to each other.<|endoftext|>
TITLE: Possible isometries of a positively curved $S^2\times S^2$
QUESTION [16 upvotes]: Just to put things in perspective, recall that the Hopf Conjecture asks whether $S^2\times S^2$ admits a metric of positive sectional curvature. By the work of Hsiang-Kleiner, it is known that, if $S^2\times S^2$ admits such a metric, then its isometry group cannot contain a circle, and is hence finite.

Q: If $S^2\times S^2$ admits a metric with $sec>0$, what is known about its isometry group $G$?


The only results I know of are:
0) [Edit suggested by Misha] The diagonal antipodal action of $\mathbb Z_2$ on $S^2\times S^2$, i.e., $\pm 1\cdot(x,y)=(\pm x,\pm y)$, cannot be isometric if $S^2\times S^2$ is equipped with a metric of positive curvature. By Weinstein's Thm, an orientation-preserving isometry of an even-dimensional positively curved manifold has a fixed point (and the antipodal map does not). Equivalently, it would induce a positively curved metric on the $2$-fold orientable cover of $\mathbb R P^2\times \mathbb R P^2$, hence on $\mathbb R P^2\times \mathbb R P^2$, but this contradicts Synge's Thm.
1) From Wilking's thesis (Prop 4.2), any simple subgroup of $G$ is either cyclic or isomorphic to a group in a finite list $F_1,\dots,F_k$ of simple groups. (This is actually true for any finitely generated subgroups of isometries of a manifold with $Ric≥0$).
2) From Fang's paper (Thm 1.2), $G$ cannot have a subgroup of sufficiently large odd order (but this lower bound is huge, since it is estimated with Gromov's universal constant for the total Betti number).

Apart from these, are there other known restrictions on what $G$ can be like?

REPLY [7 votes]: You should look at Andrew Hick's thesis, 
Andrew Hicks, Group actions and the topology of nonnegatively curved $4$-manifolds, Illinois Journal of Mathematics. Volume 41, Issue 3 (1997), 421-437. 
A corollary to his Theorem 2 shows that for a positively curved metric on $S^2\times S^2$ with $\delta$ pinching, the size of the isometry group is bounded above by a constant related to the pinching. 
There are a few other interesting results about symmetries of nonnegatively curved 4-manifolds there.<|endoftext|>
TITLE: Induced Paths of Order 4
QUESTION [13 upvotes]: In a graph $G=(V,E)$ of order $n$, what fraction of the $\binom{n}{4}$ $4$-subsets of $V$ can induce the path of order four?
I looked at this question 30 years ago and was never able to come up with a respectable upper bound.  The question has reared its head again.   The answer appears to be somewhere between $1/4$ and $1/3$, though that upper bound is almost certainly weak.  Ideas?

REPLY [12 votes]: The question appears to be difficult. The best lower bound that I am aware of is still the one provided by the question author in 1986:
$$\frac{960}{4877}\binom{n}{4}\sim 0.19684\binom{n}{4}.$$
An upper bound is referred to in the paper ``The Inducibility of Graphs on Four Vertices" by James Hirst. It is 
$$\sim 0.2064 \left( \binom{n}{4} + o(n^4)\right).$$
The bound is obtained via semi-definite programming using the flag algebra technique. This method was introduced by Razborov in 2007 and it can be used to automatically produce upper bounds on asymptotic number of induced configurations in graphs and hypergraphs. These bounds are occasionally tight. In particular, James Hirst in the paper linked above deduces asymptotically tight upper bounds on the number of induced subgraphs on $4$ vertices of any fixed type, except for the $4$ vertex path.<|endoftext|>
TITLE: Lie algebra admitting some hyperbolic automorphism is nilpotent
QUESTION [7 upvotes]: Let $\mathfrak{g}$ be a finite dimensional Lie algebra over $\mathbb{R}$ and $\phi:\mathfrak{g}\to\mathfrak{g}$ be a Lie algebra automorphism. 
Viewing $\mathfrak{g}$ as a linear space and $\phi$ a linear automorphism, we can say $\phi$ is hyperbolic if the eigenvalues of $\phi$ are disjoint from $\lbrace z\in\mathbb{C}:|z|=1\rbrace$.
Then Proposition 3.6 in Smale's paper (here) says that:

Suppose that $\phi:\mathfrak{g}\to\mathfrak{g}$ is a Lie algebra automorphism which is hyperbolic as a linear map. Then $\mathfrak{g}$ must be nilpotent.

He also mentioned the following result in (Exercise in Bourbaki with hints: Algebras de Lie, Ex. 21b, p. 124.):

Let $\mathfrak{g}$ be a finite dimensional Lie algebra having an automorphism $\phi$, 
no eigenvalue of which is a root of unity, then $\mathfrak{g}$ is nilpotent.

Do you have ideas how to prove these results? 
Thanks!

After Vladimir Dotsenko:
$$(\phi-\lambda\gamma)[u,v]=[\phi u,\phi v]-[\lambda u,\gamma
v]=[(\phi-\lambda)u,\phi v]+[\lambda u,(\phi-\gamma) v].$$
Applying above to the pair
$\hat{u}=\lambda^i\phi^j(\phi-\lambda)^{a}u$ and
$\hat{v}=\gamma^k\phi^l(\phi-\gamma)^{b}v$ we have
$$(\phi-\lambda\gamma)[\lambda^i\phi^j(\phi-\lambda)^{a}u,\gamma^k\phi^l(\phi-\gamma)^bv]=
[(\phi-\lambda)\hat{u},\phi \hat{v}]+[\lambda \hat{u},(\phi-\gamma)
\hat{v}]$$
$$=[\lambda^i\phi^j(\phi-\lambda)^{a+1}u,\gamma^k\phi^{l+1}(\phi-\gamma)^bv]
+[\lambda^{i+1}\phi^j(\phi-\lambda)^au,
\gamma^k\phi^l(\phi-\gamma)^{b+1}v].$$
Tracing the indices we get
$$(i,j,a;k,l,b)\overset{\phi-\lambda\gamma}{\to}(i,j,a+1;k,l+1,b)\cup
(i+1,j,a;k,l,b+1),$$ and in particular
$(a,b)\overset{\phi-\lambda\gamma}{\to}(a+1;b)\cup (a;b+1)$. Then
$$(\phi-\lambda\gamma)^{m+n}[u,v]
=\sum_{a+b=m+n}[\lambda^i\phi^j(\phi-\lambda)^{a}u,\gamma^k\phi^l(\phi-\gamma)^bv]=0$$
since either $a\ge m$ or $b\ge n$.

REPLY [2 votes]: This is just for answering the last question posed by Yves. In positive characteristic, a finite-dimensional Lie algebra admitting an invertible derivation need not be nilpotent. For an example, look at the paper:  
"G. Benkart - A.I. Kostrikin - M.I. Kuznetsov: Finite-Dimensional Simple Lie Algebras with a Nonsingular Derivation, J. Algebra 171 (1995), 894-916" (http://www.sciencedirect.com/science/article/pii/S0021869385710411)
However, under some assumption the conclusion is indeed true. For instance, Jacobson proved that this is just the case when $L$ is a finite-dimensional restricted Lie algebra admitting an invertible restricted derivation.<|endoftext|>
TITLE: Tor sheaves: what do they tell us about geometry
QUESTION [12 upvotes]: Hi!
I fear that I am up to ask a very vague question, but more than an answer I need a suggestion of references I should look up.
I need to know everything about Tor sheaves and what do they tell about geometry. For example if $X$ is a smooth variety and $Z$ and $B$ are subvarieties,where are the sheaves $\mathbf{Tor}_i(O_Z, O_B)$ supported?  Does the vanishing of (some) of the higher Tor sheaves have some reflection on the mutual geometry of $Z$ and $B$ (for example can we evince the dimension of the intersection of $B$ and $Z$).
As I already explain you do not need to answer this (probably very silly) questions. I just need someone to point the right book to me, since right know I have not the slightest idea of where to find these informations. 
Thank you very much for your time and your attention,
Best

REPLY [11 votes]: For example if $X$ is a smooth variety and $Z$ and $B$ are subvarieties,where are the sheaves $\mathbf{Tor}_i(O_Z, O_B)$ supported?  Does the vanishing of (some) of the higher Tor sheaves have some reflection on the mutual geometry of $Z$ and $B$?

In fact just the support or vanishing of the higher Tor sheaves can tell you quite a bit about the geometry. When $Z,B$ intersect properly and $Tor_1=0$ (which is actually equivalent to all the higher $Tor$ vanish) then you can even say that $Z,B$ are Cohen-Macaulay! 
See Serre's Local Algebra book, V.6, Theorem 4, p 110 (this great book also contains the intersection formula that Karl mentioned). 
More precisely, Auslander's ICM 1962 note (available on this page, first in Section 2) describes the support of the $Tor_i(Z,B)$ for any two sheaves over a regular Noetherian scheme (see Theorem 2). He only stated it for the unramified local case, but we now know it in full generality. A summary of his result: the support of Tor only depends on the depths of $Z,B$ at the stalks.<|endoftext|>
TITLE: Fermat's Last Theorem for Gaussian Integers ( excluding $\mathbb{Z}$ or $i\mathbb{Z}$ )
QUESTION [29 upvotes]: I am investigating solutions to Fermat's equation
$$x^n+y^n=z^n$$
with $x,y,z$ in the Gaussian integers, excluding solutions in  excluding $\mathbb{Z}$ or $i\mathbb{Z}$ .
I have found out that there are only trivial solutions for the n=3 and n=4 cases, e.g. here.
I would be grateful if you let me know of the current status or if it is already a theorem.
P.S.: This same question was asked on Math.SE but it has now drowned under the fold and I thought I will have better chances of getting answers here.

REPLY [52 votes]: This is still way open, I should think. "Elementary" methods won't even solve the analogous problem over $\mathbf{Z}$, so you need to use "modular form" methods. The problem is that even if the result were to follow from a Frey curve argument and a potential theorem of the form "all sufficiently nice Galois representations come from automorphic forms", we're a long way from establishing such a theorem.
A few more details: a key problem is that there are two natural candidates for where the automorphic forms will come from, and neither is good enough. The first is the group $GL(2)/\mathbf{Q}(i)$. This is a very natural place to look, but the problem is that this group does not admit Shimura varieties, so it's very hard to even go the "easy" way and to attach a Galois representation to an algebraic automorphic representation (i.e. do what Deligne did), let alone to do what Wiles did. There are some theorems of this nature, but they all have a self-duality hypothesis built into them, which will not hold for the Tate module of the Frey curve in general.
The second place to look is rank 2 unitary groups for the extension $\mathbf{Q}(i)/\mathbf{Q}$. These do have Shimura varieties and their geometry is understood fairly well nowadays, and there are are very strong theorems attaching Galois representations to these automorphic forms (there are even very strong theorems for rank $n$ unitary groups nowadays -- see for example the book by Harris and Taylor, but things have moved on even further since then, e.g. because of recent work of Sug-Woo Shin). However we then run into the same problem -- the unitary groups have some extra symmetry and this means that the associated Galois representations have some extra symmetry (they need to be essentially conjugate self-dual), and this extra symmetry will not in general be true for the Galois representations attached to the Tate module of the Frey curve.
So one needs a good new idea before we can push forward what one might call "Wiles' strategy" in this situation.
This is in marked contrast to the case of totally real fields, where a lot of the machinery works fine and it would not surprise me nowadays if FLT could be proved for several totally real fields. As has been implicitly mentioned in the comments, Jarvis and Meekin did this for $\mathbf{Q}(\sqrt{2})$, and this was a few years ago now, and modularity lifting theorems have moved on tremendously since then, so it would not surprise me if the experts could prove these results for other totally real number fields now. However somehow, after the Jarvis-Meekin work, which is a proof that the machine can be made to work in other cases, perhaps the interesting question now is not something like "is FLT true for $\mathbf{Q}(\sqrt{5})$?" (which one could perhaps hope to answer, perhaps with a lot of work, but with basically existing methods and a lot of hard graft to deal with the small $n$ cases) but more like "is some slightly weakened version of FLT true for all totally real fields?" or some such thing. You need to weaken it a bit because the case $n=3$ is an elliptic curve and it has positive rank for lots of totally real $F$, so now you need to deal with $n=4$ and $n=9$ and $n=6$ by hand, or just declare that you're only interested in $x^p+y^p=z^p$ with $p\geq5$ prime; and then you'll sometimes get reducible mod $p$ Galois representations -- so perhaps the correct weakening is something like "$x^p+y^p=z^p$ has no solutions for $p$ sufficiently large (depending on $F$)". Even then there may be problems in "case 2".
My impression is that the machinery being developed now is not really being developed with generalisations of FLT in mind, but it's generalising this "Wiles machine" in different directions, for example to prove things like the Fontaine-Mazur conjecture and the Sato-Tate conjecture. This is the direction deemed "trendy" -- and in a sense I can see why, because is FLT over a specific number field other than the rationals really a "natural" question? And FLT over a general number field is obviously false, so now you have to weaken things etc etc. On the other hand hard conjectures like Sato-Tate do look to me like natural questions.
IMPORTANT EDIT: I wrote this answer a long time ago -- what is it, 9 days now? -- but life moves on, and I hear from my spies in Toronto that Richard Taylor yesterday announced some results joint with Harris, Lan and Thorne, where they claim that they can attach Galois representations to (not necessarily self-dual) cohomological cuspidal automorphic representations on $GL(n)$ over totally real and CM fields. In particular apparently the theory for $GL(2)/\mathbf{Q}(i)$ is now up to about the state that the theory for $GL(2)/\mathbf{Q}$ was in the late 60s. So give it another 23 or so years and we should have FLT for $\mathbf{Q}(i)$!

REPLY [7 votes]: I suppose this is an open problem.
Exponents 5 and 7 might follow from these papers:
Algebraic points of low degree on the Fermat quintic
describes all algebraic points of degree at most 2 for exponent 5.
The non-trivial points are $(\eta,\bar{\eta},1)$ and $(\bar{\eta},\eta,1)$ where $\eta$ is primitive 6th root of unity.
For exponent 7 Algebraic points on some Fermat curves and some quotients of
Fermat curves: Progress gives all non-trivial points of degree at most 3:
$(\eta,\bar{\eta},-1)$ and $(\bar{\eta},\eta,-1)$.
Regarding Vladimir Dotsenko's comment that this will folow from Conjecture 1.2

Conjecture 1.2 (Debarre-Klassen) Let $K$ be a number field of degree $d$
  over Q. Then the equation $x^n + y^n = z^n$ has only trivial solutions over K
  when $n \geq d + 2$. Here, Debarre and Klassen define trivial solutions to mean points $(a, b, c)$ on
  $x^n + y^ n = z^ n$ where $a + b = c$.

If one considers $x+y=z$ a valid solution the conjecture won't apply and in addition this appears a counterexample to the conjecture for $n=4$
$$ (\sqrt{-7} + 3)^4 + 4^4 = (\sqrt{-7} - 1)^4 $$<|endoftext|>
TITLE: Does direct limit commute with functor of smooth sections?
QUESTION [7 upvotes]: Consider a countable family of finite-rank vector bundles $V_k$ over a finite-dimensional smooth manifold $M$. The direct limit of such a family is still a topological vector bundle even though it may have infinite rank. 
The first question is: Is there a natural structure of smooth manifold on the total space of $\varinjlim V_k$? 
And the second one: Is it true, that the functor of smooth sections commutes with direct limit? I.e, is it true that
$$\mathcal{C}^\infty(M,\varinjlim V_k) = \varinjlim \mathcal{C}^\infty(M,V_k)$$
Or turning things upside down -- is there a choice of smooth structure such that the above equation holds? If so, how does it look like?

REPLY [7 votes]: The answers are: Yes and No-but-yes-if-M-is-compact.

Kriegl and Michor's A Convenient Setting for Global Analysis describes how to put a smooth structure on an arbitrary locally convex topological vector space, say $V$, by looking first at the smooth curves in $V$ (these can be unambiguously defined).  This works for $V = \lim V_k$ where the $V_k$ are finite dimensional.  As your family is countable, this is just $\sum_{\mathbb{N}} \mathbb{R}$.  Smooth curves are continuous, and therefore an important property of this structure is that if $c \colon \mathbb{R} \to \sum_{\mathbb{N}} \mathbb{R}$ is smooth then $c([a,b])$ is contained in a finite dimensional subspace.
And that's really the key to the second answer.  If $M$ is compact, then its image lies in a finite dimensional subspace of $\lim V_k$, whence inside one of the $V_k$.  If $M$ is not compact then it will be possible to find a smooth map $M \to \lim_k V_k$ which does not lie in any subspace and so the limit will not commute with the mapping space construction.  Note that it is enough to show that such a map exists for $M = \mathbb{R}$, whereupon you take curves $\alpha_k \colon [0,1] \to V_k$ which map to $0$ on their endpoints, are infinitely slow there, and such that the image of $\alpha_k$ contains a basis for $V_k$.  Then concatenate these paths together to get a smooth path $\alpha \colon \mathbb{R} \to \lim V_k$ with image not contained in any finite dimensional subspace.<|endoftext|>
TITLE: coarse moduli space and $\pi_0$
QUESTION [11 upvotes]: I've been reading this really nice paper by Alper http://math.columbia.edu/~jarod/good_moduli_spaces.pdf, and there's a question that doesn't seem to be answered (perhaps it's not relevant).
Any stack F has a corresponding `sheaf of connected components' (or sheaf of isomorphism classes), by taking $\pi_0^{pr}(F)(S) = \pi_0 (F(S))$ and then sheafifying. (where $\pi_0$ of a groupoid, or more generally a category, is the set of isomorphism classes)
If $X$ is an Artin stack (although I'm currently more interested in DM stacks) and $X$ admits a good moduli space, then is $X \to \pi_0(X)$ a good moduli space?
Also, when is the good moduli space a scheme (and not just an algebraic space)?
And finally, if $X$ = $Spec R$ is affine and $G$ acts on it (I'm mainly interested in the $G$ finite case), is $\pi_0([X/G]) = Spec R^G$?

REPLY [7 votes]: You have probably already come up with the answer yourself, but I just thought the question shouldn't
hang around unanswered in the forum.
What you call "the sheaf of connected components", I would call the coarse sheaf of the stack or the 
sheaf associated to the stack. It is usually not representable by an algebraic space. When it is,
the stack is called a gerbe and the map to the coarse sheaf is called the structure morphism of the gerbe.
A gerbe with structure morphism $X \to Y$ is fppf locally on $Y$ of the form $B_YG := [Y/G]$, where $G$ is
a group-algebraic space which is fppf over $Y$, and the action on $Y$ is trivial.
(In fact, it is even étale locally on this form, since the structure morphism $X \to Y$ of a gerbe is smooth.)
Although gerbes certainly have at lot of good properties, the structure morphisms of a gerbe need not be a
good moduli space in the sense of Alper. It is, exactly when $G$ above is linearly reductive.
The property of being a gerbe as a very strong property. Thus good (or a coarse) moduli spaces
are seldom coarse sheaves. In particular $Spec\ R^G$ will usually not be the coarse sheaf of $[Spec\ R / G]$.
Taking the stack quotient and then taking the associated sheaf, is the same as taking the sheaf quotient
directly. The result is seldom (never?) representable unless the action of $G/N$ is free, where $N$ is the kernel
of the action.<|endoftext|>
TITLE: Interesting result on the Euler-Maschroni constant - what is the background?
QUESTION [11 upvotes]: Today I entered the following expression in maple:
$$a_i = H_{10^i} - ln(10^i) - \gamma$$
Here $H_j$ equals $\sum_{k=1}^{j} 1/k$ and $\gamma$ is the Euler-Mascheroni constant. 
When I computed $a_n$ for $i = 0$ to $10$ I obtained the following results:

 $i = 0$;    4.227843350984671393934879099175975689578406640600764011942327651151323 * $10^{-1}$

 $i=1$;   4.9167496072675423629464709201487329610707429399557393414873118115813 * $10^{-2}$

 $i=2$;   4.991666749996032162622676207122311664609813510982102304110919767206 * $10^{-3}$

 $i=3;$    4.999166666749999960317501984051226762153678825611388678758121701133 * $10^{-4}$

 $i=4$;     4.99991666666674999999960317460734126976551226762154503821179264423 * $10^{-5}$

 $i=5$;     4.9999916666666667499999999960317460321626984126226551226762154523 * $10^{-6}$

 $i=6$;     4.999999166666666666749999999999960317460317501984126984051226523 * $10^{-7}$

 $i=7$;     4.99999991666666666666674999999999999960317460317460734126984123 * $10^{-8}$

 $i=8$;                                                                            
4.9999999916666666666666667499999999999999960317460317460321623 * $10^{-9}$

 $i=9$;                                                                            
4.999999999166666666666666666749999999999999999960317460317423 * $10^{-10}$  

 $i=10$;
4.99999999991666666666666666666674999999999999999999960317423 * $10^{-11}$

So we see that the periodic strips of ...99999..., of ...66666... and ...99999... an many other periods increase for even larger $i$. The question is now: Is there any rule behind it that the remainder term $a_i$ behaves that way?

REPLY [2 votes]: Adding an example to @quid's answer:      
Using Pari/GP the harmonic numbers minus Euler-gamma can be obtained using the psi-function.
If we subtract 1 from the $\small a_i $-values the composition of the result by i-digits long blocks of decimal expansion of the bernoulli-numbers becomes then immediately visible. With $\small i \gt 20 $ or so it becomes even more impressive:
fmt(200,60)   \\ user function: internal prec 200, display prec 60 digits
i=6
psi(10^i)-1 - i*log(10)
 %681 = -1.00000050000008333333333332500000000000396825396824980158730
(psi(10^i)-1 - i*log(10))*6
 %683 = -6.00000300000049999999999995000000000002380952380949880952381
(psi(10^i)-1 - i*log(10))*42
 %684 = -42.0000210000034999999999996500000000001666666666664916666667
(psi(10^i)-1 - i*log(10))*42*30
 %685 = -1260.00063000010499999999998950000000000499999999999475000000

etc...<|endoftext|>
TITLE: What is the dimension of the product ring  $\prod \mathbb Z/2^n\mathbb Z$ ?
QUESTION [21 upvotes]: In an anwswer to a question on our sister site  here   I mentioned that a reduced  commutative ring $R$ has zero Krull dimension if and only if it is von Neumann regular i.e. if and only if for any $r\in R $ the equation $r=r^2x$ has a solution  $x\in R$.
A user asked in a comment whether this implies that an arbitrary product of zero-dimensional rings is zero dimensional.
I answered that indeed this is true and follows from  von Neumann regularity if the rings are all reduced but I gave the following counterexample in the non reduced case:    
Let $R$ be the product ring $R=\prod_{n=1}^\infty \mathbb Z/2^n\mathbb Z$.
Every $\mathbb Z/2^n\mathbb Z$ is zero dimensional but $R$ has $\gt 0$ dimension.  
My argument was that its Jacobson radical  $Jac(R)=\prod_{n=1}^\infty Jac (\mathbb Z/2^n\mathbb Z)=\prod_{n=1}^\infty2\mathbb Z/2^n\mathbb Z$ contains the non-nilpotent element $(2,2,\cdots,2,\cdots)$.
However in a zero dimensional ring the Jacobson  radical and the nilpotent radical  coincide and thus   $R$ must have positive dimension.    
My question is then simply: we know that $dim(R)\gt 0$, but what is the exact Krull dimension of $R$ ?  
Edit
Many thanks To Fred and Francesco who simultaneously (half an hour after I posted the question!) referred to an  article by Gilmer and Heinzer answering my question .
Here is a non-gated link to that paper.
Interestingly the authors, who wrote their article in 1992, explain that already in 1983 Hochster and Wiegand had outlined (but not published) a proof that $R$ was infinite dimensional.
Already after superficial browsing I can recommend this article, which contains  many interesting results like for example infinite-dimensionality of $\mathbb Z^{\mathbb N}$.
New Edit
As I tried to read Hochster and Wieland's article, I realized that it refers to an article of Maroscia to which I have no access. Here is a more self-contained account of some of  Hochster and Wieland's results.

REPLY [26 votes]: The ring $R$ is infinite-dimensional. More generally, the product of a family of zero-dimensional rings has dimension $0$ if and only if it has finite dimension. This is proven as Theorem 3.4 in R. Gilmer, W. Heinzer, Products of commutative rings and zero-dimensionality, Trans. Amer. Math. Soc. 331 (1992), 663--680.

REPLY [23 votes]: According to the paper Products of Commutative Rings and Zero-Dimensionality, your ring must be infinite-dimensional.<|endoftext|>
TITLE: Finitely many spaces generated by eta-products
QUESTION [7 upvotes]: In page 3 of Kilford's paper generating spaces of modular forms with $\eta$-products, he mentions that there are only finitely many spaces of modular forms that can be completely generated by $\eta$-products. 
My question why is this true?
The paper can be found in following arXiv link:
http://arxiv.org/pdf/math/0701478.pdf
Thanks

REPLY [7 votes]: There's a brief answer to this on page 3 of the paper; there are only finitely many eta-products with q-valuation 1 (one can write them all down), these all have a given level, so other levels will have a form with q-valuation 1 which can't be written as an eta-product. Note that I defined eta-products to be products of eta-functions with non-negative exponents.<|endoftext|>
TITLE: $A_\infty$ structure on Ext-algebras well defined?
QUESTION [14 upvotes]: Let $M$ be an object in an $k$-linear abelian category with enough projectives. Then one can construct an $A_\infty$-structure on the Ext algebra
$$Ext^\bullet(M,M)$$
as follows: One chooses projective resolution $P\rightarrow M$ and forms the Hom complex
$$Hom^\bullet (P,P)$$
Now the cohomology of this complex is the  Ext algebra and in the case where $k$ is a field, one can choose a "homotopy retraction" $Ext^\bullet(M,M)\rightarrow Hom^\bullet (P,P)$ and transfer the dg-algebra structure on $Hom^\bullet (P,P)$ along it.
My questions are:

Why does this construction (up to $A_\infty$-isomorphism) not depend on a choice of projective resolution?
One could try the same thing with an injective resolution, why is the result still the same?

REPLY [9 votes]: Let $P\to M$ be a projective resolution of $M$ and $M\to J$ be an injective resolution.  Consider the composition of the morphisms of complexes $P\to M\to J$ and set $C$ to be the cone of the morphism $P\to J$.  Then the complex $C$ has a subcomplex isomorphic to $J$ with the quotient complex isomorphic to $P[1]$. Moreover, the short exact sequence of complexes $J\to C\to P[1]$ splits as a short exact sequence of graded objects in your abelian category (i.e., after the differentials are forgotten).
Consider the subcomplex $D$ in the complex $Hom^\bullet(C,C)$ formed by all the homogeneous morphisms of graded objects $C\to C$ taking $J\subset C$ to $J\subset C$.  This condition on morphisms $C\to C$ is preserved by the composition, so $D$ is a DG-algebra over $k$. The restriction of morphisms $C\to C$ to the subcomplex $J$ defines a DG-algebra morphism $D\to Hom^\bullet(J,J)$; and the passage to the induced morphism of the quotient complexes defines a DG-algebra morphism $D\to Hom^\bullet(P[1],P[1])\simeq Hom^\bullet(P,P)$.
It is claimed that both these DG-algebra morphisms are quasi-isomorphisms.  E.g., the morphism $D\to Hom^\bullet(J,J)$ is surjective and its kernel is the complex $Hom^\bullet(P[1],C)$, which is acyclic as $P$ is a complex of projectives (bounded from above) and $C$ is acyclic.  The argument for the second DG-algebra morphism is similar.
Now, being quasi-isomorphic DG-algebras, $Hom^\bullet(P,P)$ and $Hom^\bullet(J,J)$ have $A_\infty$-isomorphic minimal $A_\infty$-models.  The proof of the independence of these from the choice of the resolution $P$ and/or $J$ is analogous.<|endoftext|>
TITLE: $\mathfrak{c}$-universal linear order
QUESTION [8 upvotes]: I've been told once or twice that the following holds:

There is a model of $ZFC+MA+\neg CH$ in which there is a $\mathfrak{c}$-universal linear order embedded in $(\omega^\omega, \le^\ast)$ 

Moreover, several people have attributed the construction of such a model to Hugh Woodin. The context that seems the most natural for this construction to appear is in his work on automatic-continuity, however for the life of me, I can't seem to find it.
It might also be the case that the embedding is into $\mathcal{P}(\omega)/fin$.
Does anybody know of a reference for this? Or if Woodin did indeed construct it (I only ask because my searches have yielded little fruit)?
Edit: 
In this context a $\mathfrak{c}$-universal linear order $(L, \prec)$ has the following property: $\vert L \vert \le \mathfrak{c}$ and for every linear order $(\ell, <) $, with $\vert \ell \vert \le \mathfrak{c}$ there is an embedding $\varphi_\ell:\ell \rightarrow L$ respecting the linear order of $\ell$.

Thought I should share what I can find/know:
So far I've been able to find models for the following

$ZFC+ \neg CH$ and there is such a $\mathfrak{c}$-universal linear order in $(\omega^\omega,\le^\ast)$.

This is due to Laver, http://www.sciencedirect.com/science/article/pii/S0049237X08716306
In addition, there is also 

$ZFC+MA+\neg CH$ and no such order is embedded in $(\omega^\omega,\le^\ast)$

There seem to be several models for this one, Laver cites two, one from Solovay and one from Kunen. I can only assume the Solovay model is the same constructed in Theorem 5.7 (p. 201, "Hausdroff Gaps and Limits", by Frankiewicz and Zbierski)
In addition there is

$CH\implies $ $(\omega^\omega,\le^\ast)$ contains a $\mathfrak{c}$-universal linear order

Which should be contained in (or atleast hinted at in) "Model Theory", by Chang and Keisler. If not there, the lack of $(\omega,\omega)$-gaps in $(\omega^\omega,\le^\ast)$ should produce it rather quickly.
So yeah, it would seem that the only link missing in this puzzle is the one I can't find the reference for.

REPLY [4 votes]: Francisco Kibedi has written down a version of this for the $c$ universal linear order consistent with MA for $\sigma$ linked partial orders. His thesis is concerned with maximal such linear orders in the sense of Hausdorff's pantachies.<|endoftext|>
TITLE: Infinitely Noetherian Rings
QUESTION [17 upvotes]: One equivalent definition of a noetherian ring is that it satisfies the ascending chain condition.  That is, for every ascending chain of ideals 
$$ I_{0} \subseteq I_{1} \subseteq I_{2} \subseteq \dots \subseteq I_{k} \subseteq I_{k+1} \dots $$
there is a j such that for all $k \ge j$, $I_{k} = I_{k+1}$.
What if this definition is extended into the ordinals?  Could there be a ring which satisfies the $j = \omega^2 + \omega\cdot3 + 5$ noetherian condition, for example? That is, could there be an infinite $j$ such that there are no ascending chains of length greater than $j$.  In this example, we would say that the ring is $\omega^2 +\omega\cdot3 + 5$ noetherian.  
Most importantly, what kind of rings would be created? Same question with a variety of definitions of noetherian.  
Edit:  The way I am trying to define such rings seems to be problematic.  I'd like to conceive a clearer definition, but I need to become a bit more familiar with noetherian rings.

REPLY [6 votes]: Possibly this is already too far away to qualify as 'answer', but it is long for a comment and might be of interest. 
There is a classical construction of descending (not ascending), so rather matching artinian than noetherian, subgroups of abelian groups (or, to artificial stay a bit closer let's say submodules of Z-modules, but then this got generalized to modules over other rings so it is not totally artificial), where indexing by ordinals is done and relevant. They are called Ulm subgroups. 
Let $G$ be an abelian group and $p$ some (fixed) prime. Than the (classical) p-height of an element $g$ is the supremum of all naturals such that the  equation $p^nx = g$ has a solution $x$ in $G$. 
One can express this differently by defining 
$p^nG$ to be the set, in fact it is a subgroup, of all $h\in G$ of the form $p^ng$ and define the p-height as the supremum of all $n$ such that $g \in p^nG$.
Now, one can continue and say $p^{\omega}G$ is the inersection of all the $p^nG$ and so on. 
Or formally, $p^0G = G$, $p^{\alpha+1}G= p(p^{\alpha}G)$ and $p^{\beta}G =\bigcap_{\alpha \lt \beta}$ for $\beta$ a limit ordinal.
This forms a descending chain of subgroups that (thus) eventually stabilzes; the ordinal where it stabilzes is called Ulm length. One can now also generalize the notion of p-height of g by defnining it as the ordinal $\sigma$ such that $g \in p^{\sigma}G$ yet not in $g \in p^{\sigma + 1}G$  and $\infty$ if such an ordinal does not exist.
And this notion is quite useful. For example, a classical result of Ulm (1930s) then says that two countable abelian $p$-groups (ie, order of each element a prime power)  $G$ and $H$ are isomophic if and only if $p^{\alpha}/p^{\alpha+1}G$ and $p^{\alpha}H/p^{\alpha+1}H$ are isomorphic for each $\alpha$ (these are called the Ulm factors) and the stable parts are isomorphic. 
And conversely for any ordinal one can construct a group of that Ulm length (and in addition one can prescribe to a considerable extent the Ulm factors at each point until there).<|endoftext|>
TITLE: Can ZFC prove "false theorems", and still be consistent? (was "Junk Theorems" follow up)
QUESTION [16 upvotes]: In Set theories without "junk" theorems? Jacques Carette consider's "junk theorems" of ZFC - theorems which are  artifacts of our means of encoding standard mathematical objects into set theory, but don't respect types.  For instance, given one encoding of the integers, it is a theorem that the only natural numbers which are functions are 0 and 3.  
The examples of junk theorems presented so far all immediately strike you as "junk" - basically there is a type error, and you can immediately recognizer that the junk theorem is the result of mixing types in a way you shouldn't.  The theorems are junk, not so much because they are wrong, but because they are not meaningful.
Is it possible for some false and meaningful statement in "normal mathematical language" to be a true statement of ZFC when it is unpackaged?
I am imagining some mundane mathematical question, like is the constant function 3 an element of the zero set of some functional F.  Unpackaging the set theoretic definitions of all of the terms involved shows that yes, 3 is a member of that set, but in fact F(3) is not zero.  Is this possible?
In other words, should we trust automated theorem provers?

REPLY [3 votes]: To Michael Blackmon (in comment on Can ZFC prove "false theorems", and still be consistent? (was "Junk Theorems" follow up) , sorry I don't have the cookie to add a new comment to that post, maybe someone with enough points can transfer this): 
Yes, the idea is that if there are two models of PA (one that affirms Goldbach's conjecture and one that refutes it), it's conceivable that ZFC disproves Goldbach's conjecture even though the conjecture is true in the standard integers.  That just means that the $\omega$ in every model of ZFC turns out to be nonstandard, i.e. ZFC itself is unsound (though still consistent) and proves theorems that are false for the standard integers.  This seems like an unlikely situation, but I don't see how it's nonsensical.  Am I missing something?<|endoftext|>
TITLE: Example of an algebraic number of degree 4 that is not constructible 
QUESTION [8 upvotes]: The number $b:=\frac{\sqrt{2a}+\sqrt{4\sqrt{a^2-3}-2a}}{2}$ with 
    $a:=\frac{\sqrt[3]{18+2\cdot\sqrt{65}}}{2}+\frac{2}{\sqrt[3]{18+2\cdot\sqrt{65}}}$ is a root of the irreducible polynomial $x^4-6x+3$.The algebraic number $a$ is a root of the irreducible polynomial $2x^3-6x-9$ and therefore not constructible with ruler and compass. 
I claim that $b$ isn't constructible either. If this is the case, how can this be shown?

REPLY [7 votes]: As explained in the comments, one needs to check if the splitting field of $x^4-6x+3$ has degree equal to a power of $2$ or not. The resolvent cubic of this polynomial equals $x^3-12x+36$ which is irreducible (over $\mathbb{Q}$), because it has no root modulo $7$. Hence the splitting field in question contains a subfield of degree $3$, so it cannot have degree equal to a power of $2$. This shows that $b$ is not constructible. You can read about the resolvent cubic here.<|endoftext|>
TITLE: Growth rate of signed sum of divisor function
QUESTION [5 upvotes]: Let $d(n)$ be the number of divisors function, i.e., $d(n)=\sum_{k|n} 1$ of the positive integer $n$. I know about some of the "gross" averages for this function, such as the estimate
$$
\sum_{n\leq x} d(n)=x \log x + (2 \gamma -1) x +{\cal O}(\sqrt{x})
$$
as well as its variability, e.g., the lim sup of the fraction
$$
\frac{\log d(n)}{\log n/\log \log n}
$$
is $\log 2$ while the lim inf of $d(n)$ is $2$, achieved whenever $n$ is prime.
How much is known about the statistics of $d(n)$? In particular, if we let $N$ grow to infinity, is there any way
to bound a sum of the form
$$
\left| \sum_{n=1}^N \varepsilon_n d(n) \right|
$$
from below for all or almost all $(\varepsilon_1,\cdots,\varepsilon_N)\in \{\pm 1\}^N$?

REPLY [3 votes]: An answer to Greg's comment. It's true about the parity. However the sum can take on any value between $-\sum_{n=1}^N d(n)$ to $\sum_{n=1}^N d(n)$ if it has the right parity. It's because if $k$ is any number between $0$ and $D(N)=\sum_{n=1}^N d(n),$ then there exists some subset $A_{k,N}$ of $\{1,\dots, N\}$ such that $\sum_{n\in A_{k,N}} d(n)=k.$
It follows from the relatively easy fact that $2D(N)+1 \geq D(N+1)$ or equivalently $D(N)+1 \geq d(n+1).$ Then the result follows by induction.
It's true for $N=1$ since $A_{0,1}$ the empty set and $A_{1,1}={1}.$
Suppose it's true for $N$, then if $0\le k \le D(N)$, let $A_{k,N+1}=A_{k,N}$
If $D(N+1)-D(N) \le k \le D(N+1),$ then let $A_{k,N+1}=A_{K,N} \cup N+1,$ where $K=D(N+1)-k.$ 
$2D(N)+1 \geq D(N+1),$ this covers all $k$ between $0$ and $D(N+1)$ finishing the induction.
Note that this idea shows the same can be done for any arithmetic function $f: \mathbb{N} \to \mathbb{N}$ with the same inequality $2F(N)+1 \geq F(N+1).$ Examples include the Euler totient function $\phi,$ the Carmichael lambda function $\lambda$ and many others.<|endoftext|>
TITLE: Calculating chern numbers yields a contradiction, why?
QUESTION [7 upvotes]: I am really stuck on this one. Let $Y=\mathbb{P}^n$ be the complex projective space and let $\tilde Y$ be the blow-up of $Y$ along a linear subvariety $X$ of codimension $d$. We get the following blow-up diagram:
$$\begin{matrix} E & \xrightarrow{\;j\;} & \tilde{Y} \\
\hphantom{\scriptstyle g}\downarrow {\scriptstyle g} &&
\hphantom{\scriptstyle f}\downarrow {\scriptstyle f} \\
X &\xrightarrow{\;i\;} & Y
 \end{matrix}$$
Denote by $P$ the proper transform, under $f$, of a hyperplane in $Y$.
I am trying to calculate $c_2(\tilde Y)$, and with the help of a post here at MO, I thought that I had figured it out. However, I wanted to do a quick check if nothing went wrong, but something did go wrong. Assume $n=4$ and $d=2$, then we can use the formula from Fulton's book in Example 15.4.3 to get
$$c_2(\tilde Y)=f^\ast c_2(Y) - j_\ast g^\ast c_1(X) - E^2.$$
By the answer to my question by Johannes Nordström, we can write $j_\ast g^\ast c_1(X)=3(E^2 + EP)$ and this yields
$$c_2(\tilde Y)=10P^2 - 3EP -  4E^2.$$
Now, we also know from the same example in Fulton's book that $c_1(\tilde Y)=f^\ast c_1(Y) - E=5P-E$. Since the blow-up map $f$ is finite of degree one, the degrees of $c_1^2(\tilde Y)c_2(\tilde Y)$ and $c_1^2(Y)c_2(Y)$ should coincide. Because I was unsure of the calculation, I asked a second question and obtained the answer that
$$P^{n-b} E^b = (-1)^{b-1+n-d} \cdot \binom{b-1}{n-d}$$
Now, we can put this all together and obtain
$$\begin{align*}
c_1^2(\tilde Y)c_2(\tilde Y) &= (5P-E)^2(10P^2 - 3EP - 4E^2)
\\&= 250P^4 - 175P^3E - 60P^2E + 37PE^3 - 4E^4
\\&= 250 + 37 + 12 = 299,
\end{align*}$$
but $c_1^2(Y)c_2(Y)=(4+1)^2\cdot\frac{4(4+1)}{2} = 250$. 
I do not know where the mistake is, since I find both of the answers I received very convincing, but I cannot find a flaw in my calculation either, nor do I doubt Fulton.

REPLY [3 votes]: Note that $P^2 = 0$, since we blow up the self-intersection of a hyperplane.
The pull-back of the hyperplane class is $P+E$, so $c_1(\tilde Y) = 5P + 4E$, and $c_2(\tilde Y) = 10(P+E)^2 - 3EP - 4E^2 = 17EP + 6E^2$.
This still does not yield $c_1(\tilde Y)^2 c_2(\tilde Y) = 250$ (I think it's $512 - 3\cdot 96 = 224$), but I don't see why this Chern number should be preserved by the blow-up.<|endoftext|>
TITLE: Forgetting and tensoring up for very connective maps of $E_{\infty}$-rings
QUESTION [6 upvotes]: Does anything happen if I forget and tensor back up along a highly connective map of $E_{\infty}$-rings?
Here's what I mean precisely: Let $f \colon A \to B$ be a $n$-connective map between connective $E_{\infty}$-rings, with $n \geq 1$. Here $n$-connective means that $\pi_i (fib(f))=0$ for $i < n$. Roughly, $f$ is a very surjective map, and if we let $fib(f)=I$, then $B$ is roughly $A/I$.
Now let's take a $B$-module $M$. We can forget the $B$-module structure and view it as an $A$-module. Then we can tensor it back up again to get $B \otimes _A M$. By adjunction there is a canonical map $m \colon B \otimes _A M \to M$, which is the multiplication map. How far is this map from being an equivalence?
The case I am really interested is when $f \colon A \to B$ is a square-zero extension obtained from a $n$-connective derivation $\eta \colon L_B \to M[1]$. Then $fib(f)$ can be identified with $M$, and so additonally has the structure of a $B$-module. Does anything special happen in this case?

REPLY [4 votes]: (Derived) tensoring preserves fiber sequences.  In particular, there is a fiber sequence
$$
fib(f) \otimes_A M \to M \to B \otimes_A M.
$$
If you allow additional assumptions, you can say more about this.  For instance, if $A$ and $B$ are connective, then the Kunneth/hypertor spectral sequence
$$
Tor^{\pi_*A} (\pi_* fib(f), \pi_* M) \Rightarrow \pi_*(fib(f) \otimes_A M)
$$
will tell you that the connectivity of the fiber is the sum of the connectivity of $M$ and the connectivity of $f$ (in the terminology that you're using).  Roughly, then, you find that the map $M \to B \otimes_A M$ is an isomorphism on the same number of homotopy groups that $A \to B$ is.
In the special case where $fib(f)$ is $M$ itself, you're getting a fiber sequence
$$
M \otimes_A M \to M \to B \otimes_A M.
$$
However, the map on the left is the multiplication map, which is zero by the square-zero assumption, and so this degenerates to an equivalence
$$
B \otimes_A M \simeq M \vee (M \otimes_A M)[1].
$$<|endoftext|>
TITLE: Form a $\mathbb{Z}^d$ lattice cycle from given lengths
QUESTION [6 upvotes]: Suppose you are given a list of integer lengths,
e.g.,
$(5,3,2,2,1,1,2,1,1)$.
The task is to decide if they can form a closed cycle
in $\mathbb{Z}^d$ by connecting segments of those
lengths in order, each parallel to a coordinate axis,
each connecting two lattice points,
each orthogonal to its neighbors.
This latter condition forbids connecting two segments collinearly:
a $90^\circ$ turn is forced between each pair of segments.
In $\mathbb{Z}^2$, the decision problem is NP-complete,
because both the horizontal, and the vertical lengths must
partition equally to close: $5+1=2+2+2$ and $3+1=2+1+1$
in the example left below.
          


Thus given an even number of integers, interleaving 1's results in
a list that can form a cycle in $\mathbb{Z}^2$ iff the given integers
may be partitioned into two equal halves.
My question is:

Q.
  Is the decision problem NP-complete in $\mathbb{Z}^d$ for $d \ge 3$?

The choices of which direction to follow at each juncture
seem to complicate a reduction from Partition.
I feel like I am missing a simple argument here ...
I'd appreciate it if anyone could supply it—Thanks!
Addendum.
A potentially more interesting question lay behind what I posed above: When might the list of lengths
be formed to realize the unknot in $\mathbb{Z}^3$?
I'll post that as a separate question if I can see how to formulate it sharply.

REPLY [6 votes]: Suppose you have a machine that solves the problem in $\mathbb{Z}^{d+1}$ quickly, and you want to use it to solve the problem in $\mathbb{Z}^d$. Suppose that your input is a list $(a_1,\ldots,a_k)$ of $k$ integers each of $n$ bits, so that your total input size is $kn$. Feed the list $(a_1,k2^n,a_2,k2^n,a_3,\ldots k2^n,a_k,k(k-1)2^n)$ to your machine. The new input size is $kn + (k-1)(\log k + n) + 2\log k + n$. All the entries $k2^n,k2^n,\ldots,k2^n,k(k-1)2^n$ must apear along the same dimension, so solving this new problem in $\mathbb{Z}^{d+1}$ amounts to solving the original problem in $\mathbb{Z}^d$. Does something like this work?

REPLY [6 votes]: Suppose we are given a sequence $a_1,\dots, a_n$ (input to a partition problem). It seems that this can be encoded as a cycle problem in $\mathbb{Z}^3$ by finding two large numbers $M$ and $N$ and taking the sequence to be $$a_1, M, N, a_2, M, N, \dots, a_n, (n-1)M, (n-1)N.$$
The numbers $M$ and $N$ should be chosen so that the two last terms cannot be canceled in any other way than the obvious one. Then by the orthogonality rule, the only remaining possibility is to let the terms $a_1,\dots,a_n$ go in the third direction, and we are back to the partition problem.
This ought to work also for $d\geq 4$, although admittedly some details remain to be filled in.<|endoftext|>
TITLE: Category-theoretic characterization of locally constant sheaves
QUESTION [7 upvotes]: Let's consider $X$ a (locally connected) topological space and $\mathcal{Sh}(X)$ the topos of sheaves over $X$.
If you see sheaves as étale spaces, locally constants sheaves correspond to covering spaces.
Is there an internal (topos-theoretic) characterization of the locally constant sheaves?
I've searched in Sheaves in Geometry and Logic, but I haven't found anything, and in the nLab there seems to be something but I do not understand it (and it does not really look like an internal characterization).

REPLY [3 votes]: One can talk about locally constant objects in a topos and these lead to covering spaces for sheaves on a space. Finite covers are already in SGA or the first Johnstone topos book. Barr and Diaconescu On locally simply connected toposes and their fundamental groups begins the modern theory. See papers by Bunge for more modern formulations.<|endoftext|>
TITLE: Subcategories of abelian categories generated by finitely many objects
QUESTION [6 upvotes]: Hello!
I am trying to understand the structure of the smallest abelian subcategory of an abelian category that contains one object $X$ and all endomorphisms of that object (or rather containing a finite number of object with all morphisms between, but this seems the same to me by just taking the sum).
Just taking the intersection over all subcategories containing my object and morphisms seems a bad idea, since sums, kernels etc are just determined up to isomorphism.
 I am not bothered about having "too many" isomorphic objects in my subcategory, though. 
I stumbled over the notion of subquotients. Since kernels, cokernels are subquotients and subquotients of subquotients are subquotients it seems to me as if I could describe the objects of my subcategory the following way:
Just take all objects isomorphic to subquotients of finite direct sums of $X$.
Does this make sense?
Thank you!
Jonas

REPLY [7 votes]: It is not true in general that the abelian subcategory (by which I mean sub-abelian category) generated by an object $X$ is all subquotients of finite sums of $X$.  It is contained in these subquotients, but it might not be all of them.  This is because, for instance, not every subobject of $X$ is the kernel of an endomorphism of $X$ (or more generally, a map from $X$ to a sum of copies of $X$).
As an example, consider the abelian subcategory of $\mathbb{Z}$-modules generated by $\mathbb{Q}$.  Because any $\mathbb{Z}$-homomorphism of $\mathbb{Q}$-vector spaces is automatically $\mathbb{Q}$-linear, you get only the finite dimensional $\mathbb{Q}$-vector spaces, and don't, for instance, get the subgroup $\mathbb{Z}$ of $\mathbb{Q}$.<|endoftext|>
TITLE: Extension of weakly compact operators from $\ell_1$ into $c_0$
QUESTION [6 upvotes]: Is every weakly compact operator from $\ell_1$ into $c_0$ extendible
to any larger space? Equivalently, is every weakly compact operator from $\ell_1$ into $c_0$ extendible to $\ell_\infty$?

REPLY [4 votes]: @Joaquin:  This one pushed me. It is, IMO, one of the nicest problems on Banach space theory asked on MO.
The answer is no.  For a counterexample, take any weakly compact operator $T:\ell_1 \to c_0$ that preserves $\ell_1^n$  uniformly for all $n$.  In fact, since $\ell_1^n$ embeds isometrically into $\ell_\infty^{2^n}$, it is easy to construct one that has norm one so that for each $n$, there is a subspace of $\ell_1$ isometric to $\ell_1^n$ on which the operator acts isometrically.
Embed  $\ell_1$ isometrically into $\ell_\infty$ and assume that $T$ extends to an operator $S$ from $\ell_\infty$ into $c_0$. The operator $S$ necessarily preserves (copies of) $\ell_\infty^n$ uniformly for all $n$.  A soft way of seeing this is to pass to an ultrapower, which will be an operator from some $C(K)$ space that is an isomorphism on a copy of $\ell_1$, hence cannot be weakly compact, whence preserves a copy of $c_0$ by Pelczynski's classical theorem.  Taking adjoints, we see that $S^*$ is an operator from $\ell_1$ into an $L_1$ space that preserves $\ell_1^n$-s uniformly, hence preserves a copy of $\ell_1$, whence is not weakly compact.  But every operator from $\ell_\infty$ into a separable space is weakly compact.<|endoftext|>
TITLE: are the smooth vectors of a Frechet space dense?
QUESTION [5 upvotes]: Given an action $\alpha$ of $V$ a Lie group on $B$ a Fréchet space with seminorms $ \{ \| \cdot \|_j \} $, let $B^\infty$ be the space of smooth vectors. Is this dense in $B$? Can I guarantee it is non-empty? Is there any requirements on $G$ or $\alpha$ or $B$? For the case I am (supposed to be) working on right now, we also have that $\alpha$ is isometric for all the seminorms and strongly continuous. 
Also, since I am almost illiterate on the subject of Lie groups, I would also be thankful for some easy to read references! Thank you so much.

REPLY [3 votes]: In full generality, the answer is "no". If you have a unitary irreducible representation of a $p$-adic Lie group on a $p$-adic Banach space, then the locally constant vectors are usually not dense. There may even be no nonzero locally algebraic vector. The locally constant vectors are dense however if you have an $\ell$-adic Lie group acting on a $p$-adic space, with $\ell \neq p$ (this is a theorem of Vigneras).<|endoftext|>
TITLE: Sheaf cohomology invariant of weak homotopy type?
QUESTION [5 upvotes]: Is sheaf cohomology an invariant of the weak homotopy type? More precisely let $R$ be a commutative ring and $f:X\rightarrow Y$ a weak homotopy equivalence. Does it follow, that the induced maps $H^n(Y,\underline R) \rightarrow H^n(X,\underline R)$ are isomorphisms?
Edit: Since any space is weakly homotopy equivalent to a CW-complex, CW-complexes are locally contractible and for locally contractible spaces sheaf and singular cohomology coincide, a positive answer to this question would imply that sheaf cohomology and singular cohomology coincide for any space. This seems unlikely, but I don't know a counter example.

REPLY [7 votes]: No. For paracompact spaces sheaf cohomology coincides with Čech cohomology. In particular it applies to the closed topologist's sine curve $C$. There is a map $C \to S^1$ inducing an isomorphism on Čech cohomology, but $C$ is weakly contractible.<|endoftext|>
TITLE: Approximation by polynomials
QUESTION [8 upvotes]: Let $f:[a,b] \rightarrow \mathbb{R}$ be of class $C^n$. Let $ x_0, ..., x_m$ be different numbers from $[a,b]$. 
Does for each $\varepsilon >0$ there exist a polynom $P$ such that $P^{(k)}(x_i)=f^{(k)}(x_i)$ for $i=0,...,m$, $k=0,...,n$ 
and $sup_{x \in [a,b]} |f(x)-P(x)|< \varepsilon$?

REPLY [13 votes]: The problem may be split into two independent and classical ones: the Hermite interpolation, and the  Weierstrass approximation. 
First, we want a polynomial $p\in \mathbb{R}[x]$ with given derivatives  at some given nodes $x _ 0,\dots, x _ m $. This is an instance of the Hermite interpolation problem; yours has exactly one solution $p$ with $ \operatorname{deg}(p) < (m+1)(n+1) $ (the degree one would expect in  terms of number of linear conditions).
So, given your $f\in C^n$, you can find a polynomial with $p^{(j)}(x _ i)=f^{(j)}(x _ i)$ for all $0 \le i \le m$ and $0 \le j \le n$. 
Second, as a consequence,
$$\frac{f(x)-p(x)}{\prod _ {i=0}^ m(x- x _ i)^n}$$
is (extends to) a continuous function on $[a,b]$ that vanishes on the points $x _ i$. By the Stone-Weierstrass approximation theorem there is a polynomial vanishing on the points $x _ i$ as well, whose uniform distance from that function on the interval $[a,b]$ is less than, say,  $\epsilon (b-a)^{-n(m+1)}$. In other words, there is a polynomial $q\in \mathbb{R}[x]$ such that
$$\bigg\| \frac{f(x)-p(x)}{\prod _ {i=0}^ m(x- x _ i)^n} - q(x) \prod _ {i=0} ^ m(x- x _ i) \bigg\|_{\infty, [a,b]}  < \epsilon (b-a)^{-n(m+1)}\ , $$
therefore the polynomial $P(x):= p(x)+ q(x) \prod _ {i=0} ^ m(x- x _ i) ^{n+1}$ fullfills the requirements, for  $P^{(j)}(x _ i)=p^{(j)}(x _ i)=f^{(j)}(x _ i)$ for all $0 \le i \le m$ and $0 \le j \le n$, and 
$$\|f- P\|_{\infty,[a,b]} < \epsilon\  .$$
btw. Incidentally, some time ago I happen to notice that one can find the solution of the Hermite interpolation problem as an application of the Chinese Remainder Theorem in the ring of polynomials, and wrote here the details. 
edit. As to why  The set $A$ of all polynomial functions on $[a,b]$ that vanish at given points $x_0,\dots, x_m$ is dense in all continuous functions on $[a,b]$ that vanish in $x_0,\dots, x_r$.  One way, a bit abstract but quite immediate is, to see it as a corollary of the Stone-Weierstrass theorem  (A separating closed algebra of real valued functions on a compact space  $X$ is either $C(X)$ or a maximal ideal $M_x\subset C(X)$, the set of all functions vanishing at  $x$). Consider $X=$ the topological quotient of $[a,b]$ obtained identifying all points $x_i$ to a point $\xi$. All functions in $A$ factor through  to the quotient map, and define a closed separating algebra of continuous functions  on $X$ that vanish on the identified point $\xi$. Thus, this algebra contains all continuous functions on $X$ that vanish on $\xi$, which is the thesis read on the quotient. 
Note that the same construction holds in general, and provides a characterization of all closed algebras $A$ of continuous functions on a compact space $X$: identifying all points that are not distinguished by the functions of $A$ (that is, under the equivalence relation $x R_A y$ iff $f(x)=f(y)$ for all $f\in A$) one gets a  Hausdorff compact quotient space (whether or not $X$ is Hausdorff), and the quotient map $\pi: X\to X/{R _ A}$ induces an isometric isomorphism of algebras $f\mapsto f\circ \pi$ of either $C(X/{R_A})$ or a maximal ideal of it onto $A$; conversely, any Hausdorff quotient of $X$ produces a closed sub-algebra of $C(X)$ this way.
Another way to see it is as a corollary of the classic Weierstrass theorem: Consider $P$ as in your comment below, then add a perturbation $L$ that makes $P+L$ vanish on the points $\{x_i\}$; this has been clearly explained in Ilya Bogdanov's answer. Here you don't have derivatives and $L$ is just a Lagrange interpolation polynomial, which is small in the uniform norm because it is small on the points  $\{x_i\}$.

REPLY [3 votes]: For the sake of simplicity, let us assume that $[a,b]=[0,1]$. $C^k$ always means $C^k[0,1]$. We will even approximate $f$ in $C^n$-norm satisfying your additional condition.

As was mentioned in the comments, you can easily approximate $f$ together with all its derivatives up to $n$th uniformly by a polynomial. In fact, it is enough to approximate $f^{(n)}$ with an adequate accuracy: if $||f'-P'||_C<\varepsilon$ and $f(0)=P(0),$ then $||f-P||_C<\varepsilon.$
Now take the polynomials $Q_{ik}(x)$ such that $Q_{ik}^{(d)}(x_j)=0$ for all $d=0,\dots,n$ 
and $j=0,\dots,m$ except that $Q_{ik}^{(k)}(x_i)=1$. Such polynomials are easy to construct: for instance, one may take 
$$
  Q_{ik}(x)=c_{ik}(x-x_i)^k\prod_{j\neq i}\left((x-x_i)^{n+1}-(x_j-x_i)^{n+1}\right)^{n+1}\;\;
$$ 
for a suitable constant $c_{ik}.$ Let $M=\max_{i,k}||Q_{ik}||_{C^n}.$ Then, let the approximation in the previous paragraph be $\delta$-accurate with $\delta=\varepsilon/(2M(m+1)(n+1)).$ To correct the values of the polynomial and its derivatives at $x_i,$ it is enough to add the polynomials $Q_{ik}$ multiplied by the coefficients with absolute values $\leq\delta,$ hence the total error will be not more that $\delta+(m+1)(n+1)M\delta<\varepsilon.$<|endoftext|>
TITLE: Bedford-Taylor theory 
QUESTION [5 upvotes]: The Dirichlet problem for the Complex Monge-Ampere equation on a bounded pseudoconvex domain in $\mathbb{C}^n$ was studied in Bedford-Taylor's seminal paper wherein they defined $(dd^{c} u)^n$ for locally bounded plurisubharmonic $u$. But, they don't seem to use it in their Perron-type method, instead using a convex-measure-theoretic construction claiming that the upper envelope is not well-behaved. Now that we know more about psh functions, have people studied the Dirichlet problem without using the measure-theoretic construction of Goffman and Serrin? (and reproved Bedford-Taylor's results)

REPLY [4 votes]: In the book Degenerate Complex Monge-Ampère Equations by V. Guedj, A. Zeriahi, they circumvent the construction of Goffman and Serrin using ideas from viscosity theory.  Let $H_n^+$ denote the set of all semi-positive Hermitian $n$ by $n$ matrices and let $\dot{H}_n^+ \subset H_n^+$ denote the matrices with determinant $n^{-n}$. For $H \in  \dot{H}_n^+$ they define
$$ \Delta_H := \sum_{i,j=1}^n h_{i,j}\frac{\partial^2}{\partial z_i \partial \bar{z}_j}$$
and show that for $u \in PSH(\Omega) \cap L^\infty(\Omega)$, $0 \leq f \in C^0(\Omega)$,
$$ (dd^cu)^n \geq f \beta^n \iff \Delta_H u \geq f^{1/n} \text{ for all $H \in \dot{H}_n^+$.} $$
The envelopes are then constructed with regards to the second statement.<|endoftext|>
TITLE: Reference for restriction of a simple module over a splitting field to a smaller field?
QUESTION [9 upvotes]: This is mainly a request for a straightforward reference (preferably at textbook level).  The question comes up while responding to a question raised by non-specialists in finite group representations.    
Let $G$ be a finite group and let $F \subset E$ be finite fields, say with $E$ a splitting field for $G$ but perhaps $F$ not.    Starting with a simple module $M$ of dimension $d$ for the group algebra $E[G]$, consider the $F[G]$-module $N$ obtained by "restriction of scalars" from $M$:  here $M$ as a vector space over $F$ typically has larger dimension related to $d$ while the representing matrices for $G$ over $F$ involve a rewriting of the matrices over $E$.   While this is elementary in principle, it's a bit complicated to write down in practice.   In the process, one could chart a precise relationship between simple modules over the field $F$ (maybe not a splitting field) and those over $E$.  

Is this written down clearly for non-specialists?

Going the other way, from a module over an arbitrary field $F$ to one over a splitting field $E$, is a more standard topic in textbooks about which much can be said in prime characteristic as well as in characteristic 0 (where the Schur index becomes a leading idea).    See for example the extensive discussion in Chapter 9 "Changing the field" in Character Theory of Finite Groups by I.M. Isaacs along with the following chapter.   But I don't recall seeing comparable detail on restriction of scalars.   The question I've been discussing with colleagues isn't especially deep or advanced, but a clear reference would help.
ADDED: To make the situation more precise, it may be enough (following Geoff) to assume that $F$ is not a splitting field for the given simple $E[G]$-module $M$ while $E$ is a minimal splitting field for it.   (Or you might just take $E$ to be a minimal splitting field for $G$.)   Then if $[E:F] = n$, the restricted $F[G]$-module $N$ has dimension $nd$ over $F$.   In this situation one expects $N$ to be simple for the smaller group algebra, which in turn yields an $nd$-dimensional module over $E$ after tensoring, etc.   The fields involved being finite, there is no Schur index complication and the Galois group of $E/F$ is easy to work with.   So the matrix version over $F$ of the given representation over $E$ should be readily described in block form.   With luck, this is all I've been asked to explain in special cases, but it would be reassuring to see a concise published version which I couldn't readily extract from books I've seen.

REPLY [2 votes]: One uses a matrix version of Hilbert's Satz 90 to answer Geoff's comment "If the field $E$ is not this minimal field, it seems less obvious to me how to realise the representation over the subfield generated by the traces, and I suppose this may be the real issue you need to resolve." An easy-to-read description is given in
S.P. Glasby and R.B. Howlett. Writing representations over minimal fields, Comm. Algebra 25 (1997), 1703--1712.<|endoftext|>
TITLE: Can a PDE constrain the degree of a $C^\infty$ map germ?
QUESTION [13 upvotes]: Let $\pi:E\to M$ be a smooth vector bundle over a smooth manifold, with $\text{rank}(E)=\text{dim}(M)$.  For a section $\sigma$ of $E$ with a zero at $p\in M$, define the degree of the zero at $p$ to be the topological degree of the induced map from a small sphere in $T_pM$ to a small sphere in $E_p$.
One motivation for studying degrees of zeros is that they contain information about the topology of $E$.  I think the following is true, although I couldn't find a good reference:
Theorem 1 (Hopf index theorem).  Suppose the zeroes of $\sigma$ are the isolated points $p_1, \ldots p_k$, with degrees $d_1,\ldots d_k$ respectively.  Then the Euler class of $E$ is $\chi(E)=\sum_{i=1}^kd_i$.
With this as motivation, my first question, the one stated in the title, is roughly (see the Example for an idea of what I'm getting at, and feel free to suggest a sharper version):

Are there conditions on [say, the symbol of] a linear differential operator $D:E\to F$, such that [some constraint] is satisfied by degree of any zero $p\in M$ of any local solution $\sigma\in\Gamma(E)$ to the PDE $D\sigma=0$?

Example. If $M$ is a Riemann surface and $E$ a holomorphic line bundle over it, the kernel of the delbar operator $\overline{\partial}:E\to T^{0,1}M\otimes E$ is precisely the holomorphic sections of $E$.  By complex analysis, zeroes of holomorphic functions have positive degree.
Theorem 1 then yields the standard result that if a line bundle admits a global holomorphic section then its Euler class (aka first Chern class) is nonnegative.

Here's an idea I had for trying to prove a theorem of the sort I ask for in Question 1.  Recall the definition of the local ring of a zero $p\in M$ of a section of E: 
Write  $\mathcal{O}_p$ for the ring of germs of smooth functions about $p$.
Definition.  Let $\sigma\in\Gamma(E)$ be a smooth section which vanishes at $p$.  The local ring of the germ $[\sigma]_p$, denoted $Q([\sigma]_p)$, is the quotient $\mathcal{O}_p/([\sigma]_p)$, where $([\sigma]_p)$ is the ideal of $\mathcal{O}_p$ generated by "components of $\sigma$": $([\sigma]_p)=\ <\{[v(\sigma)]_p:v\text{ a nonvanishing section of }E^*\}> \ \subseteq \mathcal{O}_p$.
Theorem 2 (Eisenbud-Levine-Khimshiashvili). Suppose $p$ is a zero of $\sigma$, and the local ring $Q([\sigma]_p)$ is a finite-dimensional algebra over $\mathbb{R}$.  Then there is a canonical quadratic form on $Q([\sigma]_p)$, such that the degree of the zero of $\sigma$ at $p$ can be calculated as this quadratic form's signature. 
Because a system of PDE is precisely a constraint on the local behaviour of a section, it seems plausible that local rings of zeros of solutions of a PDE might have interesting properties.

Are there conditions on [say, the symbol of] a linear differential operator $D:E\to F$, such that [some constraint] is satisfied by the signature of the local ring $Q([\sigma]_p)$ of any zero $p\in M$ of any local solution $\sigma\in\Gamma(E)$ to the PDE $D\sigma=0$?

Example. As in the previous example, let $E$ be a holomorphic line bundle over a Riemann surface $M$.  By manipulating the Cauchy-Riemann equations, one can (I think!) classify the possible local rings of zeroes of a holomorphic section, and show that all of them have positive signature. 
Theorem 2 then yields an alternative proof of the quoted result that zeroes of holomorphic functions have positive degree.

REPLY [3 votes]: This question is a bit too general, and I think that it goes beyond  the principal symbol. There clearly are constraints. Take for example, the Laplace operator $\Delta$ acting on sections of the trivial complex line bundle on the round sphere  $S^2$. The sections of this bundle are smooth  complex-valued   functions and  the functions in the kernel are  constant.  
Now replace this operator  with the operator
$$\Delta_n=\Delta-n(n+1)Id,$$
where $n$ is a nonnegative integer. The operator $\Delta_n$ has the same principal symbol as $\Delta$.  The quantity $n(n+1)$ is an eigenvalue of $\Delta$ and the eigenfunctions are the restrictions to $S^2$ of the   degree $n$ homogeneous polynomials (with complex coefficients) in $3$-variables.    The  behavior of such a polynomial $P(x,y,z)$ in a neighborhood  of  North Pole on the $2$-sphere  is equivalent  to the behavior near zero of the  polynomial map
$$\mathbb{R}^2\ni (s,t) \mapsto P(s,t,1)\in\mathbb{C}$$
Above, the affine plane $(s,t)\mapsto (s,t,1)\in\mathbb{R}^3$ is the (affine) tangent plane to the sphere at the north pole. The real and imaginary parts of this polynomial map  can be any  polynomials of degree $\leq n$.<|endoftext|>
TITLE: Maps between groups inducing isomorphisms between all nilpotent quotients
QUESTION [7 upvotes]: Let $G_1$ and $G_2$ be finitely presentable groups and let $f : G_1 \rightarrow G_2$ be a surjective homomorphism.  Denoting the kth term of the lower central series of $G_i$ by $\gamma_k(G_i)$, assume that $f$ induces an isomorphism $G_1 / \gamma_k(G_1) \rightarrow G_2 / \gamma_k(G_2)$ for all $k \geq 1$.  Is $f$ 
necessarily an isomorphism?
EDIT : I forgot the obvious assumption that the intersection of the lower central series of $G_i$ is trivial for $i=1,2$.

REPLY [2 votes]: Please allow me to give some explanation for answer of Mark Sapir for the beginners in order to avoid confusion for notation: Assume that $f$ is not an isomorphism. Since $f$ is surjective homomorphism, then $N=Ker f \ne 1$ and $G_2\cong G_1/N$. Without loss of generality, we can assume that $G_2=G_1/N$. Since the intersection of the lower central series is trivial by assumption, there exists $n$ such that $N $ is not contained in $\gamma_n(G_1)$. Now $\gamma_n(G_2)=\gamma_n(G_1)N/N$, and then $G_2/\gamma_n(G_2) \cong G_1/\gamma_n(G_1)N$, a proper homomorhic image of $G_1/\gamma_n(G_1)$. Then we get a contradiction as Mark Sapir.<|endoftext|>
TITLE: A curious definite integral
QUESTION [7 upvotes]: I was playing around with $\mathcal{I}=\int_0^1\text{frac}({\frac{1}{x^n}}) dx$, where $\text{frac(.)}$ is the fractional part function, and I discovered that 
$$
\mathcal{I} =
\begin{cases}
\frac{1}{1-n} & n \leq 0 \\
\frac{1}{1-n} - \zeta(1/n) & n \in (0,1) \\
1 - \gamma & n = 1
\end{cases}
$$
Where $\gamma$ is the Euler-Mascheroni constant. And $\zeta(s)$ is the Riemann Zeta function.
My questions are
1) Is anything similar known? Any other definite integrals relating the fractional part and the Riemann zeta function?
2) It is apparent from the above that $\zeta(1/n)
< \frac{1}{1-n}$ for $n\in(0,1)$. Now I've found out that the same inequality holds even when $n>1$. However, the same technique for evaluation for $\mathcal{I}$ doesnt work when $n>1$, as the computation depends on the sum $$\sum_{n=1}^\infty \frac{1}{n^p}$$
which diverges when $p\leq 1$. And if $n>1$, we have that $1/n<1$ so one of the sums in the process of evaluation becomes divergent. 
I'm guessing that some complex analysis is required to overcome this difficulty. But I'm not familiar with that as of now. 
I'd be grateful for any comments on this.
Thank you. :)

REPLY [8 votes]: This really isn't particularly remarkable. By definition,
$$\zeta(s) = \sum_{n = 1}^{\infty}{\frac{1}{n^s}} = s \int_{1}^{\infty}{\frac{\lfloor x \rfloor}{x^s} \frac{dx}{x}}$$
for $\Re(s) > 1$, where the second inequality follows by a partial summation argument, and $\lfloor x \rfloor$ is the floor function. We can write $\lfloor x \rfloor = x - \{x\}$, where $\{x\}$ is the fractional part of $x$, and split up the integral above in order to find that
$$\zeta(s) = \frac{1}{s - 1} + 1 - s \int_{1}^{\infty}{\frac{\{ x \}}{x^s} \frac{dx}{x}}.$$
This is now defines a meromorphic function on $\Re(s) > 0$ with just a simple pole at $s = 1$ with residue $1$. In any case, this is a well-known integral representation of $\zeta(s)$.
This should explain your result for $0 < n < 1$, by making the change of variables $n = 1/s$, and then making the change of variables $y = x^{-1/n}$ in the integral. Also, the $n = 1$ case is definitional, as the Euler--Mascheroni constant is defined to be
$$1 - \int_{1}^{\infty}{\frac{\{ x \}}{x^2} dx}$$
A partial summation argument should show why this is the same as $\lim_{x \to \infty} \left( \sum_{n \leq x}{\frac{1}{n}} - \log x\right)$.
You can also see why
$$\zeta(\sigma) < \frac{\sigma}{\sigma - 1}$$
for all $\sigma > 0$; this is simply because
$$\sigma \int_{1}^{\infty}{\frac{\{ x \}}{x^{\sigma}} \frac{dx}{x}} > 0.$$
Note that none of this really uses complex analysis, apart from the fact that $\zeta(s)$ isn't implicitly defined for $0 < \Re(s) < 1$ initially, so you need to take a leap of faith to believe that the integral representation of $\zeta(s)$ is actually valid in this strip.<|endoftext|>
TITLE: Eigenvectors and eigenvalues of a tridiagonal Toeplitz matrix
QUESTION [13 upvotes]: Is it possible to analytically evaluate the eigenvectors and eigenvalues of the following $n \times n$ tridiagonal matrix
$$
 \mathcal{T}^{a}_n(p,q) = \begin{pmatrix}
  0 & q & 0 & 0 &\cdots & 0 & 0 & 0 \\\
  p & 0 & q & 0 &\cdots & 0 & 0 & 0 \\\
  0 & p & 0 & q &\cdots & 0 & 0 & 0 \\\
  \vdots  & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots& \vdots \\\
  0 & 0 & 0 & 0 &\cdots & p & 0 & q \\\
  0 & 0 & 0 & 0 & \cdots & 0 & p & 0
 \end{pmatrix} 
$$
where $p>0$ and $q > 0$?
Furthermore, is it possible to do the same for the following tridiagonal circulant matrix?
$$
 \mathcal{T}^{b}_n(p,q) = \begin{pmatrix}
  0 & q & 0 & 0 &\cdots & 0 & 0 & p \\\
  p & 0 & q & 0 &\cdots & 0 & 0 & 0 \\\
  0 & p & 0 & q &\cdots & 0 & 0 & 0 \\\
  \vdots  & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots& \vdots \\\
  0 & 0 & 0 & 0 &\cdots & p & 0 & q \\\
  q & 0 & 0 & 0 & \cdots & 0 & p & 0
 \end{pmatrix} 
$$

REPLY [5 votes]: @ Emilio: Could you please show how you get $\det(\mathcal{T}_n(p,q)-\lambda)=2(pq)^n T_n(-\lambda/2)$?
Sorry I wanted to comment another answer. I do not know how to delete this post.
Edit
I think the first problem can be reduced to Chebyshev polynomials of the second kind  ${U_n}(x)$ because  
$$\det \left( {{T_n}(p,q) - \lambda } \right) =  - \lambda \det \left( {{T_{n - 1}}(p,q) - \lambda } \right) - pq\det \left( {{T_{n - 2}}(p,q) - \lambda } \right).$$
implies
$$\det \left( {{T_n}(p,q) - \lambda } \right) ={(\sqrt {pq} )^n}{U_n}\left( {-\frac{\lambda }{{2\sqrt {pq} }}} \right). $$
From the fact that the zeros of ${U_n}(x)$ are  $\cos \frac{{k\pi }}{{n + 1}}$ the eigenvalues are
$2\sqrt {pq} \cos \frac{{k\pi }}{{n + 1}}.$<|endoftext|>
TITLE: Integral homology circles
QUESTION [7 upvotes]: An integral homology circle is a CW-complex whose integral homology groups are isomorphic to those of the circle.
If $X$ is an integral homology circle with $\pi_1(X)=\mathbb{Z}$, must $X$ be homotopically equivalent to a circle?

REPLY [21 votes]: Example 4.35 of "Algebraic Topology" by Hatcher provides a counter-example. The space is $X=(S^1\vee S^n)\cup e^{n+1}$, where the $(n+1)$-cell is attached along the element $2t-1\in \pi_n(S^1\vee S^n)\approx \mathbb{Z}[t,t^{-1}]$. On the level of cellular homology this has the same effect as attaching the cell along a degree $2-1=1$ map, so $X$ has the integral homology of $S^1\vee D^n\simeq S^1$. But we have $\pi_{n}(X)\approx \mathbb{Z}[t,t^{-1}]/(2t-1)\approx \mathbb{Z}[\frac{1}{2}]$, so $X$ is not homotopy equivalent to $S^1$.
However if $X$ is an integral homology circle with $\pi_1(X)=\mathbb{Z}$ and $\pi_1(X)$ acts trivially on $\pi_i(X)$ for all $i\geq 2$, then $X$ is homotopy equivalent to a circle (Proposition 4.74 of AT).<|endoftext|>
TITLE: Problems with Journal Submissions due to arXiv Submission?
QUESTION [13 upvotes]: Hello, 
I wanted to ask people if they know of any journals that will not accept papers for submission if they have been already posted on arxiv. I am personally interested in Logic Journals, but people can contribute their experiences from any field. 
I guess I could read all the fine print the publishers have on their webpages, but I am trying to see if anyone had any problems.
Just to contribute my part, I didn't have problems with the Journal of Symbolic Logic and the Annals of Pure and Applied Logic. 
Thank you for your help.

REPLY [11 votes]: This recent article in the AMS notices has a handy checklist of the explicit rights given to authors in the standard publishing agreements with various publishers:
http://www.ams.org/notices/201203/rtx120300436p.pdf
Notice that there is often a distinction between the treatment of an  "author-created version" and the "publisher version" of an article.<|endoftext|>
TITLE: TP($\omega_2$) and the continuum
QUESTION [5 upvotes]: I think that in all classical models of TP($\omega_2$) we have $2^{\omega_0}=\omega_2$. Is there a known model of TP($\omega_2$) + $2^{\omega_0}>\omega_2$ at all?

REPLY [8 votes]: I think Spencer Unger in "Fragility and indestructibility of the tree property" proves that if one adds an arbitrary number of Cohen reals to Mitchell's model, then the tree property survives (see http://www.math.cmu.edu/~sunger/).<|endoftext|>
TITLE: Return probabilities for random walks on infinite Schreier graphs
QUESTION [9 upvotes]: Question:  Is there a sequence $(\delta_n)_n$ of real numbers with $\delta_n \to 0$ as $n \to \infty$, such that the following holds:
Let $F$ be a free group on two generators, let $F \curvearrowright X$ be a transitive action on an infinite set, and let $x \in X$. Then, the probability that a random word of length $n$ in $F$ fixes $x \in X$ is smaller than $\delta_n$.

I would like to consider random unreduced words (so that there are $4^n$ such words of length $n$ and each is equally likely), but probably this does not matter much. It corresponds to the nearest neighbor random walk on the Schreier graph corresponding to the action of $F$ on $X$.
It is clear that for each individual action $x \in X$, the return probability decays; and it seems plausible that this happens uniformly over all actions.

REPLY [2 votes]: While Fedja has given a concrete argument, it occured to me that it also follows from (more or less) abstract nonsense, by compactness-and-contradiction. 
Let me explain in more detail. Let $\delta_n$ be the supremum over all actions $G \curvearrowright X$ and $x \in X$ of the return probability after $n$ steps.
We need to show that $\delta_n \to 0$ as $n \to \infty$. It is clear that $\delta_{2n}$ is decreasing. This follows since the return probabilities after 2n steps for each individual action is decreasing, since they are given by the moments of positive contraction on a Hilbert space. Now, if $\inf_n \delta_{2n} = \delta >0$, then we find a sequence of actions $G \curvearrowright X_n$ and $x_n \in X_n$, such that the return probability after $2n$ (and hence $2k$ for $k<n$) steps to $x_n$ is greater or equal $ \delta/2$.
Now, the space of transitive actions with chosen point is compact! Indeed, it is just the Chabauty space of subgroups of the free group, which is obviously compact. Moreover, the return probabilities are continuous functions on the space of transitive actions with chosen point. Hence, we find a limit point in this space and hence a limit action, for which the return probability after $2n$ steps is greater or equal $\delta/2$ for all $n$. Now, the only thing we can quickly say about the limit is that it is not finite. Indeed, the finite transitive actions are all isolated in the space of actions. This is a contradiction since we know already that for each individual action the return probabilities tend to zero.<|endoftext|>
TITLE: Can the infinite von Dyck groups be subgroups of $SU(n)$?
QUESTION [5 upvotes]: I asked this over stackexchange with no answer so here we are,
I know by constructing some particular cases that I can find unitary matrices $X$, $Y$ and $Z$ such that
$X^m = Y^n = Z^p = XYZ = 1$
with 
$$
\frac{1}{m} + \frac{1}{n}+\frac{1}{p} < 1
$$
indicating an infinite von Dyck group unless the fact that the matrices are unitary implies some additional, non-trivial relations between $X$, $Y$ and $Z$. Is it possible for infinite von Dyck or triangle groups to be subgroups of $SU(n)$?
Also, can somebody point to me some references on representation theory of infinite discrete groups?

REPLY [6 votes]: The question was somewhat sloppy so it was unclear if it was about some or all von Dyck groups admitting embeddings in $U(n)$. If the question was some then Victor's answer clearly suffices. 
If the question was about all von Dyck groups, then Victor's answer almost gets there but not quite, since only finitely many of these groups are arithmetic (first observed by Takeuchi who actually listed them all, but also follows from far more powerful finiteness theorem of Borel and Prasad: Only finitely many arithmetic groups with covolume $\le const$).
For non-arithmetic groups one has to worry about all Galois conjugations yielding noncompact forms of $SO(3)$. For non-arithmetic groups, one has to use a bit more elaborate version of Victor's answer. First, observe that every von Dyck group $\Lambda$ contains a closed surface subgroup $\Gamma$ of finite index. I will consider only the case when the genus is $\ge 2$ since virtually abelian case is much easier. Then, being a closed surface group,  $\Gamma$ is isomorphic to a cocompact arithmetic subgroup $\Gamma'$ of $O(2,1)$. Now, $\Gamma'$ admits an embedding $\rho$ to some $U(n)$ as explained by Victor. To embed (unitarily) the original group $\Lambda$, use the representation induced from $\rho: \Gamma\to U(n)$ to $\Lambda$.  
A more interesting question is if every finitely generated subgroup of $SL(n, {\mathbb R})$ embeds in some orthogonal group. It seems that $SL(m, {\mathbb Z})$ is a counter-example for that. 
Update: a. Indeed, $SL(n, {\mathbb Z})$ does not embed in any $U(m)$. The same applies to all  finitely generated groups which contain distorted cyclic subgroups, see Yves' comments. The simplest example will be Baumslag-Solitary groups $BS(1,p)=\langle a, b| aba^{-1}=b^p\rangle$, $p>1$.  
b. There is no way to construct sytematically all irreducible finite dimensional unitary representations of von Dyck groups since (starting in certain dimension) there will be continuous families of such representations. (The situation is very much unlike theory of Lie groups.) One way to see this is by observing that the spaces of irreducible unitary representations (modulo conjugation) of surface groups have positive dimension. Using induction (of representations) one can then prove the same for von Dyck groups. 
Here are few links where one can read about surface group representations (including unitary ones): 
http://www.math.u-psud.fr/~labourie/preprints/pdf/surfaces.pdf
http://arxiv.org/pdf/math.GT/0509114.pdf
http://arxiv.org/abs/0710.5263
The old paper by Andre Weil "Remarks on Cohomology of Groups" 
http://www.jstor.org/stable/1970495 is particularly relevant for the discussion of von Dyck groups $\Gamma$, since Weil computes dimension of the Zariski tangent space of $Hom(\Gamma, G)$ in the end of the paper, where $G$ is an arbitrary Lie group. (He does much more, of course.) 
c. One can show that every hyperbolic von Dyck group $D(p,q,r)$, except for $D(2,6,6)$ and $D(2,4,6)$, admits a homomorphism to $PU(2)$ whose image is dense. In particular, for every von Dyck group $\Gamma$ (with the above exceptions), one gets an irreducible representation to $U(n), n\ge 3$. By working more carefully, one can probably prove the same for the two exceptions. 
d. Situation with irreducible representations to $SU(2)$ is more complicated: There will be more exceptions.<|endoftext|>
TITLE: Reference request: Martingale decompositions (positive/negative and u.i./singular)
QUESTION [7 upvotes]: For a paper I am writing, I need these two facts.  The proofs are fairly short, but I would rather just cite them. This is for martingales index by natural numbers.  Also, I call a martingale which converges to 0 "singular".  I have also seen them called "potentials".   


Is there a good reference for these
  two facts?
Do these decompositions have standard
  names?
Is there a standard term for a martingale which converges to 0?


Below, $\Vert M \Vert$ is the $L^1$-bound of the martingale $M_k$.
Decomposition 1. Let $(M_{k})$ be an $L^{1}$-bounded martingale with respect to the filtration $({\mathcal{F}}_{k})$. Then there are two nonnegative martingales $(P_k)$ and $(N_k)$ such that such that $M_{k}=P_k-N_k$   a.e. for all $k$, and $\left\Vert M\right\Vert =\left\Vert P\right\Vert +\left\Vert N\right\Vert = \Vert P_0 \Vert_1 + \Vert N_0 \Vert_1$. 
Further, this decomposition is a.e. unique;
$(P_k)=\sup_{n\geq k}E[[M_{n}]^{+}\mid\mathcal{F}_{k}]$ a.e.;
$N_k=\sup_{n \geq k}E[[M_{n}]^{-}\mid\mathcal{F}_{k}]$ a.e.;
$\lim_{k\rightarrow\infty}P_k=[\lim_{k}M_{k}]^{+}$ a.e.; and 
$\lim_{k\rightarrow\infty}N_k=[\lim_{k}M_{k}]^{-} a.e.$
Decomposition 2. Let $(M_{k})$ be an $L^{1}$-bounded martingale with respect to the
filtration $(\mathcal{F}_{k})$ and let $M_{\infty}=\lim_{n}M_{n}$.
Then there is a uniformly integrable martingale $(U_k)$ and
a singular martingale $(S_k)$ such that $M_{k}=U_k+S_k$
a.e. for all $k$. 
Further, this decomposition is a.e. unique;
$U_k=E[M_{\infty}\mid\mathcal{F}_{k}]$ a.e.;
$S_k=E[M_{k}-M_{\infty}\mid\mathcal{F}_{k}]$ a.e.; and 
$\left\Vert M\right\Vert =\left\Vert U\right\Vert +\left\Vert S\right\Vert $.

REPLY [12 votes]: These are the Krickeberg and Riesz decompositions, respectively. A good reference is section 4 of Chapter V in Probabilities and Potential B by Claude Dellacherie and Paul-Andre Meyer.<|endoftext|>
TITLE:  Is residual finiteness a profinite property?
QUESTION [6 upvotes]: Let $G,H$ be infinite finitely generated groups such that $F(G)=F(H)$. Where $F(G)$ denotes the isomorphism classes of finite quotients of $G$. Let $G$ be residually finite. can we say that $H$ is residually finite too?

REPLY [10 votes]: As Steve D observes, the answer is 'no'. This is a very general phenomenon.  Let $G$ be any finitely generated residually finite group, and let $S$ be any finitely generated group with no finite quotients (Higman gave an example, or use an infinite simple group). Then $F(G)=F(G\times S)$, but $G\times S$ is certainly not residually finite.<|endoftext|>
TITLE: Doubly covering an even lattice
QUESTION [5 upvotes]: I have read that there is a way to construct a group which is a double cover of an even lattice. The very tantalizing thing about this is that if the even lattice is chosen to be the Leech lattice, the resulting double cover is supposed to admit a natural action of the Monster group.  
(i) What is a good place to read about how to construct this double cover?
(ii) What is a good place to read about how to define the Monstrous action on the double cover of the Leech lattice?

REPLY [5 votes]: There is a brief description of the double cover of an even lattice on page 2 of Borcherds's paper Vertex algebras, Kac-Moody algebras and the monster (number 4 on the page).  It is introduced there as a set of properties that uniquely define it up to isomorphism, but without a construction.  For a detailed construction, see Chapters 5 and 7 of Vertex Operator Algebras and the Monster by Frenkel, Lepowky, and Meurman.
The automorphism group of the double cover of an even lattice $L$ is an extension of $\operatorname{Aut}(L)$ by $(\mathbb{Z}/2\mathbb{Z})^{\text{rank}(L)}$, and it is usually non-split.  For the Leech lattice, you do not get the monster, but you get something closely related to a large subgroup.  In more detail, the automorphism group of the double cover of Leech naturally acts on the Leech lattice vertex algebra, and this vertex algebra in turn can be used to construct the monster vertex algebra using a "twisted module".  The automorphism groups then yield a diagram of the following form:
$$2^{24}.Co_0 \to 2^{24}.Co_1 \leftarrow 2^{1+24}.Co_1 \hookrightarrow \text{Monster}.$$
The leftmost group is the automorphism group of the double cover of Leech, the second group is the image of its action on a fixed point subalgebra of the Leech lattice vertex algebra under an involution, the third group is a central extension that acts on a fixed point submodule of the twisted module, and it is the centralizer of an element of order 2 in the monster.  Frenkel, Lepowsky, and Meurman constructed the monster action by extracting extra symmetry from the direct sum of the fixed point subalgebra and the fixed point submodule - this required a large fraction of a book.<|endoftext|>
TITLE: Quote about errors in math writing
QUESTION [11 upvotes]: I'm searching for the original reference of a quote that went something like:
"Errors in a mathematics text add an element of surprise to an otherwise predictable plot."
I believe it may have been from the 1700s or 1800s.

REPLY [4 votes]: A resource might be
Maurice Lecat, Erreurs de mathématiciens des origines à nos jours, Bruxelles : Castaigne, 1935 
Cheers, Scott<|endoftext|>
TITLE: What are p-adic period rings?
QUESTION [14 upvotes]: I'm reading Illusie's survey on Crystalline cohomology, and I found him talking about those $p$-adic period rings like $B_{\text{dR}}, B_{\text{cris}}$. Can anybody explain what they are and give some heuristic on why they are what they are and what they are good for? So far I've only seen them appear in comparison isomorphisms. Are they just there to make the comparisons work?
I think I found a related question here: Fontaine's rings of periods but answers there simply didn't say much about it.

REPLY [18 votes]: Here are two examples where these period rings play a crucial rôle.  They represent the point of view of a spectator.  It is to be hoped that some of the actual players --- many of whom have enriched MO --- will chime in with their favourite examples.
$B_{dR}$.  
Cuspidal eigenforms $f$ (of some level and weight) give rise to galoisian representations $\rho_{f,p}:\mathrm{Gal}(\overline{\mathbf{Q}}|\mathbf{Q})\to\mathrm{GL}_2(\overline{\mathbf{Q}}_p)$ 
for each prime $p$ (Shimura, Deligne, Serre).  Such representations are called modular.
They are unramified away from finitely many primes (those which divide the level) and, crucially, derahmian (=$B_{dR}$-admissible) at $p$.
It is natural to ask :  Which galoisian representations $\rho:\mathrm{Gal}(\overline{\mathbf{Q}}|\mathbf{Q})\to\mathrm{GL}_2(\overline{\mathbf{Q}}_p)$ arise from some cuspidal eigenform ? 
Fontaine and Mazur conjectured that the necessary conditions for modularity enumerated above (along with other obvious ones) are also sufficient.  This is now almost a theorem (Kisin-Emerton); see the recent Bourbaki talk by Laurent Berger.
$B_{cris}$.  
Abelian varieties $A$ of dimension $g$ over a finite extension $K$ of $\mathbf{Q}_l$ give rise to galoisian representations 
$\rho_{A,p}:\mathrm{Gal}(\overline{K}|K)\to\mathrm{GL}_{2g}(\mathbf{Q}_p)$
coming from the galoisian action on the $p$-power torsion points of $A$ (Weil, Tate).  Does $\rho_{A,p}$ tell us whether $A$ has good reduction or not ?
The Néron-Ogg-Shafarevich theorem says that if $l\neq p$, then $A$ has good reduction if and only if the representation $\rho_{A,p}$ is unramified (see the Serre-Tate paper in the Annals).
What happens if $l=p$ ?  Fontaine proved in this case that if $A$ has good reduction, then the representation $\rho_{A,p}$ is crystalline ($=B_{cris}$-admissible).  Conversely, Coleman and Iovita have proved that if $\rho_{A,p}$ is crystalline, then $A$ has good reduction.
These are of course only two of the many things for which $B_{dR}$ (resp. $B_{cris}$) are essential.
Addendum An expert (who wants to remain anonymous) has pointed out to me that another proof of the implication "$\rho_{A,p}$ is crystalline $\Longrightarrow$ $A$ has good reduction" (in the case $l=p$) can now be given by combining an old result
(i) (Grothendieck, SGA7) If $\rho_{A,p}$ comes from a $p$-divisible group, then $A$ has good reduction,
with a conjecture of Fontaine as proved by
(ii) (Breuil, Annals 2000 for $p\neq2$, Kisin, Durham symposium 2007 for $p=2$)  If $\rho_{A,p}$ is crystalline, then it comes from a $p$-divisible group.<|endoftext|>
TITLE: Calculating Mayer-Vietoris efficiently
QUESTION [22 upvotes]: This is a question whose motivation and framing seem to involve a lot of topology, but which I suspect comes down to some simple and standard combinatorics that's probably recorded in a book somewhere. To draw in the nLab people, I'll say that I also considered entitling this "categorifying Mobius inversion".
Let $X$ be a topological space, and let $U_i$, $i \in I$, be a finite collection of open sets of $X$ such that

$X = \bigcup U_i$
For any two sets $U_i$ and $U_j$ in the collection, $U_i \cap U_j$ is also in the collection.

Suppose that I know all of the $H^{\ast}(U_i)$'s, and all of the restriction maps between them, and I would like to compute $H^{\ast}(X)$. 
One way is to compute $H^{\ast}(U_1)$, then $H^{\ast}(U_1 \cup U_2)$, then $H^{\ast}(U_1 \cup U_2 \cup U_3)$, and so forth, successively using Mayer-Vietoris to put in each new set.
I can also do it all in one go, by using the Mayer-Vietoris spectral sequence. Let $J \subseteq I$ be the set of indices $j$ such that $U_j$ is not contained in any other $U_i$. As explained here, one way to think of this is that we have an exact complex of sheaves. 
$$0 \to \mathbb{Z}(X )\to \bigoplus_{j \in J} \mathbb{Z}(U_j) \to \bigoplus_{j_1, j_2 \in J} \mathbb{Z}(U_{j_1} \cap U_{j_2}) \to \cdots \quad (\ast)$$
(See the comments on that question for issues about whether one should be using the extension by zero or the pushforward; which I'm not sure ever got resolved. I should probably get that right at some point, but it isn't what I want to focus on, so we can switch to covers by closed sets if that will avoid focusing on that point.)
It seems like sometimes one can use knowledge of the relations between the $U$'s to shorten the resolution $(\ast)$. For example, suppose that $U_1 \cap U_2 = U_1 \cap U_3 = U_2 \cap U_3 = U_4$. Then the complex $(\ast)$ looks like
$$0 \to \mathbb{Z}(X) \to \mathbb{Z}(U_1) \oplus \mathbb{Z}(U_2) \oplus \mathbb{Z}(U_3) \to \mathbb{Z}(U_4)^{\oplus 3} \to \mathbb{Z}(U_4) \to 0.$$
But there is a shorter resolution
$$0 \to \mathbb{Z}(X) \to \mathbb{Z}(U_1) \oplus \mathbb{Z}(U_2) \oplus \mathbb{Z}(U_3) \to \mathbb{Z}(U_4)^{\oplus 2} \to 0. \quad (\ast \ast)$$
Let $I$ be the poset of containment relations between the $U_i$. (Since the collection $U_i$ is closed under intersection, $I$ has joins and, if we adjoin an extra minimal element $0$ of $I$, then $I$ is a lattice.) I am looking for a recipe which would look at the poset $I$ and spit out the complex $(\ast \ast)$.
Mobius inversion tells me that the sheaf $\mathbb{Z}(U_i)$ should be used "$\mu(0,i)$ times", where $\mu$ is the Mobius function and the scare quotes are because using $U_i$ in an odd cohomological degree counts negatively. For example, the double occurrence of $U_4$ in $(\ast \ast)$ reflects that $\mu(0,2) = 2$ for this poset. So this is why I say that I want to "categorify Mobius inversion" -- I want to turn that number into a vector space (or collection of vector spaces).
Thanks!

REPLY [6 votes]: To be safe, let me assume the cohomologies are taken with coefficients in a field, like $\mathbf{C}$.
Let $I' \subset I$ be the indices for which $U_i$ is nonempty.  The incidence algebra of $I'$ is a finite-dimensional algebra that naturally acts on the vector space of $\mathbf{C}$-valued functions on $I'$.  Your "categorified Mobius inversion" amounts to finding the minimal projective resolution of this module.
Let $f:X \to I'$ be the function that carries $x$ to the index of $\bigcap_{i \in I \mid x \in U_i} U_i$.  This function is continuous for topology on $I'$ whose open subsets are order ideals.  The Mayer-Vietoris spectral sequence for the cover is also the Leray spectral sequence for the map $f$ and the constant sheaf .
$$
E_2^{st} = H^s(I';R^t f_* \mathbf{C}) \implies H^{s+t}(X)
$$
A sheaf on a finite topological space like $I'$ is the same data as a functor out of $I'$ regarded as a poset, and is also the same data as a module over the incidence algebra of $I'$.  If $\mathcal{F}$ is a sheaf, the corresponding functor $F$ is given by the formula
$$
F(i) = \Gamma(\text{minimal open neighborhood of $i$};\mathcal{F})
$$
The corresponding module $M$ is the direct sum of all the $F(i)$.  Under this correspondence:

The sheaves $R^t f_* \mathbf{C}$ take the value $H^t(U_i;\mathbf{C})$ at $i$.
Projective modules over finite dimensional algebras have a Krull-Schmidt property.  In the case of the incidence algebra the indecomposable projectives are parametrized by $i \in I'$.  The projective $P_{i}$ is given by
$$
P_i(j) = \begin{cases}
\mathbf{C} & \text{if $j \leq i$} \\
0 & \text{otherwise}
\end{cases}
$$
Homomorphisms out of $P_i$ compute the value of the functor at $i$.
The constant sheaf on $I'$ is the module $\mathbf{C}^{I'}$.

As $H^s(I';-) = \mathrm{Ext}^s(\text{constant sheaf},-)$, a projective resolution of $\mathbf{C}^{I'}$ gives a chain complex computing $H^s(I';-)$ and the $E_2$ page of the spectral sequence.  The theory of finite-dimensional algebras says that there is a unique minimal resolution (it appears as a subquotient of any other projective resolution) of $\mathbf{C}^{I'}$, or of any other finite-dimensional module $M$.  One computes it by taking the projective cover of $M$, call it $P_M \to M$, next taking the projective cover of the kernel of $P_M \to M$, and so on.<|endoftext|>
TITLE: Covering relations in $K\backslash G/B$
QUESTION [6 upvotes]: Let $G$ be a simply connected complex Lie group, $\theta$ an involution,
and $K = G^\theta$ the fixed point subgroup. Pick a $\theta$-invariant
Borel subgroup $B$. Then there is a natural
map $K\backslash G \to G$ taking $Kg \mapsto \theta(g^{-1}) g$,
inducing a map $\varphi : K\backslash G/B \to B\backslash G/B \cong W_G$.
Most of what I know about this is in papers of [Richardson-Springer],
where they study what one might call the weak Bruhat order on
$K\backslash G/B$. Unfortunately, I want the strong Bruhat order.
In particular, I seek proofs (and of course, references) for the following.

If $v \gtrdot v'$ is a covering relation in $K\backslash G/B$, then
  $\ell(\varphi(v)) - \ell(\varphi(v')) = 1,2,3$, and the difference is determined by
  whether the difference in the Cartan ranks of $v,v'$ is $-1,0$, or $1$.
If the difference is $2$, then there exists a positive root $\beta$ 
  such that $\varphi(v) = r_{\theta\cdot \beta} \varphi(v') r_\beta$.

I also have a question:

If the difference is $3$, what is the relation of $\varphi(v)$
  and $\varphi(v')$?

ADDED: The best references I have found are [Incitti] for classical groups and [Hultman] in general, but neither quite answers the above, as far as I can tell.

REPLY [4 votes]: The answer to the first question is no.
David Vogan gave me the example of $(Sp(4),GL(4))$, orbits #46 and #76 according to the Atlas numbering here: 
http://www.liegroups.org/web/atlasInput.html
Axel Hultman gave me the example of $D_4$ with generators $a,b,c,d$; $d$ being the non-leaf in the diagram, and ordinary involutions. Then, $abcdcba$ covers $abc$.<|endoftext|>
TITLE: Other Homology Theories still Count Holes?
QUESTION [14 upvotes]: This may be a naive question, but since first learning homology I considered it as a tool which counts appropriate holes in your space (on top of orientation and torsion phenomena). Then I was introduced to homology of groups and now Floer/Morse homologies.  Do these homologies still count "holes" in some fashion?
In the case of group homology, $H_\ast(G)\cong H_\ast(BG)$, so we can view this homology as a count of holes in the Milnor construction (CW-complex assembled from points in the discrete group with the group structure).
In Floer homology we're counting holomorphic curves (flow-lines in Morse homology), but it isn't viewed as having these curves "wrap around holes", so I am not sure if this hole-detecting view of homology breaks down.
[[Edit]]: I will narrow down my question. Are there instances where I can treat $HF_\ast$ as $H_\ast$ of a particular space?  For instance, I just realized that with nice conditions we have $HF^\ast(L,L)=H^\ast(L)$ in Lagrangian-Floer homology, so here it counts the holes of the Lagrangian submanifold.
Thanks to Steven Landsburg's response, we can usually find such a space (but ideally would be looking for something explicit, such as Floer homotopy type with $SH_\ast(T^\ast M)=H_\ast(\mathcal{L}M)$).

REPLY [4 votes]: In finite dimensional Floer homology, the connecting trajectory count encodes a more traditional count. Namely the unstable manifolds of the flow define,  under certain assumptions, a cellular decomposition of the manifold. The associated cellular chain complex is  isomorphic to the Floer complex. The isomorphism associates to each critical point, viewed as  an element in the Floer complex, its unstable manifold, regarded as an element of the cellular complex.  This interpretation is meaningless in infinite dimensions.<|endoftext|>
TITLE: Hartshorne's proof of the birational invariance of the geometric genus
QUESTION [6 upvotes]: I am confused about a couple of steps in the proof of the birational invariance of the geometric genus (Theorem II.8.19 in Hartshorne's Algebraic Geometry).
I shall sketch the proof and highlight my doubts.
Let $X,X'$ be two birationally equivalent nonsingular projective varieties over a field k.
Hence there is a birational map $X-->X'$ represented by a morphism $f:V\rightarrow X'$ for some largest open subset $V\subset X$.
Along taxonomic lines, the proof goes like this:

We first prove that $f$ induces an injective map $f^{\ast}:\Gamma(X',\omega_{X'})\rightarrow \Gamma(V,\omega_V)$
Then we prove that the restriction map $\Gamma(X,\omega_X)\rightarrow \Gamma(V,\omega_V)$ is bijective, using the valuative criterion of properness.

From this it follows that $\rho_g(X')\leq \rho_g(X)$, and the reverse inequality follows by simmetry.
In the proof of step 1: the map $f$ induces an isomorphism $U\cong f(U)$ for some open subset $U\subset V\subset X$ and then Hartshorne claims that this implies that f induces an isomorphism $\omega_{V|U}\cong \omega_{X'|f(U)}$. Why is that?
In the proof of step 2: from the valuative criterion of properness it follows that $\textrm{codim }(X\setminus V,X)\geq 2$. In order to prove that $\Gamma(X,\omega_X)\rightarrow \Gamma(V,\omega_V)$ is bijective it suffices to prove it on open sets $U\subset X$ trivializing the canonical sheaf $\omega_{X|U}\cong \mathcal{O}$, namely that $\Gamma(U,\mathcal{O}_U)\rightarrow  \Gamma(U\cap V,\mathcal{O}_U\cap V)$ is bijective.
Since $X$ is nonsingular, from the first remark in the previous paragraph we have that $\textrm{codim }(U\setminus U\cap V,U)\geq 2$ and then Hartsorne claims that the result (bijectivity) follows immediately from the fact that for an integrally closed Noetherian domain $A$, we have $A=\bigcap_{\textrm{ht } \mathfrak{p}=1} A_{\mathfrak{p}}$. I do not see this either.
Thanks in advance for any insight.

REPLY [2 votes]: Since $U\cong f(U)=:W$ under the isomorphism guaranteed by birationality condition, we get that there is an induced isomorphism $f^*\omega_W\to \omega_U$. This follows from the fact that a morphism of (say) varieties $f:X\to Y$ induces a map $f^*\Omega^1_{Y/k}\to \Omega^1_{X/k}$ of the cotangent sheaves. Taking wedge powers gives the morphism $f^*\omega_Y\to \omega_X$. By functoriality of the cotangent sheaf, we can conclude that an isomorphism of varieties induces an isomorphism of the cotangent sheaves and hence of the canonical bundles.
Also, the cotangent sheaf (and hence the canonical sheaf) is compatible with restrictions as in Donu's comment above. So, the birational map induces the isomorphism $f^*\omega_{X'}|_W\cong f^*\omega_W\to \omega_U\cong\omega_X|_U.$ Pretty much all of this is in Chapter 2 Section 8 of Hartshorne explicitly or implicitly.
As for the second question, the important fact is the following:

Let $X$ be a normal variety with $f\in K(X)$ such that $f\in \mathcal{O}_X(U)$ where $X\setminus U$ is codimension $\ge 2$ in $X$. Then $f$ is in fact in $\mathcal{O}_X(X)\subseteq K(X)$.

Proof: Let $V=\mathrm{spec}A$ denote an affine open. $V\cap U$ has codimension at least $2$ in $V$. So, $f\in A_{\frak{p}}$ for all $\mathfrak{p}$ of height $1$ (codimension $1$). So, $f\in \bigcap_{\frak{p}:\mathrm{ht}(\mathfrak{p})=1}\mathfrak{p}=A$ by the fact cited in Hartshorne. In particular, $f$ is actually regular on all of $V$. Noting that $X$ is covered by such affine opens, we see that $f$ is actually regular everywhere and hence in $\mathcal{O}_X(X)$.
This lets us extend rational sections of line bundles defined on open subsets with complements of codimension at least $2$. Indeed, let $\mathcal{L}$ denote a line bundle on a normal $X$ as above. Let $s\in \Gamma(U,\mathcal{L})$ for $X\setminus U$ of codimension $\ge 2$. Then choose $V$ an affine open that is trivializing for $\mathcal{L}$ so that $\mathcal{L}|_V\cong \mathcal{O}_V$. Then $s$ determines a regular function $t\in \mathcal{O}_V(U\cap V)=\mathcal{O}_X(U\cap V)$. Since $U\cap V$ has codimension $\ge 2$ complement in $V$, we can extend $t$ to $\mathcal{O}_V(V)$. Hence, we can do the same for $s$. Proceeding as before, we get that $s$ extends to all of $X$.<|endoftext|>
TITLE: Octonionic reflection groups
QUESTION [10 upvotes]: Consider Leech lattice definition provided by Wilson (Octonions and Leech lattice, 2008). 
There are 819 E8 sublattices defined by
$ (2\lambda, 0, 0);  $ 
$   (\lambda \overline{s}, (\lambda \overline{s}) j, 0);   $
$  ( (\lambda s)j, \lambda k, (\lambda j) k ) $
where $\lambda$ span 240 vectors of E8 lattice, j,k are 16 base octonions (plus, minus), and s is -1+sum of imaginary unit octonions.
(I am testing LaTeX here) See page 3, chapter 3 of Wilson paper. I wonder what is the subgroup of $Co_0$ generated by 819 reflections in 8-dim planes spanned by those E8 sublattices. They could be considered as octonion reflections. And as such they are elements of F4 Lie group being automorphism of $OP^2$.
My questions is following. Has anyone tried to extend definition of complex reflection and quaternion reflection to octonion reflection. In such definition Conway group $Co_0$ would be octonion reflection group i.e. it is generated by reflections in 8-dim planes in 24-dim Euclidean space.
In general when order 2 element in abstract group - called involution - can be considered as reflection ? I know involution is algebraic notion while reflection is geometric. But geometry is something which make group theory interesting.
Regards,
Marek

REPLY [7 votes]: Yes, such groups are interesting, see this paper by Daniel Alcock, "Reflection groups on the octave hyperbolic plane," http://www.ma.utexas.edu/users/allcock/research/oh2.pdf
If you write to Daniel, he will probably give you more references.<|endoftext|>
TITLE: Is this sum of reciprocals of zeta zeros correct?
QUESTION [5 upvotes]: I am trying to find or get a numerical approximation of
$$ \sum_{\rho \text{ non-trivial zeros of } \zeta} \frac{1}{\rho} $$
In The Riemann Hypothesis: Arithmetic and Geometry Lagarias gives the identity:
$$\hat{\zeta}(s) := \pi^{-\frac{s}{2}} \Gamma(\frac{s}{2})\zeta(s)$$
$$ \frac{\hat{\zeta}^\prime(s)} {\hat{\zeta}(s)} = \frac{d}{ds} [ \log \hat{\zeta}(s) ] = -\frac{1}{s} - \frac{1}{s-1} + {\sum_{\rho \text{ zeros of } \zeta }}^\prime \frac{1}{s-\rho} \qquad(1)$$
where the prime indicates the zeros must be summed in pairs $\rho,1-\rho$

Q1 Does the last sentence mean that the sum is over the non-trivial zeros?

Maple gives:
$$\lim_{s \to 0} {\sum_{\rho \text{ zeros of } \zeta }}^\prime \frac{1}{s-\rho} = -\gamma + \frac{1}{2}  \log\left(\pi\right) + \log\left(2\right) - 1$$

If the above result is correct, is it true that:
$$ \sum_{\rho \text{ non-trivial zeros of } \zeta} \frac{1}{\rho} = \gamma - \frac{1}{2}  \log\left(\pi\right) - \log\left(2\right) + 1 $$

EDIT As Micah Milinovich kindly answerd the above is wrong.
Trying to save the quiestion, is it true that:
$$ \sum_{\rho} \frac{1}{\rho (1{-}\rho)} =  \gamma - \frac{1}{2} \log\left(\pi\right) - \log\left(2\right) + 1  $$
Assuming RH $1-\rho = \bar{\rho}$ and the LHS is $\sum_{\rho} \frac{1}{|\rho|^2}$
According to RH Equivalence 5.3. $$\sum_{\rho} \frac{1}{\rho (1{-}\rho)}=\sum_{\rho} \frac{1}{|\rho|^2} = 2 + \gamma - \log 4\pi$$.
And the constants still don't match.

REPLY [5 votes]: These identities are not mysterious. They are simply the fact that the Riemann Zeta function has a Weierstrass product like any other meromorphic function of finite exponential order. Note here that $f'/f$ is called logarithmic derivative for a reason;)
Then it follows immediately
1) Yes, the zeros of the completed Riemann zeta function are exactly the nontrivial ones.
2) If RH hold, they come in pairs $\rho = 1/2 \pm \mathrm{i}t$ for $t>0$.
A suggestion for computing the sum: 
Let $\Omega_T$ be the boundary of $ -T \leq Im s \leq T$ and $-1/2 < Re s < 3/2$. For an approximation consider the integral
$$\frac{1}{2 \pi i} \int\limits_{\Omega_T} s^{-1} \frac{\zeta'(s)}{\zeta(s)} d \; s = \sum\limits_{-T \leq Im \rho \leq T} \frac{1}{\rho} + $$
some contribution coming from the poles, which are slightly delicate for $s=0$, since you encounter a double pole. I am pretty sure that you have missed that, and that this is why your computation fails.
Wikipedia tells you that 
$$ \zeta(s) = \frac{2^{s-1}}{s-1}-2^s \int_0^{\infty}\frac{\sin(s\arctan t)}{(1+t^2)^\frac{s}{2}(\mathrm{e}^{\pi\,t}+1)}\,\mathrm{d}t,$$
is pretty convenient for computing $\zeta$ numerically.<|endoftext|>
TITLE: Does the Mordell conjecture imply the Shafarevich conjecture
QUESTION [5 upvotes]: The base field is a number field.
It is known that the Shafarevich conjecture implies the Mordell conjecture (Kodaira-Parshin).
Is the converse also true? 
Note that both conjectures are now theorems (Faltings).
Edit: To be clear, I'm referring to the Shafarevich conjecture for curves. That is, for any number field $K$, finite set of places $S$ of $K$ and integer $g > 1$, the set of curves over $K$ of genus $g$ with good reduction outside $S$ is finite.

REPLY [6 votes]: The Shafarevich conjecture can be interpreted as stating that the set of $S$-integral points (over any number field) on the moduli space of curves of fixed genus (or principally polarized abelian varieties of fixed dimension) is finite. So, in this sense, it is a generalization of Siegel's theorem/Mordell's conjecture. But note that it is a statement about a variety of high dimension, so I very much doubt that you can reduce the Shafarevich conjecture to the Mordell conjecture. Except, of course, in the case of elliptic curves, where the Shafarevich conjecture follows directly from Siegel's theorem as Shafarevich originally remarked.<|endoftext|>
TITLE: The Haar state on compact quantum groups $A_u(Q)$ and  $A_o(Q)$
QUESTION [8 upvotes]: Let $Q\in GL_n(\mathbb{C})$. The free unitary quantum group is the universal $C^*$-algebra $A_u(Q)$ with generators $u_{ij},1\leq i,j\leq n$ and relations making $u=(u_{ij})$ as well as $Q\bar{u}Q^{-1}$ unitary, where $\bar{u}=(u_{ij}^*)$. The comultiplication is defined by
\begin{align}
\Phi(u_{ij})=\sum_k u_{ik}\otimes u_{kj}.
\end{align}
$(A_u(Q),\Phi)$ is a compact quantum group.
Let $Q\in GL_n(\mathbb{C})$ such that $Q\bar{Q}=\pm 1$. The free orthogonal quantum group is the universal $C^*$-algebra $A_o(Q)$ with generators $u_{ij},1\leq i,j\leq n$ such that $u=Q\bar{u}Q^{-1}$ is unitary. As above $(A_o(Q),\Phi)$ is a compact quantum group.
It is well-known that every compact quantum group admits a unique Haar state. My question is that 

What is the expliciit expression of the Haar state on $A_u(Q)$ and  $A_o(Q)$?
Are they faithful?

By the way, it is known for $SU_q(2)$ and $SU_q(2)\cong A_o(Q)$ for
$Q=\begin{pmatrix}
0& |q|^{1/2} \newline
-\text{sign}(q)|q|^{-1/2}&0
\end{pmatrix}.$
I just find out that the Haar state is tracial if and only if the coinverse $\kappa$ on the dense Hopf $*$-algebra satifies $\kappa^2=id$ (see "compact quantum groups" Thm 1.5 by Woronowicz).
Thanks!

REPLY [7 votes]: Have you seen arXiv:math/0511253?
INTEGRATION OVER COMPACT QUANTUM GROUPS
TEODOR BANICA AND BENOIT COLLINS 
Abstract. We find a combinatorial formula for the Haar functional of the orthogonal and unitary quantum groups. As an application, we consider diagonal coefficients of the fundamental representation, and we investigate their spectral measures.
(I do not have sufficient reputation to post this as a comment.)<|endoftext|>
TITLE: Characterization of the Lie derivative
QUESTION [14 upvotes]: The exterior differential of differential forms on a manifold can be characterized as the unique super-derivation of degree 1 on the exterior algebra of forms such that $<df,X>=X(f)$
for $f$ a $C^{\infty}$ function, $X$ a vector field. So we really only need to know how to compute $df$, and everything else follows formally.
Is there a similar characterization for the Lie derivative acting on differential forms? 
My guess: Extend $L_{X}$ from $L_{X}f=X(f)$ to an action on all forms so that it commutes with $d$ (Cartan's magic formula?) and perhaps $L_{[X,Y]}=[L_{X},L_{Y}]$. In other words, the Lie derivative should be a homomorphism of Lie algebras from vector fields to degree zero derivations of the de Rham algebra. 
If this is correct, can anyone give references for this point of view?

REPLY [2 votes]: Let me add my two cents to the great answers of Dick Palais and Tom Goodwillie. Let $A^\bullet(M)=\bigoplus_{k=0}^\infty A^k(M)$ be the graded algebra of differential forms on your manifold $M$. Then, for a vector field $X$, $\iota_X$ is the unique derivation of degree $(-1)$, satisfying $\iota_X(\alpha)=\alpha(X)$ for all $\alpha\in A^1(M)$, and $d$ is the unique derivation of degree $(+1)$, satisfying $df(X)=X(f)$ for all $f\in A^0(M)$.
If $D_1$ and $D_2$ are graded derivations of  degrees $p$ and $q$ respectively, their graded commutator
$$
D_1D_2 - (-1)^{pq}D_2D_1
$$
 is a derivation of degree $p+q$. In our case, this is
$$
L_X = d\iota_X + \iota_X d,
$$
a degree zero derivation.<|endoftext|>
TITLE: Self-dual normed spaces which are not Hilbert spaces
QUESTION [18 upvotes]: Are there any examples of non-Hilbert normed spaces which are isomorphic (in the norm sense) to their dual spaces? Or, is there any result in Functional Analysis which says that if a space is self-dual it has to be Hilbert space. 
Since, we want isomorphism in the norm sense, examples like $\mathbb{R}^{n}$ are ruled out. The norms of the space and its dual have to be equal and not just equivalent. 
Thank you.

REPLY [6 votes]: Another family of infinitely many examples: take $Y$ to be a reflexive Banach space which is not a Hilbert space, then $Y\oplus Y^*$ is isometrically isomorphic to its dual, without being a Hilbert space.
If the isomorphism verifies additional properties, then the result is true. Namely, if a Banach space is isometric to its dual, under certain conditions, it is a Hilbert space. See Theorems 2 and 4 in http://arxiv.org/pdf/0907.1813.pdf and reference therein for similar results. An example of this kind of results is the following:
Theorem: Suppose that $X$ is a Banach space and $\phi:X\to X^*$ is an antilinear isomorphism. If, for all $x\in X$, $x$ is orthogonal (in Birkhoff-James' sense) to $ker(\phi(x))$, then $X$ is a Hilbert space.<|endoftext|>
TITLE: Spaces with no topological monoid structure which are homotopy equivalent to topological monoids
QUESTION [23 upvotes]: In motivating $A_\infty$-spaces to my students I'm going to insist on the homotopy invariance of the notion, saying that "being $A_\infty$ is the homotopy invariant version of being a topological monoid" and to stress this I'd like to say that if $X$ is a topological monoid and $Y$ is a space homotopy equivalent to $X$ then $Y$ will carry an $A_\infty$-structure making it equivalent to $X$ as an $A_\infty$-space, but in general not a topological monoid structure with this property. But at this point I see to my shame that I miss an explicit example of this! 
Clearly the most dramatic example would be that of a space $Y$ which is homotopy equivalent to a topological monoid $X$, but such $Y$ carries no topological monoid structure at all, not to have to go into the equivalence issue. For a while I thought the closed interval could be an example of this (double shame: there are at least two very simple and well known topological monoid structures on $[0,1]$!), so I'm completely without examples, and I do not either know if such a space $Y$ does actually exist at all.
Any suggestion?
edit: despite I originally formulated my question in the most dramatic possible form, an example where, given a homotopy equivalence $f:Y\to X$ there is no monoid structure on $Y$ such that $\pi_0(f)$ is an isomorphism of monoids $\pi_0(Y)\to \pi_0(X)$ is even better for what I need to explain, namely that going from monoids to $A_\infty$-spaces not only $Y$ is naturally endowed with an $A_\infty$-space structure, but $f$ is promoted to an equivalence of $A_\infty$-spaces. So I will now leave the original question open as a general topology question which may have its interest in its own (despite it is admittedly an odd question), while for myself I'll be perfectly satisfied with the very nice answer by Tyler below.

REPLY [7 votes]: We can modify Neil's argument in the other thread to give an example of a contractible space with no monoid structure.
Let T be a tree with the following property: 
For each point x in T, there are at least two components of T \ x which contain an at least trivalent vertex.
In particular, the complement of x must have at least two components, so T cannot contain any 1-valent vertices.
For example, let T be the universal cover of the theta graph.
Now suppose that T has a monoid structure.  Using Neil's argument, we can get some invertibility results. Namely, let x and y be points in two different components of the complement of the identity.  Then there is a path from (e,y) to (x,y) to (x,e) in $T\times T$ which holds y fixed in the first half and x in the second half.  The image of this path must pass through e, so either x or y is invertible.  This implies that every point in all but one of the components of the complement of e is invertible.  Then there is an at least trivalent vertex g distinct from e which is invertible and a non-constant path from $g^{-1}$ to e through invertible elements.  This gives a homotopy of homeomorphisms from T to T by left multiplication.  The image of g at time 0 is e; at time 1 it is g.  This means at some time between them, g must be taken to an internal point of an edge of T; no homeomorphism can do that.
Such a T is homotopy equivalent to a point but can have no monoid structure.<|endoftext|>
TITLE: How to inject PGL (n, k) in PGL (n +1, k)
QUESTION [6 upvotes]: How to construct an injection of $PGL(n,k)$ in $PGL(n+1,k)$ if $GL(n,k)$ injects in $GL(n+1,k)$. I think it depends on the field k. For example, if we put $\varphi:GL(n,k)\longrightarrow GL(n+1,k)$ defined by : 
$$\varphi(g)=\left(
   \begin{array}{cc}
     g & 0 \cr
     0 & \chi(g) \cr
   \end{array}
 \right)$$
where $\chi$ is a caracter of $GL(n,k)$, that is a morphism of groups of $GL(n,k)$ to $k^{\times}$, then $\chi$ has the form $\chi=\phi\circ det$, where $\phi$ is an endomorphism of $k^{\times}$.
Then, $\varphi$ induces a morphisme of groups of $PGL(n,k)$ in $PGL(n+1,k)$ if and only if the morphism $\phi$ satisfies : For all $x\in k^{\times}$, $\phi(x)^{n}=x$. 
For $k=\mathbb{R}$ that is imposible.

REPLY [12 votes]: You can't do it in general. A quick computer calculation (I used Magma) shows that ${\rm PGL}(4,4)$ has no subgroup isomorphic to ${\rm PGL}(3,4)$. (It does have one isomorphic to ${\rm SL}(3,4)$.)
I suspect that there is an embedding ${\rm PGL}(2,K) \to {\rm PGL}(3,K)$, but that is coming from the irreducible orthogonal action of ${\rm GL}(2,K)$ on a 3-dimensional module.<|endoftext|>
TITLE: Which functor does the blowing up represent?
QUESTION [21 upvotes]: Let $X$ be a scheme and $I \subseteq \mathcal{O}_X$ be a quasi-coherent ideal of finite type. The blowing up $\mathrm{Bl}_I(X)$ has the following universal property: It comes with a morphism $p : \mathrm{Bl}_I(X) \to X$ such that $\langle p^*(I) \rangle \subseteq \mathcal{O}_{\mathrm{Bl}_I(X)}$ is invertible, and for every morphism $f : Y \to X$ such that $\langle f^*(I) \rangle \subseteq \mathcal{O}_Y$ is invertible, there is a unique morphism $\tilde{f} : Y \to \mathrm{Bl}_I(X)$ such that $f = p \tilde{f}$. In other words, $\mathrm{Bl}_I(X) \to X$ is a final object in the category of $X$-schemes which pull back $I$ to an invertible ideal.
First I thought that this implies that $\mathrm{Bl}_I(X)$ is a representing object of the functor
$\mathrm{Sch}^{\mathrm{op}} \to \mathrm{Set},~ Y \mapsto \{f \in \hom(Y,X) : \langle f^*(I) \rangle \subseteq \mathcal{O}_Y \text{ invertible}\},$
but this is wrong: This is not even a functor! Invertible ideals don't pull back to invertible ideals. If $H(Y)$ denotes the set above, then there is a map $\hom(Y,\tilde{X}) \to H(Y)$, which is injective, but far from being surjective.
Question. Which functor $\mathrm{Sch}^{\mathrm{op}} \to \mathrm{Set}$ does the blowing up represent? In other words, how can we simplify $\hom(Y,\mathrm{Bl}_I(X))$?
Since we have $\mathrm{Bl}_I(X) = \mathrm{Proj}_X \oplus_{n \geq 0} I^n$ , one can ask more generally what functor $\mathrm{Proj}_X \mathcal{A}$ represents when $\mathcal{A}$ is a sufficiently nice graded $\mathcal{O}_X$-algebra. This is indicated in EGA II, 3.7, but the condition that the partial morphism $r_{\mathcal{L},\psi}$ is defined everywhere is only made explicit for affine $X$ in loc. cit. Cor. 3.7.4. If $\mathcal{A}=\mathrm{Sym}(\mathcal{E})$ for some vector bundle $\mathcal{E}$, then the condition is just that $\psi$ (defined in loc. cit. 3.7.1) is surjective, so that the universal property of $\mathbb{P}(\mathcal{E})$ is very similar to the one of the usual projective space. But $\oplus_{n \geq 0} I^n$ obviously does not have this form, unless $I$ is flat or something like that.
In any case, there seems to be an injective map from
$$\{(f,\mathcal{L},\psi) : f \in \hom(Y,X) , \mathcal{L} \text{ line bundle on } Y, \psi : f^*(\mathcal{A})  \twoheadrightarrow \oplus_n \mathcal{L}^{\otimes n}\}$$
to $\hom(Y,\mathrm{Proj} \mathcal{A})$. EDIT: In Jason Starr's answer it is claimed that this is a bijection. Can someone give a reference in the literature for this observation?
In the note "Elementary Introduction To Representable Functors and Hilbert Schemes" by S. A. Stromme I have found the following amusing exercise:
"Let $X \to S$ be the blowing up of $S$ in some center $Y \subseteq S$. Try to the understand the point functor $h_{X/S}$. Then explain why it is hard to understand blow-ups."
I hope that this does not mean that we cannot understand the functor at all ...
EDIT. The universal property of Proj of a graded quasi-coherent algebra appears in section 16 of "Constructions of schemes" in the Stacks Project (here). It is the usual one when the graded algebra is generated in degree 1. I still wonder if there is any reference in the literature because I want to cite this.

REPLY [19 votes]: For a graded algebra $\mathcal{A} = \oplus_d \mathcal{A}_d$ of $\mathcal{O}_X$-algebras whose associated graded pieces $\mathcal{A}_d$ are coherent and which is generated in degree $1$, the relative Proj, $P=\text{Proj}_X \mathcal{A}$ with its projection $p:P\to X$ comes with a natural invertible quotient $q:p^*\mathcal{A}_1 \to \mathcal{O}(1)$.  Moreover, for every integer $d>0$, the induced surjection $\text{Sym}^d(q):p^*\text{Sym}^d(\mathcal{A}_1) \to \mathcal{O}(d)$ factors through the pullback of the natural surjection $\text{Sym}^d(\mathcal{A}_1) \to \mathcal{A}_d$.  In fact this is the universal property of the pair $(p:P\to X,q:p^*\mathcal{A}_1 \to \mathcal{O}(1))$, i.e., for every $X$-scheme $f:T\to X$ and for every invertible quotient $r:f^*\mathcal{A}_1 \to \mathcal{L}$ such that every induced map $\text{Sym}^d(r):f^*\text{Sym}^d(\mathcal{A}_1) \to \mathcal{L}^{\otimes d}$ factors through $f^*\mathcal{A}_d$, there exists a unique $X$-morphism $h:T\to P$ such that $p\circ h$ equals $f$ and $h^*q$ equals $r$.
Since the blowing up is Proj of the blowup algebra $\mathcal{A}$, which is generated in degree 1, the functor represented is the functor of invertible quotients $\mathcal{L}$ of the pullback $f∗I$ of $I$ such that for every integer $d>0$, the induced surjection $f^∗\text{Sym}^d(I)\to \mathcal{L}$ factors through $f^*(I_d)$.<|endoftext|>
TITLE: Montgomery's pair correlation function without RH?
QUESTION [11 upvotes]: In the theory of the Riemann zeta function, Montgomery's Pair correlation function is defined as
$$
F(\alpha) = \frac{1}{N(T)} 
\sum_{T < \gamma, \gamma' < 2T} T^{i \alpha (\gamma - \gamma')} \cdot \frac{4}{4+(\gamma - \gamma')^2}
$$
where $N(T)$ is the total number of zeroes of the $\zeta$ function at height between $T$ and $2T$, and $\gamma, \gamma'$ are ordinates of zeroes of the Riemann zeta function.
As $T \rightarrow \infty$ the dependance on $T$ vanishes and for this reason, by abuse of notation, we think of $F$ as just a function of $\alpha$. In effect, $F(\alpha)$ is the Fourier transform of the function that measures how often two distinct zeroes of the Riemann zeta function come together. 
Montgomery proved that $F(\alpha) = \alpha$ for $0<\alpha<1$ and conjectured that $F(\alpha)=1$ for $\alpha > 1$. He also proved that $F(\alpha)$ is positive and symmetric around $0$. However implicit in all this work is the assumption that the Riemann Hypothesis holds.
My question is the following: Is it possible to formulate an "unconditional" analogue of Montgomery's pair correlation function, for example, either by 1) allowing some additional large terms in the sum, corresponding to possible zeroes off the line, or 2) by taking into account only the $40$ percent or so of the zeroes of the Riemann zeta function that we know to lie on the critical line?
Once formulated, is it possible to obtain some non-trivial information about the unconditional analogue?

REPLY [13 votes]: It's a nice question. There is indeed a way to formulate an unconditional analogue. The idea is to no longer interpret $\gamma$ as ordinates, but instead label the zeroes as $\tfrac{1}{2}+i\gamma$, where $\gamma$ is a complex number with imaginary part in between -1/2 and 1/2. In this way, the Riemann hypothesis is the statement that all $\gamma$ are real, and, on RH, these $\gamma$'s correspond exactly to your $\gamma$'s. One can in this case show unconditionally that
$$
\lim_{t\rightarrow\infty} \frac{1}{N(T)}\sum_{T<\Re(\gamma),\Re(\gamma')<2T} T^{i\alpha(\gamma-\gamma')}\frac{4}{4+(\gamma-\gamma')^2} = \alpha
$$
for fixed $0 < \alpha < 1$. (If you want this uniformly for $\alpha$ you'll have to restrict $\alpha$ away from $1$ and $0$, or modify things around $0$.) Note that $\gamma-\gamma'$ may be complex.
Observations of this sort go back at least to Rudnick and Sarnak's paper "Zeros of principal L-functions and random matrix theory," where they write of the k-point correlation sums
$$
\sum_{\gamma_1,\gamma_2,...\gamma_k \textrm{ distinct}}h\Big(\frac{\gamma_1}{T}\Big)\cdots h\Big(\frac{\gamma_k}{T}\Big)f(\frac{L}{2\pi}\gamma_1,...,\frac{L}{2\pi}\gamma_k\Big)
$$
"if $h$ and $f$ are defined for complex argument and are localized, then the sums make sense even if we do not assume RH." This means that $f(x+iy) = \int e((x+iy)\xi)\hat{f}(\xi)d\xi$.
The reason one can do this here and with the slightly different 2-point sum Montgomery considers is that the explicit formulas relating zeroes to primes that Montgomery uses in his proof remain true unconditionally so long as the $\gamma$ representing an ordinate of a zero is replaced by $\gamma$ as defined above. I don't know that, for the Montgomery 2-point sum, this argument appears explicitly in print anywhere, but it's an instructive exercise to go through the standard proof seeing that everything still works and you can arrive at the formulation I have above.
Recalling the symmetry of the zeros (that is, if $\tfrac{1}{2}+i\gamma$ is a zero, then so are $\tfrac{1}{2}-i\gamma$, $\tfrac{1}{2}+i\bar{\gamma}$, and $\tfrac{1}{2}-i\bar{\gamma}$), one can see that $F(\alpha)$ (if this limit exists) is positive and symmetric using the same argument that Montgomery does. (Positivity requires a slight modification -- it is here that one needs conjugate symmetry in $\gamma$; I can elaborate if you'd like.)
I don't think anything can be said about the zeros that are already known to lie on the critical line, because in particular I don't think anything explicit can be said about the relation between these zeroes and the primes (about which we can say at least a few useful things). Using a zero-density estimate, it may be possible to show that
$$
\lim_{t\rightarrow\infty} \frac{1}{N(T)}\sum_{T<\Re(\gamma),\Re(\gamma')<2T} T^{i\alpha(\Re(\gamma-\gamma'))}\frac{4}{4+(\Re(\gamma-\gamma'))^2} = \alpha
$$
for $\alpha$ restricted as above, even if RH is false, returning our attention once again to the ordinates of zeroes. This would be interesting (to me) for a variety of reasons, but without a clever idea I don't think any known zero-density estimates are enough to get here.<|endoftext|>
TITLE: Is there a gerbe Beilinson-Bernstein Localization?
QUESTION [19 upvotes]: Suppose you want to construct a representation of an affine algebraic group $G$, you may start with a $G$-equivariant line bundle $\mathcal{L}$ on a $G$-manifold $X$ and then consider global sections, or cohomologies, for example $H^*(X, \mathcal{L})$ becomes a $G$-module. 
Suppose now you want to construct a representation of $G$ on a category $\mathcal{C}$ (the type of representations studied in the appendix of [1] or in chapter 7 of [2]). Then you may consider a $G$-equivariant gerbe $\mathcal{G}$ on $X$ and again take global sections. 
In the former case you find out that essentially all (finite dimensional, over $\mathbb{C}$, etc) representations of $G$ arise in a geometric way: there exists a manifold $X$ (ie. the flag variety $G/B$ for a choice of a Borel $B \subset G$) and an equivalence of tensor categories between certain $D_X$-modules and $G$-rep. This is known as Beilinson-Bernstein localization and I'll refer to [3] for the precise statements. 
My question is if there's a categorical analogue of this statements along the lines of 

Is there an equivalence of $2$-categories, between the category of (categorical) representations of $G$ and some category of equivariant gerbes with flat connections on a space $X$ ?

This space might be infinite dimensional. One thing that comes to mind for example is the fact that $D$-modules on the affine grassmanian $Gr_G$ carries a monoidal action of $Rep (G^{\vee})$, and actions of groups on categories are related to these monoidal actions by de-equivariantization [1]. This might be related to my question,  but  notice that this is not really what I ask up there. 
Besides the folklore that "should be true", is there anything concrete written down?  

[1] Frenkel and Gaitsgory. Local geometric Langlands correspondence and affine Kac-Moody algebras
[2] Beilinson and Drinfeld Quantization of Hitchin's integrable system and Hecke eigensheaves
[3] Milicic Localization and Representation Theory of Reductive Lie Groups

REPLY [11 votes]: [Edited to reflect Reimundo's comment]
The question addresses categorified versions of the Borel-Weil-Bott theorem (and more generally Beilinson-Bernstein localization), which states 
an equivalence between G-equivariant vector bundles on the flag variety - aka vector bundles on pt/B (modulo an action of the Weyl group - aka double cosets B\G/B - by intertwiners) and algebraic representations of G. There are two pieces of content here: first, that all representations can be realized on G/B, ie representations have highest weights, and second that irreducibles correspond to line bundles, ie their highest weight spaces are one-dimensional. The first has an analog for any representation of the Lie algebra: Beilinson-Bernstein's localization can be rephrased as simply asserting that descent holds from twisted D-modules on the flag variety to representations of the Lie algebra.
I don't know anything about the analog of the second assertion for categorified representations - ie to what extent "indecomposable" representations of some kind are induced from "one-dimensional ones" (ie from gerbes on homogeneous spaces) - except to point out a very nice paper by Ostrik (section 3.4 here) in which analogous results are proved for the case of a finite group.
As for the "descent" (first) part of BWB, it becomes completely trivial once categorified, if we consider so-called algebraic (or quasicoherent) actions of G on categories (equivalently module categories for quasicoherent sheaves on G). In fact the same assertion holds for ANY algebraic subgroup of G, not just a Borel, in sharp distinction to the classical setting: algebraic G-actions on categories are generated by their H-invariants for any H in G! More precisely we have the following theorem:
Passing to H-invariants provides an equivalence of $(\infty,2)$-categories
between (dg) categories with a G action and categories with an action 
of the "Hecke category" QC(H\G/H) of double cosets.
This is a theorem of mine with John Francis and David Nadler in a preprint that's about to appear (copies available).. it's a version of a well known result of Mueger and Ostrik in the finite group case, and is an easy application of Lurie's Barr-Beck theorem. In fact if we use a result of Lurie in DAG XI, that there is no distinction between quasicoherent sheaves of categories on stacks X with affine diagonal and simply module categories over QC(X), we can rephrase the result as follows:
G-equivariant quasicoherent sheaves of (dg-)categories on the flag variety  equipped with a "categorified Weyl group action" (module structure for QC(B\G/B) ) are equivalent (as an $(\infty,2)$-category) to (dg)-categories with algebraic G-action.
On the other hand things get much more interesting if we consider "smooth" or "infinitesimally trivialized" G-actions (module categories over D-modules on G) (also discussed in the references you provide). The fundamental example of such a category is indeed Ug-mod, the category of all representations of the Lie algebra, or equivalently (up to some W symmetry) the category D_H(G/N) of all twisted D-modules on the flag variety.
In this case my paper with Nadler "Character theory of a complex group" (and other work in progress) precisely studies the full sub(2)category of smooth G actions which ARE generated by their highest weight spaces, ie which do come via a Borel-Weil-Bott type construction (or equivalently, the full subcategory generated by the main example Ug-mod). And not all smooth G-categories are of this form, though one might hope that this is the case in some weaker sense.. in any case it seems that all G-categories of interest in representation theory do fall under this heading. But in any case I don't know a BWB type statement in this setting.<|endoftext|>
TITLE: Asymptotics of a Selberg-type integral
QUESTION [23 upvotes]: Let $\Delta(s_1,s_2,\ldots,s_n) := \prod_{i<j}(s_i-s_j)^2$. Is there a standard way to estimate the decay of the Selberg-type integral
$$ I_n:=      \frac{1}{n!^2}\int_0^1 \int_0^1\cdots\int_0^1 \frac{\Delta(s_1,s_2,\ldots,s_n) \Delta(t_1,t_2,\ldots,t_n)}{\prod_{i,j}(1-s_i t_j)^2} d s_1\ldots d s_n d t_1 \ldots d t_n$$ as $n\to\infty$ ?
Checking numerically for $n\leq 25$, the decay seems to be like $e^{-2 n^2}$.

REPLY [4 votes]: This is very late to answer this question, but let me mention a rather different approach.
You can use the Schur function expansion
$$\frac{1}{\prod_{ij}(1-s_it_j)^2}=\sum_{\lambda,\mu}S_\lambda(s)S_\lambda(t)S_\mu(s)S_\mu(t),$$
where the infinite sums are over partitions, to get
$$I_n=\frac{1}{n!^2} \sum_{\lambda,\mu}(g_{\lambda,\mu})^2,$$ with
$$g_{\lambda,\mu}=\int_0^1 \Delta(s) S_\lambda(s)S_\mu(s)ds.$$
The above integral is known (see L.K. Hua, Harmonic analysis of functions of several complex variables in the complex domains, Translations of mathematical monographs, Vol. 6) to be
$$g_{\lambda,\mu}=n!\frac{\prod_{1\le i<j\le n}(\lambda_i-\lambda_j+j-i)(\mu_i-\mu_j+j-i)}{\prod_{1\le k,r\le n}(2n+\lambda_k+\mu_r-k-r+1)}.$$
I am not sure how to extract the asymptotics from the above expression, though.<|endoftext|>
TITLE: vanishing theorems
QUESTION [7 upvotes]: I have a flat morphism $p: X \to Y$ from a smooth projective $X$ to a smooth
projective $Y$. I have a line bundle $L$ on $X$ whose restriction to every
fiber of $p$ is big and nef. I need the vanishing of $Rp_*(\Omega^b
X\otimes L^{-1})$ for b small compared to $\dim X-\dim Y$.
I could not find this in Esnault-Viehweg. Is it true? If yes, what is
the reference?

REPLY [9 votes]: First of all I suppose you meant $R^ip_*$ for $i>0$ and not $Rp_*$. Actually, neither is true, but the latter is obviously false while the former may seem believable first. Second, I suppose you are working over an algebraically closed field of characteristic zero.
In any case, unfortunately this fails even over an algebraically closed field of characteristic zero and already if $Y$ is a point. (This is not surprising as any statement you might hope for would be enough to prove in this case).

Let $X_0$ be an arbitrary smooth projective variety over an algebraically closed field of characteristic zero and $\mathscr L$ an arbitrary ample line bundle on $X_0$. Let $\pi:X\to X_0$ be the blowing up of $X_0$ at an arbitrary smooth point with exceptional divisor $E\simeq\mathbb P^{r-1}$ (i.e., $\dim X=r$). Then for any $b\in \mathbb N$, $0<b<r$ we have the following non-vanishing:
  $$
H^b(X,\Omega_X^b\otimes\pi^*\mathscr L)\neq 0
$$

Proof:
Observe that by Bott's formula for cohomology on $\mathbb P^{r-1}$ applied to $E$, we have that  $R^i\pi_*\Omega_X^b=0$ if and only if $i\neq 0, b$. By Kodaira-Akizuki-Nakano applied to $\mathscr L$ on $X_0$ we have that $H^i(X_0,\pi_*\Omega_X^b\otimes\pi^*\mathscr L)=0$ for $i>0$ (this is because $\pi_*\Omega_X^b/\Omega_{X_0}^b$ is supported at the point that was blown up). So, the Leray spectral sequence computing $H^i(X,\Omega_X^b\otimes\pi^*\mathscr L)$ degenerates and gives that for $i>0$, 
$$
H^i(X,\Omega_X^b\otimes\pi^*\mathscr L)=H^{i-b}(X,R^b\pi_*\Omega_X^b\otimes\mathscr L)
$$
For $i=b$ the latter is clearly non-zero. $\square$
Corollary: $H^{b}(X, \Omega_X^b\otimes \pi^*\mathscr L^{-1})\neq 0$ for $0<b<r$. 
In particular, if $\dim X\geq 3$, then $H^{1}(X, \Omega_X\otimes \pi^*\mathscr L^{-1})\neq 0$

The main reason vanishing fails here is that there is this big blob of a divisor, $E$, where the chosen line bundle, $\pi^*\mathscr L$, is trivial. 
There are actually still some results that do give you vanishing, but they need more assumption. Sommese proved various versions with assumptions on the fiber of the morphism induced by some high power of the line bundle, so in particular these results assume that the line bundle is actually semi-ample, however it does not always need big once you assume the restriction on the fibers. For more on this see the book by Shiffman and Sommese: Vanishing theorems on complex manifolds.
Another couple of interesting papers were written by Arapura: Check out this and this.
My result Karl mentioned in his remark is actually for singular varieties. The main idea is that since you can't expect general vanishing for not necessarily ample but big and nef line bundles, you might be able to do it on the model where they are ample, however, this usually requires working on singular varieties, which makes dealing with $\Omega$ a bit tougher. For more details see this paper. 
More recent similar vanishing results were obtained by Greb-Kebekus-Peternell and myself. The most recent relevant paper is this, but you could also look at this and this.<|endoftext|>
TITLE: Where can I find a proof of the de Rham-Weil theorem?
QUESTION [9 upvotes]: Where can I find a proof of the de Rham-Weil theorem? 
Does anyone know?

REPLY [12 votes]: Andrew Fanoe,  a former student of mine,  wrote his senior thesis on various proofs of  the DeRham theorem.  In particular he discusses in detail Weil's very elegant proof of DeRham's theorem.  Andrew's presentation  uses a more  modern language, but all the ideas are in Weil's paper. In any case, here is the link to Andrew's thesis.     It's worth having a look at it. It is well written  and may have references you might find useful.<|endoftext|>
TITLE: Exotic spectrum of Laplace operator
QUESTION [7 upvotes]: Given a closed Riemannian manifold and a generalized Laplace $\Delta$ operator, 
it is well known that $\Delta$ has discrete spectrum $(\lambda_n)_n$ (arranged in a increasing way, not counting multiplicities). 
By a (consequence of a) result of Colin de Verdière, given any finite strictly increasing sequence $a_1,\cdots,a_k$ of 
strictly positive numbers,
there exists a Riemannian manifold and a Laplace operator on it such that the first k+1 eigenvalues are 
exactly, $0,a_1,\cdots,a_k$.
My question is about what's happening at infinity. More precisely, since usually 
the spectrum of a Laplace operator is a quadratic polynomial (in the sense that 
 {$\lambda_n : {n\geq 0}$} is of the form {$P(n) : n\geq 0$} where $P$ is a quadratic polynomial),
is there a Laplace operator (on a closed manifold) such that there is no $n_0$ such that {$\lambda_n : {n\geq n_0}$} is of the form {$P(n) : n\geq 0$} where $P$ is a quadratic polynomial ?
My question could be reformulated : do you know example where the explicit (exact) eigenvalues (not asymptotics of them) are (after a certain rank) not given by a formula of the form 
$a n^2 + bn +c$ where $a,b,c$ are constants (for example some kind of 
fraction P(n)/Q(n) where the degree of P is 3 and the degree of Q is one...)
[Edit : precisions]

REPLY [5 votes]: For a generic metric on an $m$-dimensional the manifold the eigenvalues of the Laplacian are all simple.  Fix such a metric and denote the coresponding eigenvalues by
$$ \lambda_1, \lambda_2,\cdots $$
Using Weyl's asymptotic expansion we conclude that
$$\lambda_n\sim \text{const}\cdot n^{m/2}. $$
Thus for any polynomial $P$ of degree $d>m/2$ we have
$$ \lim_{n\to\infty} \lambda_n/P(n) = 0. $$<|endoftext|>
TITLE: Gaussian prime spirals
QUESTION [141 upvotes]: Imagine a particle in the complex plane, starting at $c_0$, a Gaussian integer,
moving initially $\pm$ in the horizontal
or vertical directions.  When it hits a Gaussian prime, it turns left $90^\circ$.
For example, starting at $12 - 7 i$, moving initially $+x$, this closed circuit results:
     
Instead, starting at $3+5 i$, again $+x$, this (pleasingly symmetric!) closed cycle results:
     
Here's another (added later), starting at $5+23 i$:
     
(Gaussian primes $a+bi$ off-axes have $a^2+b^2$ prime; on axis, $\pm(4n+3)$ prime.)
My question is,

Q0.What's going on?

More specifically,

Q1. Does the spiral always form a cycle?
Q2. Have these spirals been investigated previously?

(I am about to step on a plane; apologies for not acknowledging responses!)
...Later:

Q3. Under what conditions is the spiral (assuming it closes) symmetric w.r.t. reflection in a
horizontal (as is $12-7 i$), or reflection in a vertical (as is $5 + 23 i$),
or reflection in both (as is $3 + 5i$)?

REPLY [3 votes]: I wanted to report on a variation, Gaussian prime random walks.
As before
start with any point in the complex plane $c_0$. Walk in the $+x$ direction
until you hit a Gaussian prime.
Unlike the original question, which requires a $90^\circ$ counterclockwise turn,
turn left or right by $90^\circ$ with equal probability. Continue turning randomly $\pm 90^\circ$ at every Gaussian prime hit, until some Gaussian prime is hit for a
second time. Call that a Gaussian prime loop.
Below is an example:

          


     

Green: start at $c=100 + 2i$. Red: looped at $197 + 22 i$. Yellow: Gaussian prime turn.


Although this
runs into the same open-problem impediment
cited by @FrançoisBrunault,*
it appears that, for a random starting $c_0$ within some fixed distance of the origin,
the probability of eventually looping might be $1$.
Looping is much "easier" to achieve than is spiral cycling, which requires
revisiting $c_0$ from the left,
whereas looping just needs to revisit some prime twice.


*
"It's unknown whether there are infinitely many Gaussian primes of the form $n+i$ with $n \in \mathbb{Z}$. So starting at $N+i$ and moving $+x$, we cannot exclude the possibility of hitting no prime."<|endoftext|>
TITLE: Chebyshev net in 3D
QUESTION [8 upvotes]: I would like to know the reasons why the existance of Chebyshev net in 3D-case is problematic.
This question boils down to the PDE described below.
(I do not know much about PDEs, so feel free to say something trivial.)
Set
$$\mathbb W^2 =\{\,(x,y,z)\in \mathbb R^3\mid x+y+z=0\,\}.$$
Given a smooth map $f: \mathbb W^2\to \mathbb R^3$ which is $C^\infty$-close
to the identity map,
I need to extend $f$ to a neighborhood of $\mathbb W^2$ so that it satisfies the following three equations:
$$\left|\frac{\partial f}{\partial x}\right|^2=\left|\frac{\partial f}{\partial y}\right|^2=\left|\frac{\partial f}{\partial z}\right|^2=1.$$
Comments.

The 2D-case has a solution, this is due to Chebyshev and it is more than 100 years old.
In general, 3D Chebyshev net is a solution of 3 equasions as above where $f$ maps a domain in $\mathbb R^3$ in a 3-dimensional Riemannian manifold and $|{*}|$ is defined by its metric. I guess this case is just as hard as the special case described above.
Assuming that $x+y+z>0$ and $f(x,y,z)$ is defined,
it is easy to see that $f(x,y,z)$ depends only on the values
$f(x',y',z')$ for $(x',y',z')$ in the triangle $\Delta\subset\mathbb W^2$ defined by the inequalities $$x'\le x,\  y'\le y,\ z'\le z.$$

REPLY [6 votes]: The question asked is a special case of the following more general one: Let
$$
\mathbb{W} = \{x^1 + \cdots + x^n = 0\} \subset \mathbb{R}^n.
$$
Given an $n$-dimensional Riemannian manifold $M$ and a smooth map $f: \mathbb{W} \rightarrow M$, can $f$ be extended to a map $f: \mathbb{R}^n \rightarrow M$ such that
$$
|\partial_1f|^2= \cdots = |\partial_nf|^2 = 1\ ?
$$
Here, note that $\partial_if(x) \in T_{f(x)}M$, and the norm is taken with respect to the Riemannian metric on $M$.
The first remark, which I'm sure Anton and Robert already know, is that if $f$ happens to be an embedding (which is the case in the original question), then we can view it as a choice of all $n$ co-ordinate functions along a hypersurface in $M$ and the question is whether these functions can be extended to co-ordinate functions where the diagonal elements of the metric tensor written with respect to these co-ordinates are identically equal to $1$.
To determine what kind of PDE this system is, we can linearize it. Let $\dot{f}$ denote an infinitesimal variation of $f$. The linearized system is given by
$$
2\partial_if\cdot\partial_i\dot{f} = \dot{h}_i.
$$
If $f$ is an immersion, then $\partial_1f(x), \ldots, \partial_nf(x)$ are always a basis of $\mathbb{R}^n$. Therefore, solving the linearized system for $\dot{f}$ is equivalent to solving for the functions
$$
u_1 = \partial_1 f\cdot \dot{f}, \ldots, u_n = \partial_nf\cdot \dot{f}.
$$
``Differentiating by parts'', the linearized system can be rewritten as
$$
2\partial_i u_i - g^{jk}u_j\cdot\nabla_i\partial_kf = \dot{h}_i.
$$
Note that the top order terms fully uncouple into ODE's in each of the co-ordinate directions. The coupling that occurs in the zero-th order terms prevents this system from being purely a system of ODE's. However, it is easy to check that this system is indeed a symmetric hyperbolic system, where initial smooth data posed on $\mathbb{W}$ can be extended to a unique smooth solution on all of $\mathbb{R}^n$. You can find more about this in a paper I wrote with DeTurck (Duke Math. J. Vol.51, No. 2
(1984), 243-260).
Using appropriate regularity estimates and the inverse function theorem with the appropriate Banach norm, this implies existence and uniqueness of a solution to the original nonlinear system on a tubular neighbhorhood of $\mathbb{W}$. I do not know whether there is a global solution; this requires a more careful analysis of this particular system.
ADDED: Actually, since the system is fully nonlinear, you cannot use the Banach space implicit function theorem to solve the system as given. You can, however, do one of two thing: Either "prolong" (which means roughly differentiate the system and add the partial derivatives of $f$ as unknown functions) the system into a quasilinear system or just use the Nash-Moser iteration argument. Either way, you still get what I describe above.
ADDED (in response to Robert's comment): Robert is right that if you specify $f$ along $\mathbb{W}$, the extension of $f$ is not uniquely determined, due to the full nonlinearity. Robert indicates that there are exactly 2 choices when $n = 3$. You have to do the linear algebra carefully along $\mathbb{W}$ to see this and count the number of possibilities in arbitrary dimension. I haven't done this yet.
ADDED: If along $\mathbb{W}$, you set $v = \frac{1}{n}(\partial_1 f+\cdots \partial_n f)$ and $u_i = \partial_if - v$, then the $u_1, \dots, u_n$ are tangential derivatives of $f$ along $\mathbb{W}$ and therefore given by the initial data. These vectors along $\mathbb{W}$ satisfy the equations
$$
u_i\cdot v = \frac{1}{2}[1 - |u_i|^2 - \frac{1}{n}(1 - \sum_i |u_i|^2)]
$$
$$
v\cdot v = \frac{1}{n}(1 - \sum_i |u_i|^2)
$$
Clearly, a necessary condition for a solution is that
$$
|u_1|^2 + \cdots + |u_n|^2 \le 1.
$$
If strict inequality holds on $W$, I believe that there are always exactly two solutions on $W$. These translate into a corresponding inequality that the tangential derivatives of $f$ along $\mathbb{W}$ must satisfy. And if strict inequality holds, then there are exactly two ways to extend $f$ onto a tubular neighborhood of $\mathbb{W}$ such that $f$ defines a Chebyshev net.<|endoftext|>
TITLE: Two groups acting on a set.
QUESTION [6 upvotes]: Suppose we are given a set S of points on which two different groups G and G' (given by sets of generating permutations) act.  Is there an efficient algorithm for finding generators the largest pair of subgroups H and H' of G and G' whose action on S coincide?

REPLY [10 votes]: It is known that the following three computational problems for subgroups $G$ of $S_n$ are polynomially equivalent:

Computing (generators of) the centralizer $C_G(g)$ of an element $g \in S_n$ (and also testing $g,h \in S_n$ for conjugacy in $G$).
Computing (generators of) the setwise stabilizer of a subset of the set of size $n$ on which $S_n$ acts (and also testing two such subsets for being in the same orbit under $S_n$).
Computing (generators of) the intersection of $G$ with another subgroup $H \le S_n$.

As Mark says, these are all at least as difficult as graph isomorphism.
The proofs are clever but basically elementary and interesting, so I recommend them! One reference is: 
E.M. Luks, ``Permutation groups and polynomial-time computation'',
in L. Finkelstein and W.M. Kantor (eds.), Groups and Computation,
Dimacs Series in Discrete Mathematics and Theoretical Computer Science
vol. 11, American Math. Soc., 139-176, 1993.
I have just noticed that Luks has a recently published book with the same title, which I have not seen yet.
Added later: It should also be mentioned that the implementations of the above algorithms in GAP and Magma involve backtrack searches, and so are potentially exponential, but in practice they run fast for most examples of moderate degree.<|endoftext|>
TITLE: Background for classic forcing
QUESTION [5 upvotes]: When I learning forcing theory, I am surprised by various foring definition,I want to know the original purpose, why define partial order in such way and some background material which can help me gain more intuition.

Cohen forcing.It was made by Cohen when he solve CON(ZFC+nonCH),It add many new 
reals,it is natural to think this if we consider the patial order form,But I heard Cohen used the boolean value model, if condider the boolean value model,how can I understand the intuition?
Random forcing.I just know it made by solovy.But I don't know the background.
Laver forcing.I just know laver forcing used to prove CON(ZFC+BC) by laver.

How about Sacks forcing,Hechler forcing, Mathias forcing,Miller forcing?

REPLY [10 votes]: The specific notions of forcing you mention are all part of the (now) basic toolbox for getting independence results in set theory at the level of the reals. The standard reference for set theory of this flavor is Bartoszynski and Judah's text "Set theory: on the structure of the real line". If you want to go further in this area (and based on your MO questions so far, perhaps you do) after Kunen's "Set theory: an introduction to independence proofs" this is the book you want to read. It has a wealth of information on the posets you've mentioned.
Each of these forcings is used to add a real of a certain kind to the universe, and in a 'definable way'. There is a unifying view one can take of these forcings: for each there is some ideal $\mathcal{I}$ on the reals so that the forcing is equivalent to forcing with all borel sets not in $\mathcal{I}$ (and the ideal $\mathcal{I}$ has a basis consisting of Borel sets). This perspective on forcing to add reals has been extensively studied with great success by Zapletal and many elegant results characterization properties of the forcing in terms of the relevant ideal have been discovered. See Zapletal's monographs 'Descriptive set theory and definable forcing' or 'Forcing idealized'.
Let me briefly go over the forcings you've mentioned. I'll give the ideal and the original context for the forcing, although each now has many applications for a wide variety of independence results.
Cohen forcing $\mathbb{C}$ was invented by Cohen to produce a model of $\neg\mathrm{CH}$. Of course, it has found many many other uses since then. It is the unique separative countable forcing. The relevant ideal is the collection of meager subsets of $\mathbb{R}$.
Random forcing $\mathbb{B}$ was invented by Solovay. He originally used it to analyze his model where all sets of reals are Lebesgue measurable. The paper "A model of set-theory in which every set of reals is Lebesgue measurable" is still quite readable, though you can also find the proof in Jech or in BJ. The relevant ideal is the collection of null subsets of $\mathbb{R}$.
Sacks forcing $\mathbb{S}$. This was invented by Gerald Sacks in order to produce a minimal real. That is, if $g$ is a Sacks real over the constructible universe $L$, and $x$ is any real in $L[g]$ then either $x\in L$ or $g\in L[x]$. This was done in his paper "Forcing with perfect closed sets". The relevant ideal is the collection of countable subsets of $\mathbb{R}$.
Hechler forcing $\mathbb{D}$ was invented by Stephen Hechler. It is the most basic way of adding a dominating real to the universe, and often it is referred to instead as 'dominating forcing' (including in BJ). Hechler used it produce (consistently) a wide variety of possible cofinal behaviors of the structure $(\omega^\omega,\leq^*)$; this was done in his paper 'On the existence of certain cofinal subsets of $\omega^\omega$'. The relevant ideal was isolated in the paper 'Hechler reals' by Labedzki and Repicky, and is also in BJ.
Laver forcing $\mathbb{L}$ was invented by Richard Laver to produce a model of the Borel conjecture (in his paper "On the consistency of Borel's conjecture", and also in BJ). The ideal is described in Zapletal's monographs.
Mathias forcing $\mathbb{M}$ is due to Adrian Mathias (see: 'Happy families'). He used it to prove (among other things) that the infinite exponent relation $\omega\rightarrow(\omega)^\omega$ holds in the Solovay model mentioned above. For the relevant ideal see either of Zapetal's monographs.
Miller forcing $\mathbb{Q}$ was originally called 'Rational perfect set forcing' by Arnold Miller, and was introduced in his paper with the same name. The paper is on his webpage http://www.math.wisc.edu/~miller/res/rat.pdf where you can read all about it. The ideal is the ideal of $\sigma$-compact subsets of $\omega^\omega$.<|endoftext|>
TITLE: What is parameterization of the trefoil knot surface in R³?
QUESTION [9 upvotes]: What is a parameterization, say (x(u,v),y(u,v),z(u,v)), of the trefoil knot surface in R³ whose cross-section can be circular or, in general, elliptic?
Thanks!

REPLY [13 votes]: Start with the parametrization of the torus surface:
$$
x=(a+b\cos u)\cos v,
$$
$$
y=(a+b\cos u)\sin v,
$$
$$
z=b\sin u.
$$
On that surface we get a curve that sits inside the trefoil. You get a parametrization of that curve $\vec{\gamma}=(x,y,z)$ by setting $u=3s, v=2s$ and letting $s$ range over the interval $[0,2\pi]$. This way you get a curve that wraps around the hole of the donut twice, and around the tube thrice.
Then you build a tube around that curve. You need a normalized tangent vector
$$
\vec{t}(s)=\frac{\vec{\gamma}'(s)}{||\vec{\gamma}'(s)||},
$$
and a normal vector
$$
\vec{n}(s)=\frac{\vec{t}'(s)}{||\vec{t}'(s)||},
$$
and a binormal vector
$$
\vec{b}(s)=\vec{t}\times\vec{n}.
$$
Then you can parametrize the trefoil surface $(x,y,z)=\vec{r}(s,\alpha)$ as follows
$$
\vec{r}(s,\alpha)=\vec{\gamma}(s)+c\cos\alpha\vec{n}(s)+c\sin\alpha\vec{b}(s).
$$
Here $c$ is the radius of the tube of the trefoil (should be smaller than $b$
= the radius of the tube of the torus). If you want elliptical cross-sections,
then you use two constants in place of $c$ above (possibly phase-shifting $\alpha$).
Both parameters $s$ and $\alpha$ should range over $[0,2\pi]$.

Arrggh! I found my code. When producing that image I used the normal vector
$$\vec{n}_T(s)=(\cos 2s\cos 3s,\sin 2s\cos 3s,\sin3s)$$
of the torus surface at the point $\vec{\gamma}(s)$ instead of the usual normal
$\vec{n}(s)$. The reason may have been the simpler formula. Together with that I then
also used $\vec{t}\times\vec{n}_T$ in place of the usual binormal. 
I simply wanted a picture of
some kind of a tube around the trefoil curve. Therefore my recipe may not be exactly what you want. Anyway, here is the pic of the resulting tube around the trefoil on the surface of a torus:<|endoftext|>
TITLE: Simple closed curves and the coefficent of $\exp(i\theta)$ in the associated Fourier series
QUESTION [6 upvotes]: Given a continuous map $f:S^1\to \mathbb{C}$ from the unit circle to the complex numbers, one can form its Fourier series $\sum_{n=-\infty}^\infty a_n\exp(in\theta)$. I want to stick with those $f$ that give simple closed curves, bounding a closed topological disk, going round the disk in a counter-clockwise direction, and parametrized proportional to arclength. I am happy to add the hypothesis that $f'(t)$ is a continuous function of $t$ and that, for $t\in S^1$, $|f'(t)| = 1$.

Is it then true that $a_1\neq 0$?

If this is true, is $|a_1|$ bounded away from zero as $f$ varies? It may be that some other normalization might make the second question more tractable: for example, instead of normalizing the length to be $2\pi$ by a change of scale, as I have done above, one could require that a disk of unit radius be contained in the disk bounded by $f$. Any such normalization of $f$ would be highly acceptable.
I'm motivated by trying to describe the ``space of shapes'' in the plane, by using Fourier descriptors, a topic of interest both in machine vision and in microscopy in biology.

REPLY [6 votes]: OK, let's modify Sean's construction to remove any doubts (it won't look the same, but it is based on the same idea). We will consider the curves symmetric with respect to the real axis and parametrized so that $f(-\theta)=\bar f(\theta)$, so we are sure that all Fourier coefficients are real. Now take $a\in\mathbb R$ and draw any continuous family of nice symmetric counterclockwise shapes $\Gamma_a$ that visit the points $1$, $a+i$, $a-i$ in this order. Note that the shapes will be necessarily non-convex for $a\ge 1$. Take small neighborhoods of these three points and replace the quick almost straight passages that are there by some "drunken walks" without self-intersections that have huge lengths but move essentially nowhere so that the whole length of the curve becomes essentially concentrated at those 3 points and the corresponding "wasted time" intervals are close to $(-\pi/2,\pi/2)$, $(\pi/2,\pi)$, $(-\pi,-\pi/2)$. Now, $2\pi a_1$ for the corresponding function is essentially $\int_{-\pi/2}^{\pi/2}\cos\theta\\, d\theta+2\Re\left[(a+i)\int_{\pi/2}^{\pi}e^{-i\theta}\\,d\theta\right]$, which is positive for large negative $a$ and negative for large positive $a$. However, the family of curves we created is continuous and so is the family of their parametrizations, so the intermediate value theorem finishes the story.
As usual, the existence of a counterexample most likely merely means that what you asked for is not what you need. So, what's the actual goal?<|endoftext|>
TITLE: Are piecewise linear curves dense among Hölder curves?
QUESTION [5 upvotes]: Consider for some $0 < \alpha \leq 1$ the space functions $x:[0,1] \to \mathbb{R}^n$ such that $x(0) = 0$ and
$\sup_{s,t} \frac{\|f(t)-f(s)\|}{|t-s|^{\alpha}}$
is finite.
There are at least two reasonable norms defined on this space.  The first is the Hölder norm which is just the supremum above.  Another is the $1/\alpha$-variation which is the supremum over all partitions $t_0 = 0 \le t_1 \le \cdots \le t_r = 1$ of $\left(\sum_{i=0}^{r-1}|f(t_{i+1}) - f(t_i)|^{1/\alpha}\right)^\alpha$.
Let us fix $\alpha= \frac{1}{2}$ and $x(t) = \sqrt{t}$ and suppose $y:[0,1] \to \mathbb{R}$ is piecewise linear with $y(0) = 0$.  It follows easily that $\lim_{t\to 0}\frac{\|x(t)-y(t)\|}{\sqrt{t}} = 1$.
This implies that there is no sequence of piecewise linear approximations to $x$ in Hölder norm.
However, it's not too hard to show that $x$ can be approximated in $2$-variation by piecewise linear functions.
My question is the following:  Are piecewise linear functions dense among $1/2$-Hölder functions in the $2$-variation sense?
I'm also interested in the same question replacing piecewise linear functions by smooth functions.

REPLY [3 votes]: To expand on fedja's comment :
For $p>1$, a function $f$ on $[0,1]$ is in the $p$-variation closure of smooth functions $C^{0,p-var}$ iff
$$\lim_{\delta \rightarrow 0}\;\;\; \sup_{\substack{0=t_0<\ldots< t_m=1 \\\ |t_{i+1}-t_i|\leq\delta}}  \sum (f(t_{i+1})-f(t_i))^p = 0. \label{rel}$$ 
Then the function $g(x) = \sum_{i \geq 1} c^{-i/p} \sin(c^i  x)$ is $(1/p)$-Hölder, but does not satisfy this relation (for $c$ large enough).
Note that any continuous function of finite $q$-variation, for some $q< p$ (such as your square root example), is in $C^{0,p-var}$.
For $p=1$, $C^{0,1-var}$ is the space of absolutely continuous functions.
You can find these results e.g. in Subsection 5.3.3 of Friz&Victoir "Multidimensional stochastic processes as rough paths" (pdf)<|endoftext|>
TITLE: Invariants of Symmetric group
QUESTION [8 upvotes]: The celebrated Chevalley–Shephard–Todd theorem says that $\mathbb C[V]^{S_n}$ is a polynomial algebra and gives the generators of this algebra, where $V$ is the standard (or natural) representation of the symmetric group $S_n$. I am just curious to know for what other representations of $S_n$ the generators of this algebra is known ? When is this algebra a polynomial algebra ?

REPLY [17 votes]: Let $n \ge 7$. If $V$ is an irreducible representation of $S_n$ such that $\mathbb{C}[V]^{S_n}$ is a polynomial algebra then either $V$ is the trivial representation, the sign representation or the $(n-1)$-dimensional standard representation. 
Outline Proof: Let $\rho : S_n \rightarrow \mathrm{GL}(V)$ be an irreducible representation affording the irreducible character $\chi^\lambda$ where $\lambda$ is a partition of $n$. Suppose that $V$ is $d$-dimensional. By the Chevalley-Shephard-Todd theorem, $\mathbb{C}[V]^{S_n}$ is a polynomial algebra if and only if $\rho(S_n)$ is generated by pseudo-reflections. If $\rho(g)$ is a pseudo-reflection then $\rho(g)$ is similar to a diagonal matrix $\mathrm{diag}(1,1,\ldots,1,\zeta)$ where $\zeta$ is a root of unity. Hence $\chi(g) = d-1 + \zeta$. However, the irreducible characters of symmetric groups are real valued, so if $g \not= 1_{S_n}$ then $g$ is an involution and $\chi^\lambda(g) = d-2$. 
It therefore suffices to show that if $n \ge 7$ and $g \in S_n$ is an involution such that $\chi^\lambda(g) = \chi^\lambda(1)-2$ then $g$ is a transposition and either $\lambda = (1^n)$ or $\lambda = (n-1,1)$. This follows by induction using the Murnaghan-Nakayama rule. (The details are fiddly but routine.) 
Edit: Geoff Robinson shows in his answer (posted at the same time as mine) that $g$ is a product of at most $3$ transpositions. This leads to a quick inductive proof: suppose that $\lambda$ has two removable boxes, whose removal gives partitions $\mu$ and $\nu$ of $n-1$. If $\mu \not= (n-1)$ and $\nu \not= (n-1)$ then, by induction, we have $\chi^\lambda(g) \le \chi^\lambda(1)-4$. Hence $\lambda = (n-1,1)$. In the remaining case $\lambda$ is a rectangular partition, and provided $n\ge 8$, we can repeat this argument after removing two boxes from $\lambda$ in two different ways.
For smaller $n$ there are two exceptional cases, corresponding to the partitions $(2,2)$ of $4$ and $(2,2,2)$ of $6$. The former representation is $2$-dimensional and is obtained from the standard representation of $S_3$ via the quotient map $S_4 \rightarrow S_3$. The latter is $5$-dimensional, and can be obtained by applying the outer automorphism of $S_6$ to the standard $5$-dimensional representation of $S_6$; the relevant character value is $\chi^{(2,2,2)}(g) = 3$ where $g = (12)(34)(56) \in S_6$.

REPLY [10 votes]: I only concern myself with faithful representations of $S_n$ and for $n >4$. The only way to get a polynomial algebra of invariants is to represent $S_n$ as a complex reflection group (so generated by pseudo-reflections, that is elements wih a fixed-point space of codimension $1$). A complex relfection group is easily checked to always be a direct product of ireducible complex reflection groups, so from now on I consider only irreducible representations. The question then becomes: is there an irreducible representation of $S_n$ other than the usual $n-1$-dimensional irreducible constituent of the natural permutation character, where $S_n$ is represented as a complex reflection group? Note that tensoring with the sign representation of the stated $n-1$-dimensional representation does not produce a representation of $S_n$ as a complex reflection group. There is still some content to this question. It can probably be easily addressed by the Murnaghan-Nakayama  rule, or from a more detailed knowledge of the character table of $S_n,$ but I attempt a direct argument here. Since the characters of $S_n$ are rational-valued, any element represented as a pseudo reflection in the given representation must be represented as a genuine (or "real") reflection, since its trace must be rational. The generating reflections for $S_n$ in the given representation act with determinant $-1,$ so lie outside the derived group $A_n.$ Hence they are odd permutations, and expressible as a product of an odd number of (disjoint) transpositions, since they have order $2$. Since $A_n$ is simple ( as $n \geq 5$,) any single conjugacy class of these reflections generates the whole of $S_n,$ so we assume that all the generating reflections are conjugate. Note that we are not yet entitled to assume that the generating reflections are transpositions. Now we note that the product of two generating reflections has order $1,2,3,4$ or $6.$ For such a product has a fixed point space of codimension $0$ or $2,$ has determinant $1$, and has a rational trace. Now if $k >1$ is odd, a product of $k$ transpositions inverts a $2k+1$-cycle in $S_{2k+1}.$ Hence if $k > 3,$ and $n \geq 2k+1,$ then $S_{n}$ contains a pair of conjugate permutations, each a product of $k$ disjoint transpositions, whose product has order $2k+1.$ 
When $n = 2k$ we may express a $2k$-cycle as a product of two permutations, one a product of $k$ disjoint transpositions, the other a product of $k-1$ disjoint transpositions. Since $k > 3,$ if we adjoin a transposition interchanging the fixed points of the second permutation, we get a product of two permutations, each a product of $k$ disjoint transpositions, where the product is a product of two disjoint cycles of lengths (allowing a $1$-cycle) whose sum adds to $n.$ Since $n \geq 10,$ such a product never has order $1,2,3,4,5$ or $6.$ Hence we can assume that our generating reflections are products of at most three transpositions. Furthermore, if $n \geq 8,$ we can express a $5$-cycle in $S_n$ as a product of permutations, each a product of three disjoint transpositions. For we may express a $5$-cycle in $S_5$ as a product of two permutatons, each a product of two disjoint two-cycles. Affix a transposition commuting with the $5$-cycle to each of them, and this does not change the product. Hence if our reflections are products of $3$ disjoint $2$-cycles, we are left with the case $n =6.$ In fact, this case does occur. "Twist" the natural (non-unimodular) irreducible $5$-dimensional representation of $S_6$ by the exceptional outer automorphism of $S_6,$ and we obtain a reflection representation of $S_6$ in which products of $3$ disjoint $2$-cycles act as reflections. It remains to deal with reflection representations of $S_n$ in which transposition act as reflections. I presume that the Murnaghn-Nakayama rule or the character table of $S_n$ then shows that only the expected representation arises, but I leave that issue open here. (Later Edit: This now seems to be covered by Mark Wildon's answer).<|endoftext|>
TITLE: Help with a mellin-type integral 
QUESTION [5 upvotes]: greetings . i've been trying to do this integral for many days now, with no clue on how to attack it . the integral is a mellin inverse of some kind, and appears in analytic number theory . 
$$I(x)=\lim_{T\rightarrow \infty}\frac{1}{2\pi i}\int_{2-iT}^{2+iT}\frac{x^{s}}{s}\left(\sum_{k=1}^{\infty}\frac{\zeta(ks)-1}{k} \right )^{n}ds$$
$\zeta(s)$ : is the Riemann zeta function. 
$n \geqslant 2 $
any insights are more than welcome .

REPLY [2 votes]: Continued from my earlier post: I would therefore suggest approaching it in the ``counting domain'' as follows: from the explicit definition of the coefficients given by GH above, you can easily see that your Dirichlet series in the case $n=1$ is
   $$\sum_{n=2}^{\infty}\left(\sum_{n=n_i^{k_i}(n)}\log n_i(n)\right)\frac{1}{n^s\log n}=\sum_{n=2}^{\infty}\frac{\log\prod_{i}n_i(n)}{n^s\log n}.$$
From here you get 
    $$I(x)=\sum_{n\leq x}\frac{\log\prod_{i}n_i(n)}{\log n}.$$<|endoftext|>
TITLE: Busy Beaver - Proof for BB(2) = 4
QUESTION [8 upvotes]: Hi,
I need to prove the above claim.
I can show that $BB(2)\ge 4$ by building a turing machine,
but how can i show that $BB(2) \le 4$?
Searched a lot over the web, and saw that Rado proved it in 1963.
Thanks.

REPLY [2 votes]: Let allow me to write down this table of possible rules:
A0|0LA|1LA|0RA|1RA|0LB|1LB|0RB|1RB|0LH|1LH|0RH|1RH
A1|0LA|1LA|0RA|1RA|0LB|1LB|0RB|1RB|0LH|1LH|0RH|1RH
B0|0LA|1LA|0RA|1RA|0LB|1LB|0RB|1RB|0LH|1LH|0RH|1RH
B1|0LA|1LA|0RA|1RA|0LB|1LB|0RB|1RB|0LH|1LH|0RH|1RH
There will be indeed 20736 TMs, but you can easily eliminate many of them. You probably know that Robert Munafo in his site concluded that the first rule, for A0 must be 1RB, so, writing one and move to the right. So our table is:
A0|1RB
A1|0LA|1LA|0RA|1RA|0LB|1LB|0RB|1RB|0LH|1LH|0RH|1RH
B0|0LA|1LA|0RA|1RA|0LB|1LB|0RB|1RB|0LH|1LH|0RH|1RH
B1|0LA|1LA|0RA|1RA|0LB|1LB|0RB|1RB|0LH|1LH|0RH|1RH
This gives only 12^3 = 1728 TMs. We can reduce it further if we note that in all rules we can leave only one possibitily for the halting rule: 1RH (or 1LH, no matter for the result). So:
A0|1RB
A1|0LA|1LA|0RA|1RA|0LB|1LB|0RB|1RB|1RH
B0|0LA|1LA|0RA|1RA|0LB|1LB|0RB|1RB|1RH
B1|0LA|1LA|0RA|1RA|0LB|1LB|0RB|1RB|1RH
This leaves us with 729 TMs. Now we can note that in the rule B0, we can eliminate the possibilities 0RA, 1RA, 0RB, 1RB and 1RH, first four lead to infinite loop, the last one lead to 2 ones. Our table is:
A0|1RB
A1|0LA|1LA|0RA|1RA|0LB|1LB|0RB|1RB|1RH
B0|0LA|1LA|0LB|1LB
B1|0LA|1LA|0RA|1RA|0LB|1LB|0RB|1RB|1RH
324 TMs for now. Don't worry, we mustn't shall check them all, we need to explore 4 branches from 0LA, 1LA, 0LB, 1LB in rule B0. Now we can create the sub-ruleset for every branch, for example:
Branch B0|0LA
A1|0LA|1LA|0RA|1RA|0LB|1LB|0RB|1RB
B1|1RH
Note that one rule must be halting anyways, otherwise TM will works infinitely regardless to complexity of its behavior. In Branch B0|0LA, this rule must be B1|1RH, and in A1 we can eliminate the halting rule.
In every branch you will have 8 TMs, so there are 32 TMs to check. Simulate them on http://morphett.info/turing/turing.html to figure out their behavior.<|endoftext|>
TITLE: Is there a 2 component link with full symmetry?
QUESTION [16 upvotes]: If you take a (labelled, oriented) 2 component link, it has a symmetry group which is a subgroup of the 16 element group $Z2 \times (Z2 \times Z2 \ltimes S_2)$ (mirror, reverse each component, swap the components). With Parsley, Cornish, and Mastin (http://arxiv.org/abs/1201.2722) I compiled a table of the frequency with which each conjugacy class of subgroups of this group comes up as a symmetry group for links up to 14 crossings.
Interestingly, the full group never came up as a symmetry group! Can a 2 component link ever have the full 16 element group? Is there some obstruction to this? Can anyone propose an example link?
Edit: A non split example. It is easy to construct split ones!
Clarification: The traditional symmetry group is the mapping class group $MCG(S^3,L)$. The symmetry group I want is the image of that group under the homomorphism $MCG(S^3,L) \rightarrow MCG(S^3) \times MCG(L)$, which is the 16 element group above. So another way to ask the question would be: can this homomorphism ever be surjective?

REPLY [5 votes]: This is more a comment on Ryan's nice answer than an independent answer itself, but it's too long for a comment box.
Let $K$ be an oriented knot in the solid torus which is isotopic (in the solid torus) to its reverse.  [Edit: It should also be amphichiral, as well as invariant under the map of the solid torus which reverses the $S^1$ factor.]  Let $L$ be a two component unoriented framed link in $S^3$ which is isotopic to its mirror image (without exchanging the two components) and also has an isotopy which exchanges the two components.  Then replacing each component of $L$ with a copy of $K$ yields a two component oriented link in $S^3$ which has the full order 16 symmetry mentioned in the question.
In Ryan's example, $K$ is what I would call a Whitehead double with a funky figure-8 clasp, and $L$ is the Hopf link.
It's relatively easy to think of $K$ and $L$ satisfying the above conditions, so one can construct a large family of examples this way.<|endoftext|>
TITLE: Who first proved that the dimension of a vector space is unique?
QUESTION [10 upvotes]: every vector space is known to have a basis (assuming the axiom of choice). This is attributed to Georg Hamel (http://de.wikipedia.org/wiki/Georg_Hamel). Moreover, any two bases have the same cardinality. In the finite dimensional case this is often proved by the Steinitz exchange lemma. An interesting note on the history of this lemma can be found here. In particular, the author points out that the exchange lemma isn't originally due to Steinitz but was published already by Grassmann in 1862. 
However, my question is: Who has at first proved (or published a proof of) the uniqueness of  dimension in the general case ? 
I checked Hamel's paper (an online link to this paper is included in the wikipedia article), but he only proves the existence of a basis without discussing its cardinality.

REPLY [7 votes]: Since this has already been bumped, I'll copy godelian's reference into an answer:

Löwig, H. "Über die Dimension linearer Räume." Studia Mathematica, 5(1):18-23, 1934. http://matwbn.icm.edu.pl/ksiazki/sm/sm5/sm513.pdf.<|endoftext|>
TITLE: Whitehead product with identity on homotopy groups of spheres
QUESTION [17 upvotes]: For $n\geq 2$ let $(S^n,p)$ be the $n$-sphere with a base point $p$. Let $1:S^n\to S^n$ denote the identity map. Let us define the map
$Wh_1: \pi_i(S^n,p)\to \pi_{i+n-1}(S^n,p),  \alpha\mapsto [\alpha,1],$
where $[\cdot,\cdot]$ is the Whitehead bracket. 
Question 1 : What is known about $Wh_1$ ?
Question 2 : If we let $Wh_f$ denote the corresponding map on homotopy groups for $f:(S^m,p)\to (S^n,p)$ then what is known about $Wh_f$ ?

REPLY [7 votes]: As you probably know, the Whitehead product is a degree $-1$ Lie bracket on homotopy groups, i.e. it is graded anticommutative and satisfies the graded Jacobi identity, but not $[x,x]=0$. In particular your maps are homomorphisms. As for the first one, it need not be trivial, if that's what you think. Since suspensions of Whitehead products vanish, your maps are trivial when the target is stable, e.g. for $n\geq i+1$ in the first case. Also, in the first case, it is the kernel of the suspension map in the critical dimension, i.e.
$$\pi_n(S^n)\stackrel{Wh_1}\longrightarrow \pi_{2n-1}(S^n)\stackrel{\Sigma}\longrightarrow \pi_{2n}(S^{n+1})$$
is an exact sequence, e.g. for $n=2$ this looks like as follows
$$\mathbb{Z}\stackrel{2}\longrightarrow\mathbb{Z}\twoheadrightarrow \mathbb{Z}/2.$$
This uses the Blakers-Massey theorem and the fact that $1\in\pi_n(S^n)$ is a generator. Hence, the analog exact sequence works in the second case only in the special case that $f\in \pi_m(S^n)$ is a generator. Probably more things can be deduced from the elementary properties of primary homotopy operation, but this is what comes to my mind right now.<|endoftext|>
TITLE: Linear subspaces in cones over orthogonal groups
QUESTION [6 upvotes]: Consider the orthogonal group $G=O(n)$ as a subset of the vector space of $n\times n$ real matrices. Let $C=C(G)$ denote the Euclidean cone over  $G$, i.e., the space of matrices of the form $tA, A\in G, t\in {\mathbb R}$. 
Question. What are vector subspaces in $C$? For instance, are there 3-dimensional linear subspaces? (One can ask the same question for cones over other linear Lie groups $G$.) 
Update: @Dmitri noticed that in dimensions divisible by $4$ and $8$ there are 4-dimensional and 8-dimensional linear subspaces coming from quaternions and octonions. Thus, the question is: What are other linear subspaces, in particular, are there linear subspaces (which are not lines) for odd $n$? Are there linear subspaces which are not contained in cones over subgroups? 
Trivial examples of subspaces in $C(G)$ are planes, obtained as cones over $O(2)$. 
Diagonal embedding of $O(2)$ to $O(2n)$ (and its conjugates) yield examples of planes in $C(O(2n))$. Are there other examples of linear subspaces (of dimension $>1$) in $C(O(n))$? My motivation for the question comes from hyperbolic geometry (where the question is about affine subspaces of $O(n,1)$), but I find the linear algebra question about the orthogonal group intriguing, because it is so basic. The question is somewhat similar to the one in 
Spaces of matrices with same eigenvalue/Great circles in O(n)-orbits 
Alternatively, and algebraic geometers like it better, one can work over ${\mathbb C}$ and projectivize everything. Thus, I am asking about linear subspaces of the variety $X=PO(n)\subset {\mathbb P}^{N}$, $N=n^2-1$. In other words, the question is about Fano varieties $F_k(X)$ for various $k$. Algebraic geometers studied Fano varieties $F_k(X)$ of projective varieties since 19th century, so, maybe somebody looked at the case of homogeneous $X$, in particular, projective varieties $X$ which are algebraic subgroups. Unfortunately, google search did not yield anything useful.

REPLY [6 votes]: Angelo's answer essentially gives a description of all such linear spaces and also describes the only way they can happen.  To see this, suppose that you have a linear subspace $L\subset C$ that passes through $A\in O(n)$.  Then multiplying everything on the left by $A^{-1}$, you can assume that $L$ passes through the identity element $I\in O(n)$.  Suppose that $L$ has dimension $1{+}m$.  For any nonzero element $a\in L$, the matrix $a^T a$ must be a positive multiple of $I$, so there exists a positive definite quadratic function $f:L\to \mathbb{R}$ such that $a^T a = f(a) I$.  Let $a_0 = I, a_1,\ldots,a_m$ be an $f$-orthonormal basis of $L$.  Computation shows that $a_i^T\ a_j+a_j^T\ a_i = 2\delta_{ij}\ I$.  In particular, it follows that each $a_i$ for $i>0$ is skew-symmetric and hence that $a_ia_j+a_ja_i = -2\delta_{ij}\ I$ for $i,j>0$ and distinct.  Thus, the $a_i$ for $i>0$ generate the action of an orthogonal Clifford algebra on $\mathbb{R}^n$.
A similar argument can be made over the complexes, but, there, you have to watch out for degenerate quadratic forms.<|endoftext|>
TITLE: Matrices whose inverse is positive
QUESTION [11 upvotes]: I have recently come across some examples of matrices with a special structure.
I will describe these matrices here and I hope that somebody will be able 
to point out a source where I can find more information about them. Consider 
an $n\times n$ matrix $A$ with elements $a_{ij}$ having the following 
properties. The elements with $i=j$ (call them $b_i$) are negative. The 
elements with $j=i+1 {\rm mod} n$ (call then $c_i$) are positive. All other 
elements are zero. The determinant of a matrix of this type is 
$\prod_i b_i+(-1)^{n+1}\prod_i c_i$. A property of these matrices which I 
found surprising is that $(-1)^{n+1}(\det A)A^{-1}$ is a positive matrix, i.e. 
all its entries are positive. I found this by playing around with some 
examples. Can anybody point out to me some general theory which explains this
observation? I met these matrices repeatedly when looking at certain chemical
reaction networks. In that context the positivity statement is valuable because
it allows the Perron-Frobenius theorem to be applied.

REPLY [3 votes]: Matrices whose inverses are nonnegative are also called monotone. There are a number of equivalent characterizations in Theorem 6.2.3 of the wonderful book by Berman and Plemmons:
http://books.google.ie/books/about/Nonnegative_Matrices_in_the_Mathematical.html?id=MRB7SUc_u6YC&redir_esc=y
In the case of this question, the matrix might not be an $M$-matrix. It depends on the actual entries.<|endoftext|>
TITLE: Probability to be the winner in a tournament
QUESTION [14 upvotes]: In a project in Game Theory we (Ayala Arad and Ariel Rubinstein) are stuck with the following "simple" question. We are sure of the conjecture but we failed to find a (hopefully simple) proof: 
Let $A$ and $B$ be two non-empty finite disjoint sets of players.  Any two players in $A$ are "matched" and $\$2$ are transferred from one to the other.  Any player in $A$ is also matched with any player in $B$ and $\$1$ is transferred from one to the other.  The two possible directions of each transfer are equally likely and independent. No transfers are carried out between players in $B$.
    The winner is the player with the highest net transfers. In the case of a tie, the winner is selected randomly from among the highest scoring players. 
(For example if $|A|=1$ and $|B|=2$ the probability of winning for the player in $A$ is $1/4$ and the probability for the player in $B$ is $3/8$.  If $|A|=|B|=2$ the corresponding numbers are $21/64$ and $11/64$).

Claim:  If $|AUB|>3$ then the probability of winning for any player in $A$ is strictly larger than that of any player in $B$.

REPLY [3 votes]: This was intended to be a comment to Ori's post but it is too long, so I'm posting it as an answer. First of all, let us modify the game a bit by initially giving each player a random score between $0$ and $\varepsilon$. That will break the ties just as needed but will allow us to talk about the winner. 
Now the case $|A|=|B|$ is trivial. Let's do all transactions between $A$ and $B$ first and look at the resulting configurations. They split into natural pairs (swapping $A$ and $B$). Now let $a$ be the top score in $A$ and $b$ be the top score in $B$. Arrange the pair so that $a>b$. Then we need to show that for every configuration the probability that the top score in $A$ will become less than $b$ after transactions in $A$ is less than that the probability that the top score in $B$ will become larger than $a$ if we do the transactions in $B$. Identify $A$ with $B$ in some way so that the top scorers are identified. Any way to do the transactions in $A$ that moves the winner to $B$ should bring the score $a$ of the top scorer in $A$ below $b$ at the very least and that may be insufficient in some configurations. On the other hand, if have one such way and do the inverse transactions in $B$ instead, they'll bring the top scorer in $B$ above $a$ and it is not necessary to move the winner to $B$. That's all one needs to say about the equal cardinalities case. 
Now, like Ori, I have to say that I'll try to give a more complete answer later.<|endoftext|>
TITLE: Arithmetic geometry examples
QUESTION [33 upvotes]: (This is inspired by Algebraic geometry examples.)
I want to collect here (counter)examples in arithmetic geometry.

Curves violating the Hasse principle: The Selmer curve $3X^3 + 4Y^3 + 5Z^3 = 0$. It is a nontrivial element of the Tate–Shafarevich group of the elliptic curve $3\cdot4\cdot5\cdot X^3 + Y^3 + Z^3 = 0$. It is also an example of an abelian variety for which finiteness of Sha is known. In fact, $|\mathrm{III}(E/\mathbf{Q})| = 3^2$.
Non-isogeneous elliptic curves having the same Hasse–Weil $L$-series: $y^2 = x^3 \pm ix + 3$ over $K = \mathbf{Q}(i)$ (cf. Cornell–Silverman–Stevens, p. 32).
Counterexample to the Hasse norm theorem for non-cyclic extensions: $L = \mathbf{Q}(\sqrt{13},\sqrt{17})$ Galois with $G =\mathbf{Z}/2 \times \mathbf{Z}/2$, see Cassels–Fröhlich, p. 360, Exercise 5.3.

tbc

REPLY [11 votes]: It is a widely-used fact that a complex torus with "complex multiplication" is algebraic (i.e., an abelian variety) in the sense of the GAGA theorem; the proof goes via Riemann forms.  But the analogue over $p$-adic fields is false: there exists 
a non-algebraizable formal CM abelian scheme over a $p$-adic discrete valuation ring, so (using Neron-Ogg-Shafarevich) its generic fiber in the sense of Raynaud is a rigid-analytic smooth connected proper group with "complex multiplication" yet is not algebraic (in the sense of $p$-adic GAGA). 
More specifically, for $p \equiv \pm 2 \bmod 5$, consider the simple abelian surface over $\kappa = {\mathbf{F}}_{p^2}$ with Weil number $\pm p \zeta_5$ and endomorphism ring $\mathbf{Z}[\zeta_5]$ (this exists by Honda-Tate theory, and for each sign it is unique up to $\mathbf{Z}[\zeta_5]$-linear isogeny). This lifts to a formal abelian scheme $A$ with action by $\mathbf{Z}[\zeta_5]$ over $W(\kappa)$. But for $K := W(\kappa)[1/p]$ the induced $K$-linear action of $\mathbf{Q}(\zeta_5)$ on the 2-dimensional $K$-vector space ${\rm{Lie}}(A)[1/p]$ is given by a pair of embeddings $\mathbf{Q}(\zeta_5) \rightrightarrows K$ related through complex conjugation, so it is not a CM type and hence $A$ is not algebraic. 
For details, see 4.1.2 (up through 4.1.2.3) in this link.<|endoftext|>
TITLE: Does Nori's fundamental group scheme appear in Kim's work
QUESTION [10 upvotes]: This is a very vague question. I was just reading the introduction to M. Kim's article on motivic fundamental groups and the theorem of Siegel and noticed that there are essentially three fundamental groups appearing in his work: the De Rham fundamental group, the cristalline fundamental group and the etale fundamental group.
Now, does Nori's fundamental group scheme also appear somewhere in his work? If no, why not?

REPLY [11 votes]: It depends on what you call Nori's fundamental group scheme, of course. Nori himself has given several versions of his fundamental group scheme, and it has been vastly generalized.
If you think of the classical definition (the tannaka group of the category of essentially finite vector bundles) I don't think it does. It is of a somewhat different nature of the tannaka groups you list above : it is pro-finite, whereas the ones appearing in your list are all pro-unipotent. They are not designed for the same purposes: Nori's fundamental group was built to take into account positive characteristic phenomenons, especially torsors under finite group schemes that are not necessarily étale, whereas in Kim's fundamental groups both the de Rham version and the étale one are defined for a variety over a field of characteristic zero, and my vague understanding of the subject is that all of them are algebraic incarnations of the unipotent envelope of the non existing topological fundamental group.
Of course, this is not so simple. Nori has defined in his Phd an unipotent version of his fundamental group scheme which is related to Kim's de Rham. Also, a version of Nori's fundamental group scheme (rather, groupoid) was recently used in characteristic zero by Esnault and Hai to study the section conjecture, which is deeply interconnected with Kim's work.<|endoftext|>
TITLE: Complexity of surfaces bounding knots in 4-ball and 3-sphere respectively
QUESTION [9 upvotes]: I'm interested in a complexity question related to problems like the slice-ribbon problem. 
To be specific, if $K \subset S^3$ is a knot, it might be non-trivial yet still bound a smoothly-embedded compact 2-dimensional ball $B \subset D^4$, where we consider $S^3 = \partial D^4$, and the ball is required to be properly embedded, meaning $B \cap D^4 = \partial B = K$.  This is called a (smoothly) slice knot. 
The slice-ribbon conjecture states that the only way this can happen is if you can find $B$ so that the distance function $d : D^4 \to \mathbb R$ given by $d(x) = |x|$ has no local maxima on $B$, i.e. it is Morse with only saddle points and local minima on $B$, and no other critical points. 

Q1: Do people have computations describing how many such saddle points you need for knots?  In a ribbon diagram for the knot this is the number of components in the self-intersection set.  I'm particularly interested in how this number would relate to a minimal-complexity planar knot diagram for the knot -- where you minimize crossing number or some other similar feature of a knot diagram.  I want this complexity to be related to the number of tetrahedra you would need in a triangulation of $S^3$ so that the knot is a "normal curve" in the triangulation, meaning transverse to the triangulation and linear inside each tetrahedron. 

More generally, I'm interested in the computational problem of finding candidate knots that are smoothly slice yet not ribbon.  In 3-manifold theory there is a "normalization" process for normal surfaces in a triangulated manifold.  This involves some isotopy, and embedded surgery.   There is a much more primitive type of "normalization" process that works in any triangulated manifold (of any dimension) with respect to any submanifold -- simply subdivide the ambient triangulation and perturb the submanifold to be transverse.  After some number of subdivisions, the submanifold will have to appear to be linear in each top-dimensional simplex.  This is the key idea in Whitehead's proof that smooth manifolds admit triangulations.

Q2: Are there any good estimates for how many subdivisions it takes to "normalize" a submanifold in the above Whitehead sense?   

Q2 can be taken to be quite general. For example, for curves in surfaces, you can always normalize without any subdividing at all, all it takes is an isotopy.  After minimizing the number of intersection points with the 1-skeleton of the triangulation, the only curve that needs more than a small isotopy (in the sense of the mapping space topology) would be the boundary of a disc embedded in the interior of a triangle. 
For example, is there an interesting topological invariant of a knot that gives good lower-bounds on the number of tetrahedra you need to triangulate $S^3$ to make it appear to be locally linear?   Things like the maximum of the curvature aren't topological, and I consider crossing number in a planar diagram to be not computationally-friendly enough.

REPLY [8 votes]: This is an answer to the first question.
Let's indicate with $r(K)$ the ribbon number of a knot, i.e.the minimum
number of ribbon singularities needed to realize a ribbon disc spanning $K$.
We have
$$
r(K)\geq g(K)
$$
This is shown by Fox here:http://ir.library.osaka-u.ac.jp/metadb/up/LIBOJMK01/ojm10_01_08.pdf
Mizuma has shown that under certain conditions on the Alexander and Jones 
polynomial you can assume that $r(K)\geq 3$. This is Theorem 1.5 here:http://ir.library.osaka-u.ac.jp/metadb/up/LIBOJMK01/1782ojm.pdf 
It is a very special situation and I don't think that much more is known in the general case.
Maybe it is worth noting that given a band diagram for a ribbon disc one can add a fake
ribbon singularity near each singularity and then eliminate both with a tubing operation. This produces a Seifert surface whose genus equals the number of the original ribbon singularities in the band diagram. The definition of a band diagram and a picture of this trick can be found here:http://etd.adm.unipi.it/theses/available/etd-07062011-061816/unrestricted/Polynomial_invariants_of_ribbon_links_and_symmetric_unions.pdf
(pag. 27)<|endoftext|>
TITLE: Rank of isogenous elliptic curves
QUESTION [6 upvotes]: I think that k-isogenous elliptic curves have the same rank as I think rank is an isogeny invariant. However, I am not sure. Does anyone know where could I find a proof? Thanks!

REPLY [12 votes]: An isogeny $A \to B$ is a map $A \to B$ with finite kernel. Choose a splitting of $MW(A)$ into torsion-free and torsion summands. This kernel cannot include any of the torsion-free part of $MW(A)$ and so is injective on the torsion-free part so the rank of $MW(B)$ is at least the rank of $MW(A)$. Since whenever there is an isogeny $ f\colon A \to B$ (say of degree $n$), there is an isogeny $B \to A$ (for example, the one whose composition with $f$ is multiplication by $n$) , this gives you equality.<|endoftext|>
TITLE: Probability of a set of random vectors over finite field being a spanning set
QUESTION [6 upvotes]: Suppose I have a set of random vectors $f(a_1, \ldots, a_\ell) :=  (v_1, \ldots, v_m)  \subset F_p^n$, $m \ge n$, given by a matrix valued polynomial function $f$, where the $a_i$'s are independent, uniformly distributed in $F_p \setminus \{0\}$, and each component of $f$ consists of some polynomial in the $a_i$'s whose coefficients are all in $\{0,1\}$, and such that the degree of each variable is at most $1$. For example, the following set of random vectors fits the discription:
$$ v_1 = (a_1, a_2 a_1, a_3); v_2 = (a_2, a_1 a_2 a_3 + a_2, 0); v_3 = (a_1 + a_2, a_2 + a_3, a_3 a_1)$$.
Now consider the quantity $\pi_p = \mathbb{P}(f(a_1, \ldots, a_\ell) \text{ spans } F_p^n)$. My question is, is $\pi_p$ monotone non-decreasing in $p$? If not can one give a counterexample? The motivation comes from a recent result of Yuval Peres and Allan Sly (Arxiv preprint arXiv:1105.4402, 201) giving the right order of mixing time for the most natural random walk on uni-upper triangular matrices over $F_p$. Knowing the above will extend their result of $\mathcal{O}(n^2)$ to $p$ that grows with $n$, which is highly anticipated. 
Edit: Will Sawin below essentially solved an earlier version of this problem, where I forgot to state the degree condition on the $a_i$'s.

REPLY [4 votes]: One vector $v=a_1^2+1$. The probability that it forms a spanning set is $1$ or less than $1$ depending on if $-1$ is a quadratic residue mod $p$.
Edit: In the linear case, you can just set $v_1=(a_1,a_2)$, $v_2=(a_2,-a_1)$, determinant of the matrix $=a_1^2+a_2^2$. They span if and only if the determinant is nonzero, which happens if and only if $(a_1/a_2)^2\neq-1$.<|endoftext|>
TITLE: The rank of a symmetric space
QUESTION [21 upvotes]: I would like to work a theorem on a article who deals with the rank one symmetric spaces.
i looked up the definition of symmetric spaces of rank one, but I did not find a satisfactory definition then what is the meaning of rank, intuitively and mathematically? please if anybody already worked with rank one symmetric spaces..

REPLY [3 votes]: If $M=G/K$ has noncompact type and ${\rm rank}(M)=r\ge 1$, then the space of invariant differential operators on $M$ is generated by a $r$-tuple. One can choose the first element of this $r$-tuple to be the Laplace/Casimir operator, so for rank $1$ symmetric spaces it is always generated by the Laplacian.<|endoftext|>
TITLE: strict/effective epimorphism and faithfully flat
QUESTION [8 upvotes]: In SGA 1, Expose 5, there is a proposition (3.5), which states that for any two objects $X$ and $X'$ in $\textbf{Et}/S$, all $S$ morphisms between them factor into a surjective etale morphism $X\to X''$ and a canonical immersion $X''\to X'$.  If I am understanding this proof correctly, this is by the fact that to be in $\textbf{Et}/S$, any morphism from an object in the category to $S$ is finite etale, and therefore both open and closed, and so any morphism between objects in the category is both open and closed, so our image will be open and closed, and thus a finite union of connected components of $X'$.
There is a corollary following this proposition, rephrasing it in terms of an effective epimorphism followed by a monomorphism.  What are the advantages of this?  The corollary follows by the fact that the first map is faithfully flat, and thus a strict epimorphism, and a similar result for the monomorphism.  Perhaps this is just my lack of familiarity with category theory, but would every strict epimorphism be a faithfully flat morphism in this category?  Would the reason to introduce these terms just be to define Galois categories later, and should we always just think about strict epimorphisms/monomorphisms in the sense of Proposition 3.5?
(Also, is there a typo in the statement, switching the roles of $X'$ and $X''$?)

REPLY [4 votes]: I can offer two answers, but I am not sure either is what you are looking for.
First, any category has the so called ``canonical topology'', which is the finest topology in which all representable pre-sheaves are sheaves. The covers in this topology can be described (essentially) as precisely the strict epimorphisms (I think this is in SGA4, but it is also easy). The statement you mention says that the flat topology is sub-canonical, and I believe it is at least rather close to the canonical one (but I have nothing to back it up). For etale maps, this just says that $X\rightarrow X''$ from the proposition is a cover in a sub-canonical topology.
From another point of view, one often thinks of a finite etale scheme over $S$ as a "finite set" over $S$ (as opposed to a finite scheme over $S$, that could, for instance, have nilpotents). The corollary then provides evidence that this category indeed behaves in that manner: like for sets, any map is a surjection onto a subset. The fact that strict epimorphisms are reasonable analogues of surjections is discussed (for instance) in a book of Makkai and Reyes, ``First order categorical logic'' (for example, section 3.3), which also discusses some other notions from SGA4 from this point of view.<|endoftext|>
TITLE: Automorphism of Smooth manifolds
QUESTION [6 upvotes]: Suppose $M$ is a smooth manifold and $x,y \in M$ are two points. Is there always a
diffeomorphism $\phi: M \rightarrow M$ with $\phi(x)= y$ ?

REPLY [9 votes]: Here's another argument, exploiting the differentiable structure a bit more (in the sense that I don't see a straightforward way to adapt it to the topological/PL case). It might be the standard argument to show that things work locally, but it works globally as well.
Take a smooth, embedded path $\gamma: [0,1]\to M$ so that $\gamma(0)=p$ and $\gamma(1)=q$.  Since $\gamma$ is embedded, we can push forward $\partial/\partial t$ and get a vector field $W$ on the image of $\gamma$. Let $V$ be a compactly supported extension of $W$ to all of $M$.
Then it's easy to check that the flow of $V$ at time 1 is a diffeomorphism of $M$ that sends $p$ to $q$ (and is isotopic to the identity).
This argument readily generalises to $k$-transitivity, too: if $\dim M\ge 3$, we can choose generic paths $\gamma_i$ simultaneously, and use the same exact argument. If $M$ is a surface, we need to use the fact that a single embedded path doesn't disconnect $M$.<|endoftext|>
TITLE: About unpublished lecture notes of Philip Hall
QUESTION [25 upvotes]: When I study group theory, I find that there are some mysterious things. For example, Daniel Gorenstein, in his paper On a Theorem of Philip Hall,  mentioned the unpublished lecture notes of Philip Hall. Many other famous group theorists also confirmed that these notes are important to their work. Since the notes are not published, I can not find way to see them. But I am very curious about what's in these notes. In these notes, are there any theorems about groups that are not otherwise known or do not appear in published books or papers?

REPLY [18 votes]: Apparently the original notes have been published online by Washington University here 
I must add that I'm an admirer of Hall's math, and its an emotion to read his handwritten notes.<|endoftext|>
TITLE: Elementary end extension of a countable model for ZF
QUESTION [5 upvotes]: Theorem 2.2.18 in Chang and Kiesler uses omitting types to show that any countable model of ZF has an elementary end extension.
Can we control the countable order type of such a model? for example, if $ X \prec H_ {\omega_2}$ 
can we have an elementary extension $Y \prec H_{\omega_2}$ such the $order type(Y)$ is bounded by some ordinal?
Any help or reference would be appreciated.

REPLY [9 votes]: There are several interesting issues arising in your question.
First, you ask about models of ZF, but then mention elementary
substructures of $H_{\omega_2}$, which of course is not a model of ZF,
because it does not satisfy the power set axiom. In the context of
$H_{\omega_2}$, you probably intend to discuss the theory
$\text{ZFC}^-$, meaning the theory of ZFC without the power set
axiom, and this makes for a very interesting question. (Meanwhile,
please see my recent paper with Gitman and Johnstone What is the
theory of ZFC without power set? for
reasons to be careful about how this theory is described---in
particular, one should be sure to include collection rather than
merely replacement; they are no longer equivalent without power
set.)
Second, in the context of set theory, there are distinct concepts
of end-extension.

One model $(M,E)$ end-extends to another $(N,F)$ if $(M,E)$
is a substructure of $(N,F)$, meaning that $M\subset N$ and
$E=F\cap M\times M$, and also $a\mathrel{F}b\in M$ implies $a\in
 M$. That is, the sets of $M$ do not gain new elements in $N$. In
other words, $M$ is a transitive subclass of $N$.
One model $M$ top-extends to another model $N$, if $M$ is a
rank initial segment of $N$. In other words, $M$ end-extends to
$N$ and also all new objects of $N$ have ordinal rank
larger than any ordinal of $M$. For example, this is the situation
when $V_\alpha\prec V_\beta$, but in the general case,
top-extensions needn't have that the height of $M$ is an ordinal
of $N$.

Set theorists often use the term "end-extension" to refer to
top-extensions, so one must take care, as the concepts are distinct
for models of set theory. For example, a nontrivial forcing
extension is an end-extension but not a top-extension. When it
comes to elementary extensions of models of ZF, however, even in
the case of nonstandard models, then the two notions coincide. This
is because if $M\prec N$ is an elementary end-extension, then by
elementarity, the $V_\alpha^M$ remains the $V_\alpha$ of $N$ for
any ordinal $\alpha$ in $M$, and furthermore $N$ cannot add any new
ordinals below an ordinal of $M$. Thus, all new sets of $N$ must
have rank larger than any ordinal of $M$ and so $N$ is a
top-extension. So for models of ZF, the order type of an elementary
end-extension must be strictly larger.
The corresponding fact is not true for models of $\text{ZFC}^-$,
where the rank initial segments $V_\alpha$ of the von Neumann hierarchy do not exist as sets. To see this,
start with GCH in $V$ and then force to add $\aleph_2$ many Cohen
reals to form the extension $V[G]$, and consider the model
$M=H_{\omega_2}^{V[G]}$. Now add one more Cohen real $c$ over
$V[G]$, and consider $N=M[c]=H_{\omega_2}^{V[G][c]}$. I claim that
$N$ is an elementary end-extension of $M$. First, it is clearly an
end-extension, since both are transitive sets in $V[G][c]$. Second,
for elementarity, observe that the two-step forcing
$\text{Add}(\omega,\omega_2)*\text{Add}(\omega,1)$ is isomorphic to
$\text{Add}(\omega,\omega_2)$ by absorbing the final factor into an
earlier stage. Furthermore, this isomorphism can be arranged to fix
any $\omega_1$ sized initial segment of the forcing. Now, any fact
true in $N$ about some parameters in $V$ and an $\aleph_1$ sized
piece of $G$, but not involving $c$, is forced by a condition, and
thus will be true whether as computed in $H_{\omega_2}^{V[G]}$ or
in $H_{\omega_2}^{V[G][c]}$, by using that isomorphism. And so the
inclusion $M\subset N$ is elementary.
The following theorem shows that the question of whether one must
go to strictly larger order types is actually independent of ZFC.
Theorem.

It is relatively consistent with ZFC that every countable $X\prec
 H_{\omega_2}$ has a nontrivial end-extension $X\prec Y\prec H_{\omega_2}$
with exactly the same ordinals.
It is also relatively consistent with ZFC that whenever $X\prec
 Y\prec H_{\omega_2}$ and $X\neq Y$, then $Y$ has strictly higher order type than $X$.

Proof. For statement 1, use the model $V[G]$ as described above.
The point is that for any countable $X\prec H_{\omega_2}^{V[G]}$,
we can find an $X$-generic Cohen real $c$ in $V$, and it will
follow that $X\prec X[c]$ by the argument given above.
For statement 2, use $L$ and the fact that
$H_{\omega_2}^L=L_{\omega_2}$. If $X\prec Y\prec L_{\omega_2}$ and
these are end-extensions, then because there is a definable
well-ordering, each new object of $Y$ not in $X$ has an ordinal
position $\alpha$ in the canonical $L$ order, and this ordinal is
in $Y$ but not in $X$. So by the end-extension property, it cannot
be smaller than any ordinal of $X$, and so the order type of $Y$ is
strictly higher than the order type of $X$. QED
I've written too much already, but one can say much more about
bounding the order-type of $Y$ over $X$, by fixing suitable Skolem
functions and then arguing that there is a club of closure points. And there are also interesting things to say for the case of extensions of models of full ZF rather than $\text{ZF}^-$.<|endoftext|>
TITLE: Isomorphic but non-conjugate subgroups of $GL(n,\mathbb{Z})$ ?
QUESTION [6 upvotes]: I've been asked some questions by a friend interested in crystallography, and the following questions (I'm not an expert) came spontaneous to me:

1) Are there two finite subgroups $P,P'\subset\mathrm{GL}(n,\mathbb{Z})$ that are abstractly isomorphic but not conjugate in $\mathrm{GL}(n,\mathbb{Z})$?

--

2) Are there two finite subgroups $P,P'\subset\mathrm{GL}(n,\mathbb{Z})$ that are abstractly isomorphic but not conjugate in $\mathrm{GL}(n,\mathbb{R})$?

--

3) What if we don't assume $P$ and $P'$ to be finite? (ok, this has nothing to do with crystallography) 

(They may well be classical and well known results, hence the tag "reference request")

REPLY [8 votes]: Since no-one has explicitly mentioned character theory and representation theory, I will say a few words. If we have a finite group $G$, and two (faithful, ie with trivial kernel) representations $\sigma, \tau : G \to {\rm GL}(n,F)$ where $F$ is a field of characteristic zero, asking whether $G\sigma$ and $G\tau$ are conjugate in ${\rm G}(n,F)$ is the same as asking whether the representations are equivalent over $F.$ An obvious necessary condition is that $g \sigma$ and $g\tau$ have the same trace for all $g \in G,$ but character theory (and some Schur index theory, etc.) tell us that this condition is sufficient if the representation is irreducible, and then, with some work, in general (the field $F$ need not be algebraically closed for this conclusion. The (well-known) point is is that in the irreducible case, if the two representations can be intertwined over a n extension of $F,$ they already can be intertwined over $F.$ The question for integral representations is more difficult. It may be that representations $\sigma, \tau : G \to {\rm GL}(n,\mathbb{Z})$ are equivalent after extending the ground ring to $\mathbb{Q},$ but are not equivalent as representations over $\mathbb{Z}.$ A relevant theorem putting some control on the situation is one of Jordan-Zassenhaus. An example is that ${\rm GL}(2,\mathbb{Z})$ has two subgroups isomorphic to the dihedral group $D$ with $8$ elements which are not conjugate within ${\rm GL}(2,\mathbb{Z}),$ but are conjugate as subgroups of ${\rm GL}(2,\mathbb{Q}).$ The group $D$ had two normal Klein $4$-subgroups
$U$ and $V$. If we induce a non-trivial $1$-dimensional representation of $U$ to $D$ we get a subgroup $E$ of ${\rm GL}(2,\mathbb{Z})$ isomorphic to $D,$ and we can to the same for $V$ to get another subgroup $E^{\prime}.$ The subgroups $E,E^{\prime}$ are not conjugate within   ${\rm GL}(2,\mathbb{Z}),$ but the representations afford the same character, so are equivalent as rational representations. (The technical reason is that if we pass to an appropriate local ring with residue field of characteristic $2,$ we have two indecomposable modules which have non-conjugate vertices, so are not isomorphic).

REPLY [3 votes]: If two homomorphisms $h,h':G\to GL_n(\mathbb Q)$ are conjugate in $GL_n(\mathbb R)$ then they are conjugate in $GL_n(\mathbb Q)$. To see this, consider the set of all rational matrices $X$ such that $Xh(g)=h'(g)X$ for all $g$. This is a rational vector space, and a basis for it is a real basis for the corresponding space of real matrices. It follows easily that the existence of such a real matrix with non-zero determinant implies the existence of such a rational matrix with non-zero determinant. This argument does not require $G$ to be finite (or, for that matter, to be a group).
But you can easily find two non-conjugate elements of order $2$ in $GL_2(\mathbb Z)$ that are conjugate in $GL_2(\mathbb Q)$.<|endoftext|>
TITLE: Lucas numbers modulo m
QUESTION [7 upvotes]: The Lucas numbers $L(n)$ are defined by $L(0)=2$, $L(1)=1$, and $L(n)=L(n-1)+L(n-2)$, for $n\ge2$.  Looking at the sequence $\{L(n)\}$ modulo various numbers, we are lead to conjecture that $\{L(n) \mod m\}$ contains a complete residue system modulo $m$ if and only if $m$ is one of the following:  $2, 4, 6, 7, 14, 3^k$, $k\ge1$.  Example:  Modulo 5 we have the sequence $2,1,3,4,2,1,\dots$, and since the sequence repeats we never obtain $0 \mod 5$.  Example:  Modulo 6 we have $2,1,3,4,1,5,0,\dots$ and we obtain a complete residue system $\mod 6$.
The corresponding problem for the Fibonacci sequence was solved by S. A. Burr in "On Moduli for Which the Fibonacci Sequence Contains a Complete System of Residues," Fibonacci Quarterly, December 1971, pp. 497-504.  The sequence $\{F(n) \mod m\}$ contains a complete residue systems modulo $m$ if and only if $m$ is one of $5^k, 2\cdot5^k, 4\cdot5^k, 3^j\cdot5^k, 6\cdot5^k, 7\cdot5^k, 14\cdot5^k$, $k\ge0$, $j\ge1$.
Although 40 years later, I don't find that anyone has looked at the Lucas version.  If someone supplies a reference or a proof, I would appreciate it.  Thank you.

REPLY [8 votes]: Note that if the Lucas sequence modulo m contains a complete set of residues then the Fibonacci sequence must also. (If the Lucas sequence contains 0 followed by d, then it continues as d times the Fibonacci sequence.)
As 5 does not work this rules out all m divisible by 5, checking 2,4,6,7,14 by hand only leaves the powers of 3 undecided.<|endoftext|>
TITLE: Computing a determinantal representation of a bivariate polynomial
QUESTION [7 upvotes]: Let $p \in \mathbb R [x,y,z]$ be a homogeneous irreducible polynomial of degree $d$.  From Dickson in 1920 we know that there exists $A$, $B$ and $C$ such that 
$$\det (Ax + By + Cz) = c p(x,y,z)$$
where $c$ is some constant. 
Vinnikov in 1988 was able to describe all the non-equivalent determinantal representations as points on the Jacobian variety that are not on the exceptional sub variety.  The theoretical work in this paper is relatively constructive, but is still a long way from a numerically stable constructive algorithm for $A$, $B$ and $C$. 
Given any polynomial $p(x,y,z)$, can one triple $(A, B,C)$ be constructed in a numerically stable way?
Thanks in advance.

REPLY [5 votes]: This is discussed in a recent work of Plaumann, Sturmfels and Vinzant:
http://arxiv.org/abs/1011.6057<|endoftext|>
TITLE: The induced representations of $SL(2, F)$.
QUESTION [5 upvotes]: Let $G=SL(2,F)$ and $I=J_{0}\cap J_{1}$ be the Iwahori subgroup of $SL(2, F)$, where $J_{0}=\left(
             \begin{array}{cc}
               \mathcal{O}_{\mathbb{F}} & \mathcal{O}_{\mathbb{F}} \\
               \mathcal{O}_{\mathbb{F}} & \mathcal{O}_{\mathbb{F}} \\
             \end{array}
           \right)\cap SL(2)$ and $J_{1}=\left(
             \begin{array}{cc}
               \mathcal{O}_{\mathbb{F}}& \varpi_{\mathbb{F}}^{-1}\mathcal{O}_{\mathbb{F}} \\
                \varpi_{\mathbb{F}} \mathcal{O}_{\mathbb{F}}& \mathcal{O}_{\mathbb{F}} \\
              \end{array}
             \right)\cap SL(2)$ are the maximal compact subgroups of $SL(2, F)$. Here $F$ is a local non-Archimedean $p$-adic field, $\mathcal{O}_{F}$ is the valuation ring and $\varpi$ is the uniformizer.
Now, we know that if $\lambda^{2}= 1$, then $Ind_{I}^{G}\lambda=\lambda^{-}\oplus\lambda^{+}$. Also , we can write  $Ind_{I}^{G}\lambda=Ind_{J_{0}}^{G}(Ind_{I}^{J_{0}}\lambda)$. 
Does the following equality hold?
$Ind_{I}^{G}\lambda=Ind_{J_{0}}^{G}\lambda\oplus Ind_{J_{1}}^{G}\lambda$.
I do apologize for my bad English and my bad temper to create a question.

REPLY [2 votes]: I assume that $\lambda$ is a character!? How is your $\lambda$ a representation of $J_0$ or $J_1$, do you mean the induction $Ind_I^{J_k} \lambda$ instead, but the induction is never irreducible.
On some related matters on $GL(2)$:
The question whether 
$ Ind_{I}^{J_i} \lambda $
splits has been answered by Silberger for $PGL(2)$ for $J_0$ and $J_1$ in odd residue characteristic, and by Casselman for $PGL(2)$ for $J_0$ in all residue characteristics.
Silberger, Allan J. Irreducible representations of a maximal compact subgroup of ${\rm pgl}_{2}$pgl2 over the $p$p-adics. Math. Ann. 229 (1977), no. 1, 1–12.
Silberger, Allan J. ${\rm PGL}_{2}$PGL2 over the $p$p-adics: its representations, spherical functions, and Fourier analysis. Lecture Notes in Mathematics, Vol. 166
Casselman, William The restriction of a representation of ${\rm GL}_{2}(k)$GL2(k) to ${\rm GL}_{2}({\germ o})$GL2(o). Math. Ann. 206 (1973), 311–318.
In these cases, the restriction induction formula $Res_{SL(2,F)} Ind^{GL(2,F)}_{J_i} \lambda =Ind_{J_i}^{SL(2,F)} \lambda$! Induction by steps imply the results, if it $ Ind_{I}^{J_i} \lambda $ splits. 
I am not sure, if that is necessary, but you can look at the induction restriction formula for $Res_{J_i} Ind^{SL(2,F)}_{J_i} \lambda$ and see what happens.<|endoftext|>
TITLE: simplicial spaces without degeneracies
QUESTION [10 upvotes]: Suppose I have a simplicial space $X_{\bullet}$ without degeneracies (sometimes called semi-simplicial space or incomplete simplicial space). There still is a geometric realization $\lVert X \rVert$ of $X_{\bullet}$, which only uses the face maps. What properties does this realization have?

Does it still preserve products, i.e. is $\lVert X \times Y \rVert$ still (weakly?) homotopy equivalent to $\lVert X \rVert \times \lVert Y \rVert$? 
Do levelwise (weak) equivalences still induce a (weak) homotopy equivalence of the geometric realizations?

I know of the paper "Categories and Cohomology Theories" by Segal, where he defines this "fat" geometric realization in the appendix. Unfortunately, he proves the above properties by a comparison with another construction that takes the degeneracies into account. Is this the only way to prove this? Do the properties fail in case there are no degeneracies?

REPLY [12 votes]: In brief:
For your first question, no. Let $X_\bullet$ be any semi-simplicial space and $Y_\bullet$ have a point in degree zero and be empty in every other degree. Then $\vert X_\bullet \times Y_\bullet \vert = X_0$, which will not usually be equivalent to $\vert X_\bullet \vert$.
For your second question, yes. This is always true for semi-simplicial spaces, and is not always true for simplicial spaces (It is in the case where the thick and thin realisations are equivalent, of course). To see this you prove that the maps on $k$-skeleta $\vert X_\bullet \vert^{(k)} \to \vert Y_\bullet \vert^{(k)}$ are equivalences by induction on $k$, using the push-out description of the $k$-skeleton from the $(k-1)$-skeleton, and the fact that
$$X_k \times \partial \Delta^k \to X_k \times \Delta^k$$
is a cofibration so that it is a homotopy push-out. Then you use that $\vert X^\bullet \vert = \mathrm{colim} \vert X^\bullet \vert^{(k)}$ and that each $\vert X^\bullet \vert^{(k-1)} \to \vert X^\bullet \vert^{(k)}$ is a cofibration so this is a homotopy colimit.<|endoftext|>
TITLE: accumulation points within Pisot numbers 
QUESTION [5 upvotes]: Recall that Pisot numbers are algebraic integers greater than $1$, whose other Galois conjugates have modulus $<1$. The set of Pisot numbers is usually denoted $S$. It is known that $S$ is denumerable and closed (Salem). Alors, the sequence of derived sets $S',S'', ...$ does not terminate. The smallest accumulation point of $S$ is the golden ratio $\phi$.

Let $\lambda$ be an accumulation point in $S$. Therefore it is the limit of a sequence of Pisot numbers $\mu_n\ne\lambda$. It is clear that the degree of $\mu_n$ tends to $+\infty$, whereas its norm remains bounded as $n\rightarrow+\infty$. My question is about the Galois conjugates of $\mu_n$; most of them must be close to the unit circle, because their product equals $N(\mu_n)/\mu_n$. Given a Galois conjugate $\tau$ of $\lambda$, does there exist a Galois conjugate $\tau_n$ of $\mu_n$ such that $\tau_n\rightarrow\tau$ ? What is the statistics of the Galois conjugates of $\mu_n$ as $n\rightarrow+\infty$ ?

Example: the multinacci number of degree $d$ is the root $a_d>1$ of the polynomial $X^d-X^{d-1}-\cdots-X-1$. When $d\rightarrow+\infty$, one has $a_d\rightarrow2$. Then I have a proof that the empirical measure
$$m_d:=\frac1d\sum_{a\sim a_d}\delta_a$$
where the sum runs over all the Galois conjugates of $a_d$ (including itself) converges vaguely towards the uniform measure over the unit circle.
Edit. The same property holds true in the situation depicted by Dufresnoy and Pisot. Let $P\in{\mathbb Z}[X]$ be the minimal polynomial of $\lambda$, unitary. Let $A\in{\mathbb Z}[X]$ be such that $|A(z)|\le|P(z)|$ on $\mathbb T$, the equality arising only at finitely many points. Define the polynomials $P_n^\pm(X)=X^nP(X)\mp A(X)$. For infinitely many pairs $(n,\pm)$, $P_n^\pm$ has only one root $\lambda_n^\pm$ away from the unit disk, and none over $\mathbb T$. This Pisot number tends to $\lambda$, and the empirical measure defined as above tends to $\frac1{2\pi}d\theta$. It is thus tempting to conjecture that in every situation, the empirical measure  tends to the uniform measure.

REPLY [10 votes]: The answer to the first question is "certainly not". Consider the polynomials $P_d(X)=X^d-4X^{d-1}-X^{d-2}-1$. They have $d-1$ roots in the unit disk by Rouche, so their positive real roots are Pisot numbers. Also, the positive real roots of $P_d$ tend to the larger root of $X^2-4X-1$, which is a Pisot number. However, any disk or radius $r<1$ is eventually free of roots of $P_d$, so nothing tends to the conjugate.<|endoftext|>
TITLE: Surreal numbers vs. non-standard analysis
QUESTION [26 upvotes]: What is the relationship between the surreal numbers and non-standard analysis?
In particular, is there a transfer principle for surreal numbers they way there is for NSA?
A specific situation in which such a transfer principle would be useful arose in the thread Uniformizing the surcomplex unit circle ; can the surjectivity of the map $t \mapsto e^{it}$ from the reals to the complex unit circle be transferred to the surreals?  Presumably, one would need a definition of the map that was in some sense first-order; what sorts of definitions count as first-order?  It is not clear to me how definitions involving the two-sided bracket operation can be fit into a first-order framework.

REPLY [6 votes]: Coming back to the first post. Most of modern mathematics is set-theoretic, that is, it studies sets of different kind, so that reals, real and complex functions, relations on reals, as well as a variety of more complex objects like the Hilbert space - are sets of this or another kind. In that sense, any mathematical definition is a 1-st order one, assuming there is no restriction on using the language of set theory within the common axiomatics.
Regarding the surs. The definition of them yields a certain ordered field, perhaps maximal in some well-defined sense, and nothing more. That surs are so attractive to some kind of mathematically-complying minds is, in my opinion, explainable that this still is a very rare domain where meaningful facts can be explored or observed, rather than proved. On the other hand, the students of surs I believe cannot care less about some transfer and about whether their omni-something does not satisfy some Peano axiom. After all, p-adic numbers do not satisfy Peano axioms either, but who cares. 
Further it happens that the surs are isomorphic (in a class theory) to a certain nonstandard universe, defined by totally different means and towards quite different goals. This allows to enrich the surs by a variety of constructions (like the sine function) beyond their native field structure. In this case, a devoted student of surs might be interested to really figure out in some strict, well defined terms, whether a consistent sine function can be defined on surs by pure sur-means. For instance, consider a version of NBG which proves the existence of surs as a class but is not strong enough to prove the mentioned isomorphism, and prove that such a theory does not imply the existence of a consistent sur-sin. This can be very complex though.<|endoftext|>
TITLE: Subgroups of GL(2,q)
QUESTION [23 upvotes]: I have been wondering about something for a while now, and the simplest incarnation of it is the following question:

Find a finite group that is not a subgroup of any $GL_2(q)$.

Here, $GL_2(q)$ is the group of nonsingular $2 \times 2$ matrices over $\mathbb{F}_q$. Maybe I am fooled by the context in which this arose, but it seems quite unlikely that "many" finite groups are subgroups of $GL_2(q)$. Still, I can't seem to exclude a single group, but I am likely being stupid. 
Motivation: This arose when I tried to do some explicit calculations related to Serre's modularity conjecture. I wanted to get my hands on some concrete Galois representations, and play around with the newforms associated to it. Call it recreational if you like (it certainly is!). In Serre's original paper [1], there is a wonderful treatment of some explicit examples. He uses the observation that the Galois group over $\mathbb{Q}$ of
$$x^7 -7x +3$$
is isomorphic to $PSL_2(7)$. This gives him a surjection $G_{\mathbb{Q}} := \mbox{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\rightarrow PSL_2(7)$, which he combines with the character associated to $\mathbb{Q}(\sqrt{-3})$ to obtain a homomorphism $G_{\mathbb{Q}} \rightarrow PSL_2(7) \times C_2$. This homomorphism he can then lift to $GL_2(49)$ using a very clever calculation in the Brauer group. 
The way I see it (without claiming this is justified) is that it is unclear or even provably impossible to embed $PSL_2(7)$ directly into some $GL_2(q)$, and for that reason we need a nice lifting.
This raises the natural question of when this little trick with the Brauer group is "necessary". What can be said in general about subgroups of $GL_2(q)$? Can we exclude any particular finite group or families of finite groups? Can we perhaps even classify the possible isomorphism types of such subgroups? Of course this is all becomes even more natural to ask in the light of the inverse Galois problem. How many Galois groups can we "see" in the two-dimensional representations?
Remark. Of course, the question could be approached the other way. If one has the (perhaps not unrealistic?) hope of realising every $GL_2(q)$ as a Galois group over $\mathbb{Q}$, how many groups do you automatically realise over $\mathbb{Q}$ by taking quotients of $GL_2(q)$?

Which groups $G$ admit a surjection $GL_2(q) \rightarrow G \rightarrow 1$ for some $q$?

Reference:
[1] Serre, Jean-Pierre, "Sur les représentations modulaires de degré 2 de Gal(Q/Q)", Duke Mathematical Journal 54.1, (1987): 179–230

REPLY [17 votes]: We can generalize Ralph's answer to find the abelian groups that are $p$-groups for odd $p$ that cannot be subgroups of $GL_2$.
if $p|(q^2-1)(q^2-q)$ then exactly one of $p|(q+1)$, $p|(q-1)$ and $p|q$. If $p|q$ then $q=p^n$ and upper-triangular unipotent matrices provide enough, so they're a Sylow subgroup, $(\mathbb Z/p)^n$. If $p|(q-1)$ then the group of diagonal matrices contains a Sylow subgroup, which looks like $\mathbb Z/p^k \times \mathbb Z/p^k$. If $p|(q+1)$ then $\mathbb F_{p^2}^\times$ contains a Sylow subgroup, which looks like $\mathbb Z/p^l$.
Therefore $\mathbb Z/p^2 \times \mathbb Z/p \times \mathbb Z/p$ cannot be contained in any $GL_2(\mathbb F_q)$, while every abelian $p$-group not containing it can be.
Combined with Geoff's answer this gives a complete characterization of the solution to this problem for odd $p$-groups.<|endoftext|>
TITLE: Is it known that $(F_p^{\times} \ltimes F_p, F_p)$ is not a relative expander family?
QUESTION [10 upvotes]: Shalom [edit: originally M. Burger] showed that the pair $(\mathrm{SL}_2(\mathbb{Z}) \ltimes \mathbb{Z}^2, \mathbb{Z}^2)$ has Relative Property (T) with respect to standard generating sets.
(The action of $\mathrm{SL}_2(\mathbb{Z})$ on $\mathbb{Z}^2$ is the usual one, i.e. the semidirect product can be thought of as a group of affine transformations $x \mapsto A x + b$ where $A \in \mathrm{SL}_2(\mathbb{Z})$ and $b \in \mathbb{Z}^2$.
If we reduce $\mathrm{mod}\ p$, we can think of this as giving an "efficient" way of generating the translations $x \mapsto x + b$ for $b \in F_p^2$.)
A one-dimensional variant in the finite case is whether there exist bounded size subsets $S_p \subset F_p^{\times} \ltimes F_p$ and $\delta > 0$ such that the relative Kazhdan constant:
$\kappa\ (F_p^{\times} \ltimes F_p, F_p, S_p) \ge \delta$
i.e. whether the pairs $(F_p^{\times} \ltimes F_p, F_p)$ can form a relative expander family.
An equivalent formulation: do there exist bounded size sets $S_p$ of affine transformations on $F_p$, such that no non-empty subset $U \subset F_p$, $|U| \leq p/2$ is almost invariant with respect to all of them, i.e.
$\neg \exists U: \forall s \in S: |s(U) \cap U| > \frac{99}{100} |U|$
I believe the answer is no if one uses standard "generating" sets (they needn't actually generate) such as $x \mapsto x + 1,\ x \mapsto ax$, even if $a$ is allowed to vary with $p$.  This is very slightly surprising, as these do generate all translations "efficiently" in the weaker sense of logarithmic diameter.
Is there a good argument as to why this should fail in general?  Or might there be cunning sets $S_p$ such that relative expansion occurs?

REPLY [7 votes]: I think that one can show that $(F_p^\times \ltimes F_p,F_p)$ does not have relative property (T), by using the combinatorial proof that semidirect products of amenable groups are again amenable (see Proposition 5 of my notes at http://terrytao.wordpress.com/2009/04/14/some-notes-on-amenability/ ).  A bit more specifically, let $S_p$ be a bounded set of affine transformations on $F_p$, and let $D_p$ be the associated set of dilations on $F_p$, which is then a bounded subset of $F_p^\times$.  By the abelian nature of the dilation group, we can then construct a nontrivial subset $F$ of the dilation group $F_p^\times$ which is $99.9\%$-invariant with respect to the dilations of $D_p$ (thus $|dF \Delta F| \leq 10^{-3} |F|$ for all $d \in D_p$), basically by constructing a medium-sized generalised geometric progression using the dilations in $D_p$ as generators.  Note that one can make the set $F$ of bounded size (i.e. independent of $p$.)  Let $T$ be the set of all translations $t$ such that $SF$ intersects $Ft$ (i.e. the translations in $F^{-1} S_p F$).  This is a set whose size is controlled by the size of $S_p$ and of $F$, and in particular is still bounded uniformly in $p$.  We can then construct a moderately large (but still of size uniformly bounded in $p$) set $E$ of translations which is $99.9\%$-invariant with respect to any of the translations of $T$.  The set $U := EF$ will then (for $p$ large enough) be a non-trivial $99\%$-invariant set with respect to $S_p$, by the argument given in my notes above.<|endoftext|>
TITLE: Gröbner basis for Sudoku
QUESTION [14 upvotes]: I'm trying to write a program that solves Sudokus using Gröbner bases. I introduced $81$ variables, $x_1$ to $x_{81}$, this is a linearisation of the Sudoku board.
The space of valid Sudokus is defined by:

for $i=1,\ldots,81$: $$F_i = (x_i - 1)(x_i - 2)\cdots(x_i - 9)$$
This represents the fact that all squares have integer values between 1 and 9.
for all $x_i$ and $x_j$ which are not equal but in the same row, column or block: $$G_{ij} = (F_i - F_j)/(x_i - x_j)$$ This represents that the variables $x_i$ and $x_j$ can not be equal.

All these $F_i$ and $G_{ij}$ together define the space of valid Sudokus. This consists of $891$ polynomials. 
Now to solve a Sudoku we can add the clues to the space, so by example if the clue of a Sudoku is the first square is a $5$, then we add $(x_1 - 5)$ to the space. If we now take the Gröbner basis of this space we can directly see the solution for it.
I understand what I am doing this far. But I have trouble finding a computable manner for finding the gröbner bases. I have succesfully done everything for $4 \times 4$ Sudokus (or so-called Shidokus). But neither Maple nor Singular are giving me a result for the Gröbner basis of the $9 \times 9$ Sudoku space.
You can see the commands I gave to Maple here. First I define the $891$ polynomials, then I ask for a basis of it. 
I read papers saying it's feasible although non-performant to do what I strive for but I don't see how to find the solution, as they don't include many implementation details. Can anyone point me to a direction, making this problem easier for Maple or other software?

REPLY [10 votes]: Here is a Singular Code that works quite well:
ring A = 0,(t,x(1..9)),lp;
/* Characteristic 0 works suprisingly well for this problem. */
/* We choose a lexicographic ordering since we will compute an
   elimination ideal. */

poly p = (t-x(1))*(t-x(2))*(t-x(3))*(t-x(4))*(t-x(5))*(t-x(6))*(t-x(7))*(t-x(8))*(t-x(9))-(t-1)*(t-2)*(t-3)*(t-4)*(t-5)*(t-6)*(t-7)*(t-8)*(t-9);
/* p(x)=0 in Q[t] implies that x is a permutation of the numbers 1 to 9. */

matrix c = coeffs(p,t);
ideal J = (c[1..9,1]);
/* J expresses that x is a permutation of the numbers 1 to 9. However,
   surprisingly, it is better to use only constraints saying that x(8)
   and x(9) are distinct integers between 1 and 9. This is done by
   computing an elimination ideal. */
ideal JG = groebner(J);
ideal J2 = (JG[1],JG[2]);
/* J2 is the ideal expressing that x(1) and x(2) are distinct integers
   between 1 and 9. */

ring R=0,(x(1..81)),dp;
ideal I;
map psi;
proc f(k,l,m,n,o,p,q,r,s)
{intvec v = k,l,m,n,o,p,q,r,s;
 int i,j;
 for (i=1; i<=8; i++) {for (j=i+1; j<=9; j++)
 {psi = A,0,1,2,3,4,5,6,7,x(v[i]),x(v[j]); I = I + psi(J2);}}}

/* Code the rules into the ideal. */
f(1,2,3,4,5,6,7,8,9);
f(10,11,12,13,14,15,16,17,18);
f(19,20,21,22,23,24,25,26,27);
f(28,29,30,31,32,33,34,35,36);
f(37,38,39,40,41,42,43,44,45);
f(46,47,48,49,50,51,52,53,54);
f(55,56,57,58,59,60,61,62,63);
f(64,65,66,67,68,69,70,71,72);
f(73,74,75,76,77,78,79,80,81);
f(1,10,19,28,37,46,55,64,73);
f(2,11,20,29,38,47,56,65,74);
f(3,12,21,30,39,48,57,66,75);
f(4,13,22,31,40,49,58,67,76);
f(5,14,23,32,41,50,59,68,77);
f(6,15,24,33,42,51,60,69,78);
f(7,16,25,34,43,52,61,70,79);
f(8,17,26,35,44,53,62,71,80);
f(9,18,27,36,45,54,63,72,81);
f(1,2,3,10,11,12,19,20,21);
f(4,5,6,13,14,15,22,23,24);
f(7,8,9,16,17,18,25,26,27);
f(28,29,30,37,38,39,46,47,48);
f(31,32,33,40,41,42,49,50,51);
f(34,35,36,43,44,45,52,53,54);
f(55,56,57,64,65,66,73,74,75);
f(58,59,60,67,68,69,76,77,78);
f(61,62,63,70,71,72,79,80,81);

/* Code a uniquely solvable Sudoku problem into the ideal.  */
I=I+(x(3)-4);
I=I+(x(6)-3);
I=I+(x(7)-6);
I=I+(x(9)-9);
I=I+(x(12)-8);
I=I+(x(13)-9);
I=I+(x(16)-2);
I=I+(x(18)-1);
I=I+(x(22)-8);
I=I+(x(23)-1);
I=I+(x(27)-7);
I=I+(x(28)-6);
I=I+(x(33)-7);
I=I+(x(37)-8);
I=I+(x(40)-3);
I=I+(x(42)-9);
I=I+(x(45)-5);
I=I+(x(49)-6);
I=I+(x(54)-3);
I=I+(x(55)-4);
I=I+(x(59)-7);
I=I+(x(60)-6);
I=I+(x(64)-2);
I=I+(x(66)-7);
I=I+(x(69)-5);
I=I+(x(70)-1);
I=I+(x(73)-9);
I=I+(x(75)-1);
I=I+(x(76)-2);
I=I+(x(79)-4);

option(redSB);
groebner(I,30);
/* You get the solution quickly. */<|endoftext|>
TITLE: Is ramification of number fields first order?
QUESTION [11 upvotes]: Fix a prime number $p$. Is there a first order sentence $\phi_p$ in the language of fields such that $\phi_p$ holds in a number field $K$ if and only if the prime $p$ is unramified in the field extension $K/\mathbb{Q}$?

REPLY [6 votes]: I THINK, the answer is YES by a paper of Robert Rumely:
Undecidability and definability of the theory of global fields, 1980,
where he showed that "a great variety of number-theoretic objects, from rings of integers and valuations, to zeta-functions and adele rings" are definable. Many of them are actually uniformally definable. 
For the ramification issue, see the bottom of Page 210.<|endoftext|>
TITLE: A question related to ultrapower embeddings.
QUESTION [5 upvotes]: In Joel David Hamkins' "Forcing and Large Cardinals" a definition of $extender$ embeddings:
"An embedding $j:V \to M$ is an $extender$ embedding if every element of $M$ can be represented in the form $j(f)(\alpha)$ for some $f:\kappa \to V$ and $\alpha < j(\kappa)$, where $\kappa$ is the critical point of $j$."
Every ultra-power embedding by a measure $\mu$ on a measurable cardinal $\kappa$ is an extender embedding since $\kappa = [id]_{\mu}$ can be the seed for such an embedding and therefore generate $M$ by representing every element as a $j(f)(\kappa)$ for an $f:\kappa \to V$.  Such an infinite cardinal $\kappa$ is, of course, the critical point of $j$. 
However, not every extender embedding is an ultra-power embedding.  For such embeddings j: M $\to$ N, how are the images of such embeddings different from ultra-power images?  Or, how are the embeddings different? And, what mathematics can be done with extender embeddings which are not generated by it's critical point?

REPLY [7 votes]: I'm glad you're reading my book (in preparation).
There are a variety of large cardinal notions and large cardinal
embedding types that are witnessed by extender embeddings, but
which cannot be witnessed by ultrapower embeddings by an
ultrafilter on a measurable cardinal $\kappa$.

Perhaps the easiest example arises from the iterations of a normal
measure $\mu$ on a measurable cardinal. If $j:V\to M$ is the
ultrapower by $\mu$, then in $M$ the cardinal $j(\kappa)$ is a
measurable cardinal with normal measure $j(\kappa)$, and one can
take the ultrapower of $M$ by $j(\mu)$. Eventually, one builds the
system of embeddings $$V\to M_1\to M_2\to M_3\to\cdots,$$ where
$j_{n,n+1}:M_n\to M_{n+1}$ is the ultrapower of $M_n$ by
$j_{0,n}(\mu)$, and where $j_{i,j}$ is defined by composing the
maps at each step. Taking the direct limit of this system, one gets
a map $j_{0,\omega}:V\to M_\omega$, and one can show that this
limit is well-founded. This is an elementary embedding, but it is
not an ultrapower embedding by any ultrafilter on $\kappa$, because
$M_\omega$ is well-founded, but is not closed under
$\kappa$-sequences, indeed, not even under $\omega$-sequences, as
the value $j_{0,\omega}(\kappa)=\kappa_\omega=\sup_n \kappa_n$,
where $\kappa_n=j_{0,n}(\kappa)$ is the critical sequence. That is,
$\kappa_\omega$ has cofinality $\omega$ in $V$, but is a measurable
cardinal in $M_\omega$. Meanwhile, however, one can prove that
every element of $M_\omega$ has the form
$j(f)(\kappa_0,\kappa_1,\ldots,\kappa_n)$ for some function
$f:[\kappa]^{\lt\omega}\to V$, and this observation can be used to
show that $j_{0,\omega}$ has an extender representation. One can of
course continue the iterations through the ordinals, and these will
all be extender embeddings, and none of them is an ultrapower
embedding by an ultrafilter on $\kappa$. 
One can prove more generally that every extender embedding is the direct limit of the induced system of ordinary ultrapower embeddings. For any $\alpha\lt j(\kappa)$, one forms the hull $X=\{j(f)(\alpha)\mid f:\kappa\to V\}$, and this is an elementary substructure of $M$ containing the range of $j$. Following $j$ with the Mostowski collapse of that structure gives rise to a factor embeding $j_0:V\to M_0$, with $k:M_0\to M$ the inverse collapse of $X$. These embeddings fit together into a directed system of embeddings, by means of Goedel pairing of the ordinals, and the extender embedding $j$ is the direct limit of the system. 
A cardinal $\kappa$ is $\theta$-strong if there is an
embedding $j:V\to M$ with critical point $\kappa$ and
$V_\theta\subset M$. Such kind of embeddings, for
$\kappa+2\leq\theta$, cannot arise from ultrapowers by a measure
on $\kappa$, since $\mu\notin M_\mu$ for any such ultrapower
$j_\mu:V\to M_\mu$. But they can be represented by extender
embeddings, since one simply takes the Mostowski collapse of the
seed hull $\{j(f)(s)\mid f:V_\kappa\to V, s\in V_\theta\}\prec M$, and
then observes that an enumeration of $V_\theta$ turns this into
the form you stated.
A cardinal $\kappa$ is $\theta$-tall if there is an
embedding $j:V\to M$ with critical point $\kappa$ and $\theta\lt j(\kappa)$ and
$M^\kappa\subset M$. Ultrapower embeddings by a measure on
$\kappa$ cannot achieve this when $\theta$ is larger than
$(2^\kappa)^+$, since one can count the number of functions to
get a bound on the size of $j(\kappa)$. But meanwhile, one can
realize tallness with extender embeddings.<|endoftext|>
TITLE: 5/8 bound in group theory
QUESTION [73 upvotes]: The odds of two random elements of a group commuting is the number of conjugacy classes of the group
$$  \frac{ \{ (g,h): ghg^{-1}h^{-1} = 1 \} }{ |G|^2}  = \frac{c(G)}{|G|}$$
If this number exceeds 5/8, the group is Abelian (I forget which groups realize this bound).  
Is there a character-theoretic proof of this fact?  What is a generalization of this result... maybe it's a result about semisimple-algebras rather than groups?

REPLY [20 votes]: One elementary result using character theory, but going in the other direction, which is proved in the paper of R. Guralnick and myself mentioned in my comment above is that if 
$\{\chi_1, \chi_2, \ldots, \chi_c \}$ are the complex irreducible characters of $G$, where $c = c(G)$ is the numberof conjugacy classes of $G,$ then by Cauchy-Schwarz, we have $\sum_{i=1}^{c} \chi_i(1) \leq \sqrt{c}\sqrt{|G|}$, so that $\frac{c(G)}{|G|} \geq \left( \frac{\sum_{i=1}^{c} \chi_i(1)}{|G|} \right)^{2}.$.<|endoftext|>
TITLE: quadratic forms over fields of characteristic 2
QUESTION [7 upvotes]: I was wondering if anyone knows any good sources for the theory of quadratic forms over fields of characteristic 2 which are written in English?

REPLY [6 votes]: The Algebraic and Geometric Theory of Quadratic Forms by Elman, Karpenko and Merkurjev is a standard recent reference for the theory of quadratic forms, paying special attention to the differences between the theory of bilinear forms and the theory of quadratic forms in characteristic 2.<|endoftext|>
TITLE: Is Euler characteristic of a simplicial complex upper bounded by a polynomial in the number of its facets ?
QUESTION [16 upvotes]: What is the best upper bound known on the (absolute value of) the 
Euler characteristic of a simplicial complex
in terms of the number of its facets ?
In particular, I am interested in proving or disproving the following:
If $K$ is a simplicial complex with $N$ facets then 
$|\chi(K)| \leq N^{O(1)}.$  
If $K$ is "shellable" then one can show that $|\chi(K)| \leq N.$
As a partial answer, I would be interested in any other subclasses 
of simplicial complexes where the polynomial upper bound holds.

REPLY [5 votes]: If you only care about Cohen-Macaulay complexes (in particular, shellable complexes are Cohen-Macaulay) then the answer is yes.  Let $\Delta$ be a $(d-1)$-dimensional CM complex.  The key is that we should use the $h$-numbers of $\Delta$ instead of its $f$-numbers.  Most importantly: 

The number of facets in $\Delta$ is the sum of its $h$-numbers (for any complex),
$h_d(\Delta) = (-1)^{d-1}\widetilde{\chi}(\Delta)$ (also for any complex), and
$h_j(\Delta) \geq 0$ for all $j$ (for any CM complex).

Thus $$|\widetilde{\chi}(\Delta)| = h_d(\Delta) \leq \sum_{j=0}^dh_j(\Delta) = f_{d-1}(\Delta).$$<|endoftext|>
TITLE: Mathematical modeling of voting/rating (e.g. political elections,  questions on MO, gadgets on amazon,...) 
QUESTION [6 upvotes]: We need to make choices in our life and so we need to compare(=rate) things what is good what is bad. Question: are there some mathematical models which may capture features of some kind of voting/ratings systems ?
More precisely I want to have some model where there is some "ideal choice" (e.g. experts estimation of quality of question on MO) and result of real voting - which might be different. 
The model should help us to understand how to make ratings more close to "ideal choice".
The possible application which I keep in mind is user's ratings of gadgets at amazon (or other sites), with the hope to improve it taking into account something like user's "reputation"...
Another application - current science journals play the role of "rating agencies"
- papers published at "Annals" = stamp of great quality.
Assume journals will disappear. Can we create a kind of crowd-sourcing rating which
will be close to current journal based rating system ? 
There are many votings systems - political elections, ratings of questions on stackoverflow, user's ratings of gadgets sold at amazon.com,...
Of course, all of them have different features, however it might be that there are certain
simple ideas which might be relevant for understanding how all these things work... 
To understand this: are there simple ideas or there are not?  if there are - what are them ?  is the gist of this question.
The question is very vague and so any comment is welcome.

REPLY [3 votes]: If there are no incentive problems (i.e. if everyone is honestly seeking the truth) then Aumann's Agree-to-Disagree Theorem tells you that if the voters are allowed to revise their votes after seeing the outcome, they can't ultimately disagree (at least if there is some objectively correct answer).  (And various results by people like Scott Aaronson suggest that the convergence to unanimity should be fast.)  
So if we see persistent non-unanimity in these votes (which, obviously, we do), and if we want to understand what's going on, we've got to ask exactly why Aumann's Theorem doesn't apply, and that's harder than it looks (see various papers of Robin Hanson for why the most obvious solutions fail).<|endoftext|>
TITLE: Subgroups of $GL(k,q)$ for bounded $k$
QUESTION [7 upvotes]: This question on subgroups of $GL(2,q)$ asked by Jan, and especially wonderful answers to it given by Geoff Robinson, Ralph, and Will Sawin showing that "almost no finite groups" inject in $GL(2,q)$ made me wonder (completely recreationally, I have to admit) whether there exists $N$ such that every finite group, or "most finite groups" inject in $GL(N,q)$. 
Probably no such $N$ exists, but the ideas I had when thinking about the $N=2$ case use the specifics of the $2\times2$-situation way too much. Is it true, for instance that, along the lines of Ralph's and Will's answer, an abelian $p$-subgroup of $GL(N,q)$ may only have a bounder number of cyclic factors?

REPLY [12 votes]: If $p$ is prime, the least dimension for a faithful representation of $(\mathbf{Z}/p)^d$ over any field of characteristic $\neq p$ is $d$. The argument is very easy, as you can assume the field algebraically closed and diagonalize. 
It follows that if $G$ is a group containing isomorphic copies of $(\mathbf{Z}/p)^d$ for two distinct primes $p$ (e.g. $(\mathbf{Z}/6)^d$) then its smaller faithful representation over some field is of dimension $d$.
Note: for $(\mathbf{Z}/6)^d$, on the other hand there is a faithful representation in dimension two over a commutative ring (the product of finite fields $F_{2^d}\times F_{3^d}$), and looking at representations in arbitrary commutative rings seems more natural for obvious stability reasons.
 On the other hand, if $S$ is finite, simple non-abelian and has a faithful rep. in dimension $d$ over a commutative ring, then it has one over a finite field.

REPLY [9 votes]: Yes, the idea that works for ${\rm GL}(2,q)$ with Abelian subgroups works with higher derived lengths for other ${\rm GL}(N, q).$ If $p$ is a prime which does not divide $q$ then the Sylow $p$-subgroups of ${\rm GL}(n,q)$ are monomial (up to equivalence) over some extension field. An easy indction argument (no pun intended) show that a monomial $p$ goup with a faithful representation of degree $p^{k}$ or less has derived length at most $k+1.$ Hence the Sylow $p$-subgroups of ${\rm GL}(N,q)$ have derived length at most $1 +\log_{p}(N)$ for all such $p.$ If $q$ is a power of $p,$ the bound on the derived length of a Sylow $p$-subgroup of ${\rm GL}(N,q)$ is similar. The nilpotence class of the upper unitriangular group is at most $N-1$, so the derived length is at most $\log_{2}(N),$ because for any nilpotent group $U,$ we have $U^{(k)} \leq L_{2^{k}}(U),$ where $U^{(k)}$ is the $k$-the term of the derived series and $L_{m}(U)$ is the $m$-th term of the lower central series. Hence  in particular, no $p$-group of derived length $2 + \log_2(N)$ or more is a subgroup of any ${\rm GL}(N,q)$.<|endoftext|>
TITLE: Are there Hamilton paths in Cayley graphs of Coxeter groups?
QUESTION [19 upvotes]: Hi everyone.
I want to optimize certain computation on finite Coxeter groups $(W,S)$. Basically I compute the matrices $\rho(T_w)$ for all $w\in W$ of a matrix representation $H\to K^{d\times d}$ of the Hecke algebra $H=\mathcal{H}(W,S)$ and do some stuff with these matrices. The representation is given as a list of the matrices $\rho(T_s)$ for $s\in S$. The obvious way to do such a computation is to use the property $l(ws)=l(w)+1 \implies T_{ws}=T_w T_s$ of the standard basis of $H$ to move from layer to layer in the group ("layer" meaning sets of the form $\lbrace w\in W | l(w)=k\rbrace$ for fixed $k$) and by multiplying the matrices $\rho(T_s)$ to the existing ones.
Since I'm also interested in big examples, I quickly run into trouble with my memory in this way because to compute the $\rho(T_w)$ with $l(w)=l$ one has to store all the $\rho(T_y)$ with $l(y)=l-1$ which can be quite a big number if $l$ is around $\frac{1}{2}l_{max}$. Even though I have access to a machine with 128GB RAM, this is too much if $W$ and the dimension of $\rho$ are big.
A few days ago I read about Hamilton paths in Cayley graphs. This would solve my memory problem, because if I knew a Hamilton path $w_1,\ldots,w_n$ I would only need to store the single matrix $\rho(T_{w_i})$ to compute $\rho(T_{w_{i+1}})$ and forget about it afterwards. If I had access to a Hamilton path in the Cayley graph $\Gamma(W,S)$ I could carry out my calculations with using only little more memory than I already need for the input itself.
Googling showed my that in general it is not even clear if such hamilton paths always exists. That's rather unfortunate, but on the positive side I also found out that there is an easy algorithm in case of the symmetric group and its Coxeter generating set. So I'm hoping that there is a result in the case of Coxeter groups.
So my questions are:

If $(W,S)$ is a finite Coxeter system, does there exists a Hamilton path in the Cayley graph $\Gamma(W,S)$?
If this is indeed the case, is there an easy algorithm to traverse a Hamilton path?

REPLY [6 votes]: The answer to both questions is yes. See the paper of Conway, Sloane and Wilks called "Gray codes for reflection groups":
http://link.springer.com/article/10.1007/BF01788686<|endoftext|>
TITLE: "Uncertainty principle" for self-adjoint operators in a finite von Neumann algebra
QUESTION [5 upvotes]: Let $M\subset B(\mathcal H)$ be a finite von Neumann algebra of bounded operators on a Hilbert space $\mathcal H$., let $P\in M$ be a self-adjoint operator with a pure-point spectrum (for example a projection), and let $T\in M$ be another self-adjoint operator. 
The question will be formulated in the case of $P$ being a projection. Let $\mathcal H_0$ and $\mathcal H_1$ be the eigenspaces of $P$.

Question: Is there a well-studied condition on $P$ and $T$ which would imply that there exists $\epsilon$ such that whenever $U\subset \mathcal H$ is a "generalized eigenspace" of $T$ (i.e. the image of a spectral projection of $T$) then there exists $u\in U$ of norm $1$ such that $u=h_1+h_2$, $h_1\in \mathcal H_1$, $h_2\in \mathcal H_2$, and the norm of $h_2$ is at least $\epsilon$?

In particular (EDIT: the previous version had a weeker ergodicity assumption, in which case the answer is negative, by the answer of Steven Deprez below) 

Question: If a discrete group $\Gamma$ acts freely and in a measure preserving way  on a probablity measure space $X$, each element of $\Gamma$ acts ergodically, $M=L^\infty(X) \rtimes \Gamma$, $P$  is the characteristic function of some subset of $X$ of positive measure, and $T = \sum_{\gamma\in \Gamma} f_\gamma \gamma$, where $f_\gamma$ are real-valued positive functions on $X$ such that $\sum f_\gamma=1$; do $T$ and $P$ fulfill the property from the question?

REPLY [2 votes]: A refinement of the argument from my previous answer:
For any free, pmp, ergodic action of any non-trivial countable group, $\Gamma\curvearrowright (X,\mu)$, there is a projection $P\in L^\infty(X,\mu)$ and a $T=\sum_{\gamma\in\Gamma}f_\gamma\gamma$ such that the $f_\gamma$ are positive functions adding to $1$, and such that $T$ and $P$ do not fulfill the property from the question.
We can assume that every nontrivial element of $\Gamma$ acts ergodically on $X$. In particular, $\Gamma$ does not have any finite subgroups and sice the action is free, $X$ is non-atomic.
Let $e\not=g\in \Gamma$ be any element. Take any non-trivial subset $U_0\subset X$. Since $gU_0\not=U_0$, we see that $U_1=U_0\setminus gU_0$ is a non-null set. Now we know that $U_1\cap g^{-1}U_1=\emptyset$. Choose a subset $U\subset U_1$ with measure $\mu(U)<\frac14$. This set still satisfies $U\cap g^{-1}U=\emptyset$. Denote $V=X\setminus (U\cup g^{-1}U)$ and $W=V\cap gV$. Observe that $\mu(W)\geq 1-4\mu(U)>0$.
Now, set $T=\chi_V + \chi_U u_g + \chi_{g^{-1}}u_g^\ast$ and $P=1-\chi_W$. Observe that $T\chi_W=\chi_W$. It follows that every vector $\xi\in Ran(\chi_W)$ is an eigenvector of $T$ with eigenvalue $1$. For any subset $A$ of the spectrum of $T$, denote the corresponding spectral projection by $p_{T,A}$. It follows that $p_{T,A}\leq 1-\chi_W=P$ for every subset $1\not\in A\subset \sigma(T)$ of the spectrum, not containing $1$. This contradicts the property from the question.<|endoftext|>
TITLE: whitehead group of product of groups
QUESTION [8 upvotes]: I am wondering is there a formula for the whitehead group of product of groups. In other words, if we know the whitehead group of two groups, are we able to calculate the whitehead group of their products. If not, is there a example such that the whitehead group of the product is nontrivial while the whitehead group of both groups vanishes.

REPLY [2 votes]: I just realized that the Farrell-Jones conjecture can be treated as a generalization of the Bass-Heller-swan theorem, hence can answer my question. Note that the conjecture has already been verified for a large class of groups.
Given two groups, $H$ and $G$, The Group Ring $\mathbb{Z} (H \times G)$ can be considered as $(\mathbb{Z}H)G$. Denote the Ring $\mathbb{Z}H$ as $R$, then the K-theoretic Farrell-Jones conjecture predicted the following assembly map is an isomorphism
$$ H_n^G(E_{\mathcal{VYC}}(G); \mathbf{K}_R) \rightarrow K_n(RG);$$
 where $E_{\mathcal{VYC}}(G)$ is the classifying of G respect to the family of virtually cyclic subgroups, $\mathbf{K}$ is the non-connective K-theory spectrum.
Basically, the conjecture says if we know the homology theory of $G$, the K-theory of $R$, then we know the K-theory of $RG$. But in order to calculate $K_1(RG)$, we definitely need a lot K-group information of $R$, not just $K_1(R)$.  So only known the whitehead group of $G$ and $H$ is not enough to recover $Wh(G\times H)$. This also explains why there exists two groups with trivial whitehead torsion while the whitehead torsion of their product is non-zero.<|endoftext|>
TITLE: Permutations with restriction
QUESTION [6 upvotes]: We have $n$ types of objects, the number of objects of type $i$ being $a_i$, $1\leq i\leq n$.
What is the number of permutations of the 
$\sum_{i=1}^n a_i$ objects, if no two objects of the same type may be next to each other?
A simple example: If we have the objects $\{a,a,a,b,b,c,c\}$, then we allow $abcabac$ but not $aaabbcc$.
I have searched the web. The only results I found were about permutations of (pairwise) different objects avoiding certain patterns. 
I put this question on math.stackexchange.com but haven't received an answer.

REPLY [11 votes]: There's a nice result, I think due to Carlitz, Scoville, and Vaughan (I learned it as the "Carlitz-Scoville-Vaughan theorem", but I'm not sure how common that is) that the o.g.f. for words in which no pair of consecutive letters is among a restricted set is the multiplicative inverse of the o.g.f. for words in which every pair of consecutive letters is in that restricted set, evaluated at $(-a,-b,-c,\dots)$.
So for your problem, the o.g.f. for words in $a,b,c$ where all adjacent letters are the same is $$g(a,b,c) = 1 + a + b + c + a^2 + b^2 + c^2 + \cdots = 1 + \frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c},$$ which means the generating function for words in $a,b,c$ where no adjacent letters are the same is $$f(a,b,c) = \frac{1}{g(-a,-b,-c)} = \frac{1}{1-\frac{a}{1+a}-\frac{b}{1+b}-\frac{c}{1+c}}.$$<|endoftext|>
TITLE: Where can I find a full proof of the Chern-Gauss-Bonnet theorem ?
QUESTION [8 upvotes]: Hello, 
I am looking for a proof for the Chern-Gauss-Bonnet theorem. All I have found so far that I find satisfactory is a proof that the euler class defined via Chern-Weil theory is equal to the pullback of the Thom class by the zero section, but I would like a proof of the fact that this class gives the Euler characteristic when coupled to the fundamental class. Thanks in advance.

REPLY [10 votes]: For a complete proof of the Gauss-Bonnet-Chern for arbitrary vector bundles (not just tangent bundles) see Section 8.3.2  of  these notes. The proof is Chern's original proof,  based on Chern-Weil theory, but the language is more modern.
For a purely topological proof, see  Section 5.3 of   these  notes.<|endoftext|>
TITLE: symmetric difference of languages - both are in NP and coNP
QUESTION [5 upvotes]: I have this problem,
Let $L_1,L_2$ be languages in $NP \cap co-NP$. I want to show that their symmetric difference is also in $NP \cap co-NP$. Like: 
$L_1 \oplus L_2$ = {x | x is in exactly one of $L_1, L_2$}
I do not have a clue how to show it. We know that $L_1 \cap L_2 \in NP$ is unknown. So for that reason it is reasonable to ask only that  instance of the problem. From my point of view, if $L_1 \in NP$ there is some verifier for that language which runs in polynomial time. We have such verifier for the second language $L_2$ too. My proposal of the machine M which decides $L_1 \oplus L_2$ is as follows: 
Let's have 
M = "for the input x:
1. copy x on the second tape
2. run x on M_1 on the first tape
3. if M_1 accepts, (otherwise go to 4.)
    3a) run x on M_2
    3b) if  M_2 accepts, M rejects 
4. run x on M_2, if M_2 accepts, M accepts, otherwise M rejects.

I do not know the relation of this with the co-NP class ... Is my reasoning right? This machine works like a charm for languages $L_1,L_2 \in P$. Does it hold also for that intersection?
Thank you a lot

REPLY [3 votes]: NP is closed under (finite) unions and intersections, therefore using De Morgan laws, $\mathrm{NP}\cap\mathrm{coNP}$ is closed under unions and intersections. Since it is also trivially closed under complement, it is closed under all Boolean operations, including the symmetric difference.
A more careful argument shows that $\mathrm{NP}\cap\mathrm{coNP}$ is even closed under polynomial-time Turing reductions. (Note that symmetric difference can be computed by an essentially constant-time algorithm querying oracles for the two languages.) In fact, $(\mathrm{NP}\cap\mathrm{coNP})^{\mathrm{NP}\cap\mathrm{coNP}}=\mathrm{NP}\cap\mathrm{coNP}$.<|endoftext|>
TITLE: Derived Functors Versus Spectral Sequences
QUESTION [70 upvotes]: Let $A{\buildrel F\over\rightarrow}B{\buildrel G\over\rightarrow}C$ be additive functors between abelian categories.
Hartshorne, in Proposition 5.4 of Residues and Duality, constructs the obvious natural transformation $\zeta_{G,F}:R(GF)\Rightarrow (RG)(RF)$ and notes its obvious properties.  He then remarks:


This proposition shows the convenience of derived functors in the context of derived categories.  What used to be a spectral sequence becomes now simply a composition of functors.  (And of course one can recover the old spectral sequence from this proposition by taking cohomology and using the spectral sequence of a double complex).  


I've always felt like I'm missing something here.  In what sense does the composition of functors (together with the natural transformation $\zeta_{G,F}$) replace the old spectral sequence?  I understand that there's a conceptual insight here, but my questions are these:

What (if anything) is an example of a statement that used to be proved by invoking the spectral sequence but can now be proved more succinctly using the composition of functors?
In what sense does the recovery of the spectral sequence actually use the derived formalism?  Isn't the prescribed double complex exactly the same one I'd have written down if I'd never heard of derived categories?

REPLY [84 votes]: 1 Easy
Proposition Let $f:X\to Y$ be a continuous map of topological spaces, $\mathscr F$ a sheaf of abelian groups on $X$ such that $R^jf_*\mathscr F=0$ for $j>0$. Then for all $i\geq 0$ there exists a natural isomorphism 
$$
H^i(Y, f_*\mathscr F)\simeq H^i(X,\mathscr F)
$$
Proof
Apply the composition rule for the derived functors of $G=\Gamma(Y, \_ )$ and $F=f_*({\_})$. By definition, $G\circ F = \Gamma(X, \_ )$. Then
$$
R\Gamma(Y, f_*\mathscr F) \simeq R\Gamma(Y, Rf_*\mathscr F) \simeq R\Gamma (X, \mathscr F).
$$
Taking cohomology shows the result. $\square$
(edit to please anon, see the comments below)
This is usually exhibited as an example of how to use the Leray spectral sequence. Doing it that way is not much harder than the above, but perhaps a bit less "automatic". 
Furthermore, this proof shows more: Not only the cohomologies of these sheaves are isomorphic, but they come from the same complex! That's a much stronger statement. It is easy to give examples when the cohomologies of two complexes are isomorphic, but the complexes are not. I suppose one may argue that the word "natural" in the statement means exactly this, but then I'd say that proving naturality with the Leray spectral sequence is certainly possible, but it definitely needs more care.
This last point is actually an important one regarding the derived category language. You get a higher level notion. The fact that you can work with the complex whose cohomologies give the derived functors of your original functor is very very useful. 

2 Less Easy
In case you were not convinced by the above example, here is one that should do the trick:
A special case of Grothendieck duality says that if $f:X\to Y$ is a proper morphism between not too horrible schemes, let's say finite type over a field $k$ (let me not try to make a precise statement, this is in Residues and Duality that you mentioned) and $\mathscr F$ is a coherent sheaf on $X$, then 
$$
 Rf_*R\mathscr Hom_X(\mathscr F, \omega_X^{\bullet})\simeq R\mathscr Hom_Y(Rf_*\mathscr F, \omega_Y^{\bullet}).
$$
Here $\omega_{Z}^{\bullet}=\varepsilon^!k$ is "the" dualizing complex where $\varepsilon: Z\to \mathrm{Spec}\ k$ is the structure map of $Z$. 
Now try to imagine how one could state this using spectral sequences. Both sides actually correspond to spectral sequences, so the statement would be something like "there is a natural map between this an this spectral sequences, such that they converge to the same thing". 
I would argue that already the statement of this theorem would be tiring in the language of spectral sequences, but using it would be pure pain.

3 Even Less Easy
Here is an application of Grothendieck duality where one can see how the derived category formalism makes life easier and arguments that seemed complicated are reduced to a one liner.
Theorem (a.k.a. Kempf's Criterion)
Let $Y$ be a normal variety over $\mathbb C$ with a resolution of singularities $f:X\to Y$. Then $Y$ has rational singularities (i.e., $R^if_*\mathscr O_X=0$ for $i>0$) if and only if 

$Y$ is Cohen-Macaulay
$f_*\omega_X\simeq \omega_Y$

Proof Let $n=\dim Y=\dim X$ and suppose $Y$ has rational singularities. Then
$$
\omega_Y^{\bullet}\simeq R\mathscr Hom_Y(\mathscr O_Y, \omega_Y^{\bullet})\simeq 
R\mathscr Hom_Y(Rf_*\mathscr O_X, \omega_Y^{\bullet})\simeq
Rf_*R\mathscr Hom_X(\mathscr O_X, \omega_X^{\bullet})\simeq Rf_*\omega_X[n]\simeq f_*\omega_X[n].
$$
(The isomorphisms follow by the assumptions, Grothendieck duality, and the last one is the Grauert-Riemenschneider vanishing theorem).
This implies that $\omega_Y=h^{-n}(\omega_Y^{\bullet})\simeq f_*\omega_X$, which is the second condition to prove and also that $h^i(\omega_Y^{\bullet})=0$ for $i\neq -n$ which is equivalent to $Y$ being Cohen-Macaulay. 
The other direction goes essentially in the same fashion. $\square$
Now try to do this with spectral sequences. 

To answer your second question, I think you are right. In order to get the spectral sequences you do not need to go through the derived category formalism. However, if you are indeed "recovering" the spectral sequence, then you start with the derived category formalism. In other words, if you've never heard of derived categories, why (or perhaps more importantly how) would you want to recover anything from a derived category statement? (Since written word lacks intonation, let me add that I'm not trying to be confrontational, but I feel that this question is somehow off target.)<|endoftext|>
TITLE: Cartan-Hadamard's Theorem, Completeness
QUESTION [13 upvotes]: Theorem: Let $(M^n,g)$ complete, simply connected Riemannian manifold with non-positive sectional curvature. Then $M$ is diffeomorphic to $\mathbb{R}^n$.
Question: Can we change complete to "every two points can be joined by a minimizing geodesic"?

REPLY [2 votes]: Let $X$ be completion of $(M,g)$.
Note that $X$ is simply connected.
It follows since any loop in $X$ is a limit of a broken geodesic in $(M,g)$.
Therefore $X$ is CAT(0), in particular any two points are joined by unique geodesic.
Therefore $M$ is diffeomorphic to $\mathbb R^n$ as any star-shaped domain (thanks to alvarezpaiva).<|endoftext|>
TITLE: In What Way are Set Theorists' 'Experiences' in the CH Worlds Flawed, if Any?
QUESTION [21 upvotes]: This is in regards to Joel David Hamkins' new paper "IS THE DREAM SOLUTION OF THE CONTINUUM HYPOTHESIS ATTAINABLE?" (look under title in arXiv). I quote from the last paragraph of his paper:
"My challenge to anyone who proposes to give a particular, definite answer to CH is that they must not only argue for their particular answer, mustering whatever philosophical or intuitive support for their answer as they can, but also they must explain away the illusion of our experience with the contrary hypothesis.  Only by doing so will they overcome the response I have described, rejection of the argument from extensive evidence of the contrary.  Before we will be able to accept CH as true, we must come to know that our experience of the not-CH worlds was somehow flawed; we must come to see our experience in those lands as illusory."
Let me make a slight variation in the last sentence of his I quoted:
Before we will be able to accept not-CH as true, we must come to know that our experience of the CH worlds was somehow flawed; we must come to see our experience in those lands as illusory.
Since the goal of set theory (at least from Hamkins' perspective of the orthodox view (the set-theoretical universe as unique--it is the universe of all sets, "The set-Theoretical Multiverse: a model-theoretic philosophy of set theory")) is to have V (for ZFC, for example) to contain all possible sets short of inconsistency, it would seem that from this perspective that the CH worlds are already flawed and that to defend CH against not-CH one would have to say that the existence of 'Cohen reals' in the not-CH worlds is somehow illusory (or at least the belief that one can add sufficient number of Cohen reals to make CH false from the Naturalist View of Forcing perspective).  Can one make the view showing that either Cohen reals are illusory, or that the ability to add sufficient number of Cohen reals so as to make not-CH true is illusory, coherent?

REPLY [11 votes]: I would like to add to the thoughtful answers to this question and also respond to a couple points made in the answers.  Please note that this is mostly a philosophical answer.  
The part of the question that I would like to address is

. . . it would seem that from this perspective that the CH worlds are already flawed and that to defend CH against not-CH one would have to say that the existence of 'Cohen reals' in the not-CH worlds is somehow illusory

It might seem like since there are a variety of universes, that some of them must be flawed. In particular that since there are universes which are models of $CH$ and $\neg CH$ that some of these universes must be flawed. The accepted answer to this question has shown that it is coherent to resolve this difficulty in thinking by calling conflicting universes or sets illusions.  However, one can overcome this apparent cognitive difficulty by living locally in the multiverse.  For the purpose of understanding how to see through the proper perspective in the set-theoretic multiverse, let me create some characters, if you will tolerate, which live in the multiverse. Suppose in the multiverse there are travelers and inhabitants. Travelers like to travel far and wide and often, while inhabitants like to stay home.  Let's go to a universe $V$ which models $GCH$.  Inhabitants in this universe know that the continuum hypothesis is true.  However, they also know that there are many travelers coming to visit their universe.  An inhabitant meets a traveler who has just arrived from another universe $W$ where the continuum is $\aleph_{10}$.  The traveler sees that in universe $V$ the continuum is $\aleph_1$, and so accepts that the continuum is now a different size because she is in a new place.  The inhabitant is having a more difficult time understanding since he has not traveled beyond his native universe.  The traveler tells him that in universe $W$ the size of the real numbers is $\aleph_{10}$, but he doesn't understand because the size of the real numbers is most definitely $\aleph_1$, and he has no "experience of the contrary".  So, the traveler decides to take him with her to another universe $X$ where the continuum is $\aleph_2$.  Via a generic filter and forcing relation, they travel to universe $X$ where the inhabitant of universe $V$ can now see with his own eyes that in this world the continuum is indeed the second uncountable cardinal!  So, truth in the multiverse depends on location.  This way of understanding is discussed in detail in Hamkins' paper on the set-theoretic multiverse, and in the paper regarding a dream solution to CH which is quoted below:

Part of my goal in the multiverse article was to tease apart two often-blurred aspects of set-theoretic Platonism, namely, to separate the claim that the set-theoretic universe has a real mathematical existence from the claim that it is unique.

Second, I would like to respond to Andrej's perspective that all universes are created equal.  I think that there are places in the multiverse which are more pleasant than others depending on one's preferences.  For example, the inhabitants of a universe where $GCH$ holds might believe that their universe is the best since there are so many travelers who have a lay-over there.  I like this type of universe very much and indeed in my experience with forcing, it is very helpful (in order to count in the ground model) to have the $GCH$ hold. I have also been to a universe where Martin's Axiom holds and I really liked doing mathematics in that universe.  However, some regions of the multiverse may be less appealing, say a universe without the axiom of choice.
Finally, I would like to address the second to last paragraph of Joel's response.  He says that it is crippling to have to consider the rich set-theoretic multiverse as a mere simulation, an illusion we experience in the universe.  I agree, but I think it is important to discuss and distiguish the difference between the dreams of the universe and the reality of the multiverse.  In every universe, there are the classes of names of other universes with a variety of sizes of the continuum, for example.  But these classes of names are not themselves the universes to which they point since if they were then the universe which dreams of them would not be coherent.  It is only in the presensce of a generic filter that this dream class can become a real universe.  What I am saying is that even if someone holds the view that the multiverse is really an illusion experienced in the universe then that person could not be talking about the actual multiverse, but only a reflection.<|endoftext|>
TITLE: In the classical construction of conic sections, where does the axis of the cone intersect the plane?
QUESTION [5 upvotes]: Everybody knows that if I take the intersection of a right circular cone with a plane, I get a conic section. My question is, where does the symmetry axis of the cone intersect the plane? Does this point relative to the conic have a name, or a simple description? For example, for an ellipse I first guessed that it was one focus of the ellipse, but that is false.

REPLY [3 votes]: This question and accepted answer are almost ten years old, but in case anybody stumbles upon this question, here's some more information on the topic.  (For cone terminology and background, see wikipedia)
First, an edited version of one of the comments:

All [right] circular cones whose section is a given ellipse produce
[different] points. The more narrow [...] the cone, the closer to the
center is the point [of intersection]. In general, [...] the point is
between the foci, at certain distances from them, whose ratio is equal
to the ratio of the radii of the Dandelin spheres. – Pietro Majer

The apex of each cone in this family is on a hyperbola $h$ passing through the foci of the ellipse. The foci of $h$ are the points of intersection of the ellipse and its major axis.  This is easily proven in the same context as the proofs for Dandelin spheres, and uses the property that the difference of distances from a point on the hyperbola to its foci is constant for all points.  See Salmon, Conic Sections, Art. 367.
For a much more detailed treatment see Armstrong, Where is the Cone?.  From the abstract:

Real quadric curves are often referred to as “conic sections,”
implying that they can be realized as plane sections of circular
cones. However, it seems that the details of this equivalence have
been partially forgotten by the mathematical community. The definitive
analytic treatment was given by Otto Staude in the 1880s and a
non-technical description was given in the first chapter of Hilbert
and Cohn-Vossen’s Geometry and the Imagination (1932). [...] Our hope is to revive the
lost knowledge of “conic sections” by providing the slickest possible
modern treatment, by using standard linear algebra that was not
standard in 1932.<|endoftext|>
TITLE: Does bundle with torsion Chern classes admit flat connection?
QUESTION [5 upvotes]: I want to know something about torsion in topological k-theory. So, consider complex bundle with chern classes lying in torsion part of integer homologies and my question is : does it admit a flat connection? In the case of line bundles (for wich flat structure does't implies triviality) this obviously true, all you need is just add to your connection with curvature $\Omega$ 1-form A with $dA=\Omega$, so it gives curvature-free connection.

REPLY [12 votes]: No. A complex vector bundle on $S^5$ must have Chern classes zero, and in this case a flat bundle would have to be trivial, but there is a nontrivial bundle because $\pi_4U(2)$ is nontrivial.<|endoftext|>
TITLE: Application of polynomials with non-negative coefficients
QUESTION [12 upvotes]: Question 1: Are there any deeper applications (in any field of mathematics) of polynomials (with possibly more than one variable) over the real numbers whose coefficients are non-negative? So far I have found only some vague mentions of control theory and maybe some approximations of functions, but don't know the details (e.g. www.math.ttu.edu/~barnard/poly.pdf). 
Question 2: There is a similar notion, namely "positive polynomials", but they are defined as "polynomial functions that are non-negative". Does anybody know some connection between these two kinds of polynomials? Thanks.
All comments and suggested answers here (thanks for them:)) have certainly something to do with the mentioned type of polynomials. To specify more what was my intention - a desired answer (to Question 1) should fullfil the following:
"Is the application in you proposed area of such a kind that it forces an (independent) research of polynomials with non-negative coefficients on themselves?"

REPLY [2 votes]: Thanks to all for your comments and suggestions. The original motivation for this question was a connection between polynomials with non-negative coefficients and commutative semirings. After some search we found some papers asking for determining the least degree of a polynomial with non-negative coefficients that is divisible by a given (general) polynomial. Suprisingly similar questions were investigated repeatedly and independetly but without any deeper motivation. We deceided hence to make an overview, improvements of some results and suggested a few conjectures. The result (made before the latest updates of this webpage) can be found here http://arxiv.org/abs/1210.6868. (Suggestions and comments are welcome.)<|endoftext|>
TITLE: Which spaces are characterized by functions with compact support ? 
QUESTION [20 upvotes]: It's well known that two locally compact Hausdorff spaces $X, Y$ are homeomorphic iff the rings $C_0(X), C_0(Y)$ (continuous functions vanishing at infinity) are isomorphic. 
Is there a class $\mathcal{C}$ of topological spaces such that $X, Y \in \mathcal{C}$ are homeomorphic, iff the rings $C_c(X), C_c(Y)$ are isomorphic ? 
Here $C_c(X) = \lbrace f:X \to \mathbb R\mid \operatorname{cl}_X \lbrace x \in X \mid f(x) \neq 0\rbrace\text{ is compact}\;\rbrace$ denotes the ring of continuous functions with compact support.

REPLY [11 votes]: My naive feeling is that the answer is simply the class of locally compact Hausdorff spaces, for the following reasons. First, for a locally compact Hausdorff space $X$, one can recover $C_0(X)$ from $C_c(X)$ by completion in the uniform norm; and the uniform norm is an algebraic feature because it can be derived from the characters on the ring $C_c(X)$. So for $X$ and $Y$ locally compact Hausdorff, if $C_c(X)$ and $C_c(Y)$ are isomorphic then $X$ and $Y$ are homeomorphic.
Now suppose that $X$ is any completely regular (Hausdorff) space (and surely it is topological spaces in this class that we are interested in). Then $X$ is the disjoint union of an open locally compact subset $X_0$ consisting of points which have a compact neighbourhood and a closed subset $X_1$ consisting of points which do not. Each $f\in C_c(X)$ vanishes on $X_1$ so we are not going to get any information about $X_1$ from $C_c(X)$. 
So my feeling is that $C_c(X)$ determines $X_0$ up to isomorphism but gives no information about $X_1$.<|endoftext|>
TITLE: the largest eigenvalue of the matrix A with A_{ij}=(i \times j) mod p for p is a prime.
QUESTION [7 upvotes]: For a prime p, consider the $(p-1) \times (p-1)$ matrix A with entry to be $A_{ij}=(i \times j) mod$ $p$. every row (column) is permutation of 1 to p-1, such a permutation is useful in one version of proof of Fermat's little theorem. Here the question is if the largest eigenvalue is always p(p-1)/2. also anything happens for the rank of it. except for p=2,3, the rank might always be p-2.
the student taking a course (general intro to math) I am TAing asked me it. but it is embarrassing to say I don't know how to work it out. I know almost zero about primes and I believe this might be a standard result. I am grateful if anyone suggests a hint or any reference.

REPLY [9 votes]: Anthony Quas shows that your matrix has rank $(p+1)/2$ if and only if the sum
$$\sum_{j=0}^{p-2} (a^j \bmod p) \beta^{jk}$$
is nonzero for all odd $k$. Here $a$ is a primitive root of unity modulo $p$ and $\beta$ is a primitive $(p-1)$-th root of unity.
Take a look at Section 6.5 of Edward's book Fermat's Last Theorem. He establishes the formula (equation 8)
$$L(1, \chi) = \frac{ i \pi m_k}{p} \sum_{j=0}^{p-2} (a^j \bmod p) \beta^{jk}$$
where $k$ is odd, $\chi$ is the character $a^j \mapsto \beta^{jk}$ from $\mathbb{F}_p^{\times}$ to $\mathbb{C}^{\times}$ and  $m_k = \frac{1}{p} \sum_{j=0}^{p-2} \beta^{jk} \zeta^{a^j}$, for $\zeta$ a primitive $p$-th root of unity. (Warning: I have tried to change Edward's notation to match Quas's. In particular, Edwards uses $k$ and $j$ to mean the exact opposite of what Quas does! Hope I haven't introduced any errors.)
The key point is that $L(1,\chi)$ is known to be nonzero! So the result on the rank of your matrix is true, and for quite a nontrivial reason.<|endoftext|>
TITLE: What is the smallest variety of algebras containing all fields?
QUESTION [10 upvotes]: A field is a ring whose nonzero elements form a commutative group under multiplication. A field is also a commutative inverse semigroup with respect to multiplication. The unique multiplicative inverse $y$ of an element $x$ (in the sense that $xyx=x$ and $yxy=y$) is $y=x^{-1}$ if $x \neq 0$ and $y=0$ if $x = 0$.
To simplify the discussion, define an inverse ring to be a ring which is an inverse semigroup with respect to multiplication. Denote the multiplicative inverse operation by $()^{-1}$. (Warning: The notion of an inverse ring doesn't exist outside of this question.) Both rings and inverse rings form a variety of algebras, i.e. they can be defined by a set of operations ($+$, $*$, $-()$, $()^{-1}$, $0$, $1$ in this case) together with set of identities satisfied by these operations. I think that the commutative inverse rings are the smallest variety of algebras containing all fields.
Question

A direct product of a family of fields is no longer a field. However, it is still a commutative inverse ring. My question is whether every commutative inverse ring is a subdirect product of a family of fields.

(Note that subdirect product here must refer to either rings or inverse rings, because the notion of subalgebra isn't defined otherwise. The answer to my question should be independent of which one we choose, but referring to inverse rings would make more sense to me.)
Note This question is identical to this question at math.stackexchange.com.

REPLY [2 votes]: There is a notion of von Neumann regular rings. For commutative rings $A$, this notion has many equivalent definitions (for proofs see David F. Anderson, Zero-Dimensional Commutative Rings):
1) $A$ is zero-dimensional and reduced.
2) Every localization of $A$ is a field.
3) For every $x \in A$ we have $x^2 | x$.
4) For every $x \in A$ there is a unique $y \in A$ with $x = x^2 y$ and $y = y^2 x$; this $y$ is called the weak inverse of $x$.
So this is what you have called an inverse ring. The resulting variety is the smallest one containing all fields, since every reduced ring embeds into a product of fields.<|endoftext|>
TITLE: Fastest algorithm to compute (a^(2^N))%m?
QUESTION [5 upvotes]: Hi. 
There are well-known algorithms for cryptography to compute modular exponentiation $a^b\%c$ (like Right-to-left binary method here : http://en.wikipedia.org/wiki/Modular_exponentiation).
But do algorithms exist to compute modular exponentiation of the form $a^{\left(2^N\right)}\%m$ faster than with "classical" algorithms ?
Thank you very much !
Notes :
1) $m$ has no particular property
2) $N < 2^{32}$

REPLY [5 votes]: I am not sure I understand the question, but if $m \ll 2^N,$ the obvious thing to do is to compute $x = 2^N \mod \phi(m)$ [by repeated squaring], and then compute $a^x \mod m.$ If $2^N$ is not huge compared to $m$ then factoring $m$ might dominate.<|endoftext|>
TITLE: Moser regularity proof avoiding John-Nirenberg lemma
QUESTION [7 upvotes]: I heard a rumor that there exists a proof by Moser-style iteration of the $C^{0,\alpha}$-regularity for $W^{1,2}$-solutions $u$ to elliptic equations with measurable coefficients which does not rely on the John-Nirenberg lemma. 
I was wondering if somebody can point out a reference for that proof, or a reason why the John-Nirenberg lemma cannot be truly avoided (or both!).

REPLY [2 votes]: For the homogeneous equation, I have seen a proof of $C^{\alpha}$ regularity using an oscillation estimate based only on local boundedness and a Poincare-Sobolev inequality. Specifically:
Let u be a subsolution in $B_2$ satisfying $|(u \leq 0) \cap B_1| \geq \frac{1}{2}|B_1|$. Then $\sup_{B_{1/2}}u^{+} \leq \gamma \sup_{B_1}u^{+}$, where $\gamma < 1$ depends only on the ellipticity constants and $n$. (|.| denotes Lebesgue measure).
From there, one concludes that the oscillation of a solution decays by a fixed proportion each time we localize, which gives Holder regularity.<|endoftext|>
TITLE: Is it possible to make an autoequivalence of categories an automorphism?
QUESTION [7 upvotes]: In studying triangulated categories, some authors require the shift functor $T: \mathcal D \rightarrow \mathcal D$ to be an autoequivalence, whereas others require it to be an automorphism (i.e. strictly invertible). Unfortunately, I couldn't find any reference which clarifies whether the two requirements are actually equivalent or not. Indeed, I think they are, so, abstracting a little, here is my question: given an autoequivalence $T: \mathcal C \rightarrow \mathcal C$ of an arbitrary category $\mathcal C$, is it possible to find a functor $T': \mathcal C \rightarrow \mathcal C$ which is isomorphic to $T$ and is also an automorphism of $\mathcal C$?
More generally, one could ask if a similar result is true for functors between categories whose object sets are of the same cardinality...
Thanks in advance!

REPLY [13 votes]: In practice one may always assume that such a shift is an automorphism instead of an auto-equivalence. But for that one also has to modify the category with an equivalence (but this is OK for applications):
Let $F : C \to C$ be an equivalence of categories. Define the following category $C'$: Objects are sequences of objects $(X_n)_{n \in \mathbb{Z}}$ with $X_n \in C$, together with isomorphisms $\alpha_n : F(X_n) \cong X_{n+1}$. It is clear how to define the morphisms. There is a canonical automorphism $F' : C' \to C'$, which is just a shift $(X_n,\alpha_n)_n \mapsto (X_{n+1},\alpha_{n+1})_n$. Besides the evaluation at $0$ yields a functor $C' \to C$. It is easy to check that this is an equivalence of categories which makes the diagram
$$\begin{matrix} C' & \stackrel{F'}{\rightarrow} & C' \\\\ \downarrow && \downarrow \\\\ C & \stackrel{F}{\rightarrow} & C \end{matrix}$$
commute up to natural isomorphism of functors.
To put this into more global terms: Let $\mathcal{A}$ be the $2$-category of all categories equipped with an auto-equivalence, and $\mathcal{A}'$ the full $2$-subcategory of $\mathcal{A}$ consisting of those categories equipped with an automorphism. Then $\mathcal{A}' \hookrightarrow \mathcal{A}$ is an $2$-equivalence of $2$-categories.<|endoftext|>
TITLE: A curve with bad reduction for which the jacobian has good reduction
QUESTION [13 upvotes]: Let $K$ be a number field. If $X$ is a curve over $K$ with good reduction at a place $v$ of $K$, then the Jacobian of $X$ also has good reduction at $v$. This follows from the functoriality of the Jacobian.
The converse is not true, but I don't know of any examples.
Can one provide an example for all number fields $K$? 
If not, I would also be pleased with just a counterexample for some place $v$ of some number field $K$

REPLY [13 votes]: Let $E,E'$ be elliptic curves over the residue field.  Then $E \times E'$
has good reduction but is not a Jacobian.  However, any principally
polarized abelian surface that reduces to $E \times E'$ but is not
a product of elliptic curves is a Jacobian of some genus-$2$ curve $C$
that cannot have good reduction.
Explicitly (when the residue characteristic is odd), $C$ can have the form $y^2 = P(x)$ where $P$ is a sextic
with roots $x_1,x_2,x_3$, $1/x'_1,1/x'_2,1/x'_3$ 
such that each $x_i,x'_i$ reduces to zero and each
$x_i-x_j$ and $x'_i-x'_j$ ($i\neq j$) has valuation $1$.
If I remember right, the Jacobian reduces to $E \times E'$
where $E: Y^2 = (X-\bar x_1) (X-\bar x_2) (X-\bar x_3)$
and $E': Y^2 = (X-\bar x'_1) (X-\bar x'_2) (X-\bar x'_3)$.
As I recall I learned this construction from Joe harris.<|endoftext|>
TITLE: Why is this a lattice?
QUESTION [5 upvotes]: Why is $\text{SL}_3(\mathbf{Z}[1/2])$ a lattice in $\text{SL}_3(\mathbf{R})\times\text{SL}_3(\mathbf{Q}_2)$? Discreteness is pretty clear, but why finite covolume? I understand why $\text{SL}_3(\mathbf{Z})$ has finite covolume in $\text{SL}_3(\mathbf{R})$, but I'm having trouble seeing this extension.
For that matter, why does $\mathbf{Z}[1/2]$ have finite covolume in $\mathbf{R}\times\mathbf{Q}_2$?
Am I missing a simple argument?

REPLY [7 votes]: As to the second question, this is pretty elementary : use the "fractional part" in $\mathbf{Z}[1/2]$ to put any element of $\mathbf{R}\times\mathbf{Q}_2$ in $\mathbf{R}\times\mathbf{Z}_2$ (by subtraction), and the "remaining" $\mathbf{Z}$ to put the  $\mathbf{R}$ component in $[0,1]$. This way you see that the quotient is a so-called solenoid, the compact quotient $[0,1]\times\mathbf{Z}_2/(1,x)\sim (0,x+1)$ (this is a compact connected topological group).
Informally, elements of $\mathbf{R}$ have dyadic expansion that is infinite to the right and finite to the left, and $2$-adic numbers have the opposite situation. Then $\mathbf{Z}[1/2]$ is their "intersection". The (diagonal) quotient just "blends" them.
As to the first question, you might first use that $\text{SL}_3(\mathbf{Z}[1/2])$ is dense in  $\text{SL}_3(\mathbf{Q}_2)$ to put the $2$-adic component of any element of $\text{SL}_3(\mathbf{R})\times\text{SL}_3(\mathbf{Q}_2)$ in the open subgroup $\text{SL}_3(\mathbf{Z}_2)$, then $\text{SL}_3(\mathbf{Z})$ to put the $\text{SL}_3(\mathbf{R})$ component in a finite volume fundamental domain $D$. Then $D\times\text{SL}_3(\mathbf{Z}_2)$ is a finite volume fundamental domain.<|endoftext|>
TITLE: Compatibility of the KZ connection with operadic composition
QUESTION [12 upvotes]: In what sense is the Knizhnik-Zamolodchikov connection compatible with the operadic composition in the little discs operad and/or the operad of $\overline{M}_{0,n}$'s?
Here are (some) details, motivation, and a more precise question.
The KZ connection $A_n\in\Omega^1(C_n)\otimes\mathfrak{t}^n$ is given by
$$A_n =\sum_{i< j} t^{ij}\,d\log(z_i-z_j).$$
Here $C_n$ is the configuration space of $n$ different points in $\mathbb{C}$ and $\mathfrak{t}^n$ is the Lie algebra with generators $t^{ij}$ ($1\leq i,j\leq n$, $i\neq j$, $t^{ij}=t^{ji}$) and relations $[t^{ij},t^{kl}]=0$ if all $i,j,k,l$ are different and $[t^{ij},t^{ik}+t^{jk}]=0$.
The "regularized holonomy" of $A_3$, when $z_1$ stays at $0$, $z_3$ at $1$, and $z_2$ moves from $0$ to $1$, is the KZ Drinfeld associator $\Phi_{KZ}$.
In general, an associator $\Phi$ (together with its coupling constant $\mu$) is equivalent to a morphism of operads of groupoids $F:PaB\to T$. Here $PaB_n$ is the groupoid of parenthesized braids, and $T_n=\exp \mathfrak{t}^n$ ($\mathfrak{t}^n$'s form an operad of Lie algebras). In the case of $\Phi_{KZ}$, $F_{KZ}$ is given by the "regularized holonomy" of $A_n$'s. The natural explanation of the fact that $F_{KZ}$ is a morphism of operads would be that $A$ is compatible with the operad structure. In what sense it is true? (there are two questions - in what sense is $A$ compatible with operad, and how it implies that $F_{KZ}$ is a morphism of operads)
I'll mention some random stuff that might appear in the answer, but you can ignore it.
There is a flat connection on moduli spaces of rational curves with marked points, closely related to $A$. If we define $\hat{\mathfrak{t}}^n$ by imposing one more relation $\sum_i t^{ij}=0$ ($[t^{ij},t^{ik}+t^{jk}]=0$ is then a consequence), the KZ connection descends to $M_{0,n}$; let us call it $\hat{A}_n\in\Omega^1(M_{0,n})\otimes\hat{\mathfrak{t}}^n$. By putting one of the points to $\infty$ we can identify $M_{0,n+1}$ with $C_n/\{az+b\}$ ($a\in\mathbb{C}^*,b\in\mathbb{C}$), and via the isomorphism $\hat{\mathfrak{t}}^{n+1}\cong\mathfrak{t}^n/\text{center}$ we can identify $\hat A_{n+1}$ with $A_{n}/\text{center}$. The center of $\mathfrak{t}$ is lost in this way. The Lie algebras $\hat{\mathfrak{t}}^{n}$ form a cyclic operad. The compactified moduli spaces $\overline{M}_{0,n}$ also form a cyclic operad.
As $F_{KZ}$ is (modulo the center) given by the parallel transport of $\hat{A}$ between certain tangential base points of $M_{0,n+1}\subset\overline{M}_{0,n+1}$, I would imagine the operadic compositions to be (roughly) maps
$$M_{0,n+1}\times M_{0,m+1}\times\text{formal punctured disc}\to M_{0,n+m}$$
coming from the operadic composition of $\overline{M}_{0,k}$'s. But I'm not sure in what category it would be an operad (of nice families over the punctured discs?) and how to make it technically work (something should be also said about maps of the trivial $\hat T$-bundles over these spaces).
In fact, to get a parametrization of the punctured disc, it would be better to consider the moduli spaces $M_{0,n}'$ of rational curves with $n$ marked points and non-zero tangent vectors at those points.
There is a flat connection on $M_{0,n}'$ which sees also the center. Let us replace the relations $\sum_i t^{ij}=0$ with $s^j:=\sum_i t^{ij}\textit{ is central}$. The Lie algebras $\check{\mathfrak{t}}^{n}$ that we obtain in this way still form a cyclic operad, and the connection
$$\check{A}_n=  \sum t^{ij}\,d\log(z_i-z_j)+ \sum s^i\,d\log(v_i)$$
is a flat connection on the configuration space of $n$ different points with chosen non-zero tangent vectors; this 1-form is again $SL(2,\mathbb{C})$-basic, and so it descends to the moduli space $M_{0,n}'$. By putting one of the points to $\infty$ and also normalizing its tangent vector, we can identify $M_{0,n+1}'$ with $(C_n/\text{translations})\times (\mathbb{C}^*)^n$. We have an isomorphism $\check{\mathfrak{t}}^{n+1}\cong\mathfrak{t}^n\oplus \mathbb{C}^n$, and so $\check{A}_{n+1}$ gets identified with $A_n$ plus a central part corresponding to the tangent vectors. It is the framed version of the KZ connection (and is my favorite).
Edit: There is another (Alekseev-Torossian) connection $A^{AT}_n$ on $C_n$ with values is $\mathfrak{t}^n$. $A^{AT}_n$ is in fact on $FM_2(n)$ ($FM_2(n)$ is a compactification of $C_n/\{az+b\}$, where this time $a\in\mathbb{R}_+$). $FM_2(n)$'s form an operad (a version of the little discs operad) and $A^{AT}_n$'s are compatible with the operad structure in the obvious way: if $o_i:FM_2(m)\times FM_2(n)\to FM_2(m+n-1)$ is one of the compositions then $o_i^*A^{AT}_{m+n-1}$ is equal to the connection $A^{AT}_m\oplus A^{AT}_n$ on $FM_2(m)\times FM_2(n)$, after we apply the corresponding $o_i:\mathfrak{t}^m\oplus\mathfrak{t}^n\to\mathfrak{t}^{m+n-1}$. The operad $PaB$ is a sub-operad of the fundamental groupoid of $FM_2$. $\Phi_{AT}$ and $F_{AT}$ are defined as (ordinary, not regularized) parallel transport of $A^{AT}$, and $F_{AT}$ is obviously a morphism of operads. I would like to understand the corresponding picture for $A_{KZ}$.

REPLY [3 votes]: Edit: Answer reformulated
Hi Pavol,
While it's not true that you can extend the KZ connection on the compactified space, there is a well defined procedure to "specialize" it to get a connection on each boundary component. My claim is that these specializations are precisely obtained through application of operadic composition.
The precise story is as follows: De Concini--Procesi introduced a compactification $Y_n$ of the configuration space satisfying the following properties:

The complementary $K=Y_n\backslash C_n$ is a divisor with normal crossings
The KZ connection induces a meromorphic connection $\nabla_n$ on $Y_n$ with logarithmic singularities on $K$.

These properties means the following: let $D$ be a smooth irreducible divisor. Then there exists some local coordinates $z_1,\dots,z_{n-1},t$ such that $D$ is defined by $t=0$, and such that
$$\nabla_n=d-\omega_D(z_1,\dots,z_{n-1},t)+A \frac{dt}t$$
where $\omega_D$ is holomorphic at $t=0$ and $A$ is a constant (the residue of the defining 1-form of $\nabla_n$). Now for any choice of a tangential basepoint inside $D$ the monodromy of $\nabla_n$ around $D$ is then given by $e^A$.
Now you can specialize $\nabla_n$ to a flat, meromorphic connection on $D$ by setting
$$\nabla_D:=d-\omega_D(z_1,\dots,z_{n-1},0)$$
It turns out that there is a natural identification $D=Y_{n-1}$, and that $D$ is the image of some operadic composition
$$\mu:Y_{n-1}\times Y_2 \rightarrow Y_n$$
Let
$$\mu:\mathfrak t_{n-1}\times \mathfrak t_{2}\rightarrow \mathfrak t_{n}$$
be the "same" operadic composition.
Now my precise attempt to answer your question is the following

Claim: $$\nabla_D=\mu(\nabla_{n-1})$$

Now you can repeat this process on $D=Y_{n-1}$ and get the compatibility with other operadic maps.
Example
Consider the KZ connection on $Y_4$ (obtained from the original KZ connection by setting $z_1=0,z_2=x,z_3=y,z_4=1$):
$$
\nabla_4=t_{12}d\log x+t_{23}d\log(x-y)+t_{24}d\log(x-1)+t_{13}d\log y+t_{34}d\log(y-1)$$
Let $D \subset Y_4$ be the divisor associated to the hyperplane $x=y$ (that is, to the hyperplane $z_2=z_3$ in $C_n$). Setting $t=x-y$ one gets:
$$\nabla_4=t_{12}d\log x+t_{13}d\log(x-t)+t_{24}d\log(x-1)+t_{34}d\log(x-t-1)+t_{23}d \log t$$
Then we can define a connection on $D$ it by setting $t=0$ in the holomorphic part of $\nabla_4$:
$$\nabla_D= t_{12}d\log x+t_{13}d\log x+t_{24}d\log(x-1)+t_{34}d\log(x-1)$$
i.e. 
$$\nabla_D= (t_{12}+t_{13})d\log x+(t_{24}+t_{34})d\log(x-1)=\nabla_3^{1,23,4}$$<|endoftext|>
TITLE: Double coset decomposition of symplectic group over a quadratic extension
QUESTION [8 upvotes]: I'm trying to understand the double coset decomposition of $G(F)\setminus G(E)/K_E$ , where $G = \mathrm{GSp}_{2n}$ is the rank $n$ group of symplectic similitudes, $E/F$ is a quadratic extension of $p$-adic fields and $K_E=G(\mathcal{O}_E)$ is the ring of integers in $E$.
My knowledge of buildings is limited at present, but as I understand it, $G(E)$ acts transitively on the vertices of the affine building over $E$, and $K_E$ is the stabilizer of a hyperspecial point, so the quotient $X(E) = G(E)/K_E$ can be identified with the vertices of this building.
If this is correct, then the double coset decomposition becomes a question about orbits of $G(F)$ on $X(E)$.  

What is known about the orbits of $G(F)$ on $X(E)$?

Specifically, I would like to find representatives that give some idea as to the geometry of the situation.  (I.e., Sage or Matlab might be able to find coset representatives, but I would still have no idea why these are the representatives.)
(I'm reading through Garrett's book on buildings and trying to work through Tits' article in Corvallis, but any other direction for sources would be appreciated as well.)

REPLY [4 votes]: This question is answered for a large number of reductive groups (including the symplectic)  in P. Delorme and V. Sécherre, "An analogue of the Cartan decomposition for p-adic reductive symmetric spaces" (arXiv, MSN, published).<|endoftext|>
TITLE: Perverse vs real formality?
QUESTION [5 upvotes]: Let $\cal A$ be an abelian category, say linear over a field, with enough injectives and $\cal P$ be the heart of a t-structure on the bounded derived category $D^b(\cal A )$. Assume that $\cal P$ also has enough injectives. Suppose that the realization functor 
$$real:D^b(\cal P)\rightarrow D^b(\cal A)$$
is an equivalence. Now given two objects corresponding through this equivalence, are their Ext-algebras $A_\infty$-quasi-isomorphic?
Edit: If I understand it correctly the realization functor is constructed in BBD as follows:

They take the homotopy category of injective complexes of objects in $\cal A$, equipped with descending filtration, whose filtration steps lie in $\cal P$. This category is called $DF_{bete}$. 
Using the boundary map of the triangles $gr^{i+1}F\rightarrow F^{i-1}/F^{i+1} \rightarrow gr^{i} F$ they construct a functor to the category of complexes $DF_{bete}\rightarrow C^b(\cal P)$ and show that it is an equivalence. 
Forgetting of the filtration on $DF_{bete}$ translates to a functor $C^b(\cal P)\rightarrow D^b(\cal A)$. The derived functor of this is the realization functor.

REPLY [3 votes]: OK, I still doubt that if you divide out chain homotopies in (1) in the category of filtered complexes, you can get the category of complexes (and not the homotopy category) as a target in (2). I also doubt that you get an equivalence in this way unless you invert something in source and target. Sorry that I'm too lazy to look now at BBD.
Suppose that for any bounded complex $X$ in $\mathcal{P}$ you can construct a filtered complex of injectives $F$ in $\mathcal{A}$ such that each $F^{n}/F^{n+1}$ is an injective object in $\mathcal{P}$ and $gr^{\ast}(F)=F^{\ast}/F^{\ast+1}$ equipped with the differential $\delta$ obtained as in (2) is a bounded below complex quasi-isomorphic to $X$ in $C(\mathcal{P})$. You can probably check in BBD that all this is possible (at leas under reasonable assumptions).
Consider the following zig-zag of DG-algebra morphisms:
$\operatorname{End}_{C(\mathcal{P})}(gr^{\ast}(F),\delta)\leftarrow 
\operatorname{End}_{\mbox{filtered}}(F)\rightarrow
\operatorname{End}_{C(\mathcal{A})}(F)$
By definition, the $\operatorname{Ext}$ algebra of $X$ is the cohomology of the DG-algebra on the left. According to (3), the $\operatorname{Ext}$ algebra of $real(X)$ is the cohomology of the DG-algebra on the right. Moreover, the $A$-infinity structures on these $\operatorname{Ext}$ algebras are obtained from these DG-algebras by the standard transfer procedure, choosing bases of cohomology vector spaces and representing cocycles for the elements in these bases. By definition, the transfer maps are $A$-infinity quasi-isomorphisms between the $\operatorname{Ext}$ $A$-infinity algebras and these DG-algebras.
Your claim in (2) should instead say, I think, that the functor you sketch induces an equivalence after inverting quasi-isomorphisms on the right, and something appropriate on the left (filtered quasi-isos?). Therefore $\leftarrow$ above is a quasi-isomorphism. Moreover, since you assume that $real$ is fully faithful, $\rightarrow$ above is also a quasi-isomorphism. Now you can get your desired $A$-infinity quasi-isomorphism between the $\operatorname{Ext}$ $A$-fininity algebras by inverting and composing.<|endoftext|>
TITLE: Higher computability : Constructive ordinal and $\Delta^1_1$ predicates
QUESTION [5 upvotes]: Everything I know on this subject comes from Sacks book : "Higher recursion theory"
Let $\mathcal{O^Y}$ be the set of codes for ordinals constructive in $Y$.
We should have the result that $A \subseteq \omega \times 2^\omega$ is $\Delta^1_1$ iff $\exists a \in \mathcal{O}\ \ \exists e \in \omega$ such that $A(n, Y) \leftrightarrow \varphi_e^{H_a^Y}(n) \downarrow$ where $H_a^Y$ is the $|a|$-th iteration of the turing jump of $Y$.
Two things are now in contradiction in my mind :
The first one :
$X$ is $\Delta^1_1(Y)$ iff $\exists a \in \mathcal{O^Y}\ \ \exists e \in \omega$ such that $n \in X \leftrightarrow \varphi_e^{H_a^Y}(n) \downarrow$. Potentially we can have $|a| \geq \omega_1^{ck}$ if $\omega_1^{Y} \geq \omega_1^{ck}$
The second one :
$X$ is $\Delta^1_1(Y)$ iff then there exists a $\Delta^1_1$ predicate $A \subseteq \omega \times 2^\omega$ and $\exists a \in \mathcal{O}\ \ \exists e \in \omega$ such that 
$n \in X \leftrightarrow A(n, Y) \leftrightarrow \varphi_e^{H_a^Y}(n) \downarrow$
This time, the code $a$ for the constructive ordinal is always smaller than $\omega_1^{ck}$.
Can anyone see where I made a mistake ?
Thanks in advance

REPLY [2 votes]: I finaly got an answer from another forum.
The answer is simple, I was assuming that if $A \subseteq \omega$ is $\Delta^1_1(Y)$ it means that there is a $\Pi^1_1$ predicate $F \subseteq \omega \times 2^\omega$ and a $\Sigma^1_1$ predicate $E \subseteq \omega \times 2^\omega$ such that $\forall n\ \ A(n) \leftrightarrow F(n, Y) \leftrightarrow E(n, Y)$ and $\forall X\ \ \forall n\ \ F(n, X) \leftrightarrow E(n, X)$. But it does just mean that $\forall n\ \ F(n, Y) \leftrightarrow E(n, Y)$ which is indeed very different.<|endoftext|>
TITLE: "generic" in elementary submodels
QUESTION [10 upvotes]: Suppose you have a Suslin tree $T$ and you have a countable elementary submodel $M$ containing the usual "enough stuff" (including $T$). A comment in Todorcevic's Partition Problems in Topology indicates that a branch of $T$ of height $M \cap \omega_1$ will be generic, i.e., will meet every dense set of $T$ that's an element of $M$. Why? And is the result more general than for Suslin trees? I.e., does it use the fact that their branches are countable (so $M$ knows of no branch of height $M \cap \omega_1$)? Or does it hold for more general partial orders?

REPLY [5 votes]: In answer to Judy's second question, the fact that branches are all countable is not enough to get genericity of any branch of the same height as the elementary submodel. For example, consider the classic Aronszajn tree consisting of one-to-one functions from the ordinals to the integers. For any integer k the set of all functions with k in their range is an antichain. However, it is easy to construct a branch going to the top of an elementary submodel without k in its range, hence missing this definable antichain.<|endoftext|>
TITLE: Algebraic Geometry - Definition of a Morphism
QUESTION [8 upvotes]: What's the best (or your favourite) definition of a morphism of a quasi projective variety? I've seen a huge number of equivalent definitions, and I'd like to know which is the best to memorise! 
I'd prefer to have a definition that doesn't mention sheaves/schemes or rational maps (since I normally define a rational map as a morphism on an open set, and don't want my definitions to be circular).
Many thanks in advance!

REPLY [8 votes]: I confess I really like this question, because it troubled me a lot when trying to teach (and understand) beginning algebraic geometry.  There seem to be two questions here: 1) which is the more fundamental notion, morphism or rational map; and 2) what special definition is  preferred in the case of quasi projective varieties?  I finessed these difficulties for years by working largely in the category of complex analytic varieties and appealing to Chow's theorems, so I am not expert, but I have thought about it.  I will share my necessarily naive conclusions, so I can learn from the responses.
As to 1), the notion of rational map seems more fundamental except in the one case of polynomial maps of affine varieties.  I.e. if one wants the concept of regularity to be local, one must define it using rational functions.  This arises as soon as one defines a regular function on even a quasi affine variety.  So it appears as if regular functions are naturally a subclass of rational ones.  I.e. first one must understand rational functions, and the set on which such a function is regular.
If the varieties are embedded in projective space as closed sets, one could use the standard affine open cover and stick to polynomials, but if they are abstract or even quasi projective, it seems to require a notion of isomorphism, or of the sheaf of regular functions to define an affine open set.
Note Ruadhai's definition above requires the use of rational functions to define that of regular ones, after which the definition of morphism appears to involve only the concept "regular".
In case 2) we are dealing with projective space, which has two representations, as a space with a natural affine open cover, but also as a space which is a group quotient of an open subset of affine space.  Hence there are two natural definitions of regular map, locally regular in terms of the affine cover, but also as a regular map of the big affine space which is constant on orbits. the latter definition yields a map by homogeneous polynomials of the same degree.  This definition is the one most convenient for writing down examples.
Again, Ruadhai's definition of regular locally rational function uses the homogeneous representation, so that definition of morphism is an amalgam of abstract sheaf theoretic concepts and concrete projective ones.
The fully concrete version of Ruadhai's definition of morphism is thus: a map locally defined by homogeneous polynomials of the same degree with non vanishing denominators. This is the definition actually used in practice to write down morphisms of projective and quasi projective varieties.
Several basic books give both definitions (locally rational and regular for the affine open cover, and local homogeneous representation) in the quasi projective case, Shafarevich, Harris, and Reid, I believe.  This seems to me a helpful practice.
Bill Fulton's lovely little book on curves, which is intended to prepare one for schemes, uses only the abstract open cover approach as definition, but then leaves as exercise to the reader to check that such innocent maps as  [x:y]-->[x:y:z] from P^1 to P^2 are regular.
To summarize, it seems one cannot avoid local definitions, and if one wants to avoid fractions altogether, the only definition I know of a morphism of quasi projective varieties is as a map given locally by homogeneous polynomials of the same degree without common zeroes.
In my example above of a map from the intersection of two quadrics to a plane cubic the map [x:y:z:w] to [x:y:z] is undefined at [0:0:0:1].  By using the locally rational regular map definition in affine coordinates, it seems that point goes to [-1:0:1], as given by the hopefully equivalent map [w(y-w) : (x-z)(y-w) : w^2 ]. 
Edit:This is my attempt to understand Donu’s definition above.
Now I think I understand it, as well as why it is a benchmark, and yet not his favorite.  Of course I cannot speak for him.  One of its beauties as he said is that it does mimic closely the familiar chart/coordinate system definition from analytic and differential geometry.
Its shortcoming is perhaps its cumbersome nature.  E.g. it requires an infinite family of coordinate charts even to discuss regular functions on (open subsets) of affine space.  As to whether it enables one to dispense with rational as opposed to polynomial maps, well yes and no.  It is true that all regular maps are locally polynomial in terms of the charts, but the trick is that the charts themselves are defined by rational functions.
E.g. Suppose I want to glue two copies of A^1, along the complement of their origins, to get P^1.  Can this gluing be done by polynomial functions?  Well yes, as Karl said, but it depends somewhat on your point of view.  I may not think the regular function 1/z is a polynomial function on the set {z≠0} in A^1.  But to understand Karl’s point, I must realize that z is not an affine coordinate system for that subset, so I should not expect all regular functions there to be polynomials in z.  But the pair of functions (z,1/z) = (z,w) is an affine coordinate system there, and then 1/y is a polynomial in w, i.e. 1/y is a polynomial in the variable 1/y!  In this way one can make any regular locally rational function look locally polynomial.  Here are some details as I understand them.
Define an abstract variety as a topological space with a basis of open sets {Uj} such that each Uj is equipped with a homeomorphism fj:Uj-->Vj where Vj is a Zariski closed subset of some affine space.  Then require that every inclusion map Ui into Uj becomes a polynomial map of the corresponding affine varieties (fj)o(fi)^(-1):Vi-->Vj.
In this category Donu’s definition of morphism makes perfect sense.  I.e. a continuous map of abstract varieties is a morphism if it is locally polynomial in some collection of coordinate systems.  In particular a continuous k valued function is regular if and only if it is locally defined by polynomials in some coordinate cover. Nothing could be conceptually simpler or more natural.
What is the catch?  With this definition it is not immediately obvious that any familiar (non finite) example at all is an abstract variety, not even k ≈ A^1.  I.e. it takes an infinite number of affine coordinate charts even to give affine space itself the structure of an abstract variety.  Moreover these charts are defined by rational functions.
If we define a coordinate system in U as a finite set of regular functions such that every regular function in U is a polynomial in terms of those functions, then it is sufficient but not necessary to be affine in order to possess a coordinate system.  Fortunately every affine variety has a topological basis of affine open sets, thus we can put a structure of abstract variety on every quasi projective variety.  
The difference between the algebraic case and the analytic and differential cases is that the restriction of an affine coordinate system may not be an affine coordinate system.  Thus we cannot use the same coordinate system on every open subset of affine space, as we would in differential geometry.
Lemma:  The principal open subsets Uf = {f≠0} define a structure of abstract affine variety on affine space A^n.
Proof:  It suffices to use polynomials f which have no repeated prime factors.  Define the coordinate map Uf-->A^(n+1) by sending x-->(x,1/f(x)) = (x,T).  Then the image Vf = {1-f.T = 0} is a closed affine set.   The coordinate map itself is defined by regular rational functons on Uf, and is a homeomorphism. 
Moreover, Ug is contained in Uf if and only if  g = fh, for some polynomial h.  If we map Ug-->Vg by x-->(x,1/g(x)) = (x,W), then Vg = {1-g.W=0}.  Hence the inclusion map from Ug to Uf becomes in affine coordinates, the map (x,W)-->(x,W.h(x)) from Vg to Vf, a polynomial map in the coordinates (x,W).  QED.
With this lemma it seems one can use restrictions to define a structure of abstract variety on every quasi affine and quasi projective variety.
Finally, as Donu remarked, it is not obvious that one can check regularity using any coordinate cover.  I.e. there might be one cover by affine coordinate systems in which a given map is locally polynomial, and yet another cover by different affine coordinate systems in which it is not.  One must prove the usual theorem, via the nullstellensatz, that a locally polynomial function on an affine variety is globally polynomial.
This is a beautiful, conceptually natural point of view on what it means to be a morphism of varieties.  I would advocate, after giving this definition, proving that a map of quasi projective varieties is a morphism in this sense if and only if it has the property in the accepted answer above, if and only if it can be defined locally by sequences of homogeneous polynomials with no common zeroes.  I.e. it is hard to be prepared for all situations with just one characterization.
The last (homogeneous polynomial) point of view can then lead to the important idea that a morphism of a variety to projective space is also defined by a line bundle and a sequence of regular sections without common zeroes, an approach not yet mentioned.  I.e. maps of a variety to projective space are much more special than maps to arbitrary varieties, and this special case is well worth understanding.
If we want to discuss the meaning of the common zeroes of sections, as in the example of the intersection curve of two quadrics, note the restriction of a linear polynomial to this curve must vanish 4 times, so the image of the map defined by linear polynomials should have degree 4.  Since the image curve has degree three there must be a point where the rational map is not defined.  I.e. the line bundle on the intersection curve defining the morphism is O(1) restricted to the curve tensored with O(-p) where p is the point [0:0:0:1].  Since we extended the map using quadratic polynomials, each vanishing 8 times on the curve, presumably those polynomials have 5 common zeroes on our curve.<|endoftext|>
TITLE: Primes that are the sum of three squares
QUESTION [9 upvotes]: This is in some sense an extension of the earlier MO question, "Gaussian prime spirals."
Gaussian primes in the complex plane, $a+b i$, require $a^2 +b^2$ prime off the axes.
The generalization to quaternions leads to Hurwitz primes $a + b i + c j + d k$, which
require $a^2 + b^2 + c^2 + d^2$ prime (here I am ignoring the Hurwitz integers whose
components are in $\mathbb{Z}+\frac{1}{2}$).
So this suggests exploring points in $\mathbb{Z}^3$ with $a^2+b^2+c^2$ prime.
(I realize this ignores the nice algebraic properties of complex numbers and quaternions.)
The Gaussian prime spirals were created by walking along a lattice direction and turning 
left at Gaussian primes.  The generalization is to start at a point $p \in \mathbb{Z}^3$, walk
along a lattice direction until a point $(a,b,c)$ is hit with $a^2 +b^2 + c^2$ prime, and then 
turn.  It makes sense to iterate through the six lattice directions,
$(+x,+y,+z,-x,-y,-z)$, in that order.
The result seems again to be a closed cycle.
Here is one example, with $p=(30,40,10)$ (marked in red), 
which cycles after encountering 739 "primes" (marked in yellow):
         

It appears that whether or not every Gaussian prime spiral cycles runs up against
long-unsolved problems.
What is the situation with these 3D "prime" spirals?
Is there much known about the density and distribution of primes that are
are sum of three squares?
Are there theorems or conjectures that would either imply all spirals are closed,
or the opposite, that there should exist infinite unclosed spirals?

REPLY [9 votes]: From one direction, everything is known about the resulting primes. All primes 
$$  p \equiv 1,3,5 \pmod 8  $$ are the sum of three squares, so is $p=2,$ while no numbers
$$  n \equiv 7 \pmod 8  $$ are ever the sum of three squares.
However, your construction involves fixing two coordinates, say $x=a, y=b,$ then varying $z$ in either direction and hoping to find another prime. This is still unsure, for example if 
$a^2 + b^2 = 1,$ nobody knows for sure that there are infinitely many primes of the form $1 + z^2.$
For what it may be worth, Siegel's theorem gives an exact answer for the number of representations of a given number $n = x^2 + y^2 + z^2$ because the form is alone in its genus. I also gave a version in the style of Guass at Is there a simple way to compute the number of ways to write a positive integer as the sum of three squares?  Meanwhile, following Linnik, the main result in Duke and Schulze-Pillot (1990) is that the lattice points on the sphere are asymptotically equidistributed.  
About Guass, I had a girlfriend who had taken Latin, and expressed a desire to write a murder mystery with hero Dexter, while only in the last few pages do we find out about the evil twin Sinister.  The class had a motto, Semper Ubi Sub Ubi or Always Wear Underwear.<|endoftext|>
TITLE: Counterexample in cohomology for symmetric spectra?
QUESTION [6 upvotes]: Symmetric spectra are a particular model for spectra, introduced by Hovey, Shipley and Smith. They have the nice property that they have a well-behaved smash product. Our interest in spectra comes from homotopy theory and hence we want to define a model category structure on it. Several model category structures are given in Hovey, Shipley and Smith. One model category structure that has the stable homotopy category as its homotopy category uses stable equivalences instead of the level equivalences, which one might naively guess are a good choice. 
These stable equivalences are supposed to be maps that induce isomorphisms on "all" generalized cohomology theories. However, they are defined using cohomology with coefficients in a particular class of spectra, the injective $\Omega$-spectra. The reason for picking this particular class is given earlier in the article:

It would be nice if the 0th cohomology group of the symmetric spectrum $X$ with coefficients in the symmetric $\Omega$-spectrum $E$ could be defined as $\pi_0 Map_{Sp^\Sigma}(X,E)$, the set of homotopy classes of maps from $X$ to $E$. But, even though the contravariant functor $E^0 = \pi_0 Map_{Sp^\Sigma}(-,E)$ takes simplicial homotopy equivalences to isomorphisms, $E^0$ may not take level equivalences to isomorphisms. ... We introduce injective spectra as a class of spectra $E$ for which the functor $E^0$ behaves correctly.

I'm trying to understand this problem and I think it would be helpful to see an example. So my question is: what is an example of a level equivalence $f: X \to Y$ of symmetric spectra and a symmetric $\Omega$-spectrum $E$ such that $f$ does not induce an isomorphism on $E^0$?

REPLY [7 votes]: The core problem is: Maps of symmetric spectra are, levelwise, maps equivariant with respect to the symmetric group, but the notion of weak equivalence ignores that.
The injectivity notion essentially works because there is a (simplicial) model structure on symmetric spectra where

weak equivalences are levelwise weak equivalences,
cofibrations are levelwise monomorphisms, and
fibrant objects are "injective" spectra.

This is sometimes called the injective level model structure.  This ensures that $Map_{Sp^\Sigma}(X,E)$ has a sensible homotopy type, making $\pi_0$ invariant under levelwise weak equivalence.  The $\Omega$-spectrum condition is not as relevant.
However, you've asked for an example, and so we have to come up with one.  The problems with doing so are as follows.
There is more than one way to skin this particular cat.  There are at least three different level model structures on symmetric spectra (see Stefan Schwede's book project, chapter III, for coverage of them), and if we try to come up with an example we need to have one where the mapping space has an incorrect homotopy type for all of them.

We need $E$ to be an $\Omega$-spectrum.  Since this implicitly involves loop spaces, I want to stick to an example where $E$ is levelwise fibrant (and hence is fibrant in the projective model structure).  So we need an example where the domains are not projective-cofibrant, which boils down to a condition about the symmetric groups not acting sufficiently freely.
I am lazy, and want to construct the domain spaces using a "free" functor so that I can analyze the mapping spaces easily.  There are the functors $G_n$ described in Hovey-Shipley-Smith's paper, which are left adjoint to the forgetful functor $X \mapsto X_n$ from symmetric spectra to $\Sigma_n$-spaces.  Objects constructed in this way are always cofibrant in the flat level model structure, so we need the target $E$ to not be fibrant in the flat level model structure.  This is a specific condition on what the fixed-point subspaces of $E_n$ look like.
Again in the "laziness" department, you've asked for an example where $E$ is an $\Omega$-spectrum, and these are more difficult to write down.


Thus, here is a recipe for taking some injective $\Omega$-spectrum $E$ and constructing a new one, $E'$, in a way ensuring that $E^0$ doesn't take weak equivalences to isomorphisms.
Let $X \to Y$ be the map
$$G_n((E\Sigma_n)_+ \wedge S^k) \to G_n(S^k),$$
induced by the projection, where $S^k$ is given the trivial action of the symmetric group.  One checks by the explicit formulas for $G_n$ that the map $X \to Y$ is a levelwise equivalence.  For any symmetric spectrum $E$, the map $E^0(Y) \to E^0(X)$ is
$$\pi_0 Map_{\Sigma_n,*}(S^k, E_n) \to \pi_0 Map_{\Sigma_n,*}((E\Sigma_n)_+ \wedge S^k, E_n)$$
which is
$$\pi_k ({E_n}^{\Sigma_n}) \to \pi_k Map_{\Sigma_n}(E\Sigma_n, E_n).$$
If $E$ is injective, this is an isomorphism by assumption.
Let $$E_n' = E_n \wedge (E\Sigma_n)_+.$$
Here the symmetric groups act diagonally, and the structure maps are induced by the structure maps of $E$ and the functorial maps $E\Sigma_n \to E\Sigma_{1+n}$.  Then $E'$, levelwise, has no fixed points, and the map $(E')^0 (Y) \to (E')^0 (X)$ is
$$
0 \to \pi_k Map_{\Sigma_n} (E\Sigma_n, (E\Sigma_n)_+ \wedge E_n) \cong \pi_k Map_{\Sigma_n} (E\Sigma_n, E_n)
$$
by freeness.  If $E$ was injective in the first place, the right-hand side is equivalent to $\pi_k E_n$.  If $E^*$ is not trivial, we can pick some $k$ and $n$ making this nonzero.<|endoftext|>
TITLE: Where does the canonical basis differ from the KLR basis?
QUESTION [14 upvotes]: The question implicitly asked in Ben Webster's question is: Does the canonical basis of Uq(n+) agree with the basis coming from categorification via Khovanov-Lauda-Rouqier algebras?
Thanks to Shunsuke Tsuchioka's answer to the very same question, the answer is No.
One can now ask for more, and ask for an explicit example where these two bases disagree. So that's what I am going to do here. 
In case someone brings it up, I'm disqualifying Example 3.25 of Khovanov-Lauda I on the grounds that it doesn't count (the algebra isn't quite defined "correctly" in characteristic zero (it doesn't match the geometry), and weird stuff is to be expected outside of finite type for KLR algebras in positive characteristic). I'd prefer to work in finite type please.
Of course Tsuchioka's answer provides us with an upper bound on where to look.

REPLY [3 votes]: Since this question was first asked, nontrivial decomposition numbers for Quiver Hecke algebras (=KLR algebras) have been discovered in finite type. Thus there is an alternative family of examples where the canonical and KLR basis disagree, if one is willing to accept ones algebras living over a field of positive characteristic.
Known examples are not always explicated in the literature right now. The historically first example is due to Geordie Williamson - see http://arxiv.org/abs/1212.0794 or http://arxiv.org/abs/1210.6900.<|endoftext|>
TITLE: Can the set of iso classes of G-equivariant H-bundles be given by ordinary homotopy classes of non-equivariant maps?
QUESTION [5 upvotes]: Let $G$ be a (nice enough) topological group (actually a filtered colimit of compact Lie groups), and let $X$ be a manifold with an action (a proper one in fact) by a Lie group $H$. Let $X//H := (X\times EH)/H$ be the Borel construction. 
Does the following claim appear in the literature:

Claim: The set of isomorphism classes of $H$-equivariant principal $G$-bundles is given by $[X//H,BG]$

For the case when $H$ is a compact Lie group and $G$ is an abelian Lie group, then this is proved in a 1983 paper by Lashof, May and Segal,
Equivariant bundles with Abelian structure group, Contemporary Math 19 (1983) 167--176. (pdf)
And some related cases are treated by May in
Some remarks on equivariant bundles and classifying spaces, Asterisque 191 (1990) 239--263 (pdf)
The claim seems intuitively true, but I can't seem to be able to find it, so my intuition may be off. If it is true, then surely May would have mentioned it (perhaps he does and for some reason I missed it).
EDIT: May also treats the topic in chapter VII of the book Equivariant homotopy and cohomology theory, but it seems a compilation of older work.
Another way to view the question is to ask if the categories of H-equivariant G-bundles on X and ordinary G-bundles of X//H are equivalent. There is some work in Lashof-May-Segal where they refrain from assuming G is abelian (there called A, however), I'm digesting it at the moment.
Also note I'm not asking for a classifying space for equivariant bundles - this is a much different question.

REPLY [9 votes]: That guy that keeps getting mentioned here never claimed it in general because he does not
believe it in general.   Think about $G=U(n)$ and about equivariant $K$-theory.  This
is very close to the Atiyah-Segal completion theorem.   The result for Abelian structure
groups still seems somewhat surprising to him.<|endoftext|>
TITLE: Recognizing the 4-sphere and the Adjan--Rabin theorem
QUESTION [15 upvotes]: The problem of recognizing the standard $S^n$ is the following:
Given some simplicial complex $M$ with rational vertices representing a closed manifold,
can one decide (in finite time) if $M$ is homeomorphic to $S^n$. 
For $n=1$, this is obvious,
and for $n=2$,
one can solve it by computing $\chi(M)$.
A solution for $n=3$ is due to 
J.H. Rubinstein. An algorithm to recognize the 3-sphere. In Pro-
ceedings of the International Congress of Mathematicians, vol-
ume 1, 2, pages pp. 601–611, Basel, 1995. Birkhäuser.
By a theorem of S.P. Novikov,
the problem is unsolvable if $n\geq 5$.
The idea is the following: By the Adjan--Rabin theorem,
there is a sequence of super-perfect groups $\pi_i$ for which the triviality problem is unsolvable. 
Now construct homology spheres $\Sigma_i$ with fundamental groups $\pi_i$.
If one can decide which of the $\Sigma_i$ are standard spheres,
then one can solve the triviality problem for the fundamental groups.
Question: Is the recognition problem for $S^4$ solvable?
The problem with this proof of S.P. Novikov's theorem is that there is no result that asserts that for any given super-perfect group $\pi$ there is a homology $4$-sphere satisfying $\pi_1(\Sigma) = \pi$.
However,
Kervaire has proved that every perfect group with the same amount of generators and relators may be realized as the fundamental group of a homology $4$-sphere.
Thus the question:
Is there an improved Adjan--Rabin theorem that asserts the existence of a sequence of perfect groups $\pi_i$ with the same amount of generators and relators,
the triviality problem of which is unsolvable?

REPLY [8 votes]: As mentioned algorithmic 4-sphere recognition is an open problem.    Since Rubinstein's solution to the 3-sphere recognition problem is so simple and elegant, perhaps the first thing you might guess is, why not try those techniques in dimension 4?  Normal surfaces, crushing normal 3-spheres, searching for almost-normal 3-spheres.  
That theory is still in its infancy.  Rubinstein and his former student Bell Foozwell have been developing normal co-dimension one manifold theory in triangulated manifolds.   They have a "normalization" process that follows Rubinstein's general normal/almost-normal schema but it appears to do a fair bit of damage to the manifolds, so it's not clear to me if anything like this could eventually be used for 4-sphere recognition, but maybe some creative variant of the idea will work-out. 
Another closely-related problem would be an algorithmic Schoenflies theorem, to determine if a normal 3-sphere bounds a ball.<|endoftext|>
TITLE: Etale topos as a classifyng topos ?
QUESTION [7 upvotes]: Hello !
If $X$ is a scheme, we can consider the etale topos of $X$ whose object are etale scheme above $X$ with the etale topology.
My question is : is there a know way to express this topos as the classifying topos of some geometric theory ? Of course it is possible, just because it's a grothendieck topos, but I'm looking for an explicit theory at least on some particular case (like when $X$ is affine, or when $X$ is the spectrum of the ring of integer of a number field, or when $X$ is a projective curve over a finite field... )
For example, if $A$ is a ring, then the Zariski topos of $Spec A$ (topos of finite presentation scheme above $Spec A$ with the Zariski topology) is the classifying topos of the theory of local $A$ algebra. (the universal local $A$ algebra being the structural sheaf).

REPLY [7 votes]: I cannot give the details, but my guess is that the etale topos should be the classifying topos of the theory of strictly local $A$ algebras. By a strictly local $A$ algebra I mean a henselian local algebra with separably closed residue field. I don't know if this is a honest algebraic theory.
Bonus: in this vein the Nisnevich topos should be the classifying topos of the theory of henselian local $A$ algebras. The proof should follow similar lines to the previous one.<|endoftext|>
TITLE: Willmore minimizers for genus $\geq 2$
QUESTION [7 upvotes]: For an immersed closed surface $f: \Sigma \rightarrow \mathbb R^3$ the Willmore functional is defined as
$$
\cal W(f) = \int _{\Sigma}  \frac{1}{4}  |\vec H|^2   d \mu_g,
$$
where $\vec H$ is the mean curvature vector in $\mathbb R^3$and $g$ is the induced metric.
If $\Sigma$ is closed we have the estimate
$$
\cal W(f) \geq 4 \pi
$$
with equality only for $f$ parametrizing a round sphere. 
Recently, the Willmore conjecture was proved (the paper can be found on arxiv), which states that for closed surfaces $\Sigma$ of genus $g \geq 1$ this estimate can be improved:
$$
\cal W(f) \geq 2 \pi^2
$$
with equality only for the Cilfford torus. 
Are there any conjectures about the minimizers in the case of genus $g \geq 2$? And what happens if we consider surfaces immersed in some $\mathbb R^n$ instead of $\mathbb R ^3$?

REPLY [3 votes]: I remember there is a paper by Kusner named: comparison surfaces for the Willmore problem in which the author 
conjectured that the Lawson surface(see Sebastian's answer) minimizes the Willmore energy of genus g surface. For surfaces immersed in R^n, it is also conjectured the Clifford torus should be the minimizer, but it seems to me that this is still an open question.<|endoftext|>
TITLE: Two versions of "absolutely ccc"
QUESTION [17 upvotes]: I have recently been slogging my way through Shelah's "Large continuum, oracles".  Essentially from the start there has been a question needling me which I cannot seem to answer.

In the paper, Shelah says that a forcing notion $\mathcal{P}$ is absolutely ccc if it remains ccc after forcing with any ccc notion.
Elsewhere, I have seen it defined that a forcing notion $\mathcal{P}$ is absolutely ccc if it remains ccc after any forcing.  (This would be indestructibly ccc from Bartoszyński-Judah.)

Any forcing having the Knaster property is absolutely ccc (in the strong sense), and MA$_{\aleph_1}$ implies that all ccc forcings have the Knaster property.  Thus, it is consistent that the two are equivalent.
Do these two versions of absolute ccc-ness provably coincide?

REPLY [7 votes]: Bartoszyński-Judah's definition is NOT what you say, it is in fact the same as Shelah's notion.  In Set Theory: On the Structure of the Real Line, they define "indestructibly c.c.c." on page 177:

"Recall that a forcing notion $\mathcal{P}$ is indestructibly ccc if for every forcing notion $\mathcal{Q}$ satisfying ccc, $V^\mathcal{Q} \vDash$ "$\mathcal{P}^V$ satisfies ccc."
 
Then they prove (theorem 3.5.26) that if $\mathcal{P}$ has the Knaster property, then it is indestructibly ccc, per this definition.  (One can also show that having the Knaster property is itself indestructible by ccc forcing.)
I have not seen elsewhere the claim that the Knaster property implies that the ccc is indestructible by any forcing.  Do you have a proof of this?
Tangentially, while I was looking in Bartoszyński and Judah's book, I found a notable mistake.  They claim (lemma 1.5.14) that $\mathcal{P} * \dot{\mathcal{Q}}$ has precaliber $\aleph_1$ iff $\mathcal{P}$ has precaliber $\aleph_1$ and $\Vdash_{\mathcal{P}} \dot{\mathcal{Q}}$ has precaliber $\aleph_1$.  This is false.  As noted in the comments on the original post, MA$_{\aleph_1}$ implies the ccc is equivalent to having precaliber $\aleph_1$.  So assume MA$_{\aleph_1}$, and let $\mathcal{C}$ be Cohen forcing, and $\dot{\mathcal{T}}$ be a $\mathcal{C}$-name for a Suslin tree (by Shelah).  So in $V$, the iteration $\mathcal{C} * \dot{\mathcal{T}}$ has precaliber $\aleph_1$, but $\Vdash_{\mathcal{P}} \dot{\mathcal{T}}$ does not have precaliber $\aleph_1$.<|endoftext|>
TITLE: Quantum group Uq(sl(2))
QUESTION [15 upvotes]: I'm looking at motivating the standard deformation of $U(\mathfrak{sl}(2))$.  As an algebra $U(\mathfrak{sl}(2))$ is generated by $X,Y$ and $H$ and subject to the relations $[X,Y] = H$, $[H,X] = 2X$ and $[H,Y] = -2Y$.  From $q$-analysis I have that an integer $n$ is deformed according to $[n] = \frac{q^{n}-q^{-n}}{q-q^{-1}}$.
This seems to motivate, at least partially, the means of deforming $U(\mathfrak{sl}(2))$.  That is, set $[X,Y] = [H] = \frac{q^{H}-q^{-H}}{q-q^{-1}}$ and then set $K = q^{H}$ and $K^{-1} = q^{-H}$.  But what motivates the relations $KX = q^{2}XK$ and $KY = q^{-2}YK$?

REPLY [11 votes]: Allow me the excuse to mention a cute way of constructing Uq(sl2) as a q-analogue. Presumably this is ahistorical, since I suspect that initially Uq(sl2) was constructed first over the formal disc, then later it was realised that it was definable over $\mathbb{Q}(q)$.
Anyhow, let S be a finite set. Consider a vector space with basis $\{[I]\mid I\subset S\}$. Define operators e,f,h on this vector space by
$$e[I]=\sum_{i\notin I}[I\cup\{i\}]$$
$$f[I]=\sum_{i\in I}[I\setminus \{i\}]$$
$$h[I]=(2|I|-|S|)[I].$$
You can check that this is a representation of sl2 (it's just a tensor power of the standard represenation in disguise).
Now q-analogise! How? Well we should replace the idea of a subset of a set with that of a subspace of a vector space (of course working over the finite field with q elements).
Let V be a finite dimensional vector space over $\mathbb{F}_q$. We should consider the space of all complex-valued functions on the Grassmannian Gr(V) of all subspaces of V. Well actually we'd really like something of dimension 2^n, so we should consider only those functions invariant under the action of a Borel subgroup.
Let F be the space of pairs (W,W') with $W\subset W'$ two subspaces of V satifying dim(W')=dim(W)+1. There are two obvious maps $p_1,p_2:F\to Gr(V)$.
Now we can define two operators on our space of functions on Gr(V), namely $E=(p_2)_!(p_1)^!$ and $F=(p_1)_!(p_2)^!$. Here p! is pullback of functions and p! means sum the function over the fibre. These are almost the generators of the quantum group we're after, to make the formulae match up nicely, we need to modify the definition of p! by multiplying it by a factor of q-d/2 where d is the dimension of the fibre summed over (a Tate twist).
Now you have two operators E and F, and when you compute their commutator, you naturally find another operator K such that $$EF-FE=\frac{K-K^{-1}}{q^{1/2}-q^{-1/2}}.$$
Computing KE=qEK and qKF=FK is now easy.
But even better, this approach gives more than just the product structure on the quantum group. You can also get the coproduct. I will be brief since I haven't checked all powers that q that appear. I haven't thought about obtaining the antipode.
Consider a short exact sequence of vector spaces 0-->V'-->V-->V''-->0. Given functions f' on Gr(V') and f'' on Gr(V''), one can define a function on Gr(V) by
$$(f'\otimes f'')(X)=f'(X\cap V')f''(X/(X\cap V')).$$
There is a reasonable chance this needs correction by a power of q. The point however is that this defines an isomorphism between Borel-equivariant functions on Gr(V) and the tensor product of Borel-equivariant functions on Gr(V') with Borel-equivariant functions on Gr(V''). (If you drop Borel-equivariant, then you're still getting a map, you're just not deforming a representation of sl2).
I leave it as an exercise to the interested reader to use this to deduce the coproduct formulae in Uq(sl2).<|endoftext|>
TITLE: Countably generated $\sigma$-algebras of ${\mathcal P}({\mathbb R})$ and choice
QUESTION [6 upvotes]: It is consistent with ${\sf ZF}$ that the reals are the countable union of countable sets. Since any countable set is Borel, it follows that in any such pathological universe, let's call it $W$, every set is Borel; this has come out here before. 
Since there is a countable basis for the Borel sets, it follows that in $W$ there is a countable family of sets of reals, such that the $\sigma$-algebra they generate is all of ${\mathcal P}({\mathbb R})$. I would like to see whether we can improve this slightly:

Suppose ${\mathbb R}=\bigcup_n A_n$, where each $A_n$ is countable. Can we further assert that, in addition, the $\sigma$-algebra the $A_n$ generate is ${\mathcal P}({\mathbb R})$? More precisely, can we modify the $A_n$ to a new family $A_n'$ of countable sets with union ${\mathbb R}$ and this generating property?

Asaf's answer shows I'm being (somewhat) sloppy in this formulation. So, does it make sense to ask for some intrinsic construction? (Something other than: We start with such and such countable family that generates the Borel sets, we code it into bits of the $A_n$, and then we are done. I confess I do not see immediately how to formalize "intrinsic", perhaps it makes no sense.)

REPLY [8 votes]: We can cheat in the following way:
Suppose $\lbrace A_n\rbrace$ is a countable collection of countable sets, let $\lbrace B_n\rbrace$ be an enumeration of the countable collection of open intervals with rational end points.
Now consider the family $\lbrace A_n\cap B_m\mid n,m\in\omega\rbrace$. It is a countable family of countable sets, and every open interval $B_k$ is the countable union $\bigcup_n(A_n\cap B_k)$. Therefore the $\sigma$-algebra generated by this family is the entire power set of the real numbers.

The reason we cannot talk much about the general case is that given a countable family of countable sets $\lbrace A_n\rbrace$ we can make them disjoint via $A'_n=A_n\setminus\bigcup_{k < n} A_k$. The $A'_n$ are still countable, disjoint and have the same union - namely $\mathbb R$.
Now the $\sigma$-algebra the disjoint family generates is exactly $\lbrace\bigcup_{i\in I} A'_i\mid I\subseteq\omega\rbrace$. Since there can only be countably many singletons in this family we cannot separate all the points of $\mathbb R$ from one another and therefore we cannot generate $\mathcal P(\mathbb R)$.
So if the $A_n$ were already disjoint we could not have generated $\mathcal P(\mathbb R)$ without some sort of modification as in the first part.<|endoftext|>
TITLE: Primes $ 1 + x^2 + y^2$
QUESTION [11 upvotes]: EDIT, Saturday 11:32 am, March 24: a complete answer for $4 + x^2 + y^2$ and generally $4 m^2 + x^2 + y^2$ for fixed $m$ is given in Friedlander and Iwaniec page 282, Theorem 14.8. We might also expect useful stuff in Harman. 
ORIGINAL: There is no inspiration involved with this. It just happened. Well, it is about a possible method of giving Joseph some prime spirals. For his birthday. (See "Primes that are the sum of three squares.")
I do not seem to know whether there are infinitely many primes of the form $1 + x^2 + y^2,$ for that matter $4 + x^2 + y^2,$ or $9 + x^2 + y^2,$ or $16 + x^2 + y^2.$ So that is the question, for fixed $a,$ are there infinitely many primes $a^2 + x^2 + y^2?$ 
It seems a very good bet. Every prime $p$ not with $p \neq 7 \pmod 8$ can be expressed as $x^2 + y^2 + z^2.$ Furthermore, we know that all $p \equiv 1 \pmod 4$ can be written as $x^2 + y^2,$ so that is one example of an infinite set. 
Furthermoremore, the count of numbers up to some large $N > 0$ that can be written as $a^2 + x^2 + y^2$ is asymptotically the same as the result with $a=0,$ namely 
$$  \frac{0.7642... \; \; N \;}{\sqrt{\log N}}   $$ 
Looking at some computer output, it does seem to be very easy to write a bunch of primes with fixed $a$ in $a^2 + x^2 + y^2.$ 
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./primes_1_x2_y2 | sort -n
           2           1           0           1
           3           1           1           1
           5           1           0           2
          11           1           1           3
          17           1           0           4
          19           1           3           3
          37           1           0           6
          41           1           2           6
          53           1           4           6
          59           1           3           7
          73           1           6           6
          83           1           1           9
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ 


jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./primes_1_x2_y2 | sort -n
           5           2           0           1
          13           2           0           3
          17           2           2           3
          29           2           0           5
          29           2           3           4
          41           2           1           6
          53           2           0           7
          89           2           2           9
          89           2           6           7
         101           2           4           9
         113           2           3          10
         149           2           1          12
         149           2           8           9
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$


jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./primes_1_x2_y2 | sort -n
          11           3           1           1
          13           3           0           2
          17           3           2           2
          19           3           1           3
          29           3           2           4
          41           3           4           4
          43           3           3           5
          59           3           1           7
          59           3           5           5
          61           3           4           6
          67           3           3           7
          73           3           0           8
          83           3           5           7
          89           3           4           8
         107           3           7           7
         109           3           0          10
         109           3           6           8
         113           3           2          10
         131           3           1          11
         137           3           8           8
         139           3           3          11
         139           3           7           9
         157           3           2          12
         173           3           8          10
         179           3           1          13
         179           3           7          11
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ 


jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./primes_1_x2_y2 | sort -n
          17           4           0           1
          29           4           2           3
          41           4           0           5
          41           4           3           4
          53           4           1           6
          61           4           3           6
          89           4           3           8
          97           4           0           9
         101           4           2           9
         101           4           6           7
         113           4           4           9
         137           4           0          11
         173           4           6          11
         197           4           9          10
         241           4           0          15
         241           4           9          12
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$

REPLY [7 votes]: It is known that there are infinitely many primes of this form; see the references in this previous thread [corrected link -- that'll teach me to post late at night!]:
Primes represented by two-variable quadratic polynomials
We still don't have an asymptotic for the number of primes $p \leq X$ of the form $1+x^2+y^2$, but the order of magnitude ($\asymp X/(\log{X})^{3/2}$) is known.<|endoftext|>
TITLE: Square of a continuous map
QUESTION [13 upvotes]: Recently a student asked me the following (elementary looking) question :
If $T$ is an invertible linear transformation of some finite-dimensional space $E$ into itself which factorizes as $T = f \circ f $ where $f : E \mapsto E$ is continuous, must $T$ have positive determinant ?
Of course this is trivially true if $f$ is itself linear. It is also an easy exercise to show that this also holds when $f$ behaves locally like a linear transformation, that is, when it is $C^1$ : $T$ then factorizes as $T = df_{f(0)} \circ df_0 $, and since $x \mapsto \det df_x$ keeps a constant sign, we're done.
When $f$ is only continuous, this certainly still holds but I suspect this requires rather deep properties of continuous maps (unless I missed something obvious ...) with which I'm not very familiar. Hence two questions :
1) Is there an "elementary" proof of this ? (in which case I apologize for this question)
2) Does this property sound obvious to experts ? That is, is there some two-lines proof of this with a sufficient background ? If yes, what would be good references (books for example) to acquire this background ?

REPLY [25 votes]: The first relevant fact about $f$ is that it is a proper map. In such a situation the topological (Brouwer) degree of $f$ is well-defined, and by the product rule $\operatorname{deg}(T)= \operatorname{deg}(f\circ f)= \operatorname{deg}(f) \operatorname{deg}(f)$. For an invertible linear transformation, the topological degree is the sign of the determinant, which proves your claim.<|endoftext|>
TITLE: Are algebraic geometry error correcting codes (Goppa codes) "good" ? 
QUESTION [7 upvotes]: Question (informal version): Are algebraic geometry error correcting codes (V.D. Goppa codes) "good" ? 
Some details. There is certain construction of error-correcting codes by means of algebraic geometry,
originating from pioneering work by Russian mathematician Valerii Denisovich Goppa
(70-ies or early 80-ies ?). 
I wonder what is known about these codes: 
a) are they "capacity-achieving" b) are there some "low-complexity" soft-decoders, like belief propogation which complexity is  linear in the length of code c) are there some practical applications of these codes in error-correcting applications, if not - why ?  
PS
It is known that they are involved in McEliece cryptosystem, but it is crypto-application,
not error-correcting.

REPLY [3 votes]: AG codes for a given length n and alphabet q will beat corresponding turbo and LDPC but only over a channel with finite field alphabets (and over a complex channel if properly mapped and decoded properly). 
Firstly hamming metric is not the true metric of gaussian channels which are the real channels (remember Trellis coded modulation which naively achieves coding gain by proper mapping of constellations that ordinary algebraic codes could not). There is no good way known to map algebraic codes to complex constellations (an analogy would be smaller distance code words should be mapped far apart in the complex constellations.. there is no subexponential way to do this). So even turbo and ldpc inspite of their bad minimum distance properties have an advantage.
Secondly even if you have a good map complex constellations (say gaussian distribution), there is no efficient soft decoder or hard decision ML decoder for AG codes and the notion of MAP decoding is hard (computationally). That is why even turbo and ldpc perform better inspite of having just a suboptimal MAP decoder over a complex channel.
My intuition is AG codes should achieve capacity faster (that is for a given rate could use much shorter codes) than inferior turbo and ldpc. However there is no proof for this. I believe the situation is due to the complexity involved in providing an argument through the first point (namely you have to show how many points are closer than distance d from each other and how should one map them over complex channels and how will the gains scale up.... these all seem to hit a wall due to the formidable computing complexity involved for any given n).<|endoftext|>
TITLE: When Cayley graphs of the symmetric group wrt generating sets of transpositions are isomorphic?
QUESTION [8 upvotes]: Dear All,
I thought the following question might be well-known, but couldn't find anywhere, so decided to ask here:
Let $A$ and $B$ be two generating sets for $S_n$, consisting  of transpositions.
Question: When the Cayley graphs of $S_n$ with respect to $A$ and $B$ are isomorphic?
Well, if $\Gamma(S_n,A)$ is isomorphic to $\Gamma(S_n,B)$, then of course $|A|=|B|$. Is it true that the answer to the question is "whenever $A$ and $B$ are conjugate"?

REPLY [8 votes]: I claim the answer to your question is yes.
This is my first time posting on mathoverflow. I hope my latex goes ok.
Given $\Gamma(S_n,A)$, build an auxiliary graph $X(\Gamma(S_n,A))$, with vertex set $\{1,\ldots,n\}$ and two vertices are adjacent if the corresponding involution is in $A$.
Build a second auxiliary graph $Y$ with vertex set the elements of $A$ with an edge between them if they commute.
Note that $Y$ is the complement of the line graph of $X$.
Let $\Gamma_1=\Gamma(S_n,A)$ and let $\Gamma_2=\Gamma(S_n,B)$.
We have to show that if $\Gamma_1$ and $\Gamma_2$ are isomorphic, then so are $X(\Gamma_1)$ and $X(\Gamma_2)$. Since $X(\Gamma_1)$ and $X(\Gamma_2)$ are connected, they are isomorphic if and only if $Y(\Gamma_1)$ and $Y(\Gamma_2)$ are (assuming they have at least 4 vertices, see http://en.wikipedia.org/wiki/Line_graph#Characterization_and_recognition). It thus suffices to show that if $\Gamma_1$ and $\Gamma_2$ are isomorphic, then so are $Y(\Gamma_1)$ and $Y(\Gamma_2)$.
I will do this by showing that, given $\Gamma_1$ without labels, I can recover $Y(\Gamma_1)$ uniquely up to conjugacy in $S_n$.
The crucial observation is that in $\Gamma(S_n,A)$, an element at distance 2 from the identity is either a 3-cycle or a product of two disjoint transpositions. If it is a product of two distinct transpositions, then there will be exactly two paths of length 2 joining it with the identity (in other words, it will be contained in a unique $4$-cycle with the identity).
If it is a 3-cycle, there will be exactly either one or three paths of length 2 joining it to the identity.
First, label one vertex of $\Gamma_1$ "1" (think of it as the identity). 
Now, labels the neighbours of 1 with $x_1,\ldots,x_k$ (where $k$ is the valency of $\Gamma$).
We think of these as being undetermined transpositions.
By the argument above, $x_i$ and $x_j$ commute if and only if they are contained in a unique $4$-cycle with the identity.
We can now construct $Y(\Gamma_1$), in a unique way up to conjugacy in $S_n$.<|endoftext|>
TITLE: Can we define homotopy groups using Tannakian categories
QUESTION [13 upvotes]: This is another vague question. Hope you guys don't mind.
Let $T$ be a Tannakian category. For any fibre functor $F$ on $T$ we define the fundamental group of $T$ at $F$, denoted by $\pi_1(T,F)$, to be the tensor-compatible automorphisms of $F$. This fundamental group is representable by an affine group scheme.
Can one give a meaningful definition of homotopy groups $\pi_n(T,F)$ using the Tannakian formalism?

REPLY [16 votes]: There certainly is a notion of higher Tannakian category which would have meaningful higher homotopy groups. I'm not sure how much of the theory has been worked out already, but higher Tannakian duality is formulated for example in Conjecture 5.13 in this 2003 paper by Bertrand Toën, and is maybe proved by Jacob Lurie in section 5 of DAGVIII.
The higher analogue of Jan's example is the following: local systems of $\infty$-groupoids on a space $X$ are equivalent to representations of the fundamental pro-$\infty$-groupoid of $X$ (which is the homotopy type of $X$ if $X$ is a paracompact space homotopy equivalent to a CW complex).<|endoftext|>
TITLE: "Functors between monads": what are these really called?
QUESTION [22 upvotes]: Let $(S,\eta,\mu)$ be a monad on a category $C$, and $(T,\eta,\mu)$ a monad on a category $D$.  The following kind of gadget is ubiquitous: a functor $F:D\to C$, together with a natural map $\sigma: SF\to FT$, satisfying the identities

$F\eta = \sigma\circ \eta F$,  and
$\sigma\circ \mu F = F\mu\circ \sigma T\circ S\sigma$.

I might call this a "functor between monads".  Question: What do people actually call these, and what are the standard references?
Some random examples:

$S$ is the free commutative ring monad on abelian groups, $T$ is the free commutative monoid monad on sets, and $F$ is the free abelian group on a set.  (In this case, $\sigma$ is an isomorphism.)
$S=T=J$, the free associative monoid on pointed spaces (where we force the basepoint to be the unit element), with $F=$ loop space.  (Enriched category theory provides a lot of examples like this.)
If $F=$ identity functor, we recover the usual notion of morphism between monads.

Some notes: 

A functor between monads as above gives you a way to turn a $T$-algebra into an $S$-algebra, so you obtain a functor $F^*: \mathrm{Alg}_T\to \mathrm{Alg}_S$.  
You can compose "functors between monads", so there is a category with these as morphisms, and monads as the objects.

REPLY [2 votes]: Pumplün, D. Eine Bemerkung über Monaden und adjungierte Funktoren.
Math. Ann. 185 (1970). 329-337.
also defines a category of monads.<|endoftext|>
TITLE: Are Vitali-type nonmeasurable sets determinate?
QUESTION [6 upvotes]: Here, by a Vitali set, I mean the following.  Call $f_1,f_2:\omega\rightarrow 2$  tail-equivalent if  {$n| f_1(n)\not=f_2(n)$}$<\infty$.  Vitali sets (existence via AC) contain one such $f$ from each tail-equivalence class.  
Since any Vitali set $V$, as a subset of $2^\omega$, has inner measure 0 and positive outer measure, $V$ gives rise to an indeterminate game roughly of the following sort: Alice must name successive binary digits of a number $v$ in $V$ while Bob names some increasingly very thin open sets $U_i$ hoping that in the end, for some $i$, $v\in U_i$ ("$U_i$ captures $v$").
My question: does any $V$, in itself, define the payoff set of an indeterminate game?  (So now, a turn for either Alice or Bob consists of naming the next digit of one common number that Alice, but not Bob, would like to steer into $V$.)
(Easy observation: each tail-equivalence class contains many functions $f$ with $f(1)=0$, and a $V$ containing only these would give Bob a win as early has his first turn, thus making the game trivially determinate.  So one can't expect all such $V$ to define indeterminate games.  But how about all dense Vitali sets?)

REPLY [4 votes]: To answer the last (parenthetical) piece of the question, which Joel seems to have ignored: Density doesn't help; a dense Vitali set $V$ can still have an easy winning strategy for Bob.  The point is that, if $C$ is a tail-equivalence class and $B$ is any basic open set (the set of all $f\in 2^\omega$ that extend a particular finite sequence $b$), then there is an element $f\in C\cap B$ for which the odd-numbered terms $f(2n+1)$ ("Bob's moves") are not all zero --- just define $f$ to agree with $b$ where $b$ is defined, to have a 1 in some later odd-numbered position, and thereafter to behave in a way that puts it into $C$.  Now associate, arbitrarily, to each basic open set $B$, one tail-equivalence class $C_B$.  For each of these, pick one $f\in C_B\cap B$ as above; for all the other tail-equivalence classes (those not associated to any $B$) pick any $f\in C$ whose odd-numbered terms are not all zero.  Then the chosen $f$'s constitute a dense Vitali set, for which Bob has the winning strategy "always play 0".<|endoftext|>
TITLE: How many models of Peano arithmetic are isomorphic to the standard model and how many models of Peano arithmetic are non-standard?
QUESTION [6 upvotes]: I am currently writing a paper on non-standard models of Peano arithmetic and I am having trouble finding references or information that discuss the relative sizes of how many models of Peano arithmetic there are in the standard and the non-standard cases. 
I see it quoted all over the place that, "It is familiar that there are continuum-many pairwise non-isomorphic countable models of $\mathsf{PA}$". From this I take it that there are $\mathcal{c}$-many ($\aleph$-many) non-standard models of Peano arithmetic. Where can I find a proof of this fact? How many models of Peano arithmetic are there that are isomorphic to the standard model?
Thank you!

REPLY [14 votes]: Here is another way to do it. 
By the Gödel-Rosser theorem, there are continuum many distinct consistent completions of PA. One can see this by building a tree of finite extensions of PA, using the Gödel-Rosser theorem at each node to branch with the Rosser sentence or its negation relative to that theory (and also deciding the $n^{\rm th}$ sentence), so that every branch through the tree is a complete consistent extension of PA. Every such consistent completion of PA has a countable model. Since different complete theories cannot have isomorphic models, you get continuum many non-isomorphic countable model of PA. 
(Meanwhile, Andreas's answer applies not just to PA, but to any fixed theory, and so in fact, the compactness argument he mentions shows that each of these continuum many extensions of PA has continuum many non-isomorphic models.)<|endoftext|>
TITLE: Calculating the Perron-Frobenius eigenvector of a positive matrix from limited information
QUESTION [6 upvotes]: In the background of this question is a matrix $A$, all of whose elements are positive.  The Perron-Frobenius theorem tells us that the eigenvalue with largest absolute value is real, and that there is an associated dominant eigenvector, all of whose elements are positive.
Suppose we don't actually observe $A$, but are told what its first row sum is.  We're also told the first row sum of $A^{2}$, $A^{3}$, $A^{4}$, ... .  In other words, writing $e$ for the vector of ones, we're told the first element of $Ae$, $A^{2}e$, $A^{3}e$, and so on.  If, for example, $A$ is a stochastic matrix then $Ae = e$ so that the information given is simply a list of ones: $(Ae)_{1}=1$, $(A^{2}e)_{1}=1$, etc.
This information is enough to work out the dominant eigenvalue of $A$ via the power method: simply compute $\lim_{n \to \infty} \left( (A^{n} e)_{1} \right)^{1/n}$.
My question is:

Can anything at all be said about the dominant eigenvector?

REPLY [2 votes]: There is some very partial information you can obtain. See this recent paper:
Das, Kinkar Ch.
A sharp upper bound on the maximal entry in the principal eigenvector of
symmetric nonnegative matrix. (English)
Linear Algebra Appl. 431, No. 8, 1340-1350 (2009). ISSN 0024-3795<|endoftext|>
TITLE: A sequence of finite groups
QUESTION [6 upvotes]: Question: does there exist a strictly ascending sequence of finite groups 
$G_0<G_1<G_2<\dots $ such that for every $i \in \mathbb{N}$ there is $a_i \in G_{i+3}$ and the following two conditions are satisfied:
(i) $a_i$ centralizes $G_{i-1}$ in $G_{i+3}$;
(ii) $G_{i+3}=\langle G_i, a_i G_i a_i^{-1} \rangle$.
Remark: taking $G_i=Sym(d_i)$ for some increasing sequence of integers $(d_i)$ with the evident embedding $Sym(d_i) \hookrightarrow Sym(d_{i+1})$ does not seem to work.

REPLY [3 votes]: Apparently such sequences do exist. For example, one can take $G_i:=Alt(5^i)$ for $i=0,1,2,\dots$, where the embedding $\gamma_i: G_i \to G_{i+1}$ is the diagonal embedding of $Alt(5^i)$ into $Alt(5^{i+1})$ (i.e., if $\sigma \in Alt(5^i)$ then $\gamma_i(\sigma)(1)=\sigma(1)$,  $\dots$, $\gamma_i(\sigma)(5^i)=\sigma(5^i)$, $\gamma_i(\sigma)(5^i+1)=\sigma(1)$, $\dots$, $\gamma_i(\sigma)(2\cdot 5^{i})=\sigma(5^i)$,  $\gamma_i(\sigma)(2\cdot 5^i+1)=\sigma(1)$, $\dots$
For $i \in \mathbb{N}$, the element $a_i \in G_{i+3}$ can be chosen as a certain permutation of blocks of length $5^{i-1}$, which are the orbits of $G_{i-1}=Alt(5^{i-1})$ when it is considered as a subgroup of $G_{i+3}=Alt(5^{i+3})$. This clearly implies the condition (i). The argument that $G_i$ and  $a_iG_ia_i^{-1}$ generate $G_{i+3}$ is somewhat technical. Hopefully the details will appear in the second version of http://arxiv.org/abs/1203.3317.<|endoftext|>
TITLE: Duality of eta product identities: a new idea?
QUESTION [7 upvotes]: Looking at the collection of Eta Function Product Identities by Michael Somos, it seems like generally those identities come in pairs:
let's call two eta product identities $\sum\limits_{i=1}^r a_iP_i=0$ and $\sum\limits_{i=1}^r b_iQ_i=0$ dual if the eta products $Q_i$ can be re-labeled and the factors of each product re-arranged such that for all $i$, we can transform $P_i$ into $Q_i$ by replacing each argument $q^k$ or $k\tau$ of an eta function ($k\in\mathbb N$) by $q^{n/k}$ or $(n/k)\tau$. Note that the scalar coefficients $a_i$ and $b_i$ need not be the same, but they should be non-zero of course.
Example: The first two identities on the n=14 page, both of degree $12$, are
q14_12_44 = +1*u1^7*u2*u7^3*u14 +7*q*u1^3*u2*u7^7*u14 -1*u2^8*u7^4 -7*q^3*u1^4*u14^8 ; 
q14_12_64 = +1*u1^8*u14^4 +7*u2^4*u7^8 -56*q^2*u1*u2^3*u7*u14^7 -8*u1*u2^7*u7*u14^3 ;

re-arranged and written in a more humanly-readable form, putting $s_k=\eta(k\tau)$, we have
q14_12_44$\iff\qquad s_1^7s_2^{\ }s_7^3s_{14}^{\ } +7s_1^3s_2^{\ }s_7^7s_{14}^{\ } -s_2^8s_7^4 -7s_1^4s_{14}^8=0$
q14_12_64$\iff -56s_{14}^7s_7^{\ }s_2^3s_1^{\ }-8s_{14}^3s_7s_2^7s_1^{\ } +7s_7^8s_2^4+s_{14}^4s_1^8\,=0$
so both identities are duals of each other.

It should be easy to see that an eta product identity (edit: unless it is dual to itself) cannot have two (linearly independent) duals. Is that really easy to see?

An eta product identity can also be self-dual, e.g. the known linear identities mentioned in my recent MO thread or the one equivalent to the well-known theta identity $\theta_3^4(q) = \theta_4^4(q) + \theta_2^4(q)$, which becomes, when written in terms of $\eta$'s, $$s_2^{24}=s_1^{16}s_4^8 +16s_4^{16}s_1^8 .$$

This concept of duality seems so logical that I wonder: Has there already been any research on that? Has each eta product identity a dual counterpart, unless it is self-dual ?

In the collection mentioned above, for $n=22$ there is only one identity listed, and that one is not self dual, it is even highly asymmetrical. Its dual should have the label x22_19_124. Maybe that one is missing just because the computer search had to stop at some point.

Given that self-dual identities have a higher degree of symmetry: can they always be expressed in terms of Ramanujan's psi, phi, and/or chi functions?

I would think the answer is yes, as there are even some not self-dual ones that can be expressed in that way.
Edit concerning the first question: Even this may be trickier than expected. For example, there are two self-dual identities, labeled q6_14_36a and q6_14_36b. They are:
$s_1^9s_3^{\ }s_6^4 +\ \ 6s_1^4s_2^{\ }s_6^9    +2s_2^9s_3^4s_6^{\ }    -3s_1^{\ }s_2^4s_3^9=0$
$s_1^9s_3^{\ }s_6^4     +12s_1^4s_2^{\ }s_6^9   -4s_2^9s_3^4s_6^{\ }    +3s_1^{\ }s_2^4s_3^9=0$.
They contain exactly the same $\eta$-products and so can also be considered as dual to each other (and to linear combinations of them). Moreover, by linear combinations we may eliminate any one of the four terms, resulting in four 3-term identities with coefficients
 0  1 -1  1
-1  0 -8  9
 1  8  0 -1
-1 -9  1  0. 

Note that each of those has exactly one dual counterpart again, as they come in two dual pairs.
So the above conjecture of a unique dual doesn't necessarily hold if the given identity is already self-dual.

REPLY [5 votes]: This looks like the Fricke involution to me.  Given any positive integer $M$, define $W_M$ to be the operator given by slashing with the matrix 
$$\left(\begin{array}{cc} 0 & -1 \newline M & 0 \end{array}\right),$$
or, equivalently, if $f(z)$ is weight $k$ modular, by $f(z)\mid_k W_M = (-iz\sqrt{M})^{-k}f\left(\frac{-1}{Mz}\right)$.  This is the weight $k$, level $M$ Fricke involution.  It preserves $M_k(\Gamma_0(M))$, albeit changing the character.
Now, using the transformation properties of $\eta(z)$, in particular that $\eta(-1/z)=(-iz)^{1/2}\eta(z)$, it's possible to see that hitting any $\eta$-product with the appropriate Fricke involution takes it to another $\eta$-product, perhaps after scaling.  Thus, hitting an $\eta$-product identity with the Fricke involution produces another identity, which is moreover dual to the original.  At least the first example you give has this property, i.e., it follows from the Fricke involution.
As to the question of unique duals, you've already seen that things get wonky if you allow for self-dual identities.  But even if you require the identities to not be self-dual, things are still bad (although perhaps unsatisfyingly so).  Let $I_1$ and $I_2$ be two linearly independent, not dual to each other, and not self-dual identities.  Let $I_1^\prime$ and $I_2^\prime$ denote their duals, respectively (if you want, just take them to be images under the Fricke involution).  Consider now the identity $I=I_1+I_2$.  By your definition of dual, any of the identities $I^\prime=aI_1^\prime+bI_2^\prime$ will be dual to $I$, and so the dual-space is at least two-dimensional.  Modifying this in the obvious way suggests that the dimension of the dual-space can be arbitrarily large, with the requisite $I_1, \dots, I_n$ being of the form $I_1^aI_2^b$ varying over $a+b=n-1$, say.
I think the right way to phrase the uniqueness question is by asking whether there is a set of primitive $\eta$-product identities from which all other identities can be obtained, with the property that every dual of an identity is generated by Fricke involutions of the constituent primitive elements of the original identity.  In particular, if an identity is primitive, is its dual-space one-dimensional?  I don't know the answer to this question.<|endoftext|>
TITLE: Double log ratio of number of conjugacy classes to order of group in infinite simple group families (or their corresponding Schur covers, automorphism groups)
QUESTION [10 upvotes]: For a group G, denote by c(G) the number of conjugacy classes in G.
If $S_n$ denotes the symmetric group on n letters, then:
$$\lim_{n \to \infty} \frac{\log \log c(S_n)}{\log \log |S_n|} = \frac{1}{2}$$
[NOTE: Assume all logs are to the same base. It doesn't matter what the base is, as long as we use the same one in the numerator and denominator.]
[Proof that I know of: $c(S_n)$ equals the number of unordered integer partitions of n, and there exist asymptotic formulas for that which show it to be exponential in $\sqrt{n}$. $|S_n|$ is exponential in $n \log n$. Taking double logs, we get roughly $(1/2)\log n$ and $\log n + \log(\log n)$ respectively, and the quotient goes to $1/2$.]
Obviously, the result also holds for alternating groups instead of symmetric groups.
Similarly, if $GL(n,q)$ denotes the general linear group of degree n over a field of q elements, then for fixed q:
$$\lim_{n \to \infty} \frac{\log \log c(GL(n,q))}{\log \log |GL(n,q)|} = \frac{1}{2}$$
[Proof that I know of: the number of conjugacy classes can be expressed as a polynomial in q of degree n, and the group order is a polynomial in $q$ of degree $n^2$, so we get the result.]
The result holds if we replace the general linear group by the special linear group, projective general linear group, or projective special linear group (which is the Chevalley group of type A).
My question:

Does the result also hold for the groups in the other three infinite Chevalley families B, C, D, and in the other infinite families of simple groups?
If the answer to 1 is yes, is there some deeper reason why the result holds both in the symmetric/alternating group case and in the linear groups case? I imagine that the linear groups cases can be tied together using some general Lie group or algebraic group principles. If so, what are they? How do the alternating groups fit into the picture? May be something to do with Iwahori-Hecke algebras or the field of one element? All explanations are welcome.

REPLY [3 votes]: (1) is true, and follows from a paper of Liebeck and Shalev ("Character degrees and random walks in finite groups of Lie type" on Liebeck's webpage). They count representations according to their degrees rather than conjugacy classes, but it follows from their results that if $\Phi$ is a (reduced, irreducible) root system and G is a corresponding algebraic group (possibly twisted), then the number of conjugacy classes of $G(F_q)$ is roughly $q^{rk \Phi}$ (the lower bound is easy, just take conjugacy classes of elements in the torus). The limit of $$ \frac{\log \log q ^{rk \Phi}}{\log \log |G(\mathbb{F}_q)|}$$ as the rank of $\Phi$ tends to infinity is 1/2 also for types B,C, and D.<|endoftext|>
TITLE: Kazhdan's property T for Kahler surfaces
QUESTION [10 upvotes]: Is it true that the fundamental groups of compact Kahler surfaces have property T if and only if it they are finite? I am having trouble finding counterexamples to this, but maybe that's just me...

REPLY [8 votes]: According to this survey by Donu Arapura, Toledo proved that many arithmetic lattices in higher rank algebraic $\mathbb{Q}$-groups (with hermitian symmetric space) are fundamental groups of smooth projective surfaces.
In particular $Sp(2n,\mathbb{Z})$ for $n>2$, is such a group, and has property (T).
Note that once you get a group as fundamental group of a smooth projective variety you obtain a smooth projective surface with the same fundamental group by intersecting with some generic hyperplanes.<|endoftext|>
TITLE: When are all centralizers in a Lie group connected?
QUESTION [15 upvotes]: Let $G$ be a compact connected Lie group acting on itself by conjugation,
$$ G\times G\to G,\qquad (\sigma,h)\mapsto \sigma h \sigma^{-1}.$$
The fixed point set of a closed subgroup $H\le G$ equals the centralizer $C_G(H)$ of $H$ in $G$.
For a current project I am searching for examples of actions $G\times X\to X$ such that every fixed point set $X^H$ is connected, and I would like to fit the above conjugation action into this framework. Hence my questions:

Are there any conditions on a connected Lie group $G$ which ensure that the centralizer $C_G(H)$ of every closed subgroup $H\le G$ is connected?
Can someone give examples where this is true? I was thinking the unitary groups $U(n)$ might be candidates, since at least their centre $U(1)$ (which clearly is contained in every centralizer) is connected.

Thanks, and apologies if this is too vague or elementary.

REPLY [6 votes]: Centralizers of various types of subgroups, and their relationship to invariants like the cohomology of the classifying space $BG$ has a long history of study, at least going back to the papers from the 50s by Borel and others.
EDIT: Changed what's below to a more limited statement.
I don't know a reference to your exact question, but the following more limited statement is true:
Theorem: A compact connected Lie group G has the property (**) "For every elementary abelian $p$-subgroup $H \leq G$, $C_G(H)$ is connected" if and only if $G$ has torsion-free fundamental group and has a finite cover isomorphic to a product of copies of $SU(n)$, $Sp(n)$, $S^1$ for various $n \geq 1$.
In particular, if you in addition assume $G$ to be simply connected, the only examples are products of $SU(n)$ and $Sp(n)$ for various $n$.
Sketch of proof:
1) The condition is necessary: If (**) holds holds for all elementary abelian p-subgroups $H$ for all p, then every elementary abelian p-subgroup has to be contained in a maximal torus. By a result of Borel this implies that the fundamental group is torsion free. One easily checks case-by-case that the exceptional groups and Spin(n), $n \geq 6$ cannot be involved in $G$ since they have non-toral elementary abelian p-subgroup (which remain so even after modding out by a central subgroup). This leaves the cases above.
2) The condition is sufficient: We just have to check that the centralizer of an element of finite order of a group as above will again be a group of the same form, which I think you can do by hand.
--
Further info: The groups in the list above are exactly the ones whose classifying spaces have integral cohomology rings a finitely generated polynomial algebra. In fact any polynomial ring which occur as the cohomology ring of the space has to be of this form ---  this is the so-called Steenrod problem, whose solution is described in Andersen-Grodal: The Steenrod problem of realizing polynomial cohomology rings. J. Topology 1 (2008), 747-760. (DOI: 10.1112/jtopol/jtn021). 
Also, see Andersen-Grodal-Møller-Viruel The classification of p-compact groups for p odd. Annals of Math. 167 (2008), 95--210, in particular section 10+12 for a summary of the relationship between subgroups of $G$, their centralizers, and cohomology. Our setup is more general in that paper (we study p-compact groups), but it contains references to the papers of Borel etc.<|endoftext|>
TITLE: Direct summands of direct sum of line bundles on projective varieties
QUESTION [7 upvotes]: Serre-Swan's theorem (see the MO discussion) says that any locally free sheaf over an affine variety is a direct summand of a free sheaf. However, this is not true on projective varieties. It is not hard to check that a non-trivial line bundle with non-zero global sections can not be a direct summand of a free sheaf. The reason is that, being a direct summand of a free sheaf implies that the dual line bundle also has non-zero global sections. But that implies the line bundle is trivial. 
I am wondering if the same holds for vector bundles, i.e. if a vector bundle and its dual over a projective variety both have non-zero global sections, then the vector bundle is trivial. 
Another I think related question is the following:
Is a direct summand of a direct sum of line bundles on a projective variety also a direct sum of line bundles?

REPLY [4 votes]: 1) The vector bundle $E=\mathcal O(1) \oplus \mathcal O(-1)$ over $\mathbb P^1$ has non trivial sections and so has its dual (which happens to be the same bundle $E$ ).
 However  $E$ is non trivial  because each of its global sections has a zero.  
2) Yes, a direct summand of a direct sum of line bundles on a complete variety is also a direct sum of line bundles.
This follows from Atiyah's general version of the Krull-Schmidt theorem .    
Edit
Since Fei asks, let me remark that the trick in 1) works also for bundles which are not direct sums of line bundles:
If $T$ is the tangent bundle to $\mathbb P^2$, then $E=T \oplus T^* $ has non-trivial global sections , and so has  $E^*=E$
However, since $T$ is indecomposable   $E$  is not a sum of line bundles by Atiyah's result and is thus a fortiori not trivial.<|endoftext|>
TITLE: Interesting conjectures "discovered" by computers and proved by humans?
QUESTION [36 upvotes]: There are notable examples of computers "proving" results discovered by mathematicians, what about the opposite:

Are there interesting conjectures "discovered" by computers and proved by humans?

Possible example in graph theory is "Some Conjectures of Graffiti.pc (2004-07)," suggested by Joseph O'Rourke in another answer.
The question might not be well defined because "discovered" is controversial.
Added This question may be a duplicate (or refinement) of (2) in Experimental Mathematics as Kristal Cantwell pointed out.

I am mainly interested in examples where the program is designed to make conjectures which are not known identities to the program and later proved.

REPLY [7 votes]: In Automated conjecture making in number theory using HR, Otter and Maple, the HR program discovered the following conjecture:

Paraphrased,... if you take an integer and add up the divisors, then if the result is a prime, the number of divisors you have just added up will also be prime

And formally:

all a (isprime(sigma(a)) -> isprime(tau(a)))

Where sigma(a) and tau(a) are the sum of the divisors of a, and the number of divisors of a, respectively.
They then proved that this conjecture is true. To me this seems very interesting and non-obvious. But I have no idea if this theorem is significant, or if it was totally unknown before this program discovered it.
EDIT: I'm told that specific conjecture might not be particularly interesting to mathematicians. Here are some more conjectures the program discovered, from that paper:
It conjectured (and then the author proved) that sum of the divisors of square numbers is always odd, i.e.:

issquare(a) → odd(σ (a))

Probably not new either, but still cool that it was automatically discovered.
The program also discovered (invented?) a new concept called "refactorable numbers", and then made some surprising and interesting conjectures about this new concept:

HR invented the concept of refactorable numbers, which are such that the number of divisors is itself a divisor, e.g., 9 is refactorable, because 9 has three divisors (1, 3 and 9) and 3 divides 9.
...
This left us with 26 open conjectures, which we present in Appendix C. We have not yet fully investigated these remaining conjectures, and it seems likely that the majority may be false. Of particular interest to us are the conjectures about refactorable numbers: amongst others, HR made the conjectures that: (i) for even numbers, if σ (a) is refactorable, then τ (a) and σ (a) will be even, (ii) for odd numbers, if σ (a) is even and refactorable, then τ (τ (a)) and σ (τ (a)) will both be prime, (iii) if τ (a) is refactorable and τ (τ (a)) is prime, then σ (τ (a)) will also be prime, and (iv) if both σ (a) and σ (σ (a)) are refactorable, then
τ (σ (a)) will be refactorable and σ (τ (a)) will be odd.

The author has a number of other papers on this program and things it's discovered. I'm just now reading through Automatic Invention of Integer Sequences, where they claim it invented 17 novel integer sequences that were accepted into OEIS. Including the refactorable numbers mentioned above.<|endoftext|>
TITLE: Hot-topics in error correcting coding related to interesting math. ? 
QUESTION [12 upvotes]: What are topics in error-correcting coding which are related to interesting math. ?
I am primarely  interested in nowdays hot topics, but old days topics are also welcome. 
Let me try to mention what I heard about.
1) Hot topic in error-correction is finding LDPC codes with very low "error-floor" for code lengths dozens thoursands bits, this might be useful for optic transmission. However it is not clear for me what kind of math playing role here ? ("Error-floor" is related with codewords with small Hamming weight. So the code might be quite good - means majority of codewords have big Hamming weight, 
so in most case code performs well, but very small number having small Hamming weight will cause small number of  errors - it can be seen on the BER/SNR plot as a "floor".)
2) There is certain number of papers applying number theory (lattices in algebraic number fields) to consruct good codes.
One may see papers by F. Oggier, G. Rekaya-Ben Othman, J.-C. Belfiore, E. Viterbo:
e.g. this one : http://arxiv.org/abs/cs/0604093.
I am not aware how "hot" is this topic and how far it is from practical applications...
3) Polar codes is a hot topic. What kind of math is playing role here ?
4) Probably most classical example is the Golay code (1948) and sporadic simple  Mathieu groups.
Let me quote Wikipedia: http://en.wikipedia.org/wiki/Binary_Golay_code :
"The automorphism group of the binary Golay code is the Mathieu group . The automorphism group of the extended binary Golay code is the Mathieu group . The other Mathieu groups occur as stabilizers of one or several elements of W." By the way - is it occasional coincidence of there is something behind it ?

REPLY [4 votes]: For an additional answer, you might be interested in "permutation codes".
This is a warm/hot topic in coding theory with strong mathematical roots.<|endoftext|>
TITLE: Family of subsets such that there are at most two sets containing two given elements
QUESTION [8 upvotes]: We have a set $S$ with $k$ elements, a positive integer $n$, and subsets $S_1, S_2, \dots, S_n,$ each with $n$ elements. For any two elements $a,b$ of $S$, there are at most two sets $S_i$ containing both $a$ and $b$. Must $k$ be $\Omega(n^2)?$

If we require instead that, for any two $a,b$ there are at most one set $S_i$ containing both $a$ and $b$, then we have $k \ge \frac{(n+1)n}{2},$ because $S_i$ contains at least $n+1-i$ elements that are not in $S_1, S_2, \cdots, S_{i-1}.$ In this case, a special case of this is that if we have $n$ pairwise disjoint lines in the plane and a set of points such that each line contains at least $n$ points, we have at least $\frac{n(n+1)}{2}$ points.
One motivation for this question is the generalization of Szemerédi-Trotter to a family of curves satisfying 1) two curves intersect in at most m points and 2) for every two points, at most n curves go through both of those points. 
I'm trying to solve a graph theoretical problem assuming only an analogue of condition 2, and this is the easiest case of it. I expect that an extra condition is necessary, but I cannot find any obvious counterexamples.

REPLY [8 votes]: I think I may have an answer. Let $p$ be a prime (for simplicity). Let $r=p^3$ and let $T_1,\dots,T_r$ be all possible graphs of quadratic functions defined on the integers mod $p$. These graphs live in a set of size $n=p^2$, so $r=n^{3/2}$. Note that no two quadratic functions agree in more than two places, so $|T_i\cap T_j|\leq 2$ for every $i\ne j$. 
Now for each $k,l\leq p$ let $S_{kl}=\{i:(k,l)\in T_i\}$. If $q_i$ is the quadratic function corresponding to $T_i$, then $S_{kl}=\{i:q_i(k)=l\}$. Then $|S_{kl}|=p^2=n$ and there are $n$ of these sets. Each $S_{kl}$ is a subset of $\{1,2,\dots,r\}$ so lives inside a set of size $n^{3/2}$. And finally, the number of $S_{kl}$ that contain $i$ and $j$ is the number of pairs $(k,l)$ such that $q_i(k)=l$ and $q_j(k)=l$. But two distinct quadratic functions can agree in at most two places (since their difference has at most two roots).
Thus, it seems to me that the $n^{3/2}$ bound is the correct one and not $\Omega(n^2)$.<|endoftext|>
TITLE: What properties make $[0,1]$ a good candidate for defining fundamental groups?
QUESTION [109 upvotes]: The title essentially says it all. Consider the category $\mathfrak{Top}_2$ of triples $(J,e_0,e_1)$ where $J$ is a topological space, and $e_i \in  J$. There is an obvious generalization of the definition of homotopic maps. Suppose we have selected $(J,e_0,e_1)\in \mathfrak{Top}_2$. We could say that two continuous maps $f,g:X\to Y$ are "$J$-homotopic"  if there is a continuous map $h:X\times J\to Y$ such that $h(x,e_0) = f(x)$ and $h(x,e_1) = g(x)$. We could then define $\pi_1 (X,x)$ to be the set of continuous maps $f:J\to X$ satisfying $f(e_0)=f(e_1)=x$, with $J$-homotopic maps identified. Here  in order to define composition of paths in the naive way, we need to have picked some homeomorphism continuous map from $J$ to $(\{0\}\times J\cup \{1\}\times J)/((\{0\},e_1) = (1,e_0))$, taking $e_0$ to $e_0\times 0$ and $e_1$ to $e_1\times 1$. I have two questions:

Can $([0,1],0,1)$ be characterized as an object in $\mathfrak{Top}_2$ in a purely categorical manner?
When is $\pi_1 (X,x)$ a group? For that matter, when is $\pi_1 (X,x)$ associative?

Essentially, the question comes down to: what properties of $[0,1]$ are needed in order to do homotopy theory?

REPLY [4 votes]: You may be interested in a paper entitled "The Big Fundamental Group, Big Hawaiian Earrings, and Big Free Groups" by J. Cannon and G. Conner.  In this paper, they talk about "big intervals" and use them to define a "big fundamental group".  A "big interval" is a totally-ordered set (with the order topology) which is compact and connected (equivalently, a linear continuum with first and last point).  It is true that [0,1] is terminal in the category of (non-degenerate) big intervals.  (Which also means that it is initial since for big intervals I and J, a monomorphism $I\to J$ exists precisely when an epimorphism $J\to I$ exists).<|endoftext|>
TITLE: Atkin–Lehner operator for GL(3)?
QUESTION [9 upvotes]: Let $f$ be an automorphic form for $\Gamma_0(N)\subset SL(3,\mathbb{Z})$.
$\Gamma_0(N)=(a,b,c;d,e,f;g,h,i)\in SL(3,\mathbb{Z})|g=h=0(mod N)$
Is there any Atkin-Lehner operator for $\Gamma_0(N)$ which will give a functional equation for L-function of $f$?

REPLY [2 votes]: You will have to decompose
$$ Endo_{SL(3, \mathbb{Z})} ( Ind_{\Gamma_0(N)}^{SL(3, \mathbb{Z})} 1 ) .$$ 
This will give you the analog of the Atkin-Lehner theory. However, I have some doubts that this exists in the literature. This question of mine gives you the resaon why it has not been done for $d,g,h = 0 \bmod N$:
Parabolic induction GL(n,Zp)
I am happy, if somebody proves me wrong though.
Edit: I forgot to mention, that the case for $N$ square free is in general possible, since you can rely on the representation theory of reductive groups over residue fields.<|endoftext|>
TITLE: Good introduction to Morse-Novikov theory?
QUESTION [14 upvotes]: Morse theory investigates the topology of compact manifolds using critical points of real-valued functions $f\colon\, M\to \mathbb{R}$. Motivated by problems in dynamical systems, Novikov (Multivalued functions and functionals: An analogue of the Morse Theory) began to study Morse theory of closed $1$-forms, a special case of which is circle-valued Morse Theory (the subject of a very nice book by Pajitnov). The key new idea was to use the Novikov ring, a certain ring of formal power series, as a tool to count gradient flow lines of an induced Morse function on a covering space of $M$.
Novikov's ideas have been quite influential, but I have not been able to find a good introductory text on general Morse-Novikov theory just by googling keywords. My interests are in classical differential topology, in low dimensions. The reason that I want such a reference is that I work quite a bit with finite covering spaces (not necessarily abelian covers), and I'm curious as to whether or not Morse-Novikov theory gives me a better intuitive handle on what is happening than just plain old garden-variety Morse theory.

Reference Request: Can anyone recommend/ Does there exist a good introductory text on Morse-Novikov Theory?

REPLY [2 votes]: S. P. Novikov, An analogue of Morse theory for many-valued functions. Certain properties of Poisson brackets. Appendix I in Dubrovin, Novikov, and Fomenko, Modern Geometry-Methods and Applications Part III. Introduction to Homology Theory. This short paper was written for the student working through their textbook treatment of differential geometry.<|endoftext|>
TITLE: Intrinsic characterization of Soergel bimodules?
QUESTION [13 upvotes]: A Soergel bimodule (for $S_n$) is a bimodule over $R = \mathbb{Q}[x_1,\dots,x_n]$ which appears as a summand/grading shift of tensor products of the basic bimodules
$$B_{i,i+1} = R \otimes_{i,i+1} R$$
where $\otimes_{i,i+1}$ means the tensor product over the subring of polynomials invariant under permuting $i$ and $i+1$.  It follows immediately that every Soergel bimodule $M$ has the following properties:
(1) $M$ is free as a left module or as a right module, although not necessarily as a bimodule.
(2) $M$ commutes with invariant polynomials, in the sense that for every invariant polynomial $p \in \mathbb{Q}[x_1,\dots,x_n]$ and $m \in M$, we have
$$
pm = mp.
$$
I think they also have the following property:
(3) There is an invariant vector, an element $m_0 \in M$ so that
$$
x_i m_0 = m_0 x_i
$$
for every $i=1,\dots,n$.
Do these properties characterize Soergel bimodules?  Without the third condition, you could have, for instance, a bimodule that just permuted the $x_i$: a one-dimensional module with a single generator $a$ as a right module, so that
$
x_i a = a x_{\sigma(i)}
$
for some permutation $\sigma$.
Edit: The natural generalization for a general Weyl group $W$ would be to replace the invariant polynomials in (2) by the polynomials that are invariant under $W$.  Clearly all Soergel bimodules would still satisfy this generalization of (2).
Any references are welcome.  If it's not known, I'll try to prove it.
Edit: Ben Webster gave a counterexample below.  More generally, I'm still interested in some sort of intrinsic, elementary characterization.

REPLY [12 votes]: There is an intrinsic characterisation which is probably more complicated than what you are looking for. As Ben says, Soergel bimodules are pretty subtle things ...
Because Soergel bimodules are (finitely generated) $R$-bimodules one can think about them as coherent sheaves on $V \times V$ (where $V = Spec R$). Inside $V \times V$ one has for any $w \in S_n$ its reversed graph:
$Gr_w = \{ (wv, v) \;| \;v \in V \}$
Hence, given any subset of $U \subset S_n$ one can talk about "sections of an $R$-bimodule $M$ with support in $U$": those sections of $M$ which have support in the union of the graphs of all elements of $U$. In this way, for any subset $I$ of $W$ one can consider $\Gamma_I M \subset M$.
Your point (2) means that Soergel any Soergel bimodule satisfies $\Gamma_{S_n} M = M$ (that is, every element is supported on the union of all the graphs of elements of $S_n$). It follows that any Soergel bimodule has a canonical filtration indexed by the ideals of the poset $S_n$. A basic fact is that if one considers the quotient
$\Gamma_{\le w / < w} (M) := \Gamma_{\le w} M / \Gamma_{< w} M$
this is a free left $R$-module, isomorphic as a bimodule to a direct sum of copies of $R_w$ (the ``standard'' bimodule with normal left action and right action twisted by $w$). This is proved in Soergel's "Kazhdan-Lusztig-Polynome und unzerlegbare Bimoduln über Polynomringen" and is also discussed and generalised in my "Singular Soergel bimodules".
So now one can consider all bimodules which satisfy the above property. One this category one can put an exact structure: a sequence is exact if whenever one applies the functor $\Gamma_{\le w / < w}$ one obtains a split exact sequence of $R$-bimodules (necessarily isomorphic to direct sums of shifts of $R_w$'s).
Then the claim is that Soergel bimodules are the injective objects in this exact structure. I don't think this is written down anywhere. In the very similar language of moment graphs it is proved by Peter Fiebig in "Sheaves on moment graphs and a localization of Verma flags" here:
arxiv.org/abs/math.RT/0505108
(I might be mixing things up a bit. I think Peter considers the opposite filtration, which is why he gets projective objects. Anyway, if this is really what you're looking for then I can try to provide some more detail.)
By the way, the condition that the subsequent quotients in this filtration be split has other applications. In this paper
http://arxiv.org/abs/1205.4206
we examine when Rouquier complexes satisfy this property. It turns out that this is the case if (probably: and only if) the braid is a positive lift of an element of the Weyl group.<|endoftext|>
TITLE: Can there be an almost-special not-fully-special Aronszajn tree?
QUESTION [13 upvotes]: Question. Can there be an Aronszajn tree $T$, such that no
c.c.c. forcing extension adds a cofinal branch to $T$, but there is an
$\omega_1$-preserving forcing extension adding a cofinal branch to
$T$?
To give some background, an Aronszajn tree is an
$\omega_1$-tree with no $\omega_1$-branches, and it is special,
if it is the union of countably many antichains. Any forcing
extension in which a special Aronszajn tree $T$ comes to have a cofinal branch must collapse
$\omega_1$, since the branch has size $\omega_1^V$ and will contain
at most one node from each of the antichains witnessing that $T$ is
special. In other words, a special Aronszajn tree must remain Aronszajn in any
$\omega_1$-preserving forcing extension, and this includes all
c.c.c. extensions. Meanwhile, any Aronszajn tree can be forced to
be special, and the specializing forcing is c.c.c. and absolute
(consisting of finite partial specializing functions). Thus, this
forcing remains c.c.c. in any extension in which $T$ remains
Aronszajn.
Concerning the question,

A special Aronszajn tree satisfies the first requirement of the question, since we
cannot add a branch to it by c.c.c. forcing, but it doesn't
satisfy the second requirement, since no $\omega_1$-preserving extension
can add a cofinal branch. So special Aronszajn trees won't work.
A Souslin tree satisfies the second requirement, but not the first, precisely because it is
itself c.c.c. and adds a branch through itself. So Souslin trees
won't work.

What is needed is a tree that is almost special, in the sense that
it remains Aronszajn in every c.c.c. extension, but not fully
special, in the sense that we can add a cofinal branch by some
(non-c.c.c.) $\omega_1$-preserving forcing.
I am interested in this question in part because the existence of
such a tree $T$ would provide an answer Arthur Fischer's question
on two
versions of absolutely ccc, because the forcing $\mathbb{P}$ to specialize $T$ is c.c.c. and would remain
c.c.c. in any c.c.c. extensions, since the tree would still be Aronszajn
there, but it would not be c.c.c. in the $\omega_1$-preserving
extension in which $T$ gains a cofinal branch.

REPLY [2 votes]: To supplement this with another example, it is also possible to construct a tree that is $\omega$-distributive and $S$-st-special (in Shelah's terminology) from $\Diamond^*(S^c)$ for some $S$ bi-stationary (in fact I think $\Diamond(S^c)$ works). It appears in Shelah's proper forcing book Chapter IX Lemma 3.9. This tree satisfies your requirement since forcing with the tree itself adds no countable sequences of ordinals, and no forcing preserving stationary of $S$ (and preserving $\omega_1$) can add a branch by $S$-st-specialness. 
$T$ is $S$-st-special if there exists $f: T\restriction S\to \omega_1$ such that $f(x)\in ht(x)$ and $x<_T y \rightarrow f(x)\neq f(y)$.<|endoftext|>
TITLE: Embedding number fields in fields with class number 1
QUESTION [16 upvotes]: (Apologies if this question isn't quite research-level: a colleague came across it while preparing a non-examinable bonus lecture on class field theory for an undergraduate algebraic number theory course.)

Let $K$ be a number field. Is there always a finite extension $L / K$ such that $L$ has class number 1? 

If $K$ has finite class field tower (i.e. the tower of fields $(K_n)_{n \ge 0}$, where $K_0 = K$ and $K_{n+1}$ is the Hilbert class field of $K_n$, eventually terminates) then that solves the problem. But it's a well-known theorem of Golod and Shafarevich that the class field tower of $K$ doesn't terminate if $K$ is an imaginary quadratic field with enough primes ramified.
The textbook my colleague has been using claims that it follows from Golod-Shafarevich that these fields $K$ cannot be embedded in any number field with class number 1, but this implication isn't clear to me. Golod-Shafarevich shows that $K$ has no finite, solvable, everywhere-unramified extension with class number 1, but that's a much weaker statement, isn't it?

REPLY [15 votes]: See Proposition 1 on p.231 of Cassels and Frohlich for a proof of the claim in the textbook:
The point is that if such an $L$ exists then $K_1L$ is abelian and unramified over $L$ so it is contained in the Hilbert classfield of $L$ which is $L$ itself. By induction, this implies that $K_i \subset L$ for all $i$ so the class field tower of $K$ must be finite.<|endoftext|>
TITLE: Weak Bott periodicity vs. strong Bott periodicity
QUESTION [5 upvotes]: I asked this also here, but maybe it's also appropriate to ask it here.
Bruno Harris' proof (or I guess also Bott's original proof) of Bott periodicity (see here for instance) shows that there is a homotopy equivalence $h\colon\mathbb{Z}\times BU \rightarrow \Omega^2 (\mathbb{Z}\times BU)$, hence there is a natural isomorphism of set-valued functors $KU\rightarrow KU^{-2}$. Karoubi calls this "weak Bott periodicity" in his book.
Now, is it possible to deduce "strong Bott periodicity" from that?
I.e. that external multiplication by the Bott element is an isomorphism?
It would suffice to show that the maps $KU(X)\rightarrow KU^{-2}(X)$ are $KU(X)$-module homomorphisms, i.e. I think one has to show that a diagram like
$\begin{matrix}
\mathbb{Z}\times BU\wedge\mathbb{Z}\times BU &\stackrel{\otimes}{\rightarrow}&\mathbb{Z}\times BU\\\
\downarrow{h\wedge id}&&\downarrow{h}\\\
\Omega^2(\mathbb{Z}\times BU)\wedge \mathbb{Z}\times BU&\stackrel{\otimes}{\rightarrow}&\Omega^2(\mathbb{Z}\times BU)
\end{matrix}
$
commutes up to homotopy (the lower $\otimes$ should be something like pointwise multiplication of maps $S^2\rightarrow \mathbb{Z}\times BU$ by elements of $\mathbb{Z}\times BU$). Of course, the problem is that the map $h$ is not very explicit.
Is it at all possible to do that? Many thanks.

REPLY [14 votes]: As a matter of history, the original Bott maps are very explicit, and in fact they are $E_{\infty}$ maps with respect to the actions of the linear isometries operad (as shown in the first chapter of
$E_{\infty}$ ring spaces and $E_{\infty}$ ring spectra).  Bott himself, in his paper Raoul Bott.
Quelques remarques sur les théorèmes de périodicité. Bull. Soc. Math. France 87 1959 293–310,
showed how to derive the strong form from the weak form by showing that his original maps are
homotopic to the adjoints of the evident maps obtained by tensoring with the Bott classes.
He does this for the real and quaternionic cases as well.<|endoftext|>
TITLE: Relationship between monodromy representations and isomorphism of flat vector bundles
QUESTION [12 upvotes]: This question is somehow related to this one.
Let $M$ be a smooth (compact, if you wish) connected manifold. 
Then, it is well known that there is an equivalence between the isomorphism classes of pairs $(E,\nabla)$, where $E\to M$ is a complex vector bundle of rank $r$ and $\nabla$ a flat connection on $E$, and the isomorphism classes of complex linear representations of dimension $r$ of the fundamental group $\pi_1(M)$ of $M$.
Here, $(E,\nabla)\simeq (F,\nabla')$ iff there exists an isomorphism of differentiable vector bundles $f\colon E\to F$ over $M$ such that $f^*\nabla'=\nabla$.  
Question. Are there necessary and/or sufficient conditions such that given two representations $\rho,\rho'\colon\pi_1(M)\to\operatorname{Gl}(r,\mathbb C)$, the associated complex vector bundles $E_\rho,E_{\rho'}\to M$ are isomorphic? 
Of course, in the question we are forgetting the flat connections naturally given by these two representations, and we are asking the two complex vector bundles to be isomorphic just as complex vector bundles.
More specifically, can we for instance give an answer in terms of connected components of the space of the representations - or, if you want, in terms of (some well-chosen notion of) homotopies between the representations?

REPLY [11 votes]: I have two comments:
(1). It is a trivial remark, but one should note that representations $\rho, \rho'$ which are in the same component of $Hom(\pi,G)$ (in your case, $G=GL(r, {\mathbb C}))$ yield flat bundles $E_{\rho}, E_{\rho'}$ which are isomorphic as vector bundles. Here and below $\pi=\pi_1(M)$.  
Proof. It suffices to work with principal $G$-bundles. Let $\rho_t$ be a path of representations between $\rho, \rho'$. Then $\rho_t$ determines a (flat) principal $G$-bundle $P$ over $M\times [0,1]$, whose restrictions $P_0, P_1$ to $M\times 0, M\times 1$ are isomorphic to the principal $G$-bundles associated with $\rho, \rho'$. I claim that this bundle is just the product $P_0\times I$. Indeed, construct a section of the bundle $Hom(P, P_0\times I)$ starting with the identity on $M\times 0$ and then extend it to the rest of $M\times I$ (path-lifting property for Serre fibrations). QED.  
This gives a sufficient, but, of course, far from necessary, condition for an isomorphism of vector bundles. In particular, you get finiteness of the number of isomorphism classes of vector bundles. As  a simple example consider the case $r=1$ and $H_1(M)$ torsion-free. Then $Hom(\pi, {\mathbb C}^\times)= 
({\mathbb C}^\times)^n$ is connected (here $n=rank(H_1(M))$). In particular, all flat line bundles in this case are trivial. 
(2). A far less trivial is the result of Deligne and Sullivan "Fibres vectoriels complexes a groupe structural discret", C. R. Acad.  Sci. Paris 281 (1975),  1081-1083. They prove that for every finite cell complex $M$ and every $\rho: \pi=\pi_1(M)\to GL(r, {\mathbb C})$ there exists a finite cover $\tilde{M}\to M$ 
so that the pull-back of the associated flat bundle $E_\rho$ to $\tilde{M}$ is trivial (as a bundle). Their proof is constructive and (from what I can tell by reading Math Review of their paper since I lost my copy) gives a sufficient condition for triviality of $E_\rho$: Let $A$ denote the subring of ${\mathbb C}$ generated by the matrix entries of $\rho(\pi)$. Suppose that there are two maximal ideals $m_1, m_2$ in $A$ so that:
a. $A/m_i$ have distinct characteristics and
b. $\rho(\pi)$ maps trivially to $GL(r, A/m_i), i=1,2$. 
Then $E_\rho$ is trivial as a vector bundle.  
From this one can extract a sufficient condition for an isomorphism of bundles $E_\rho, E_{\rho'}$ (by considering the flat bundle $E_\rho^*\otimes E_{\rho'}$).  
If I remember correctly, a proof of this theorem by Deligne and Sullivan was redone by Eric
Friedlander, in "Étale homotopy of simplicial schemes". Annals of Mathematics Studies, 104. Princeton University Press, 1982. I cannot tell if it is easier to read than the original.<|endoftext|>
TITLE: Can any numerical polynomial be a Hilbert polynomial of something?
QUESTION [12 upvotes]: is it true that any numerical polynomial , i.e. $f(t)\in \mathbb Q[t], f(n)\in\mathbb Z\  \forall n\in\mathbb Z\ $ can be presented as Hilbert polynomial of some variety?

REPLY [12 votes]: Steven Landsburg's answer  completely solves the problem for schemes. I notice that you asked for a variety, which is more restrictive. For example, $2t+2$ is the Hilbert polynomial of a pair of skew lines, but it is not the Hilbert polynomial of any irreducible variety. Such a variety would have to be a degree $2$ curve, by looking at the leading coefficient, and this curve would have to have genus $-1$ by Riemman-Roch, a contradiction. Similarly, $2 \binom{t+2}{2} -1 = t^2+3t+1$  is the Hilbert polynomial of two $\mathbb{P}^2$'s which meet transversely as a point and I believe that it is not the Hilbert polynomial of any irreducible variety.<|endoftext|>
TITLE: A space of ideals
QUESTION [15 upvotes]: Definition: Let $R$ be a commutative ring with 1.  Endow the power set $2^R$ with the product topology.  The ideal space $\mathcal{I}(R)$ is defined to be subset of $2^R$ consisting of ideals, equipped with the induced topology.
This is the ring-theoretic analogue of the Gromov--Grigorchuk space of marked groups, which can be used to give nice proofs of simple facts about algebraic geometry over groups (cf. this paper by Champetier and Guirardel).  My question is:

Is the 'ideal space' of a ring a standard construction in commutative algebra or algebraic geometry?  If so, what's it called and where can I read more about it?

My motivation is to strengthen the analogy between algebraic geometry over groups and classical algebraic geometry.  It would be nice, when doing algebraic geometry over groups using the space of marked groups, to be able to say 'This is analogous to the foobar widget in classical algebraic geometry.'
To demonstrate that this concept has some use, I'll give a very simple application.  But first, here are a few basic facts.

$\mathcal{I}(R)$ is compact (because it is a closed subset of $2^R$, which is itself compact by Tychonoff's Theorem), Hausdorff and totally disconnected.
Each point $I\in \mathcal{I}(R)$ is contained in a canonical closed subset
$U_I=\{J\in \mathcal{I}(R)\mid I\subseteq J \}$
(which is in fact isomorphic to $\mathcal{I}(R/I)$).
If $R$ is Noetherian then each $U_I$ is equal to the set of ideals that contain a (finite) generating set for $I$, and hence is open.
The subset of prime ideals in $\mathcal{I}(R)$ is closed: indeed, for each pair of non-units $x,y$, the subset $N(x,y)=\{x\notin I, y\notin I, xy\in I\}$ is open, and the union of these sets is the complement of the set of prime ideals.

Now here's the promised application -  a proof of a well known lemma.
Lemma: If $R$ is a Noetherian ring then there is a finite set of prime ideals $\mathfrak{p}_1,\ldots,\mathfrak{p}_k\subseteq R$ with the property that every prime ideal contains one of the $\mathfrak{p}_i$.
Proof: The set of $U_{\mathfrak{p}}$ for $\mathfrak{p}$ prime is an open cover of the set of prime ideals.  Since the set of prime ideals is compact, there is a finite subcover. QED
By the way, my research concerns, among other things, algebraic geometry over groups, but I have never seriously studied algebraic geometry or commutative algebra.  This question was first posted at math.stackexchange.

REPLY [6 votes]: I wrote a paper about the space of ideals in a commutative ring (or more generally submodules of a module over a commutative ring):
http://arxiv.org/abs/0904.4216
I focused on the case of finitely generated commutative rings, because, in the general case, the pointwise convergence seems not to be the most natural one, see the discussion in Section 5 of the paper, and also because my motivation was to deal with f.g. metabelian groups, that I consider in
http://arxiv.org/abs/0904.4230
As Benjamin mentions, these constructions are particular cases of general abstract constructions known by people in model theory and others, and the contribution of the above papers is to obtain precise description of the homeomorphism type of certain of these objects.<|endoftext|>
TITLE: Punctured spectrums of local rings
QUESTION [8 upvotes]: Let $A$ be a local ring with the unique maximal ideal $\mathfrak{m}$. The punctured spectrum of $A$ is the open subset $\text{Spec}(A)\setminus \{\mathfrak{m}\}$.  I have seen many papers (for instance Horrocks' papers) studying vector bundles over algebraic varieties (in particular, projective spaces) by putting them over a punctured spectrum of a local ring. 
However, I am wondering why punctured spectrum is better than varieties in this satiation. I feel that geometric pictures of varieties are clearer than that of a punctured spectrum. More essential, are the categories of coherent sheaves over a variety and its punctured spectrum equivalent? How much information about coherent sheave (in particular, vector bundles) can be recovered from punctured spectrums? For second question, I am thinking examples that reflect some relations between those two gadgets.

REPLY [6 votes]: Here is a general principle. If a problem is formulated over a general (say noetherian) base, by localization arguments one can often reduce to the case of a local base. Arguing by induction on the dimension of the base, we can exploit the fact that the punctured spectrum, while often not affine, is of smaller dimension!  Then we can try to "extend" the solution over the punctured spectrum to a solution over the local spectrum (by taking into account local cohomology or other tools).  I think this brilliant idea is due to Grothendieck (truly exploiting the concept of a scheme).
For example, observe that the punctured spectrum of a 2-dimensional regular local ring is a (typically non-affine!) Dedekind scheme, over which flatness is a hands-on concept. Much deeper, the proof of Zariski's Main Theorem in EGA IV$_3$ goes via dimension induction on the base using the above trick with a punctured spectrum having smaller dimension.
For the study of coherent sheaves, the categories on the spectrum and punctured spectrum are not at all equivalent, though the situation for vector bundles is more accessible; the problem is governed by local cohomology and depth considerations (so the situation is more tractable over a regular base).  This is studied in depth (no pun intended) in the middle of SGA2.<|endoftext|>
TITLE: Q-factorial and rational singularities on surfaces
QUESTION [6 upvotes]: Let $X$ be a normal surface. Is any rational singularity $\mathbf{Q}$-factorial? I've seen this somewhere for surfaces over fields, but what about the general case of an integral 2-dimensional excellent normal scheme?
In this generality it might not hold so what if we assume that $X$ is fibered ( = flat projective) over a Dedekind scheme?  When can we hope for such a result to hold. Probably there are some problems depending on the characteristic.
What about the converse?
I know that every surface fibered over $\mathrm{Spec} \mathbf{Z}$ is $\mathbf{Q}$-factorial. Are all its singularities rational?
I know that one has to be careful with the base scheme. Probably if the base scheme is a smooth projective curve over a field things might not work so well, but maybe if the base is $\mathrm{Spec} \mathbf{Z}$ things might become better.

REPLY [5 votes]: About the converse: one does need all the assumptions Karl mentioned in his answer. There are $2$-dim. complete local rings which are UFD but does not have rational singularity. One such example (due to Salmon) is $k(u)[[x,y,z]]/(x^2+y^3+uz^6)$ which is factorial for any field $k$.  
Removing the hensenlian condition is also a problem:  $R=k[x,y,z]_{(x,y,z)}/(x^r+y^s+z^t)$ where $r,s,t$ are pairwise prime, is factorial over any field $k$!<|endoftext|>
TITLE: If a polyhedron is homeomorphic to a simplex, is it piecewise-linear homeomorphic?
QUESTION [7 upvotes]: If a polyhedron is homeomorphic to a simplex, is it piecewise-linear homeomorphic?
In particular, is this true in $R^{4}$? In 2 and 3 dimensions any two polyhedra
that are homeomorphic are PL-homeomorphic, by theorems of Rado and Moise. In dimension $\geq 5$, this is a trivial special case of theorem 1.1 in 
M.A. Armstrong "The Hauptvermutung According to Lashof and Rothenberg" in 
The Hauptvermutung Book. But I have not found a statement that covers it for dimension 4; and I am not confident that dimension 4 can easily be reduced to
dimension 5. 
Also, if anyone can suggest a reference for this particular case that does not go through these very high-powered, difficult, general theorems, I would be interested on stylistic grounds.

REPLY [5 votes]: It is not true in dimension 5 or above. Edwards' double suspension theorem implies this. See this Wikipedia article section.<|endoftext|>
TITLE: About Tate's computation of $K_2^{\rm M}(\mathbb Q)$
QUESTION [7 upvotes]: For any field $F$, there is a natural group homomorphism $K_n^{\rm M}(F) \to K_n(F)$ from Milnor's $K$-theory to Quillen's $K$-theory. If $n=2$, it is an isomorphism, by Matsumoto's theorem. It is a well known theorem of Quillen that if $F$ is a number field, then the groups $K_n(F)$ are finitely generated for $n\geq 2$. It is known in particular, that the group $K_2(\mathbb Q)$ is finite, isomorphic to $\mathbb Z/2\mathbb Z$.
I am now confused about the following result of Tate (I got it from Milnors "Introduction to Algebraic K-theory", Theorem 11.6): There is a canonical, split exact sequence of commutative groups
$$0 \to \{\pm1\} \to K_2^{\rm M}(\mathbb Q) \to \bigoplus_p (\mathbb Z/p\mathbb Z)^\ast \to 0 $$
and in particular, $K_2^{\rm M}(\mathbb Q)$ is an infinite torsion group. The first term in the sequence should be interpreted as $K_2^{\rm M}(\mathbb Z)$, and each $(\mathbb Z/p\mathbb Z)^\ast$ should be seen as $K_1$ of the finite field with $p$ elements. 
What did I misunderstand? Is it not true that $K_2^{\rm M}(\mathbb Q) \cong K_2(\mathbb Q)$ -- or is it not true that $K_2(\mathbb Q)$ is finitely generated --- or did I misquote Tate's theorem?

REPLY [9 votes]: The mistake is your assumption $K_2(\mathbb Q) = \mathbb Z/2$. 
For any number field $F$ with ring of integers $\mathcal{O}$, the isomorphism $K_n(F) \cong K_n(\mathcal{O})$ is true only for $n > 1$ odd. If $n > 1$ is even, there is a short exact sequence 
$$0 \to K_n(\mathcal{O}) \to K_n(F) \to \oplus_P K_{n-1}(\mathcal{O}/P) \to 0$$
where $P$ runs through the primes of $\mathcal{O}$. 
Moreover, it's known that $K_n(\mathcal{O})$ is finite for $n>1$ even (for instance $K_2(\mathbb Z) = \mathbb Z/2$). So in this case, $K_n(F)$ is an infinite torsion group. 
An excellent survey on the K-theory of number fields can be found in this paper of Weibel.<|endoftext|>
TITLE: Best known Margulis constants?
QUESTION [12 upvotes]: A Margulis number for a hyperbolic $n$-manifold $M=\mathbb{H}^n/\Gamma$ is a number $\epsilon>0$ such that for each $x\in\mathbb{H}^n$ the group generated by the elements in $\Gamma$ which move $x$ less than distance $\epsilon$ is elementary.
The Margulis constant for hyperbolic $n$-manifolds is the largest number $\epsilon(n)$ which is a Margulis number for every hyperbolic $n$-manifold.
Meyerhoff showed that $\epsilon(3) > 0.104$. (Robert Meyerhoff. A lower bound for the volume of hyperbolic 3-manifolds. Canad. J. Math., 39(5):1038–1056, 1987.)
Shalen proved that 0.29 is a Margulis number for all but finitely many orientable hyperbolic 3-manifolds.  He also notes that experimental evidence suggests that $\epsilon(3) <0.616$ (Peter Shalen. Topology and geometry in dimension three, 103–109, Contemp. Math., 560, Amer. Math. Soc., Providence, RI, 2011)
Question 1: Is Meyerhoff's lower bound the best known lower bound for $\epsilon(3)$?
Question 2: What is known about the Margulis constants for higher dimensional hyperbolic manifolds?

REPLY [3 votes]: For Question 2: In addition to Kellerhals' lower bounds on $\epsilon(n)$, there exists an absolute constant $C>0$ such that
$$
\epsilon(n)\le \frac{C}{\sqrt{n}},
$$
see Proposition 5.2 here.<|endoftext|>
TITLE: Massey Products vs. $A_\infty$-Structures
QUESTION [22 upvotes]: Does anyone know a good reference for a proof of the fact that given a dga $A$, an $A_\infty$-structure on $HA$ is ''the same'' as coherent choices for all of the higher Massey products of $HA$?  More concretely the fact I am looking for is something like the following.
When defining the Massey product $\langle x_1,\dots, x_n\rangle$ there are multiple non-canonical choices that need to be made, which in turn give multiple cycles that could be called the Massey product of $x_1,\dots, x_n$. If $M(x_1,\dots, x_n)$ is the set of all  possible resulting Massey products of $x_1,\dots, x_n$, and $\mu_n$ is the $n$-th $A_\infty$ structure map (on $HA$), then $$\mu_n(x_1\otimes\cdots\otimes x_n)\in M(x_1,\dots, x_n)$$
for all $n$ and $x_i$.

REPLY [16 votes]: When $n=3$, this is in Stasheff's H-spaces from a homotopy point of view, Chapter 12. For general $n$, it is in a paper of mine with Lu, Wu, and Zhang, "$A_\infty$-structures in Ext algebras, J. Pure Appl. Alg. 213 (2009), 2017--2037 (Theorem 3.1 and Corollary A.5).<|endoftext|>
TITLE: Quadrature of the Lune
QUESTION [7 upvotes]: What is a good reference for the following result which I believe is proved by Tchebotarev.
There are exactly 5 types of Lunes that are squarable. (Hippocrates produced three and then two more were given by Euler).

REPLY [6 votes]: "The Problem of Squarable Lunes."
M. M. Postnikov, translated from the Russian by Abe Shenitzer.
The American Mathematical Monthly.
Vol. 107, No. 7 (Aug.-Sep., 2000), pp. 645-651.
JSTOR link.
See also the MathPages web page entitled "The Five Squarable Lunes,"
and the Wikipedia page "Lune of Hippocrates," from which this image 
of the "lunes of Alhazen" is taken:
          


"The two blue lunes together have the same area as the green right triangle."<|endoftext|>
TITLE: Total spaces of $TS^2$ and $S^2 \times R^2$ not homeomorphic 
QUESTION [48 upvotes]: Hello,
I'm looking for an invariant to distinguish the homeomorphism types of homotopy equivalent spaces. Specifically, how does one show that the total spaces of the tangent bundle to $S^2$ and the trivial bundle $S^2 \times R^2$ are not homeomorphic? (I am not asking for a proof that $TS^2$ is not the trivial bundle.)
Also, is there a way to reduce the question, "Are the total spaces of two vector bundles homeomorphic" to "Are the associated sphere bundles homeomorphic"? In the case of $TS^2$ and $S^2\times R^2$ it's not too difficult to show that the sphere bundles are not homeomorphic, and I'm wondering if there's a way to leverage that.
Thanks,
Zygund

REPLY [4 votes]: An argument using only basic topology and fundamental group (no homology or higher $\pi_i$).
Consider the condition $\mathcal{P}$, for a topological space: for every compact subset $K\subset X$ there exists a compact subset $K'$ containing $K$, such that $X-K'$ is path-connected and has a finite fundamental group.
Then $T\mathbf{S}^2$ satisfies $\mathcal{P}$ ($K$ can be enlarged to the set $K'$ of vectors of norm $\le R$ for some $R$, and the complement of $K'$ to the product of $\mathbf{R}$ and the set of vectors of norm $2R$, so is homeomorphic to $\mathbf{R}\times\mathrm{SO}(3)$ and has $\pi_1$ with 2 elements).
While $\mathbf{S}^2\times\mathbf{R}^2$ does not satisfy $\mathcal{P}$: given $K$, for any $K'$ we can retract the complement of $K'$ onto $\mathbf{S}^2\times \mathbf{S}^1(R)$ for large enough $R$, where $\mathbf{S}^1(R)$ is the circle of radius $R$, so the $\pi_1$ complement of $K'$ has an infinite homomorphism onto the group $\mathbf{Z}$ of integers.
Informally, the fundamental groups at infinity of the 2 spaces differ. (This argument is, anyway, related to the previous ones.)<|endoftext|>
TITLE: Noether Normalization
QUESTION [9 upvotes]: I teach a course of undergraduate algebraic geometry. I noticed that students have difficulty grasping the proof of Noether normalizsation Lemma (given in Reid's undergraduate algebraic geometry). Also, the proof given in Mumford red book is also non-elementary. Does anyone know of some easier proof of the result?
 Thank you

REPLY [9 votes]: Here is the proof from my class notes last time I taught the course.  I don't remember where I borrowed it from.  Hope it helps.  It is tedious but elementary in the sense of Martin's comment.  It is the special case where the base field is infinite.  It might become clearer if you give the various steps as exercises to your class.
Noether normalization lemma:
Given any infinite field $k$, and any finitely generated $k$-algebra $B$, then we can decompose the extension $k$ in $B$, into $k$ in $A$ in $B$ where $A$ is isomorphic to a polynomial algebra over $k$, and $B$ is a module-finite extension of $A$.
Proof:  Proof by induction on the number of $k$-algebra generators of $B$ over $k$.
    First do the case of one generator.  If $B = k[c]$ and $c$ is transcendental over $k$ we are done.  If $c$ is algebraic over $k$ then $B\cong k[T]/(f(T))$ where $f$ is the monic irreducible polynomial of least degree $d$ satisfied by $c$ over $k$. Then $B = k[c]$ is generated over $k$ by the elements $1,c,…,c_{d-1}$, and in particular $B$ is a finite vector space over $k$.  So take $A = k$, and we are done.
Now assume $B = k[c_1,...,c_n]$ where $n\geq 2$.  Consider the canonical surjective $k$-algebra map $k[T_1,...,T_n]\to B$ taking $T_i\mapsto c_i$.  If this map is also injective we are done, since then we may take $A = B$.
If this map is not injective then some non-constant polynomial $f$ maps to zero.  If $f$, considered as a polynomial in $T_n$, is monic of degree $d\geq1$, then $k[c_1,\dots,c_n]$ is generated as a module over $k[c_1,\dots,c_{n-1}]$ by the finite set of monomials of degree less than $d$ in $c_n$.  In this case we are done, since by the inductive hypothesis $k[c_1,…,c_{n-1}]$ is module finite over some subalgebra isomorphic to a polynomial algebra over $k$.  Then by transitivity of finiteness,  $B = k[c_1,...,c_n]$ is also finite over that same polynomial subalgebra of $B$.
We claim we can “change coordinates” linearly in $k[T_1,\dots,T_n]$, until the polynomial $f$ becomes monic in $T_n$.  That will mean there is another $k$-algebra surjection $$k[T_1,\dots,T_n]\to B=k[e_1,\dots,e_n],$$taking $T_i\mapsto e_i$ such that some element of the kernel is monic in $T_n$.  Then we are done by the same inductive argument.  So all we have to do is the change of variables argument, sometimes known as the Weierstrass preparation lemma.
Let $f(T_1,\dots,T_n)$ be any non constant polynomial, and consider the linear change of coordinates 
$$T_i\mapsto(T_i+a_iT_n), 1≤i≤n-1,\quad \text{and}\quad T_n\mapsto T_n,$$ where the $a_i$’s are elements of $k$.  Write $f$ as $f^d + $ lower degree terms, where $d =$ degree $f$, and $f^d$ is homogeneous of degree $d$.  
Under this transformation, $f\mapsto f(T_1+a_1T_n,\dots,T_{n-1}+a_{n-1}T_n, T_n)$.  We claim it is possible to choose the coefficients $a_i$ so that this new polynomial has non zero coefficient of $T_n^d$.  Just expand, and get $$f(T_1+a_1T_n,\dots,T_{n-1}+a_{n-1}T_n, T_n)=f^d(T_1+a_1T_n,\dots,T_{n-1}+a_{n-1}T_n, T_n) + \text{lower degree terms}.$$
Then expand the top degree term and get
$$f^d(T_1+a_1T_n,\dots,T_{n-1}+a_{n-1}T_n, T_n) = f^d(a_1,\dots,a_{n-1}, 1)T_n^d + \text{lower degree terms in }T_n.$$
Adding gives $$f(T_1+a_1T_n,\dots,T_{n-1}+a_{n-1}T_n, T_n) = f^d(a_1,\dots,a_{n-1}, 1)T_n^d + \text{lower degree terms in }T_n.$$
Thus we only have to choose the $a_i$’s so that $f^d(a_1,\dots,a_{n-1}, 1)\neq0$.  Since $f^d$ is a non-zero homogeneous polynomial of degree $d\geq 1$, $f^d(a_1,\dots,a_{n-1}, 1)$ is a non-zero polynomial of degree less than or equal to $d$ in the $a_i$’s.  By induction on $d$, since our field $k$ is infinite, it does not vanish for all choices of the $a_i$’s. QED.<|endoftext|>
TITLE: What is a self-dual category?
QUESTION [24 upvotes]: $\newcommand{\C}{\mathcal{C}}\newcommand{\D}{\mathcal{D}}\newcommand{\op}{\mathrm{op}}$I would like to define the notion of a self-dual category, which should mean a category isomorphic to its opposite in a natural way, and the notion a self-dual functor between such categories. For a category $\C$, I denote by $\C^\op$ its opposite category; for a functor $F \colon \C \to \D$ its opposite functor is $F^\op \colon \C^\op \to \D^\op$; for a natural transformation $\eta \colon F \Rightarrow G$ I denote $\eta^\op \colon G^\op \Rightarrow F^\op$ its opposite natural transformation.  
Definition. A self-dual category is a category $\C$, a functor $i_\C \colon \C \to \C^\op$, and a natural isomorphism $$\epsilon_\C \colon i_\C^\op \circ i_\C \Rightarrow \mathrm{id}_\C.$$
I think this is the "correct" definition. (Not quite, see below.) An example would be for $\C$ the category of finite dimensional vector spaces over a field $k$, and $i_\C = \mathrm{Hom}(-,k)$. We would now like to talk about functors being self-dual, which should mean that they commute with taking duals.
Definition. A self-dual functor $\C \to \D$ is a functor $F \colon \C \to \D$ and a natural isomorphism 
$$ \eta \colon i_\D \circ F  \Rightarrow  F^\op \circ i_\C $$
satisfying the following coherence condition: The diagram of natural isomorphisms
$$\begin{matrix} F \circ i_\C^{\op} & \stackrel{\eta^{\op}}{\Rightarrow} & i_\D^{\op} \circ F^{\op} \\\\ \epsilon_D \Uparrow ~  ~ ~ & &  ~ ~ ~ \Downarrow \epsilon_C^{\op} \\\\ i_\D^{\op} \circ i_\D \circ F \circ i_\C^{\op} & \stackrel{\eta}{\Rightarrow} & i_\D^{\op} \circ F^{\op} \circ i_\C \circ i_\C^{\op} \end{matrix}$$
commutes.
This whole definition is quite a mouthful and it feels like someone ought to have defined this carefully somewhere. Googling for self-dual or autodual categories produces several hits where people use this term for categories isomorphic to their opposite, but I haven't seen anyone discuss categories which have such an isomorphism in a coherent way. Does anyone know whether there is such a reference? Maybe this is a special case of a more general construction?

REPLY [6 votes]: The question has essentially been answered in the comments, I am recording this here so that the question does not get bumped back to the top. Theo Buehler and Buschi Sergio both gave nice references, and it seems that this notion is well known in K-theory under the name "category with duality". Martin's remarks were also helpful.<|endoftext|>
TITLE: $A_\infty$-categories basic reference
QUESTION [14 upvotes]: Can anyone provide me with a basic reference on $A_\infty$ categories?

REPLY [3 votes]: Here's a list of references:

BLM,              483 Pages, Pretriangulated _∞-Categories;
COS,               25 Pages, Localizations of the Category of _∞-Categories and Internal Homs;
Faonte,           156 Pages, Nerve construction, _∞-functors and homotopy theory of differential graded categories;
Faonte,            47 Pages, Simplicial nerve of an _∞-category;
Faonte,            53 Pages, _∞-functors and homotopy theory of dg-categories;
Fukaya,            97 Pages, Floer Homology and Mirror Symmetry;
Horel,             10 Pages, The homotopy theory of _∞-categories;
Keller,            31 Pages, Introduction to _∞-algebras and modules;
KS,                70 Pages, Notes on _∞-algebras, _∞-categories and non-commutative geometry I;
Lefèvre-Hasegawa, 230 Pages, Sur les _∞-Catégories;
Ornaghi,          113 Pages, Comparison Results About DG-Categories, _∞-Categories, Stable ∞-Categories and Noncommutative Motives;
Seidel,           334 Pages, Fukaya Categories and Picard–Lefschetz Theory.<|endoftext|>
TITLE: Teaching Experience for Graduate Students.
QUESTION [11 upvotes]: I am currently a graduate student, who will (hopefully!) graduate in the next year (or two..).  I have slowly come to realize that I enjoy teaching, and consequently want to do more of it!  My main reasons are such:

to gain experience
to bolster my CV
to learn to be a better teacher

That last point is particularly important to me, because I feel I have received insufficient training in how to teach mathematics well.
Now, I have done the usual TA thing.  For the past year, I have also been an adjunct instructor at a local four-year college, so I pretty much know what it's like to be "fully responsible" for a course.  What I am looking for are challenging opportunities which allow me to do some - or all - of the following:

teach fairly sophisticated math to bright high school students / undergraduates
engage them in innovative thinking/research
be consistently mentored/evaluated throughout the duration of this experience

Again, the last point is rather important to me.  So my question is:

Do such programs exist?

I am sure they do, but when Googling, I invariably come up with graduate summer schools, or "local" opportunities. (By "local" here, I mean those which are only eligible to students in that particular school's graduate program.) So I am hoping someone (or several people) here know more about it than Google does.
I know that most opportunities for this summer have probably already expired, but I want to stress I am not only focused on summer sessions.  In particular, I would gladly forego my usual TA appointment for a semester, to be a part of a more difficult and rewarding experience somewhere else.
Finally, I am on the fence about whether this question should be CW or not; as of now, it isn't.  If people feel it should be, I will change it.
Thanks in advance.

REPLY [7 votes]: Many small liberal arts undergraduate colleges in the US have temporary adjunct positions
often or usually filled by new Ph.D.s. These colleges generally focus more attention on teaching than do research universities, and consequently have infrastructure for mentoring and improving teaching.  Many have special fellowships whose goal is specifically to mentor young postdocs interested in a teaching career.  At my school, these fellowships require teaching only six courses over a three-year appointment, a generous arrangement.
The US National Postdoctoral Association maintains a list of such teaching fellowships,
although I am uncertain of its comprehensiveness.
One can also find specific school fellowships by web-searching "postdoctoral teaching fellowships in mathematics."
I suspect analogous summer teaching opportunities would be less useful in terms of development,
but perhaps(?) easier to secure. There are many summer math programs for high-schools students;
teaching at one of these would be quite enjoyable.  I can only mention examples (again, in the US) rather than point
you to a comprehensive list: Hampshire College Summer Studies in Mathematics,
Cornell Summer Math Institute Awesome Math at Cornell/UTDallas/UCSantaCruz, Berkeley Math Path Summer Camp, 
Johns Hopkins Center for Talented Youth, etc.<|endoftext|>
TITLE: Is Diagonalization worth to be taught?
QUESTION [5 upvotes]: When students come to the College (first two years of the University system in most of the developped countries) to train in mathematics, they get a linear algebra / matrix analysis course. After a few months, perhaps after one year, they are taught about diagonalization of matrices. They learn many criteria that are either necessary or sufficient or both. This seems to be a mandatory step for future engineers and other categories of scientific workers.
My question is a bit provocative:

Is diagonalization that important? Should we teach it thoroughly to people who will have to use linear algebra and matrices in the future?

Here are a few arguments why we should refrain ourselves to enter this topic, except when teaching future mathematicians:
1- The solution of this problem is not so nice, many matrices being not diagonalizable. And the set of diagonalizable matrices is neither open nor close in any sense (usual, if the field is $\mathbb R$ or $\mathbb C$, Zariski otherwise).
2- Diagonalization is not effective. As a matter of fact, we cannot compute explicitely the eigenvalues of an $n\times n$ matrix if $n\ge5$ (Abel plus companion matrix).
3- Diagonalization is not really useful. You don't use it to calculate the exponential, or to invert, ... What engineers are interested in is often stability of dynamical systems. Thus a good problem is whether the spectrum belongs to either the left half-plane or the unit disk, whether there are eigenvalues on the unit circle or with vanishing real part.
I therefore open a discussion, in which I am looking for either pro- or con- arguments about teaching diagonalization to engineers.

REPLY [3 votes]: You are talking about training in mathematics, not numerical analysis, so the answer is surely "yes". What is more, given that software packages exist, it should be taught in a more conceptual way than is done traditionally. I'm struck by how much of the "Moscow School" or Gel'fand way of looking at things depends on a good feel for the basics of linear algebra,<|endoftext|>
TITLE: Update to Shephard's "Twenty Problems on Convex Polyhedra"
QUESTION [20 upvotes]: Forty-three years ago, Geoffrey Shephard published an influential list of open problems
on convex polyhedra.
Progress has been made on several of his problems, and perhaps some have been completely solved.
I am wondering if anyone has written a survey that updates the status of all 20 problems?
Thanks for any pointers!

"Twenty Problems on Convex Polyhedra: Part I."
  G. C. Shephard.
  The Mathematical Gazette, Vol. 52, No. 380 (May, 1968), pp. 136-147. 
  (JSTOR link).
  "...: Part II."
  No. 382 (Dec., 1968), pp. 359-367.
  (JSTOR link).

REPLY [4 votes]: A listing from 1991 is given in Unsolved Problems in Geometry (Croft, Falconer, and Guy). Chapter B contains 25 open problems on polygons, polyhedra, and polytopes, including some from Shephard's 1968 list.
For updates on specific problems from Shephard's list, see

Equiprojective
polyhedra (2008).   
Moser's Shadow Problem
(2013).
(Non-)Equiprojectivity and (Non-)Biprojectivity of Simplicial Polyhedra (2010).<|endoftext|>
TITLE: Binary representation of powers of 3
QUESTION [10 upvotes]: I asked this question at Mathematics Stack Exchange but since I didn't got a satisfactory answer I decided to ask it here as well.
We write a power of 3 in bits in binary representation as follows.
For example $3=(11)$, $3^2=(1001)$ which means that we let the $k$-th bit from the right be $1$ if the binary representation of this power of 3 contains $2^{k-1}$, and $0$ otherwise.

Prove that the highest power of 3 that has a palindromic binary representation is $3^3 = (11011)$.
Prove that $3 = (11)$ is the only power of 3 with a periodic binary representation (in the sense that it consists of a finite sequence of $1$s and $0$s repeated two or more times, like "$11$" consists of two repetitions of the bitstring "$1$").

REPLY [5 votes]: Here is an alternative (and complete) proof of 2. Following Mark Sapir, we show that the only solution of
$$ \frac{2^{tm}-1}{2^m-1} = 3^r $$
in positive integer triplets $(t,m,r)$ is $(2,1,1)$ and $(2,3,2)$. I am sure this was written down before, by the way.
Our main tool is the observation that $\mathrm{ord}_{3^n}(4)=3^{n-1}$ when $n\geq 1$ (in other words $2$ is a primitive root modulo $3^n$), which implies that $v_3(4^k-1)=1+v_3(k)$. The observation follows from $4^{3^{n-2}}\equiv 1+3^{n-1}\pmod{3^n}$ when $n\geq 2$, which in turn can be proved by induction or by the binomial theorem. 
In the diophantine equation $r\geq 1$, hence $tm$ is even. If $m$ is even, then
$$ r=v_3\left (\frac{2^{tm}-1}{2^m-1}\right)=(1+v_3(tm))-(1+v_3(m))=v_3(t), $$
so that $t$ is divisible by $3^r$. This is a contradiction:
$$  3^r=\frac{2^{tm}-1}{2^m-1} > t \geq 3^r, $$
which proves that $m$ is odd and $t$ is even. In that case $4^m-1\mid 2^{tm}-1$, and
$$ \frac{2^{tm}-1}{4^m-1} = 3^s $$
for some $0\leq s\leq r$. As before
$$ s=v_3\left (\frac{2^{tm}-1}{4^m-1}\right)=(1+v_3(tm))-(1+v_3(m))=v_3(t), $$
so that $t$ is divisible by $3^s$. This is a contradiction when $t\geq 4$:
$$  3^s=\frac{2^{tm}-1}{4^m-1} > 4(t/2-1)\geq t \geq 3^s, $$
which proves that $t=2$. Now the original diophantine equation becomes
$$ 2^m+1=3^r,$$
where $m$ is odd. Then $1+v_3(m)=v_3(4^m-1)=r$, so that $m$ is divisible by $3^{r-1}$. It follows that $1+2^{3^{r-1}}\leq 3^r$. It is easy to see that this holds only for $r=1$ and $r=2$, which yields $m=1$ and $m=3$, respectively.<|endoftext|>
TITLE: Configuration spaces and non homeomorphic vector bundles
QUESTION [10 upvotes]: Given two complex line bundles over the complex projective line ${\mathbb CP}^1$, prove or disprove that their total spaces are homeomorphic if and only if their Chern numbers are equal up to sign.

This question is a generalization of this question by zygund. In my answer I recalled an idea of Wu Wen-Tsün which can be found in the introduction to his remarkable book A theory of embedding, immersion, and isotopy of polytopes in a Euclidean space: in order to distinguish, up to homeomorphism, two topological spaces that are homotopy equivalent (like any two lines bundles over the projective line), it is fruitful to consider the homotopy type of their configuration spaces. Wu Wen-Tsün's book shows that this is particularly useful in considering embedding problems in Euclidean spaces. 
So my real question is: 
Given two complex line bundles over ${\mathbb CP}^1$ such that their two-point configuration spaces are homotopy equivalent, are their Chern numbers equal up to sign?

REPLY [16 votes]: The answer is yes. The third homology group distinguishes these spaces, namely $H_3(C_2(E_k))=\mathbb{Z}_k$ (up to extension). 
This shows that configuration spaces of homotopic open manifolds are easier to tell apart than configuration spaces of homotopic closed manifolds. The latter have the same 
additive homology, and one needs finer invariants, like Massey products, to tell them apart.
Here is the argument:
The space $C_2(E_k)$ has the homotopy type of a pushout 
$$A_1 \leftarrow A_{12} \rightarrow A_2$$
The space $A_1$ corresponds to pairs $(x,y) \in C_2(E_k)$  such that $\pi(x)$ and $\pi(y)$ are close to each other in $S^2$ (say $<\pi(x),\pi(y)> \geq 0$), and fibers are distinct (use parallel transport). Up to homotopy $A_1$ is the fiberwise configuration space, a bundle $A_1 \to S^2$ with fiber $C_2(\mathbb{R}^2) \simeq S^1$, that is the sphere bundle of $E_k$. The space $A_{12}$ corresponds to the case $<\pi(x),\pi(y)>=0$ and is the pullback of $A_1$ along the projection $SO(3) \to S^2$. Finally $A_2$ corresponds to the case when $\pi(x)$ and $\pi(y)$ are close to antipodal (negative scalar product), and is  homotopically a bundle $A_2 \to S^2$ with fiber $(\mathbb{R}^2)^2$, so $A_2 \simeq S^2$.  The bundle $A_1$ has clutching function $S^1 \times S^1 \to S^1$
$(z,u) \mapsto z^ku$ and $A_{12}$ has clutching function $S^1 \times (S^1 \times S^1) \to S^1 \times S^1 $ $(z,u,v) \mapsto (z^k u, z^2 v)$. 
Apply the Mayer Vietoris sequence to compute the homology of $A_1, A_{12}$, and the induced map.  Then $H_1(A_1)=\mathbb{Z}/k, \, H_2(A_1)=0,\, H_3(A_1)=\mathbb{Z}$; in the odd case $H_1(A_{12})=\mathbb{Z}, \, H_2(A_{12})=0, \, H_3(A_{12})=\mathbb{Z}, \, H_4(A_{12})=\mathbb{Z} .$ 
In the even case $H_1(A_{12})$ and $H_2(A_{12})$ have an extra $\mathbb{Z}/2$
 summand. 
The map $A_{12} \to A_1$ induces multiplication by $k$ (odd) or $k/2$ (even) on $H_3$.
Apply again Mayer Vietoris to compute the homology of $C_2(E_k)$. 
This gives the answer. In the even case either $H_3(C_2(E_k))=\mathbb{Z}_k$ or $H_3(C_2(E_k)) = \mathbb{Z}_{k/2} \oplus \mathbb{Z}_2$  .<|endoftext|>
TITLE: What is the possible usefulness of étale topology and cohomology apart from the resolution of the Weil conjecture ?
QUESTION [5 upvotes]: My question is as stated in the title:
What is the possible usefulness of étale topology and cohomology apart from the resolution of the Weil conjecture ?
I am particularly interested to know if it's reasonably possible to deduce results in areas like 3 and 4 manifolds topology or Kähler geometry from this machinery. Has anyone ever tried ?

REPLY [2 votes]: Not sure that should count, but still: morally there is relation with Milnor's K-groups.
According to Milnor conjecture proved by Voevodsky 2-torsion in Milnor K-groups are the same as Galois cohomology mod Z/2. Etale cohomology is "globaliztion" of the Galois cohomology, and there are certain "globalizations" of Milnor's conjecture. So you get the relation... (As far as I understand Voevodsky extended this to Z/l coefficients also).<|endoftext|>
TITLE: To which extent can one recover a manifold from its group of homeomorphisms
QUESTION [48 upvotes]: Question. Suppose that $M$ is a closed connected topological manifold and $G$ is its group of homeomorphisms (with compact-open topology). Does $G$ (as a topological group) uniquely determine $M$?
One can ask the same question where we regard $G$ as an abstract group (ignoring topology), replace topological category by smooth category (here one can equip $G=\mathrm{Diff}(M)$ with a finer structure of a Frechet manifold), varying degree of smoothness, dropping compactness assumption, recovering $M$ up to homotopy, etc.
I do not know how to answer any of these questions. I do not even know if one can recover the dimension of $M$ from its group of homeomorphisms. In low dimensions, or assuming that $M$ has a locally-symmetric Riemannian metric, and if $\dim(M)$ is given, I know few things. For instance, among 2-dimensional manifolds one can recover $M$ from $G$ since $G/G_0$ is the mapping class group $Mod(M)$ of $M$ and one can tell the genus of $M$ from maximal rank of free abelian subgroups of $Mod(M)$. Same for, say, closed hyperbolic manifolds with non-isomorphic isometry groups. However, given, for instance, two closed hyperbolic 3-manifolds $M_1, M_2$ with trivial isometry groups, I do not know how to distinguish $M_i$'s by, say, $\mathrm{Homeo}(M_i)$ (the problem reduces to a question about homeomorphism groups of the unit ball commuting with $\pi_1(M_i)$, $i=1,2$, but I do not see how to solve it).
Update: Results quoted by Igor and Martin give the complete answer in topological and smooth category in the strongest possible form (much more than I expected!). Positive answer is also known in the symplectic category, but, apparently, is open for contact manifolds and their groups of contactomorphisms.
Another reference in the smooth case, sent to me by Beson Farb is
Banyaga, Augustin, The structure of classical diffeomorphism groups, Mathematics and its Applications (Dordrecht). 400. Dordrecht: Kluwer Academic Publishers. xi, 197 p. (1997). ZBL0874.58005.

REPLY [11 votes]: For the smooth case, the result is in:

Takens, F. (1979). Characterization of a differentiable structure by its group of diffeomorphisms. Bol. Soc. Brasil. Mat., 10, 17–25. MR552032

and the answer is "Yes".
For completeness, Takens' theorem is:
Theorem Let $\Phi \colon M_1 \to M_2$ be a bijection between two smooth $n$-manifolds such that $\lambda \colon M_2 \to M_2$ is a diffeomorphism iff $\Phi^{-1} \circ \lambda \circ \Phi$ is a diffeomorphism.  Then $\Phi$ is a diffeomorphism.<|endoftext|>
TITLE: Did Emmy Noether ever publish under a man's name?
QUESTION [41 upvotes]: A recent article in the New York Times, http://www.nytimes.com/2012/03/27/science/emmy-noether-the-most-significant-mathematician-youve-never-heard-of.html?pagewanted=all says, among other things, "Noether was a highly prolific mathematician, publishing groundbreaking papers, sometimes under a man’s name, in rarefied fields of abstract algebra and ring theory." This is the first I have ever heard of Emmy Noether publishing under a male pseudonym, and I ask whether anyone can confirm, or refute, the assertion in the Times. 
I wonder if the author is confusing Emmy with her mathematician father Max; or if the author has in mind times when Noether gave lectures that were advertised as Hilbert's; or if the author has in mind Sophie Germain, who wrote under the name M. LeBlanc. 
EDIT: I have an answer from the writer, and it appears that Zsban hit the nail on the head in a comment. The writer says her point was badly phrased, and she was referring to Noether's letting (male) students and colleagues publish her ideas as if those ideas were their own. My thanks to all who have contributed here.

REPLY [6 votes]: EDIT: it appears I am behind Gerry by about 16 hours. That's what comes of not  reading ALL the comments. Sigh.
There was an option to email the author by clicking on something, I sent:
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Dear Ms. Angier,
        We are having trouble substantiating your suggestion that Emmy Noether sometimes published under a man's name. She did sometimes have male co-authors, of course. Please see Did Emmy Noether ever publish under a man's name? 
        In short, we think that she never published anything under a man's name. If you know otherwise for certain, I would be interested in details.
Sincerely,
William C. Jagy
Berkeley, CA 
https://mathoverflow.net/users/3324/will-jagy 
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
NOTE this is not the same as a letter to the editor or to a corrections department.<|endoftext|>
TITLE: Relation between topos and $\infty$-topos
QUESTION [10 upvotes]: I'm currently reading The book of Jacob Lurie, 'Higher Topos Theory', and I'm a little confused by the relation between classical topos and $\infty$-topos :
to an $\infty$-topos I can attach the ordinary topos of its $0$-truncated objects. And to a classical topos I have several way to associate $\infty$-topos 'above' it. 
Jacob Lurie (in his book, section 6.4) present this relation as similar to the relation between a classical topos and its locale of sub-terminal objects. In this situation, I know I can have plenty of topoi (a proper class) that are associated to the same locale, even if this locale is just a point. But I have no idea of what happens in the case of $\infty$-Topos : I have seen that in some cases there might be several non equivalent $\infty$-topos above a same ordinary topos, but I see them more like "different ways of doing homotopy theory in the internal logic of $X$ because some classical result of homotopy theory (like Whitehead's theorem) may fail in the internal logic" rather than completely different objects that just share a small property" (like the class of topoi whose locale of subterminal objects is reduced to a point is just the class of topos whose internal logic is two-valued)
So for example : 
Are there several (non equivalent) $\infty$-topoi, whose topos of $0$-truncatued objects is the topos of set ? If it's the case, can I have an example ? Are we able to 'classify' them ?

REPLY [20 votes]: Jacob's answer to your "for example" question is a good one, but let me be so bold as to try to address the general question.  I think there are two different issues in play: the fact that the site of a topos may not be truncated, and the fact that an $(\infty,1)$-topos may not be hypercomplete.  The first is just as much the case for 1-toposes, while the second is a purely $\infty$-phenomenon, and this may be the source of some of your confusion.
Every 1-topos is a left exact localization of a category of presheaves on some 1-category.  If this 1-category is in fact a (0,1)-category (a poset), then the 1-topos is localic and equivalent to the topos of sheaves on its locale of subterminal objects, and the 2-category of localic 1-toposes is equivalent to that of locales.  But, as you note, many different 1-toposes can have equivalent locales of subterminal objects, because we can consider 1-toposes of sheaves on 1-categories that are not posets.
Similarly, every $(\infty,1)$-topos is a left exact localization of a category of presheaves on some $(\infty,1)$-category.  If this $(\infty,1)$-category is in fact a 1-category and the localization is topological, then the $(\infty,1)$-topos is 1-localic and equivalent to the $(\infty,1)$-topos of sheaves on its 1-topos of 0-truncated objects, and the $(\infty,2)$-category of 1-localic $(\infty,1)$-toposes is equivalent to that of 1-toposes.
Jacob's answer points out that even among topological localizations of presheaves, many different $(\infty,1)$-toposes can have equivalent 1-toposes of 0-truncated objects, becuse we can consider $(\infty,1)$-toposes of sheaves on $(\infty,1)$-categories that are not 1-categories.  (His examples, as you note, are categories of presheaves on $\infty$-groupoids.)  This situation is entirely analogous to the 1-topos-theoretic one.
However, in the $\infty$-case there is the separate issue that even a left exact localization of a category of presheaves on a 1-category need not be 1-localic, if the localization is not topological.  For a fixed 1-site, the lattice of corresponding localizations of its $(\infty,1)$-presheaf category, with the topological localization at one extreme and its hypercompletion at the other extreme, is probably what you are thinking of when you talk about "different ways of doing homotopy theory in the internal logic of $X$".  So in this sense, $(\infty,1)$-toposes are more of a generalization of 1-toposes than 1-toposes are of locales.
(Incidentally, I'm not sure how accurate it is to think of the various localizations of presheaves on the site of a 1-topos $X$ as "ways of doing homotopy theory in the internal logic of $X$".  As far as I know, only the hypercomplete such localization can actually be constructed "internally" to the 1-topos $X$, using the model structure on simplicial sheaves.  It seems more accurate to me to describe the others as "ways of extending the internal logic of $X$ to include homotopy theory".  But I'd be interested to hear others' thoughts on that subject.)<|endoftext|>
TITLE: Perron, Fourier
QUESTION [17 upvotes]: Perron´s formula is in some sense just Fourier inversion, but I have never seen proven it that way in a textbook. I take this must be because the conditions for the Fourier inversion formula to hold may be difficult to verify in this case. Or are they? Is it feasible to prove Perron´s formula using mainly just the fact that the Fourier transform is self-dual?

REPLY [4 votes]: I think these answers may be elaborated upon together. Assuming $\sum |a_n| n^{-c}$ converges, the formula to which BR refers explains the sense in which the "traditional" form of Perron's formula is a Fourier integral. By a change of variables, you have 
$$\frac{1}{2\pi}\int_{-T}^{T}\left(\sum_{1}^{\infty}\frac{a_n}{n^{c+iy}}\right)\frac{e^{ixy}dy}{c+iy}=e^{-cx}\sum_{n< e^x}a_n+O\left(\frac{1}{T}\sum_{1}^{\infty}\frac{|a_n|}{n^c|x-\log n|}\right)$$
(see Titchmarsch for a generalization that leads to PNT, p61-63). As $T\rightarrow\infty$, the (sharp) estimate on the r.h.s. shows that the convergence is not uniform, even though the interchange of limits on the l.h.s. is justified because the Dirichlet series converges absolutely (hence uniformly). 
This is obviously where the Schwartz function $\phi$ to which Ben refers justifies appeal to Fourier duality, instead of having to justify interchanging limits beyond the scope of dominated convergence. It also gives you more parameters to tweak for other purposes, if required. Yet, without introducing a further limit, the resulting formula is not Perron's so that doesn't resolve your question (but perhaps Perron's formula is obsolete?). 
As I understand it, this was the precisely the controversy with Riemann's statement of the Fourier expansion of the prime counting function $J(x)$ - he just appealed to Fourier duality and left it there, but that was ultimately verified by Von-Mangoldt, so perhaps there is a proof. Certainly Fourier-Stieltjes applies when the coefficients are positive, so maybe that can be deployed by someone who knows more about measure than me.<|endoftext|>
TITLE: information on an Euler product
QUESTION [5 upvotes]: The following Euler product came up in some sieving applications:
$f(z, s) = \prod_{\mbox{primes}} \left(1-\frac{z}{p^s}\right).$
What is known about this function? (Analytic continuation? Asymptotics?) This must be quite classical...

REPLY [15 votes]: This function has indeed been studied before. For example, its reciprocal is
$$
f(z,s)^{-1} = \prod_p \bigg( 1-\frac z{p^s} \bigg)^{-1} = \sum_{n=1}^\infty z^{\Omega(n)} n^{-s},
$$
where $\Omega(n)$ is the number of prime factors of $n$ counted with multiplicity. When $|z| <2$, this function can be written as
$$
f(z,s)^{-1} = \zeta(s)^z g_z(s)
$$
where the Dirichlet series $g_z$ converges for $\Re s > \frac12$. In particular, for $|z|<2$ the function $f(z,s)$ can be analytically continued to $\Re s > \frac12$ minus a branch cut on $[\frac12,1]$, say. Similar things can be said for larger $|z|$ as well, although now the individual factors in the product defining $f(z,s)$ can vanish.
Information about this function appears in Section 7.4 of Montgomery/Vaughan's Multiplicative Number Theory and Chapter 5 of Tenenbaum's Introduction to Analytic and Probabilistic Number Theory; the key words "Selberg-Delange method" might also help you find references.<|endoftext|>
TITLE: Degree of generators of irreducible components
QUESTION [10 upvotes]: Let $V$ be a Zariski-closed subset of $\mathbb{A}^n_k$, where $k$ is an algebraically closed field. Assume that $V$ may be defined by polynomials of degree at most $d$ (or to put it otherwise $V$ is an intersection of hypersurfaces of degree at most $d$). My question is the following: is it also possible to define the irreducible components of $V$ by polynomials of degree at most $d$?
This is true if $V$ is an hypersurface (the irreducible components are defined by the factors of a polynomial defining $V$), so the answer is positive when $n$ is at most 2. I have managed to prove it in a few other cases but not much and I would appreciate any advice.
There exists algorithms to compute the irreducible components. I have checked a few of them but they could let the degree of generators grow. Any algorithm using Gröbner basis for instance will not fit. For the same reasons, trying to prove the results using projections is probably hopeless, since they may increase the degree of the generators.
A few more remarks:

I am not sure how relevant the fact that $k$ is algebraically closed is, but I suspect there could be very non-trivial arithmetic issues otherwise, even when $V$ is 0-dimensional.
I do not mind to work in the projective space instead of the affine space (it implies the result anyway and makes it easier to deal with degrees).
I do not mind to get the irreducible components only as a set (i.e. the ideal up to a radical).
I tried to make a few computations but found it rather hard. If you know a way to compute the least integer $d$ such that a given Zariski-closed subset may be defined by polynomials of degree $d$, I would also be glad to know.

REPLY [8 votes]: Here's a counterexample with $n=d=3$.
Let $C$ be the rational curve $\lbrace (x,y,z) = (t,t^4,t^6) \rbrace$.
Then the space $S$ of cubics that vanish on $C$ is the span of
$\lbrace x^2 y - z, x^2 z - y^2, y^3 - z^2 \rbrace$.
But all such cubics vanish also on the line $y=z=0$.  Therefore
we can take $V$ to be the zero-locus of any two-dimensional
subspace of $S$, and $C$ will not be cut out by the space $S$
of cubics vanishing on $C$ (even though for this example we defined
$V$ by a proper subspace of $S$).<|endoftext|>
TITLE: How can one prescribe the pairwise intersection measuress of $n$ sets?
QUESTION [10 upvotes]: Take $n\geq 1$, and $m_{ij}\in [0,1], 1\leq i,j \leq n$. Under what conditions is it possible to find measurable subsets $X_1,...,X_n$ of, say, $[0,1]$, such that $leb(X_i\cap X_j)=m_{ij}$?
Some relations are necessary, like $m_{ii}\geq m_{ij}$, or the fact that the matrix $(m_{ij})$ must be semi-definite positive, but it does not appear to be sufficient.
The same question holds with $m_{ij}\in \mathbb{R}_+, X_i \subset \mathbb{R}$.

REPLY [3 votes]: Here is a partial idea  where the last step needs a bit more justification  which doesn't work as well as I had hoped, but may yet have promise. 
As noted, this is a question about adding real numbers, no specialized measure theory is involved although the language is convenient.   We will   had hoped to define $2^n$ values $t_I$, one for each $I \subseteq [n]=\lbrace1,2,\cdots,n\rbrace $ so that if they are all non-negative, then the specified values can be achieved, otherwise they can not. 
Suppose that we have a set $U$ of size (or measure) $m_{\emptyset}=|U|$ along with $n$ subsets $X_i$ for $i \in [n]$ each with complement $\overline{X_i}.$ For each $I \subseteq [n],$ let $m_I=|\cap_{i \in I}X_i|$ while $t_I=|(\cap_{i \in I}X_i) \cap (\cap_{j\notin I}\overline{X_j})|.$ As in the problem we can abbreviate $m_i=m_{i,i}$ for $m_{\lbrace i \rbrace}=|X_i|$ and $m_{i,j}$ for $m_{\lbrace i,j\rbrace}=|X_i\cap X_j|.$ 
If we know either set of values we can uniquely find the others:  $$m_I=\sum_{I \subseteq J \subseteq [n]}t_J\hspace{0.1in}  \text{ while } t_I=\sum_{I \subseteq J \subseteq [n]}(-1)^{|J|-|I|}m_J$$
If the $m_I$ are given, then the $t_I$ will be determined over the reals, but we want non-negative values.  In the given problem we have only $n+\binom{n}{2}$ of the $m$ values, perhaps along with $m_{\emptyset}=1.$
So we first define the rest of the $m_I$ by $m_I=\min_{i,j \in I}m_{i,j}$ then solve for the $t_I$ and check that they are non-negative. I think that these choices will make $t_{[n]}$ as large as possible and thought that they would give all the $t_I$ the best chance to be non-negative consistent with the given information.
BUT now I notice problems with the simple case of asking for the sides of a triangle: $|U|=3$ $|X_1|=|X_2|=|X_3|=2$ and $|X_1 \cap X_2|=|X_1 \cap X_3|=|X_2 \cap X_3|=1$ If we do add the final condition $m_{\lbrace 1,2,3 \rbrace}=|X_1 \cap X_2 \cap X_3|=0$ then we do get the desired solution $t_{\{1,2\}}=t_{\{1,3\}}=t_{\{2,3\}}=1$ with the rest of the $t_I=0.$ HOWEVER if we make my suggested choice of $m_{\lbrace 1,2,3 \rbrace}=|X_1 \cap X_2 \cap X_3|=1$ then we do get $t_{\{1,2,3\}}=1$ as large as possible but then $t_{\{1,2\}}=t_{\{1,3\}}=t_{\{2,3\}}=0$ making $t_{\{1\}}=t_{\{2\}}=t_{\{3\}}=1$ and finally $t_{}=-1$
Think of the values $t_I$ as weights to be assigned to the regions of an ideal Venn diagram for $n$ sets. The conditions on the $m_i$ and $m_i,j$ (along perhaps with $m_{\emptyset}=1$) give $\frac{n^2+n}2$ (or else  $\frac{n^2+n}2+1$ ) equations in $2^n$ variables $t_I$ along with the side condition that all the $t_I \ge 0.$ This is a linear programming problem. Perhaps there is a way to start with my assignmet and then adjust the values until success or proved failure, but I am not sure.
later The convex hull might be unworkable once the number of dimensions is $43$ or $43+\binom{43}{2}$ or $2^{43}$, but maybe not.  I found the comments on integral problems pretty convincing and asked this question. However I subsequently answered it and realized that, while there is no projective plane of order $6$, We can achieve $m_{\emptyset}=43, m_i=7$ for $1 \le i \le 43$ and all $m_{i,j}=1$ by setting $|t_I|=0$ except that $|t_J|=1/\binom{41}{5}$ for all $\binom{43}{7}$ choices of $J$ with $|J|=7.$ So now I am back to thinking that there might be a simple algorithm which achieves the constraints with least $L_1$ error (so $0$ if possible.)

REPLY [3 votes]: The discrete version of this problem, where the $X_i$ are subsets of $\{1,2,\ldots,n\}$, is extremely difficult. For example, it includes the questions of the existence of Hadamard matrices and the existence of finite projective planes, which both remain unsolved despite a huge effort. So one can't expect to have necessary and sufficient conditions that are routine to check.  It could be that the continuous problem is easier than the discrete one, though it isn't obvious to me.  Actually I think the case when the $\{m_{ij}\}$ are integers is the same as the discrete problem since then each of the sets can be written as the disjoint union of atoms of measure 1.<|endoftext|>
TITLE: Does the hyperfinite II_1 factor admit two irreducible representations that are not unitarily equivalent?
QUESTION [13 upvotes]: Regarding the hyperfinite $II_{1}$ factor $R$ as $C^{*}$-algebra, is it known whether any two irreducible representations of $R$ are unitarily equivalent? If it is known that there exists a pair of irreducible representations that are not unitarily equivalent, please provide a reference.
(Note: I am not requiring that the representations be normal!!)

REPLY [18 votes]: There are $2^c$ mutually non-equivalent irreducible representations. Since $\ell_\infty(N)$ has $2^c$ many pure states (there are $2^c$ many ultrafilters on $N$), any $C^*$-algebra containing $\ell_\infty(N)$ has at least as much pure states. Since $R$ has $c$ many unitary elements, there are $2^c$ many mutually non-equivalent pure states. This is a very old observation, but I don't know a specific reference. (It was noted in our paper too. http://xxx.lanl.gov/abs/math.OA/0110152)<|endoftext|>
TITLE: long enough interval of integers to solve a simultaneous congruence
QUESTION [26 upvotes]: Let $a$, $b$ be two coprime natural numbers. Let $A \subseteq \{0,1,\ldots, a-1\}$ and $B \subseteq \{0,1,\ldots,b-1\}$ be two nonempty sets, which we think of as sets of residues mod $a$ and $b$ respectively.
I would like to know if anyone has ever seen (or knows a proof for) the following result: that any interval of $(a - |A| + 1)(b - |B| + 1)$ consecutive integers contains a number $x$ such that $x$ mod $a$ is in $A$, while $x$ mod $b$ is in $B$.
Actually, I do have a proof of this result, but it's complicated, and I have no proof for the $k$-variate case. To be precise the $k$-variate case is the following: we have $k$ natural numbers $a_1, \ldots, a_k$ that are pairwise coprime, and nonempty sets of residues $A_1, \ldots, A_k$ where $A_i$ is a set of residues mod $a_i$. The question is to show that any interval of at least
$$
(a_1 - |A_1| + 1)(a_2 - |A_2| + 1)\cdots (a_k - |A_k| + 1)
$$
consecutive integers contains an integer $x$ such that $x$ mod $a_i$ is in $A_i$ for $i = 1, \ldots, k$. The case $k = 1$ is obvious, for the case $k = 2$ I have a proof, and for $k \geq 3$ I only have a partial result, namely that the statement holds as long as the interval length mentioned above is strictly greater than
$$
\sum_i \prod_{j\ne i} a_j.
$$
Would be grateful if people could tell me what they know about this problem, or their insights. Thanks!

REPLY [14 votes]: [Edited again  mostly to spell out why the $m_j \bmod a$ are distinct.]
Yes, the desired result is true for all $k$.  The following proof
is elementary but possibly more algebraic than expected (apparently
some kind of variant of the "polynomial method" in combinatorics, though
with no need for anything as advanced as the "combinatorial Nullstellensatz").
This would make for a good Putnam B-6 problem; indeed I wouldn't be surprised
if this question has already been used for such a competition.
Let $a_1,\ldots,a_k$ be pairwise coprime positive integers,
and set $a = \prod_{i=1}^k a_i$.
For each $i$ let $A_i$ be a nonempty subset of ${\bf Z} / a_i {\bf Z}$,
and let $Z_i$ be the complement of $A_i$ in ${\bf Z} / a_i {\bf Z}$.
Let $A \subseteq {\bf Z} / a {\bf Z}$ consist of the residues $n \bmod a$
such that $n \bmod a_i \in A_i$ for each $i$.
Let $Z$ be the complement of $A$ in ${\bf Z} / a {\bf Z}$, consisting of
the residues $n \bmod a$ such that $n \bmod a_i \in Z_i$ for some $i$.
We claim:
Proposition. This set $Z$ cannot contain a run of
$$
N := \prod_{i=1}^k (a_i - |A_i| + 1) = \prod_{i=1}^k (|Z_i| + 1)
$$
consecutive residues $\bmod a$.
Proof: Let $w \in {\bf C}$ be a primitive root of unity of order $a$,
so that $w_i := w^{a/a_i}$ is a primitive root of unity of order $a_i$
for each $i$.
Set $W_i := \lbrace w_i^n | n \in Z \rbrace$, a set of size $|Z_i|$, and
$P_i(X) := \prod_{x \in W_i} (X-x)$, which is thus a polynomial
of degree $|Z_i|$.  Then for any $n \bmod a$ we have $n \in Z$ iff
$$
0 = \prod_{i=1}^k P_i(w_i^n) = P(w^n),
$$
where $P$ is the polynomial defined by
$$
P(X) := \prod_{i=1}^k P_i(X^{a/a_i}).
$$
Because each $P_i(X^{a/a_i})$ is the sum of at most $|Z_i|+1$ monomials,
their product $P$ is the sum of at most $\prod_{i=1}^k (|Z_i|+1) = N$
monomials, say
$$
P(X) = \sum_{j=1}^N c_j X^{m_j}.
$$
The $N$ exponents $m_j$ are the integers of the form $a \sum_{i=1}^k b_i/a_i$
with $0 \leq b_i \leq |Z_i|$.  Since each $|Z_i| < a_i$ (this is where
we use the hypothesis $A_i \neq \emptyset$) and the $a_i$ are pairwise
coprime, it follows that these $m_j$ have pairwise distinct residues 
$\bmod a$.
We claim, then, that for each $n \bmod a$ at least one of 
$P(w^n), P(w^{n+1}), \ldots, P(w^{n+N-1})$ is nonzero.
Suppose not.  Then $(c_1,\ldots,c_N)$ would be a nonzero solution of
the $N \times N$ linear system
$$
\sum_{j=1}^N w^{m_j (n+k)} c_j = 0 \phantom{\infty} (k=0,1,\ldots,N-1).
$$
Hence $(w^{-nm_1^{\phantom.}} c_1^{\phantom.}, \ldots, w^{-nm_N^{\phantom.}} c_N^{\phantom.})$
would be a nonzero vector
in the kernel of the Vandermonde matrix with entries $(w^{m_j})^k$.
But then some two $w^{m_j}$ would coincide, contradicting our observation
that the residues $m_j \bmod a$ are distinct.  This completes the proof.
P.S. Note that we do not even need the formula for
the determinant of a Vandermonde matrix, only
the fact that it is invertible, which can be obtained by interpreting the
kernel of the transposed matrix as the space of polynomials of degree
less than $N$ that vanish at the $N$ distinct points $w^{m_j}$.<|endoftext|>
TITLE: A measure theory question
QUESTION [11 upvotes]: Here's an interesting problem one can formulate for a student. This problem arises when considering special ergodic theorems:
On a finite dimensional manifold $M$ with a Lebesgue measure $\mu$, does every measure zero set equal a countable union of the sets of less than full Hausdorff dimension? 
For a diffeomorphism $f$ of $M$ and a continuous function $\varphi$ on $M$, define $$\overline \varphi = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_0^{n-1} \varphi \circ f^k(x).$$ Then the Birkhoff theorem asserts that for almost all $x$,  $\overline \varphi \rightarrow \int_M \varphi, n \rightarrow \infty$. But consider the set $K_{\alpha}$ of $x$ where $$\alpha \leq |\overline \varphi - \int \varphi|.$$
So Birkhoff says $\mu(K_\alpha)=0$, but what about the Hausdorff dimension of $\mu(K_\alpha)=0$?  For some diffeomorphisms, for example hyperbolic maps,
it was proven that $\dim_H K_\alpha < \dim X$. 
That fact gives a rise to my question. I expect a negative answer, but I can not find a counterexample.

REPLY [10 votes]: Consider a function $h$ defined on the unit interval $[0,1]$ which is monotone 
nondecreasing and for which $h(0)=0$, $h$ continuous at $0$. We may define 
a Hausdorff measure $H_h$ associated to $h$ (see Donoghue, Distributions and Fourier 
Transforms Academic Press New York 1969  p. 30--35, or  C. A. Rogers, Hausdorff Measures,
Cambridge University Press, 1970). When $h(x)=x^\alpha$ you get the ordinary 
Hausdorff measures $H_\alpha$.  Consider also $f(x)=x\log(e/x)$. 
Then a set $A\subset{\bf R}$ with  $0<  H_f(A)<1$  has measure of Lebesgue $0$ but it is not 
union of a numerable set $A=\bigcup A_n$  with  $H_{\alpha_n}(A_n)=0$ and 
$0<\alpha_n<1$, because this implies  $H_f(A_n)=0$ and so $H_f(A)=0$.
As similar construction applies to each ${\bf R}^n$.<|endoftext|>
TITLE: Transporting model structures via adjunctions
QUESTION [6 upvotes]: Hello,
If $F$ is a left adjoint between $C$ and $D$, and $D$ has a model structure; We can define cofibrations and equivalences in $C$ to be those that are so after applying $F$. What are criterions for this to define a model structure? Where can I find a discussion? Also, the dual question about a right adjoint.
Thank you,
Sasha

REPLY [11 votes]: Since in practice the dual situation occurs more often than the one you stated I will dualize your question. Given categories $\bf {D}$ and $\bf {E}$ and adjoint functors $F:\bf {D}\to \bf {E}$ and $G:\bf {E}\to \bf {D}$, with $F$ left adjoint to $G$, if $D$ is a model category then define weak equivalences and fibrations in $E$ as follows. Declare an arrow $f$ in $E$ a weak equivalence (resp. a fibration) if $G(f)$ is a weak equivalence (resp. fibration). 
Then it is guaranteed that with these fibrations and weak equivalences $E$ is a cofibrantly generated model category provided the following conditions hold:
1) $D$ is cofibrantly generated. 
2) The left adjoint $F$ preserves small objects. 
3) The weak equivalences in $E$ contain any sequential colimit of pushouts of images $F(g)$, where $g$ is allowed to vary over the generating trivial cofibrations in $D$.
Of course condition 3 is difficult to check so it is good to know that it is satisfied automatically if:
$E$ has a functorial fibrant replacement and $E$ has functorial path objects for fibrant objects.
First appearance in the literature of such transfer principles seems to be S. E. Crans – Quillen closed model structures for sheaves, J. Pure Appl.
Alg. 101 (1995), 35–57.
The more recent (and where I learned of the answer) article "Axiomatic homotopy theory for operads", by Berger and Moerdijk can be found here:http://arxiv.org/abs/math/0206094 where the result is used to establish a model structure on (one-object) operads.
Of course a dual answer will provide a transfer principle for fibrantly generated model categories. However, as interesting/important fibrantly generated model categories are very rare such a transfer result is hardly useful (showing that mathematics is not self-dual).<|endoftext|>
TITLE: Maslov index and Heegard Floer homology
QUESTION [11 upvotes]: I am an undergraduate who wants to learn Knot Floer homology. I was told to start with this   expository paper which was working quite well until I reached the actual definition of the differential. The definition involves counting the holomorphic representatives of $\phi \in \pi_2(x,y)$ when $\phi$ satisfies $\mu(\phi) = 1$. 
First of all I do not understand how one might tell if $\mu(\phi) = 1$. The only definition of $\mu$ given in the paper is that it is the "expected dimension" of the moduli space of holomorphic representatives. I was wondering what is meant here? I have read other definitions of $\mu$ that define it as the index of an operator and while this may be precise I do not think it is what was intended to be used to work out the simple examples and exercises given in the paper.
Specifically, there are quite a few problems for the reader in the paper where a picture of the domain of $\phi$ namely $\mathcal{D}(\phi)$ is given and one is expected to compute $\mu(\phi)$ (see for instance page 17). I was wondering how one might even begin to do this?
For instance one can find a $\phi$ so that $\mathcal{D}(\phi)$ is the annulus given in figure $4$. We are later told in the paper that in this case $\mu(\phi) = 0$. How might one figure something like this out? In other examples $\mu$ is negative, how might one see this as well? Since the paper is for a beginner I am hoping there is an answer suitable for a beginner.
To sum up.
In what cases is it possible to just look at $\mathcal{D}(\phi)$ and use the picture to compute $\mu(\phi)$ and how does one go about it? I would be very happy if someone explained any of the examples given in the paper or simply point me to a reference. I feel like I am missing something simple and any help would be greatly appreciated.

REPLY [11 votes]: Knowing that the maslov index is equal to the "expected" dimension of the moduli space of the disks is helpful too. For example in the figure on the right side of page 17, you can see that a holomorphic disk with boundary on $\alpha$ and $\beta$ can have "cuts" along the dashed lines, i.e. can send part of the boundary of the disk to the dashed lines. Since the length of these two "cuts" are variable, the moduli space is two dimensional and therefore the Maslov index is two. This also shows you that the moduli space is not compact. (Why?)<|endoftext|>
TITLE: Is the class of additive groups of rings axiomatizable?
QUESTION [15 upvotes]: I know that it is impossible to axiomatize the multiplicative structures of rings, called $R$-semigroups. Is anything known about the first-order axiomatizability of the class of abelian groups which are additive groups of some ring? I don't want to restrict the meaning "ring" here. I would like to know whether this question is answered for rings with any subset of the set of adjectives {"associative", "unitary", "commutative"} attached.
EDIT I forgot to mention that I do want to exlude some rings, that is rings with zero multiplication.

REPLY [16 votes]: The answer to the question in the title is No. You can prove this from the work of Wanda Szmielew on the elementary properties of abelian groups. This answer works for any kind of nonzero, bi-additive, binary multiplication (associative or not, commutative or not, unital or not).
In particular, an abelian group $A$ is elementarily equivalent to our favorite  group $\mathbb Z$ iff $A$ is torsion free and of $p$-rank 1 for every prime $p$. The $p$-rank of $A$ is defined to be the minimum of $\textrm{dim}_{\mathbb Z_p}(A/pA)$ or $\omega$. An example of a torsion free abelian group of $p$-rank 1 for every $p$ is the subgroup $S\leq \mathbb Q$ consisting of rationals with square free denominator.  
So $\mathbb Z$ and $S$ are elementarily equivalent. $\mathbb Z$ is the additive subgroup of the unital ring $\mathbb Z$, while (as Tom Goodwillie has pointed out) $S$ is not the additive subgroup of any unital ring. In fact, it is impossible to equip $S$ with any nonzero bi-additive multiplication. For, if $s, t\in S$, then $s$ and $t$ are $p$-divisible for almost all primes $p$. By bi-additivity, $s*t$ is $p^2$-divisible for almost all primes $p$. But the only element of $S$ that is $p^2$-divisible for almost all $p$ is 0, so $s*t=0$.<|endoftext|>
TITLE: Program for computing group cohomology
QUESTION [15 upvotes]: Is there any computer program with which I can compute the group cohomology H^n(G,V) for a group G acting linearly on a vector space? 
I mainly care about infinite groups.

REPLY [18 votes]: GAP can be used to compute free $\mathbb ZG$-resolutions of $\mathbb Z$ for some infinite groups G. These resolutions can then be used to compute homology and cohomology of $G$. I hope the following examples give a flavour of the kind of infinite groups that can be handled at present. (A range of finite groups can also be handled.)

EXAMPLE   $H_{99}(SL_2(\mathbb Z[1/7]),\mathbb Z) = \mathbb Z_4 \oplus \mathbb Z_{12}$
gap> R:=ResolutionSL2Z(7,100);;
gap> Homology(TensorWithIntegers(R),99);
[ 4, 12 ]

EXAMPLE $H_3(SL_3(\mathbb Z),\mathbb Z) = \mathbb Z_{12} \oplus \mathbb Z_{12}$ 
gap> R:=ResolutionArithmeticGroup("SL(3,Z)",4);;
gap> Homology(TensorWithIntegers(R),3);
[ 12, 12 ]

EXAMPLE $H_3(SL_2(O_{-11}),\mathbb Z) = \mathbb Z_2 \oplus \mathbb Z_{24}$ where $O_{-11}$ is the ring of integers of $\mathbb Q(\sqrt{-11})$. 
gap> R:=ResolutionArithmeticGroup("SL(2,O-11)",4);;
gap> Homology(TensorWithIntegers(R),3);
[ 2, 24 ]

EXAMPLE $H_3(G,\mathbb Z)= \mathbb Z_2 \oplus \mathbb Z_2 \oplus \mathbb Z$ for $G$ the space group P62. 
gap> G:=Image(IsomorphismPcpGroup(SpaceGroupBBNWZ("P62")));;
gap> R:=ResolutionAlmostCrystalGroup(G,4);;
gap> Homology(TensorWithIntegers(R),3);
[ 2, 2, 0 ]

EXAMPLE  $H_3(G,\mathbb Z) = \mathbb Z_2 \oplus \mathbb Z^{110}$ for $G$ the Heisenberg group of degree 5.
gap> R:=ResolutionNilpotentGroup(HeisenbergPcpGroup(5),4);;
gap> Homology(TensorWithIntegers(R),3);
[ 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ]

EXAMPLE $H_3(G,\mathbb Z) = \mathbb Z_2$ for $G$ the braid group on eight braids (or seven generators). 
gap> Dynkin:=[[1,[2,3]],[2,[3,3]],[3,[4,3]],[4,[5,3]],[5,[6,3]],[6,[7,3]]];;
gap> R:=ResolutionArtinGroup(Dynkin,4);;
gap> Homology(TensorWithIntegers(R),3);
[ 2 ]<|endoftext|>
TITLE: Perfect matching in a vertex-transitive hypergraph
QUESTION [7 upvotes]: In connection with this MO problem, I wonder whether the hypergraph in
question was actually vertex-transitive. And so, as a natural variation (and,
perhaps, a refinement):

If the vertex set of a vertex-transitive hypergraph $H$ can be partitioned as
  $V_1\cup\cdots\cup V_r$, so that every edge of $H$ contains exactly one
  vertex from each of the partite sets $V_i$, what reasonable conditions
  guarantee that $H$ possesses a perfect matching?

(As an example of a reasonable condition: $H$ is non-empty. An unreasonable
condition would be that $H$ is (almost) complete in the sense that it
contains the edge $\{v_1,\ldots, v_r\}$ for (almost) any $v_1\in V_1,\ldots,
v_r\in V_r$.)

The case $r=2$ is easy: we are then looking at vertex-transitive bipartite
graphs, and every such graph has a perfect matching by Hall's marriage
theorem (provided it is non-empty). Indeed, it suffices that the graph be
regular. For $r=3$ vertex-transitivity is insufficient as shows, for instance,
the following construction. Let $G$ be a finite abelian group of order
divisible by $2$, but not by $4$. Let $V_1,V_2,V_3$ be (disjoint) copies of
$G$, and consider the hypergraph $H$ on the vertex set $V_1\cup V_2\cup V_3$
whose edges are all triples $(v_1,v_2,v_3)$ with $v_1+v_2+v_3=0$. If a
perfect matching in $H$ existed, then the sum of all elements of $G$,
multiplied by $3$, would be equal to $0$, which is not the case.

REPLY [2 votes]: This answer is not for vertex transitive hypergraphs (I have not noticed that condition)!
No simple necessary and sufficient condition can exists as 3DM is NP-complete:
http://en.wikipedia.org/wiki/3-dimensional_matching
Of course, if you are only looking for a sufficient condition, one can come up with several, eg. see:
http://arxiv.org/abs/1101.5830 where it is proved by Imdadullah Khan that
"A perfect matching in a 3-uniform hypergraph on $n=3k$ vertices is a subset of $\frac{n}{3}$ disjoint edges. We prove that if $H$ is a 3-uniform hypergraph on $n=3k$ vertices such that every vertex belongs to at least ${n-1\choose 2} - {2n/3\choose 2}+1$ edges then $H$ contains a perfect matching. We give a construction to show that this result is best possible."<|endoftext|>
TITLE: probability theory for combinatorialists
QUESTION [8 upvotes]: More than one combinator(ial?)ist has asked me to recommend a good book to learn probability from, and I never know what to say; the probability theory that I use in my research up was mostly learned piecemeal.  (The stuff I learned in grad school from reading Chung and Feller hasn't been as useful, and I didn't especially enjoy those books.)  Any suggestions?

REPLY [9 votes]: The Probabilistic Method by Noga Alon and Joel Spencer!
Not a probability textbook per se ---Feller or whatever for that---
but sufficiently self-contained that one can learn the tools as one 
sees them applied -- to combinatorics!<|endoftext|>
TITLE: Distribution of primes in small intervals
QUESTION [6 upvotes]: Let $\pi(x)$ be the number of primes smaller than $x$. Do there exist unconditionally universal constants $c > d$ such that 
$$
\lim_{x \rightarrow \infty} \frac{\pi(x + \log^c x) - \pi(x)}{\log^{c-d} x} \geq 1
$$
We know that by Maier Theorem, it is not possible that $c = d+1$. 
By Selberg theorem, for any function $y(x)$ grows faster than $\log^2 x$, it holds that
$$
\lim_{x \rightarrow \infty} \frac{\pi(x + y) - \pi(x)}{y/\log x} = 1
$$
for  $ \textit{almost all}\ \ x$ (assuming the Riemann hypothesis). Does it hold for  $ \textit{all}\ \ x $
if $y(x) = \log^c x$ for some constant $c$ (with Riemann hypothesis)?

REPLY [6 votes]: Assuming the Riemann Hypothesis, I believe the best known result is due to Cramer (I cannot figure out how to add the accent of the e) and it says the following:
There is a constant $C > 0$ such that if if the Riemann Hypothesis is true, then for every $x \ge2$ the interval $(x, x + C \sqrt{x} \log x)$ contains at least $\sqrt{x}$ prime numbers.
This is Theorem 13.3 in Montgomery and Vaughan's Multiplicative Number Theory.
Translating things from $\pi(x)$ to $\psi(x)$, exercise 2, pp. 430-431 of the same book outlines a proof that the Riemann Hypothesis implies that
$$\psi(x+y)-\psi(x)=y+O\left(\sqrt{x} \log x \log\left(\frac{2y}{\sqrt{x} \log x}\right) \right).$$
Thus an asymptotic holds as soon as $\frac{y}{\sqrt{x} \log x} \to \infty$. This formula simultaneously implies both Cramer's result and von Koch's well-known result that 
$$ \psi(x) = x + O(\sqrt{x}\log^2 x) \quad \text{equivalently } \quad \pi(x) = \int_2^x \frac{dt}{\log t} + O(\sqrt{x}\log x) $$
assuming the Riemann Hypothesis.<|endoftext|>
TITLE: Lengths over a local ring
QUESTION [12 upvotes]: Let $A$ be a noetherian domain, $\mathfrak{m}$ а maximal ideal, $s$ a non-zero element of $\mathfrak{m}$, $d= \dim A_\mathfrak{m}$.
Is the following claim true?
Claim:
For any $\epsilon>0$, there exists a positive integer $n$ s.t. for any ideal $I$ satisfying
1) $ I \subset\mathfrak{m^n}$
2) $\sqrt  I = \mathfrak{m}$
3) $I$ can be generated by $d$ elements,
the following holds: 
$$ \mbox{length}(A/(I+As)) /\mbox{length}(A/I) < \epsilon$$
Note: The following example shows that the claim can be false if one drops the requirement that that the number of generators of $I$ be bounded. 
Example: $A:= k[x,s]$, and let $\mathfrak{m}$ denote the ideal $(x,s)$. Let $I_{n,m}$ be an ideal of $A$ given by 
$$ I_{n,m}= s\mathfrak{m}^{n-1} + \mathfrak{m}^m$$.
We can calculate that for any $n$, 
$$\lim_{m\to \infty}  \mbox{length}(A/(I_{n,m}+As)) /\mbox{length}(A/I_{n,m}) = 1$$

REPLY [4 votes]: Here is a counterexample for you question.
Let $A = k[[s, x]]$, $\dim A = 2$
For each pair $n, m$, $n < m$, we consider the parameter ideal
$$\mathfrak{q}_{n, m} = (s^n+x^m, sx^{n-1})$$
We have $\mathfrak{q}_{n, m} + sA = (s, x^m)$. Hence 
$$\ell(A/(\mathfrak{q}_{n, m} + sA)) = m$$
On the other hand, we can check that $s^{n+1}$ and $x^{m+n-1}$ is contained in $\mathfrak{q}_{n, m}$. Thus
$$\ell(A/\mathfrak{q}_{n, m}) \leq \ell(A/(s^{n+1}, sx^{n-1},x^{m+n-1})) = m + n^2-1.$$
Therefore
$$\lim_{m \to \infty} \ell(A/(\mathfrak{q}_{n, m} + sA))/ \ell(A/\mathfrak{q}_{n, m}) = 1$$
Remark:
(i) It should be noted that, I contruct this example based thinking the minimal reduction of the ideal $I_{n,m}$ of your question.
(ii) Your question is true in the case $I = \mathfrak{m}^n$, it means
$$\lim_n \;\ell(A/(\mathfrak{m}^n + sA))/ \ell(A/(\mathfrak{m}^n) = 0.$$<|endoftext|>
TITLE: Why $\partial$ and $\bar{\partial}$ defined in that way (the Wirtinger derivatives)?
QUESTION [11 upvotes]: For $\mathbb{C}$-valued functions, why are $\frac{\partial}{\partial z}$ and  $\frac{\partial}{\partial \bar{z}}$ defined as
$$
  \frac{\partial}{\partial z}=
  \frac{1}{2}\left(
    \frac{\partial}{\partial x} - i
    \frac{\partial}{\partial y}
  \right)
  \quad \quad
  \frac{\partial}{\partial \bar{z}}=
  \frac{1}{2}\left(
    \frac{\partial}{\partial x} + i
    \frac{\partial}{\partial y}
  \right)
$$
where $x$ and $y$ are the real and imaginary parts, respectively.
It seems to be me that the inventor of this notation could have easily reverse the signs: Let $\frac{\partial}{\partial z}$ have a $+$ in the middle and let $\frac{\partial}{\partial \bar{z}}$ have a $-$ in the middle instead.  Is there any reason that they are defined the way they are now? The same question extends to the $\partial$ and $\bar{\partial}$ operators.
One possible guess is that in complex analysis, one usually works with holomorphic functions, so one operator is used much more often than the other. We probably want the more frequently used one to be the one that is easier to write. But I really doubt this is the reason. After all, it's just one extra stroke.

REPLY [6 votes]: By the chain rule,
$$
\frac{\partial}{\partial z} = \frac{\partial}{\partial x} \cdot \frac{\partial x}{\partial z} + \frac{\partial}{\partial y} \cdot \frac{\partial y}{\partial z}. 
$$
Now
$$\frac{\partial x}{\partial z} = \frac{\partial \frac{z + \bar{z}}{2}}{\partial z} = \frac{1}{2}
\quad \text{and} \quad \frac{\partial y}{\partial z} = \frac{\partial \frac{z - \bar{z}}{2 i}}{\partial z} = \frac{1}{2i} = - \frac{i}{2}.$$
Therefore
$$
\frac{\partial}{\partial z} = \frac{1}{2}\frac{\partial}{\partial x}  - \frac{i}{2} \frac{\partial}{\partial y}. 
$$
Similar derivation gives the formula for $\frac{\partial}{\partial \bar{z}}$.<|endoftext|>
TITLE: Has the Weil conjectures been proved using other (Weil) cohomology theory?
QUESTION [14 upvotes]: According to Weil, the Weil conjecture should follow once one has a sufficiently powerful cohomology machine. And it is proved using one of them, namely étale cohomology. 
My question is, has there been any attempt, after its proof using étale cohomology, to prove it using other Weil cohomology theories? i.e. a cohomology theory which has finiteness, allow Poincaré duality, Künneth formula, cycle map, weak and strong Lefschetz. After all it is a motivic thing.

REPLY [20 votes]: Yes.  See Kedlaya's Fourier transforms and p-adic Weil II.
This is a proof using Berthelot's rigid cohomology.<|endoftext|>
TITLE: Milnor exact sequence in $K(n)$ local Morava $E$-theory
QUESTION [8 upvotes]: Let $L_E$ denote Bousfield localisation with repsect to the cohomology theory $E$. I am trying to follow through some calculations in Hovey-Strickland's paper Morava $K$-theories and localisation
Claim 7.10(e) is that
$$L_{K(n)}X = \underset{\leftarrow}{\text{holim}}_I L_{E(n)}X \wedge S/I$$
where the homotopy limit is over a tower of generalised Moore spectra (the Moore spectra $S/I$ is defined in 4.12 and the tower in 4.22)
Define
$$E_*^{\vee}:=\pi_{*} (L_{K(n)} (E \wedge X))$$
the $K(n)$-local version of Morava $E$-theory (where, I believe, Morava $E$-theory here is what I might call a completed Johnson-Wilson theory, but I don't believe it really matters).
The claim (8.04) is then that we can extract a Milnor exact sequence from this:
$$0 \to \varprojlim_I {}^1 (E/I)_{\ast+1}(X) \to E_*^\vee X \to \varprojlim_I (E/I)_*X \to 0 $$
I'm not sure how to show this. I would like to think that we can get a sequence
$$0 \to \varprojlim_I {}^1 \pi_{\ast+1}(L_{E(n)}(E \wedge X) \wedge S/I)\to E_{\ast}^\vee X \to \varprojlim_I \hspace{1mm} \pi_* (L_{E(n)}(E \wedge X) \wedge S/I)  \to 0 $$
and then if you drop the $E(n)$-localisation it seems to work, but I'm not really sure about this.

REPLY [9 votes]: In this case, this follows because $E(n)$-localization is a smashing localization.  Specifically,
$$
L_{E(n)}(X) = L_{E(n)}(\mathbb{S}) \wedge X
$$
for any $X$.  In particular, this means that the smash product of any spectrum with an $E(n)$-local spectrum is already $E(n)$-local.
The fact that $E(n)$-localization is smashing was one of the Ravenel conjectures; specifically, conjecture 10.6 in his paper "Localization with respect to certain periodic homology theories".  I don't have a copy of his orange book handy and so can't give you a reference for the theorem statement.
As a result, as the Morava $E$-theories are $E(n)$-local already, so are the smash products $E \wedge X$ in your equation.<|endoftext|>
TITLE: Significance of the vanishing of $K_{-1}(A)$
QUESTION [6 upvotes]: In M. Schlichting's paper, he defines the negative $K$-theory for derived categories. In this he states that for $\mathcal{A}$ an idempotent complete (see below) triangulated category, $K_{-1}(\mathcal{A}) = 0$ is equivalent to the fact that for any full triangle embedding $\mathcal{A} \hookrightarrow \mathcal{B}$ of $\mathcal{A}$ into an idempotent complete triangulated category $\mathcal{B}$, we have that $\mathcal{B}/\mathcal{A}$ is also idempotent complete.
An additive category is said to be idempotent complete (sometimes called Karoubian) if for every idempotent $e: A \rightarrow A, \hspace{2pt} e^2 = e$ defines a splitting $A = Im(e)\oplus Ker(e)$. 
My question is, what is the importance of this quotient category $\mathcal{B}/\mathcal{A}$ being idempotent complete?
I would even be happy if the following, more specific question were answered. For a ring R, Bass defined the negative $K$-groups to be the cokernel of the map 
$K_{-(n-1)}(R[t]) \oplus K_{-(n-1)}(R[t^{-1}]) \rightarrow K_{-(n-1)}(R[t, t^{-1}])$, for $n \geq 0$. 
If $R$ is regular, by the homotopy invariance property of the $K$-groups, $K_{-1}(R) = 0$. It is proven in Schlichting's paper that the negative $K$-groups for a ring defined this way coincide with the negative $K$-groups of the category $Ch^b(Proj(R))$ of bounded chain complexes of the exact category of projective $R$-modules. The more specific question is what is the significance of the above quotient condition where $\mathcal{A}$ is the category $Ch^b(Proj(R))$?

REPLY [7 votes]: There are several questions here. I'm not sure I understand them all. I'm going to try to address the role of idempotent completion. I suspect that if you reread Schlichting, you'll see everything I say was already there, but it may be helpful to have someone else say it.
First, what is the point of $K$-theory? As I see it, the main point is $K_0$ and the other groups are mainly there to help compute it. Quillen or Waldhausen tells us that if we have a short exact sequence of nice categories $A \to B\to C$, then we get a fiber sequence of spaces (or connective spectra) and a long exact sequence of higher $K$-groups. In particular, $K_0(B)$ surjects onto $K_0(B/A)$ for the trivial reason the objects of the two categories are the same and so the equivalence relation can only get coarser; $B/A$ is defined by taking the objects of $B$ and modifying the morphisms. But $B/A$ is rarely what you really want. Most natural categories are Karoubian closed and $B/A$ might not be. In the most basic example, projective modules are the Karoubian envelope of free modules.
In particular, if you have a localization of rings, you might expect such a short exact sequence of categories. With $B=R[t]$-modules and $C=R[t,t^{-1}]$-modules, the map $B\to C$ is a localization of categories, killing off the Serre subcategory of modules supported on $(t)$. So we get an LES of $K$-groups. But this is only true if we allowed infinitely generated modules, in which case all the groups are zero. If we impose a finiteness condition (bounded chain complexes of finitely generated projectives), it is not true that the localization of $R[t]$-modules is $R[t,t^{-1}]$-modules. If it were, then LES of higher $K$-groups would end with a surjection $K_0(R[t])\to K_0(R[t,t^{-1}])$, but as we see from Bass's definition, that is not surjective. But Thomason showed (with Trobaugh, but see also his other papers) that this is the only obstruction and the Karoubian envelope of the quotient $B/A$ is the category of $R[t,t^{-1}]$-modules.
But when we combine positive and negative $K$-theory, we do get a full LES, including $K_0(B)\to K_0(C)\to K_{-1}(A)$. The purpose of $K_{-1}(A)$ is to fit in this sequence, to fix the failure of surjectivity, which stems from the fact that $C$ is not quite the same as $B/A$, but instead its Karoubian envelope. As to the characterization, if $K_{-1}(A)$ vanishes, then the LES shows that $K_0(B)$ surjects on $K_0(C)$, which is to say that $B/A$ is already closed and thus equivalent to $C$. Conversely, if $A$ had the property that $B/A$ were always closed, then no nonzero element of $K_{-1}(A)$ would be hit by the boundary map from an element of $K_0$ of some other category. Given our motivation, these elements would seem pointless, though that doesn't prove that they don't exist. To do that, embed $A$ in a category $B$ subject to a swindle; one is given, roughly, by infinite objects of $A$. This category has contractible $K$-theory, so $K_{n}(B/A)=K_{n-1}(A)$, in particular, $K_0$ (of the Karoubian closure) of $B/A$ is isomorphic to $K_{-1}(A)$. So that gives the characterization of vanishing, modulo the existence of a nice $K_{-1}$. 
What's so great about categories being Karoubian closed? As I said above, natural categories like modules are. But also, in the last paragraph, I secretly assumed that $A$ was. Swindle categories automatically are, so the swindle category containing $A$ also contained its closure. And it is only possible to take the quotient by thick subcategories, so $A$ had better be thick, hence closed. In particular, this definition of negative $K$-theory only really makes sense for closed categories. Even if we start with a closed category, if we want to iterate this procedure to define negative $K$-theory, we have to take the Karoubian envelope at each stage to continue. This is typical of how they are easier to work with.<|endoftext|>
TITLE: Generalizations/applications of a formula for the Dedekind zeta function?
QUESTION [6 upvotes]: I saw the following nice formula in an unpublished paper of H. Cohen. Let $L$ be a quartic field, whose Galois closure $\widetilde{L}$ has Galois group $S_4$. Denote the cubic resolvent field of $L$ by $K_3$, and let $K_6$ be the Galois closure of $K_3$.
Then, we have the following relation among Dedekind zeta functions:
$$\zeta_L(s) = \frac{ \zeta(s) \zeta_{K_6}(s) }{\zeta_{K_3} (s)}.$$
This is proved using the formalism of Artin $L$-functions, and reflects the relation
$$ Ind_{G_L} 1 = 1 + Ind_{G_{K_6}} 1 - Ind_{G_{K_3}} 1$$
between the characters of $S_4$ induced from the trivial characters of the subgroups of $\mathrm{Gal}(\widetilde{L}/\mathbb{Q})$ fixing $G_L$, $G_{K_6}$, $G_{K_3}$ respectively. So essentially this is a calculation in group representation theory.
One nice consequence is the simple relation $D_L = D_{K_6} / D_{K_3}$ between the discriminants of these number fields. (This may be proved by comparing the conductors in the functional equation.)
I imagine this is old hat, but this formula really struck me. Two questions:
(1) Are there other similarly nice relations between other families of Dedekind zeta functions?
(2) Have this or related formulas seen other interesting applications in number theory?

REPLY [8 votes]: As Keith says, such relations between permutation representations are often called Brauer relations, because Brauer was the first one to note that such isomorphisms of permutation representations give rise to relations between zeta functions (Kuroda noticed the same phenomenon at the same time). Any non-cyclic group has Brauer relations, so in any $G$-extension of number fields with $G$ non-cyclic, you will get relations between zeta functions of the intermediate fields. Moreover, the space of Brauer relations is a finitely generated abelian group, and its rank is equal to the number of conjugacy classes of non-cyclic subgroups of $G$ (this is a consequence of Artin's induction theorem, together with the fact that the number of irreducible rational representations of $G$ is equal to the number of conjugacy classes of cyclic subgroups). So that's precisely the number of essentially different relations between zeta functions that you get this way in any given Galois extension of number fields. The Brauer relations in an arbitrary finite group are completely classified in this joint work of mine with Tim Dokchitser.
Such relations have lots of applications. For example numerous papers by Bart de Smit, Perlis, and others investigated relations that only consist of two groups. See my answer here for a description of some of their results. In this paper I used these relations to investigate integral units as Galois modules, and in this paper Bart and I generalised those findings.
In that paper with Bart we make use of the fact that you also get similar relations between other $L$-functions, e.g. $L$-functions of elliptic curves. This had been used to great effect by Tim and Vladimir Dokchitser (see here and here) to obtain results in the direction of the Birch and Swinnerton-Dyer conjecture. Also, Bley and Boltje have used Brauer relations to obtain relations between $p$-parts of class number, of Tate-Shafarevich groups, and of other interesting number theoretic invariants.
A final remark: to get the relations between discriminants from Brauer relations, you do not need zeta functions. They follow immediately from the conductor-discriminant formula.
The references to the original papers by Brauer and Kuroda are:
R. Brauer, Beziehungen zwischen Klassenzahlen von Teilkörpern eines Galoisschen Körpers, Math. Nachr. 4 (1951), 158–174 and
S. Kuroda, Über die Klassenzahlen algebraischer Zahlkörper, Nagoya Math. J. 1 (1950), 1–10.
If you search for forward citations, you will find lots of literature.<|endoftext|>
TITLE: Diagram calculus for weak 2-functors between bicatgories
QUESTION [8 upvotes]: I have to do some messy calculations with weak 2-functors between bicategories, and I know the most efficient way to do it would be via some sort of string diagram methods. Also, it means that I can put pretty calculations in my paper, unlike some of this paper's 'ancestors' (written by others) where they simply can't put in all the diagrams. To me this seems to be a blow to the reader, because not only does the reader have to figure out what on earth is going on, they are stuck with inelegant methods for doing so.
I know exactly what I should be doing if I was working in a single bicategory (just a simple extension of string diagrams for monoidal categories) but I don't know if anyone has written out a version with weak 2-functors. I'm sure someone has. Any pointers?

REPLY [2 votes]: FWIW, not exactly sure how relevant this is, but have a look at
Peter Sellinger's - A survey of graphical languages for monoidal categories (or on his homepage)
This gives a nice overview of many constructions used in graphical languages.<|endoftext|>
TITLE: admissible group operation on etale, separated, finite type scheme
QUESTION [7 upvotes]: Grothendieck in SGA 1 introduces a proposition in expose 5 (proposition 3.1) which states:
Let $X$ be etale, separated of finite type over $Y$, locally noetherian, and let $G$ be a finite group which operates on $X$ by $Y$-automorphisms.  Then $G$ operates admissibly and the quotient scheme $X/G$ is etale over $Y$.
The hint he gives is that we may show this for $X$ quasi-projective, and to use proposition 1.8, which states that $G$ operates admissibly on $X$ iff $X$ is the union of open affines that are invariant under the action of $G$.
I am unsure how to show this.  Help please?
EDIT: so I understand how to make the reduction to the quasi-projective case (since every etale morphism is quasi-finite, and then just apply Zariski's main theorem, and we get thus that this is quasi-projective), but I am unsure how to show that if a finite group operates on a quasi-projective scheme then it operates admissibly.  I have a rough idea for how it could be done for a projective scheme, but I do not see how I could alter it for quasi-projective :(  I could post this proof if it is of any help to readers??

REPLY [8 votes]: We can reduce to the case where $Y$ is affine, and in this case (as you observe), $X$ is quasi-projective over $Y$ by Zariski's main theorem. Consequently, the key step is to show that a finite group operating on a quasi-projective scheme does so admissibly (meaning that we can form the quotient nicely). 
Admissibility is equivalent to the condition that every $G$-orbit is contained in an open affine, but any finite subset of projective space is contained in an open affine (e.g. find a hypersurface not containing any element of the subset and take the complement). 
A proof of étaleness in the case of a finite morphism is on p. 56 of Murre's "Lectures on an introduction to Grothendieck's theory of the fundamental group." It is based on a sequence of reductions, which I'll try to outline (without assuming finiteness of $X \to Y$, but I will assume that $Y$ has finite Krull dimension). First, étaleness descends under faithfully flat base change, and it can be checked on the stalks. Moreover, the process of taking the quotient by a finite group (which, on the level of rings, amounts to taking fixed points) is preserved under flat base change. (This is essentially the following observation: if $G$ acts on an $A$-algebra $B$, then if $A'$ is a flat extension of $A$, then $(A' \otimes_A B)^G \simeq B^G \otimes_A A'$; one proves this by fitting both into exact sequences.)
What all this means is that if you want to check that a map $X/G \to Y$ is étale, then you may as well base-change the whole problem to $T \to Y$ where $T$ is a faithfully flat extension of the local scheme $\mathrm{Spec} \mathcal{O}_y$ (for any $y \in Y$). 
Anyway, a lot of things simplify when you're allowed to make these kinds of base changes. So let's assume for the sake of argument that $Y$ is the $\mathrm{Spec}$ of a complete local ring, noetherian, whose residue field is algebraically closed. Then $X$ splits into two pieces: the first is a piece finite étale over $Y$ (and thus a disjoint union of copies of $Y$, since etale covers of $T$ split) and  a second piece whose image does not contain the closed point of $Y$. 
Let's consider what happens to the piece which is finite étale over $Y$, and so looks like $Y \sqcup Y \sqcup \dots \sqcup Y$. Here the group $G$ just acts by permuting the factors, and the quotient is what you get by identifying a bunch of pieces of $Y$, so in particular looks like $Y \sqcup Y \sqcup \dots \sqcup Y$ with a different number of pieces. 
Now what can we say about the piece $X'$ over $Y - \{ \ast\}$ for $\ast \in Y$ the closed point? Not that much, but we do know that it is of smaller Krull dimension, and using induction on the dimension, we can assume that  $X'/G$ is étale over $Y - \{ \ast\}$. It follows that $X/G$ is etale over $Y$, completing the (sketched) proof. 
The key lemmas that make possible this replacement of $Y$ by such a nice ring are the following: 
Lemma: Let $(A, \mathfrak{m})$ be a local noetherian ring and $L/k$ an algebraic extension. Then there exists a flat extension $(\widetilde{A}, \widetilde{m})$ of $(A, \mathfrak{m})$ such that the $\widetilde{A}/\widetilde{m} \simeq L$ of the same Krull dimension. 
This is essentially the result of EGA III.10.3. 
Corollary: If $(A, \mathfrak{m})$ is any local noetherian ring of finite Krull dimension, then there exists a flat extension $(B, \mathfrak{n})$ of the same Krull dimension and such that $B$ is complete local and such that $B/\mathfrak{n}$ is algebraically closed. 
This follows from the lemma applied to the completion of $A$. The whole point is that the etale maps to $\mathrm{Spec} B$ are a lot simpler than the maps to $\mathrm{Spec} A$ (for instance, the etale covers of $\mathrm{Spec} B$ are all trivial). In the present case, the strict henselianization would also do. 
To recap, the point of the argument is to induct on the dimension of $Y$. Then, one reduces to checking the claim of etaleness by making a very strong base change, such that $X$ splits into an easily understood finite piece and a less easily understood but inductively controlled non-finite piece. 
These types of reductions are quite useful. For instance, they're integral to the proof of Zariski's main theorem given in EGA. Also, I'm pretty sure that the hypothesis of finite Krull dimension on $Y$ can be removed by a "noetherian descent" type argument (i.e., regarding $X, Y$ as a limit of schemes of finite type over $\mathbb{Z}$), but I'm not completely sure how noetherian descent is supposed to work for étaleness.<|endoftext|>
TITLE: Smooth variety with positive point-count polynomial and odd cohomology
QUESTION [5 upvotes]: I am looking for an example of a smooth irreducible quasiprojective variety $X$ over ${\mathbb C}$, such that when reduced over finite fields ${\mathbb F_q}$, the number of its points is a polynomial $P(q)$ of $q$ with nonnegative (integer) coefficients, but $X$ has some odd cohomology. 
Background: as discussed in an answer to this question, if a variety $X$ is paved by affine spaces, then it only has $(p,p)$ cohomology, and the number of its ${\mathbb F_q}$-points equals $P(q)$, where $P(t)$ is the Poincare polynomial of (compactly supported cohomology of) $X$. Note that the coefficients of $P$ are necessarily non-negative, given by the number of affine cells in the paving of a fixed dimension. In the appendix to  this paper, N.Katz proves a kind of converse to this statement: if the number of points of $X$ over a finite field is given by a polynomial $P(q)$ of $q$, then this polynomial determines the so-called $E$-polynomial $E(x,y)$ of $X$ by the formula $E(x,y)=P(xy)$. The $E$-polynomial is a partial Euler characteristic, where we remember the weights of (compactly supported) cohomology but not the degrees. 
Of course varieties of the latter type can have odd cohomology; the typical example is $X={\mathbb C}^*$, with point count polynomial $P(q)=q-1$. $X$ of course has odd cohomology. A slighly more complicated example, due to N.Katz, shows that $X$ can also have non-$(p,p)$ cohomology. But in these examples, the polynomial $P$ has some negative coefficients. 
Hence the question: can $X$ have positive polynomial count, but still some odd cohomology (which cancels in the $E$-polynomial)? Note that $X$ can't be smooth projective, since then its cohomology would be pure so any odd cohomology would have to show up in the $E$-polynomial. 
There may of course be a trivial example which I am missing.

REPLY [7 votes]: The answer to your question is yes:
Let $X$ be the blowup of $\mathbb{A}^1 \times (\mathbb{A}^1 - \{0\})$ in a point. The number of points over a field of $q$ elements is $q(q-1) + q = q^2$ and $X$ has non-trivial $H^1$.<|endoftext|>
TITLE: Chern classes of a blow-up at a point
QUESTION [10 upvotes]: Let $X$ be a nonsingular projective variety over $\mathbb{C}$, and let $\widetilde{X}$ be the blow-up of X at a point $p\in X$. 
What relationships exist between the degrees of the Chern classes of $X$ (i.e. of the tangent bundle of $X$) and the degrees of the Chern classes of $\widetilde{X}$?
Thanks.

REPLY [9 votes]: Like Georges says, 15.4 of Fulton's Intersection Theory deals with the general theory. For this special case it's not too hard to work out the Chern classes by hand though.
Let $f : \widetilde X \to X$ be the projection and $E \cong \mathbb{C}P^{n-1}$ the exceptional divisor.
$H^*(\widetilde X) \cong f^*H^*(X) \oplus \langle \textrm{Poincare duals of planes } P_k \textrm{ in } E $ $\textrm{of dimension }k = 1,\ldots, n-1\rangle$. Note that $[P_{n-i}][P_{n-j}] = -[P_{n-i-j}]$, while $(f^*\alpha) [P_k] = 0$ for any $\alpha \in H^i(X)$ ($i, k > 0$).
$f^* c_i(X)$ and $c_i(\widetilde X)$ are equal outside the exceptional divisor, so their difference is Poincare dual to something in $E$. On the other hand the restriction of $f^*c_i(X)$ to $E$ is 0 (for $i > 0$), while the restriction of $c_i(E)$ is $c_i(\mathcal{O}(1)^n \oplus \mathcal{O}(-1)) = \left({n\choose i} - {n \choose i-1}\right)H^i$, where $H \in H^2(E)$ is the hyperplane class. For $0 < i < n$ we deduce that $c_i(\widetilde X) = f^*c_i(X) - \left({n\choose i} - {n \choose i-1}\right)[P_{n-i}]$ by comparing the evaluations on $P_i$.

REPLY [4 votes]: Assume $X$ is smooth compact of dimension $n$ and $x_0\in X$ is the point where we perform the blowup. Set  $ X_* := X \setminus x_0 $,  $ \tilde{X}_* := \tilde{X} \setminus E$. Denote by $N$ a tubular neighborhood of $E$ in $\tilde{X}_* $.  By Mayer-Vietoris, the Chern classes of $ \tilde{X} $  are determined once we know their restrictions to $ X_* $ and $ N $.
We identify $ \tilde{X}_* $ with $ X_* $ via the blowdown map $p:\tilde{X}_* \to X_* $. The restriction of $c_k( \tilde{X}) $ to $X_*$ is equal to the restriction of $c_k(X)$. The restriction of  $c_k(\tilde{X})$  to $N$ is  easy to determine since
$$TN \cong  \pi^* T\mathbb{CP}^{n-1} \oplus \pi^*  H^*, $$
where $\pi: N\to E= \mathbb{CP}^{n-1}$ is the natural projection and $H\to \mathbb{CP}^{n-1}$ is the hyperplane line bundle.  Thus, 
$$ c_k(\tilde{X})|_N = c_k( N ) = \pi^*c_k(\mathbb{CP}^{n-1} ) +\pi^* c_{k-1}(\mathbb{CP}^{n-1} ) \pi^* c_1(H^*) $$
$$ = \pi^*c_k(\mathbb{CP}^{n-1} ) - \pi^* c_{k-1}(\mathbb{CP}^{n-1} )\cup \pi^*[H].  $$
Things can be simplified a bit if we introduce the notation $h=\pi^*[H]\in H^2(N,\mathbb{Z})$ and we observe that for some integers $\nu_k$ and $\nu_{k-1}$
$$ \pi^*c_k(\mathbb{CP}^{n-1} ) =\nu_k h^k, $$
$$ \pi^* c_{k-1}(\mathbb{CP}^{n-1} )=\nu_{k-1} h^{k-1}. $$
Then
$$   c_k(N) = ( \nu_k -\nu_{k-1} ) h^k. $$
As for the  integers $\nu_k$  they are determined from the equality
$$ 1+c_1( \mathbb{CP}^{n-1} )+\cdots + c_{n-1}( \mathbb{CP}^{n-1} )= (1+H)^n - H^n $$
$$= \sum_{k=0}^{n-1}\binom{n}{k} H^k. $$<|endoftext|>
TITLE: What is the etymology of model?
QUESTION [10 upvotes]: What is the etymology of model? The answer is of course pre-WWW, but the better part of an hour in the library searching both classic model theory and modal logic textbooks turned up nothing. Every book I touched, without exception, uses the word in the usual way - a structure consistent with some theory - but of course gives no justification for it.
I say 'of course' because, given the word's common meaning of 'a representation of something', 
phrases like 'model theory', 'canonical model', etc. are quite jarring.
To my great relief, a couple of books commented on this jarring nature (it's not only me!) but I am genuinely curious as to who was the first person to use model to mean 'consistent structure', and even more curious as to why they did so.
Summary:
Thanks one and all for your comments! I've come to realize that I simply had the 'mathematical model' usage of the word etched into my brain, but the past couple of days spent reading 'model' as '(toy) model' have jolted me into agreeing that perhaps 'model' is a reasonable choice of term after all. I still feel (perhaps wrongly, I'm not an expert in model theory) that a word like cast or casting (as in a die-casting) would be better;
it conveys a sense of fitting (satisfying) some mould (theory) while still being short and a little light hearted like 'model'. But that is just me, and thanks to the comments here, I think that 'model' is quite good enough.

REPLY [6 votes]: I'm sure the origin of the term is quite complex. Indeed, as Henry and David have pointed out, it is a very natural choice in this context.
I think the first definition of 'model' (not the first use) in the exact sense currently used in model theory is due to Tarski in O pojȩciu wynikania logicznego (On the concept of following logically; English translation MR1951812; German translation from Polish by Tarski himself). This is the paper where the current definition of logical consequence first appears:

We say that the sentence $X$ follows logically from the sentences of the class
  $\mathfrak{K}$ if and only if every model of the class $\mathfrak{K}$ is at the same time a model of the sentence $X$. [Translation by M. Stroinska and D. Hitchcock.]

I don't know much about Tarski's choice of terms here, but the commentary to the English translation by M. Stroinska and D. Hitchcock could be enlightening.
It is interesting to note that Tarski published his paper in 1936, half a decade after Gödel's Die Vollständigkeit der Axiome des logischen Funktionenkalküls (The completeness of the axioms of the functional calculus of logic; MR1549799). So it appears that these ideas were already known to members of the Vienna Circle and affiliates.<|endoftext|>
TITLE: Explicit 3-cocycles for the symmetric group $S_6$
QUESTION [8 upvotes]: Hello,
this is a request for literature/a reference. I'm looking to do some calculations with the
symmetric group ($S_6$ and higher) and would be interested in explicit expressions for
3-cocycles, i.e. elements of $H^3(S_6, U(1))$.
Does anyone know whether these have already been calculated somewhere?

REPLY [2 votes]: I think Graham's answer is relevant to the question. 
A 3-cocycle on $G$ with values in an abelian group $A$ my also be described as a morphism of chain complexes from the standard free resolution $F(G)$ of $G$ to the chain complex say $(A,3)$ which is $A$ in dimension 3 and zero elsewhere. But it may be more convenient to compute another free resolution $C$ of $G$ and describe a morphism $C \to (A,3)$. A standard 3-cocycle then comes from the description of a morphism of chain complexes $F(G) \to C$,  which "in principle" is standard homological algebra. 
But the issue remains of the form required for the answer, and for this one probably needs to know  the place of the question in a research project. Why is the question asked? 
There are other ways of representing elements of $H^3(G,A)$, for example by "crossed sequences"
$$0 \to A \to C_2 \to C_1 \to G \to 1$$ 
where $C_2 \to C_1$ is a crossed module. There are good examples where $G,A, C_2,C_1$ are finite.  But intriguingly, for the current question $A$ is a topological abelian group!<|endoftext|>
TITLE: what are the conditions for the product of 2 symmetric matrices being symmetric
QUESTION [6 upvotes]: In generally, the product of two symmetric matrices is not symmetric, so I am wondering under what conditions the product is symmetric.
Likewise, over complex space, what are the conditions for the product of 2 Hermitian matrices being Hermitian?
Thanks!

REPLY [4 votes]: Incidentally, every real matrix is the product of two symmetric matrices. (If I remember correctly, I once read about this in a paper by Halmos).<|endoftext|>
TITLE: Excellent uses of induction and recursion
QUESTION [19 upvotes]: Can you make an example of a great proof by induction or construction by recursion?
Given that you already have your own idea of what "great" means, here it can also be taken to mean that the chosen technique :

is vital to the argument;
sheds new light on the result itself;
yields an elegant way to fulfill the task;
conveys a powerful and simple view of an intricate matter;
is just the only natural way to deal with the problem.

Here induction and recursion are meant in the broadest sense of the words, they can span from induction on natural numbers to well-founded recursion to transfinite induction, and so on...
Elementary examples are especially appreciated, but non-elementary ones are welcome too!

REPLY [2 votes]: There are infinifely many positive integers $n$ such that $2^n+1$ is divisible by $n$. One can show this by proving that if $n$ divides $2^n+1$, then $2^n+1$ divides $2^{2^n+1}+1$.<|endoftext|>
TITLE: Construction of Serre Spectral Sequence
QUESTION [5 upvotes]: I'm trying to follow Hopkins' construction of the Serre Spectral Sequence, but some "obvious" things are not that obvious to me.
He starts with considering a double complex $C_{\bullet,\bullet}$ with $C_{p,q}$ to be a free $\mathbb{Z}$-module generated by the maps $\Delta[p]\times\Delta[q]\rightarrow E$ ($E$ is a total space of Serre fibration over $B$) which fit into the diagram
\begin{matrix} 
\Delta[p]\times\Delta[q] & \to & E \\\ 
\downarrow & & \downarrow  \\\  
\Delta[p] & \to & B 
\end{matrix}
with obvious differentials (coordinate by coordinate differentials as in normal singular complex).
There are two filtrations, he uses the first one (by rows) to determine the homology of the total complex, ane the second one to get $E^2_{p,q}=H_p(B,\underline{H_q}(F)$. I have a problem with the second part. He fixes $p$ and a map $c$ in the bottom row and interprets the diagram as a map $\Delta[q]\rightarrow F_c$ where $F_c$ is the image of $\Delta[p]\times\Delta[q]$ in $E$ such that the diagram commutes, which basically means that $C_{p,q}=\bigoplus_c \mathbb{Z}[F_c]$ (as a module, that's true). Now he calculates the homology of the column and says that it's $\bigoplus_c H_*(F_c)$. Why can we apply the vertical differential here? Do we need to check some compatibility condition?
Next, we want to use that $E^1_{p,q}$ is a module of singular $p$-chains with coefficients in a local system $\underline{H_q}(F)$ and say that the horizontal differential is just the regular differential to get the desired output. But how do we know that this differential works in a nice way?

REPLY [5 votes]: I wanted to post the following as a comment, but it's too long. 
It might help to realize where the differential comes from. Let $p: E \to B$ be a Hurewicz fibration. Assume $B$ is a connected CW complex. 
Then $B$ has a cellular filtration $B_k \subset B_{k+1} \dots$. If we pull back $p$ along this filtration we obtain a filtration (in fact, by cofibrations!) of $E$
$$
E_0 \subset E_1 \subset \cdots 
$$
I claim that the homology spectral sequence of this filtration is then the Serre spectral sequence. The $d_1$-differential of the above filtration is given by the composition
$$
E_k/E_{k-1} \overset{\delta}\to \Sigma E_{k-1} \to \Sigma E_{k-1}/E_{k-2} \qquad (\ast)
$$
where the map $\delta$ is the Barratt-Puppe extension of the cofibration $E_{k-1} \to E_k$
and the second displayed map is given by collapsing $E_{k-2}$ to a point.
Furthermore, it's easy to check that 
$$
E_{k}/E_{k-1} \simeq F_+ \wedge B_k/B_{k-1} \qquad (\ast\ast)
$$
So I guess that your question, in the end, amounts to the following: 
With respect to the equivalence $(\ast\ast)$,
what does the map $E_k/E_{k-1} \to \Sigma E_{k-1}/E_{k-2}$ look like when considered
as a map 
$$
F_+ \wedge B_k/B_{k-1} \to F_+ \wedge \Sigma B_{k-1}/B_{k-2} \quad ?
$$
In other words, on the level of homology, is this map of the form 
$\text{id}\wedge c$, where $c:B_k/B_{k-1} \to \Sigma B_{k-1}/B_{k-2} $ is the map for the filtration $B_k \subset B_{k+1} \dots$ constructed like the one in $(\ast)$?<|endoftext|>
TITLE: Some questions about Ackermann set theory
QUESTION [5 upvotes]: In a comment  on this site Andreas Blass stated:  
"To fit this situation into my philosophical point of view, I'd say that what Ackermann's theory 
calls proper classes are really certain sets. That notion has some support in the Levy-Vaught interpretation
 of Ackermann set theory in a conservative extension of ZF, where both the sets and the classes of Ackermann 
are interpreted as certain sets in the sense of ZF.  
– Andreas Blass Feb 9 at 16:58 "  
Does it mean that the following statements (analogs of the axioms of pairing, union and powerset respectively) are consistent with Ackermann set theory:    
1) For any two classes $X, Y$ there exists the class $Z$ which contains just $X$ and $Y$.  
2) For any class $X$ there exists the class, whose members are just the members of the members of $X$;  
3) For any class $X$ there exists the class whose members are just all the subclasses of $X$.

REPLY [3 votes]: The answer to your question is "Yes."
In trying to understand why Ackermann nonetheless wanted to distinguish proper classes such as V from sets - where V is the proper class of all sets, taken here to be a natural model of ZFC - it is necessary to take seriously Ackermann's claims about the universe V being continually "under construction," so to speak, or always in the process of being "built" (see Penelope Maddy's article "Proper Classes" (Journal of Symbolic Logic, 1983), p. 122, on this point).  At some particular "time" (or "step") t in this construction process, there may exist (e.g.) classes larger than V ("superclasses") that have been obtained by iterating the power-set operation on V denumerably many times. Consider the totality T of such superclasses existing at t; i.e., T (at t) is the limit of a denumerable sequence of iterated power-set operations on V. Since T (at t) is clearly not a natural model of ZFC, it cannot be regarded as a new "universe" that supersedes or replaces V. Hence, those collections which (at t) are members of T but not of V are not members of a suitable universe or natural model; and so they are to be distinguished from the members of V, a distinction expressed by calling them "proper classes" (and calling the members of V "sets").
One glaring problem with this account is that it's unclear what we're supposed to take as the time t.  Why, in particular, should we suppose that t is such that the power set operation on V has been iterated denumerably many times (or even some nonzero number of times)? Without any constraints on t, there's simply no way to know what proper classes we're supposed to regard as existing; and, in fact, there are no theoretical constraints on what t's value is.  Hence, there's no basis for saying anything definite about proper classes at all; and so it seems best, as Andreas Blass said, to treat Ackermann's "proper classes" simply as sets.<|endoftext|>
TITLE: Squares in an Arithmetic Progression
QUESTION [11 upvotes]: Let $P(x;a,b) := \{an+b,  0\leq n \leq x \} $ denote an arithmetic progression. Further let $A(x;a,b)$ denote the number of elements of $P(x;a,b)$ that are squares.  It's an old conjecture of Rudin that $A(x;a,b) \ll x^{1/2}$.  Less ambitiously, Erdős posed the problem of showing that $A(x;a,b) = o(x)$. This was proven by Szemeredi around 1974 (amusingly the paper is only a few sentences). 
Here is Szemeredi's proof: If the theorem was false then we could find arbitrarily large arithmetic progressions composed of at least $\delta>0$ percent squares. Then invoking the 4 case of Szemeredi's (most well-known) theorem we have that there must be a length 4 arithmetic progression consisting of only squares. However, this contradicts an old theorem of Euler. 
While this proof is slick, it is natural to want to avoid having to use anything as powerful as Szemeredi's theorem.  I recently I ran across the paper "On the Number of Squares in an Arithmetic Progression" by Saburo Uchiyama (Proc. Japan Acad. 52, no. 8 (1976), 431-433). The complete paper is freely available here.  The paper claims to give a simple and self-contained solution to Erdős' question (that  $A(x;a,b) = o(x)$). In fact, the proof given is so short that I will repost it in its entirety:
       


Now I don't follow the claim that "This clearly proves (1)." (where (1) is the claim that $A(x;a,b) = o(x)$). Certainly when $a=o(x)$ this gives the desired result, but why does this work when $a$ is large? 


Since this is so short and simple (and I have never seen the argument cited anywhere) I am skeptical, however perhaps I am missing something obvious. 
(Perhaps, it should be pointed out that there is a more recent approach to the problem that yields better quantitative bounds that goes through Falting's theorem due to Bombieri, Granville and Pintz.  In fact, their analysis shows that the case when $a$ is large compared to $x$ is the 'hard case', which raises further suspicion of the above argument.)

REPLY [11 votes]: The conjecture of Rudin is about the maximum of $A(x; a,b)$, taken over all values of $a$ and $b$. Not with just keeping $a$ and $b$ fixed and letting $x$ get large. So, like you said, if $a$ is not $o(x)$, the proof of Uchida doesn't go through.
The latest news on this problem (as far as I know) can be found here: E. Bombieri, U. Zannier, A Note on squares in arithmetic progressions, II, Rend. Mat. Acc. Lincei, s. 9, v. 13:69-75 (2002) (link: Wayback Machine).<|endoftext|>
TITLE: Is the tensor product of a power series ring and a field noetherian?
QUESTION [11 upvotes]: Suppose that $k$ is an algebraically closed field.  Let $F/k$ be a (possibly non-finitely generated) field extension.  Is
$$
k[[x]] \otimes_{k} F
$$
noetherian?
If not, is the natural map $k[[x]] \otimes_{k} F \to F[[x]]$ injective?

REPLY [3 votes]: I was emailed the following argument:
We prove that $k[[x]]\otimes_{k} k((x))$ is not noetherian by showing directly that $k((x)) \otimes k((x))$ is not noetherian (as suggested by Georges Elencwajg). I will just handle the case $k=\bar{\mathbb{Q}}$ and then make a remark about the general case at the end.
The field $k((x))$ has only countably many finite separable extensions because every such extension is obtained by adjoining a root of $x$.  On the other hand, the transcendence degree of $k((x))$ over $k$ must be uncountable because $k((x))$ is uncountable and $k$ is countable.  Fix a transcendence basis $( t_i )_{i \in I}$  for $k((x))$ over $k$. 
The algebraic extension $k((x))$ of $K := k((t_i))$ is algebraic with infinite separable degree.  Indeed, if the separable degree was finite, then $K$ would admit at most countably many finite separable extensions as this is true for $k((x))$.  This is absurd because $( t_i )$ is uncountable.  
Because the separable degree of $k((x))$ over $k((t_i))$ is infinite, $(k((x)) \otimes_{K} k((x)))_{\text{red}}$ has infinitely many idempotents and so 
$$k((x)) \otimes_{K} k((x))$$
and hence
$$k((x)) \otimes_{k} k((x))$$
are non-noetherian.  This completes the proof.
With work, this proof can be modified to hold when $k$ is a finite field $\mathbb{F}$.  In this case, one must argue more carefully to show that $k((x))$ has only countably many finite separable extensions. (The email indicated that one should use local compactness together with Krasner's Lemma.)  Finally, one can deduce the case of a more general field $k$ from the case $k=\bar{\mathbb{Q}}$ or $\mathbb{F}$ by using a faithfully flat descent argument.<|endoftext|>
TITLE: derivative in the Wasserstein space
QUESTION [7 upvotes]: Villani gives the following formula to find the gradient of a function $F$ of a probability density function $\rho$ in the Wasserstein space :
$$\nabla_W F(\rho) = -\nabla.(\rho \nabla \frac{\delta F}{\delta \rho})$$
where, if $F$ is given as $F(\rho)=\int U(\rho) dx$, then $\frac{\delta F}{\delta \rho} = U'(\rho)$.
I am trying to find how much the value at a given (fixed) point $x$ of a continuous pdf varies along a displacement interpolation. I am thus trying to apply the formula above in the distributional sense where $U = \delta_x$ (here, with $\delta$ the Dirac distribution), to get $F(\rho) = \rho(x) = <\delta_x, \rho>$.
By using $\nabla.(f G) = (\nabla f) G + (\nabla.G) f$, I thus get : 
$$\nabla_W F(\rho) = - (\nabla\rho.\nabla\delta_x + \rho\triangle\delta_x)$$
which equals to $0$ after using the definition of the derivatives in the distributional sense since both terms cancel each other.
So, what I get from it, is that the value at a fixed point $x$ (ie., not a point which gets advected with the function!) does not change when a function moves in the Wasserstein space... which I really don't believe to be true ! What went wrong ? 
Thanks !

REPLY [4 votes]: The dirac distribution $\delta_x$ is not smooth enough to carry out this computation, which only works for functionals $F(\rho)=\int U(\rho)$...
One thing though: a displacement interpolation is nothing but a geodesic in the metric space of probabilities (endowed with the Wasserstein distance). If you look for example at Villani's book (chapter 8), we know that such a geodesic $(\rho_t)_{t\in[-\varepsilon,\varepsilon]}$ satisfies a continuity equation
$$
\partial_t\rho_t+\nabla\cdot(\rho_tv_t)=0
$$
for some particular vector-field $v_t$ determining the tangent vector $s=-\nabla\cdot(\rho_t v_t)\in T_{\rho_t}\mathcal{P}$. This PDE is what you are looking for, since it allows to compute the variation $\partial_t\rho_t(x)$ for a given point $x$ (forgetting about smoothness issues, of course).<|endoftext|>
TITLE: Polynomials for addition in the Witt vectors
QUESTION [19 upvotes]: The addition of $p$-typical Witt vectors ($p$ a prime number) is given by universal polynomials $S_n=S_n(X_0,\dots,X_n;Y_0,\dots,Y_n)\in\mathbb{Z}[X_0,X_1,\dots;Y_0,Y_1,\dots]$ determined by the equalities  
$\Phi_n(S_0,\dots,S_n)=\Phi_n(X_0,\dots,X_n)+\Phi_n(Y_0,\dots,Y_n)$ for all $n\ge 0$,  
where  
$\Phi_n(T_0,\dots,T_n)=(T_0)^{p^n}+p(T_1)^{p^{n-1}}+\dots+p^nT_n$.  
I guess that whoever sees the Witt vectors for the first time writes down explicitly $S_0=X_0+Y_0$, $S_1=X_1+Y_1+\frac{1}{p}((X_0)^p+(Y_0)^p-(X_0+Y_0)^p)$, maybe $S_2$ if she/he is courageous, and then stops since the computation becomes extremely messy. I think that there is no reasonable explicit expression in general, but patterns seem to exist and my question is about making these patterns more precise. Before I ask, let me illustrate with $S_2$. It is easy to see that there exists a unique sequence of polynomials $R_n\in\mathbb{Z}[X,Y]$, $n\ge 0$, such that  
$X^{p^n}+Y^{p^n}=R_0(X,Y)^{p^n}+pR_1(X,Y)^{p^{n-1}}+\dots+p^nR_n(X,Y)$.  
For example $R_0=X+Y$ and $R_1=\frac{1}{p}(X^p+Y^p-(X+Y)^p)$. Then:  
$S_1=R_0(X_1,Y_1)+R_1(X_0,Y_0)$  
$S_2=R_0(X_2,Y_2)+R_1(X_1,Y_1)+R_1(R_0(X_1,Y_1),R_1(X_0,Y_0))+R_2(X_0,Y_0)$.  

Can someone make the shape of $S_n$ more precise, e.g. in the form $S_n=P_0+\dots+P_n$ presumably with $P_0=R_0(X_n,Y_n)$, $P_n=R_n(X_0,Y_0)$? The intermediary $P_i$'s are more complicated but should be (uniquely) determined by a condition of the type
  "$P_i$ is an iterated composition of $R_i$ involving only the variables $X_0,\dots,X_i$".
  Maybe the polynomials $P_i$ should be homogeneous w.r.t. some graduation.  

Any hint or relevant reference will be appreciated. Thanks!

REPLY [10 votes]: I found a formula for $S_n$ in terms of the previous $S_i$'s and the polynomials $R_i$, which I'm quite happy about. In fact, I need the multivariate version of the $R_i$, which I will construct all at the same time. Let us consider the ring of formal power series in countably many variables $X_1,X_2,\dots$ with integer coefficients, that is $A=\mathbb{Z}[[X_1,X_2,\dots]]$. Note that there are several notions of power series in infinitely many variables; mine is that of Bourbaki, where the underlying module of $A$ is just the product of copies of $\mathbb{Z}$ indexed by (finitely supported!) multiindices. (Some people require that the homogeneous components of a power series be polynomials; this is not the case here.) Then one can see that there exists a unique sequence $(R_n)$ of elements of $A$ such that for all $n\ge 0$ we have  
$X_1^{p^n}+X_2^{p^n}+\dots=R_0^{p^n}+pR_1^{p^{n-1}}+\dots+p^nR_n$ 
where on the left is the sum of all $p^n$-th powers of the variables. This is a straightforward application of Bourbaki, Algèbre Commutative, Chapitre IX, $\S~1$, no 2, prop. 2, c) with the endomorphism $\sigma:A\to A$ defined by $\sigma(X_i)=X_i^p$ for all $i$. Now if we have finitely many (say $s$) variables, then we set $R_n(X_1,\dots,X_s)=R_n(X_1,\dots,X_s,0,0,\dots)$. Examples:  
$R_0(X_1,\dots,X_s)=X_1+\dots+X_s$ and $R_1(X_1,\dots,X_s)=\frac{X_1^p+\dots+X_s^p-(X_1+\dots+X_s)^p}{p}$.  
Assuming that the $R_n$ are computable, I have an inductive recipe for $S_n$ which is interesting because it shows that all the $p$-adic congruences implying integrality of the $S_n$ are contained in the $R_n$. The recipe goes like this. For each $i$, the polynomial $S_i$ is a sum of $2i$ terms (it will be obvious below what these terms are) and assuming $S_1,\dots,S_{n-1}$ are known then  
$S_n=R_0Z_n+R_1Z_{n-1}+\dots+R_nZ_0+R_1S_{n-1}+R_2S_{n-2}+\dots+R_{n-1}S_1$ 
where: $Z_j$ is short for the pair of variables $(X_j,Y_j)$, $R_iZ_j$ is short for $R_i(X_j,Y_j)$ (the bivariate $R_i$) and $R_iS_j$ is the ($2j$-variate) polynomial $R_i$ evaluated at the $2j$ terms of $S_j$. I hope the following examples make it clear what this means, and how efficient it is:  
$S_1 = R_0 Z_1 + R_1 Z_0$ 
$S_2=R_0 Z_2 + R_1 Z_1 + R_2 Z_0 + R_1 (R_0 Z_1 , R_1 Z_0 )$ 
$S_3 = R_0Z_3+R_1Z_2+R_2Z_1+R_3Z_0 \\ \quad
+ R_1(R_0Z_2,R_1Z_1,R_2Z_0,R_1(R_0Z_1,R_1Z_0))+ R_2(R_0Z_1,R_1Z_0)$ 
$S_4 = R_0Z_4+R_1Z_3+R_2Z_2+R_3Z_1+R_4Z_0 \\ \quad
+ R_1(R_0Z_3,R_1Z_2,R_2Z_1,R_3Z_0,R_1(R_0Z_2,R_1Z_1,R_2Z_0,R_1(R_0Z_1,R_1Z_0)),R_2(R_0Z_1,R_1Z_0)) \\ \quad
+ R_2(R_0Z_2,R_1Z_1,R_2Z_0,R_1(R_0Z_1,R_1Z_0)) \\ \quad
+ R_3(R_0Z_1,R_1Z_0)$ 
The proof that the recipe is correct is an exercise.<|endoftext|>
TITLE: Finite Field Grassmannians as Homogeneous Spaces
QUESTION [5 upvotes]: For the real Grassmannian Gr$(N,k)$ we have the well-known isomorphism 
$$
\text{Gr}(N,k) = O(N)/(O(k) \times O(N-k))
$$ 
For the complex case, we have
$$
\text{Gr}(N,k) = U(N)/(U(k) \times U(N-k))
$$ 
I would like to know if anything like this holds in the finite field setting, ie can the finite field Grassmannians be described as a homogeneous space of an algebraic group over a finite field, or something like this?

REPLY [2 votes]: There is a description of these homogeneous spaces over finite fields in terms of graphs in "Distance-Regular Graphs" by Brouwer, Cohen and Neumaier (Springer, 1989) and in "Algebraic combinatorics I: Association schemes" by E.Bannai and T.Ito (Benjamin/Cummings, 1984).<|endoftext|>
TITLE: Sums of powers mod p
QUESTION [20 upvotes]: For prime $p > 7$ with $p-1=rs$, $r>1$, $s>1$, let $A=\{x^r|x \in \mathbb{Z}_p\}$ and $B = \{x^s|x \in \mathbb{Z}_p\}$. If $g$ is a primitive root mod $p$ then $A = \{0\} \cup \{g^{ir}|0 \leq i < s \}$ and $B = \{0\} \cup \{g^{js}|0 \leq j < r \}$.  Is it always true that $\mathbb{Z}_p \neq A + B$?

REPLY [2 votes]: For fixed $r>1$, use of Weil's bound will give the result. 
    We refer to Theorem 5.1 from chapter 2 of W.M.Schmidt, Equations over Finite Fields: An Elementary Approach.
Theorem
    Let $f_1,\cdots, f_n$ be polynomials with coefficients in $\mathbb{F}_q$ and of degree $\leq m$. Put $\delta=\textrm{lcm}(d_1,\cdots,d_n)$, and $d=d_1d_2\cdots d_n$. Let X be a variable and let $\eta_1,\cdots,\eta_n$ be algebraic quantities with
    \begin{equation}
    \eta_1^{d_1}=f_1(X), \cdots, \eta_n^{d_n}=f_n(X).\
    \end{equation}
    Suppose
    \begin{equation}
    [\overline{\mathbb{F}}_q(X, \eta_1,\cdots,\eta_n):\overline{\mathbb{F}}_q(X)]=d.
    \end{equation}
    Then if $q>100\delta^3m^2n^2$, the number $N$ of solutions $(x,y_1,\cdots,y_n)\in\mathbb{F}_q^{n+1}$ of the equations $y_1^{d_1}=f_1(X),\cdots,y_n^{d_n}=f_n(X)$ satisfies
    \begin{equation}
    |N-q|<5mnd\delta^{5/2}q^{1/2}.
    \end{equation} 
    We will use this theorem to solve the problem for fixed $r>1$. We have $q=p$ prime number. Let $g$ be a primitive root modulo $p$. We put $n=r+1$, $m=1$, $d_1=\cdots=d_n=r$, $\delta=r$, $d=r^{r+1}$, $f_i(X)=g(X-g^{is})$ for $0\leq i \leq r-1$, and $f_r(X)=gX$. Then the condition is satisfied, and the number $N$ of solutions to the system $Y_i^r=f_i(X)$ ($0\leq i \leq r$) is 
    \begin{equation}
    |N-p|<5(r+1) r^{r+1+5/2}p^{1/2}.
    \end{equation}
We look for the number $N^{*}$ of solutions with no $Y_i$ being zero. 
Then we have
$$N^{*}\geq p-5(r+1) r^{r+1+5/2}p^{1/2}-(r+1)r^{r+1}.$$ 
This is in fact greater than zero for sufficiently large $p$. For any such solution $(X,Y_0,\cdots, Y_r)\in \mathbb{F}_p^{r+2}$, we obtain that
$$X\in \mathbb{Z}_p-(A+B).$$ 
This proves the result.<|endoftext|>
TITLE: Is there an algebraic approach to metric spaces?
QUESTION [48 upvotes]: It is well known that most topological spaces can be studied via their algebra of continuous real-valued (or complex-valued) functions. For instance, in the setting of compact Hausdorff spaces, there is a complete dictionary between topological properties of the space $X$ and corresponding algebraic properties of the algebra $C(X)=C(X,\mathbb{R})$, and two such spaces $X$ and $Y$ are homeomorphic if and only if their algebras $C(X)$ and $C(Y)$ are isomorphic. A similar theory is available in the locally compact Hausdorff case, by replacing $C(X)$ with the algebra of real-valued functions on $X$ which vanish at infinity. Furthermore, these algebras are precisely all the commutative C*-algebras (at least if you take complex-valued functions). Thus if one wishes to study locally compact Hausdorff spaces, he might as well study commutative C*-algebras.
Is there an analogous algebraic approach to the theory of "nice" metric spaces (e.g. compact, locally compact, connected, etc.)? More specifically, is there a natural function space one can attach to every nice metric space which essentially contains all data about the metric space, up to isometry?
I'll welcome any references related to this question.
EDIT: Many of the answers referred me to Nik Weaver's (and others') theory of "Lipschitz algebras". I took a look at his book and while very interesting (and well written), it doesn't seem to be quite what I am looking for. More specifically, it seems that his constructions of Lipschitz algebras only characterize the metric space up to bi-Lipschitz equivalence (which he calls a quasi-isometry), while I'm looking for natural function spaces that would characterize the space up to isometry. His results in section 1.8 of the book show that two (pointed) complete metric spaces have isomorphic Lipschitz function algebras if and only if they are bi-Lipschitz equivalent, rather than isometric. He does have a characterization up to isometry (using a different Lipschitz algebra) for the class of metric spaces of diameter less than 2 (section 1.7) , but his reduction of a general metric space (say complete) to a metric space of diameter $\le 2$ is of course far from preserving isometries (rather preserving Lipschitz mappings).
So it seems that I am looking for a somewhat more "rigid" construction. Any ideas about that?
Thanks.

REPLY [2 votes]: An algebraic context for metric spaces--attractive to me--can be be provided by translation lattices, introduced (in the bounded, distributive case) by Irving Kaplansky, Lattices of continuous functions II, Amer. J. Math. 70 (1948), pp. 626-634; see also Włodzimierz Holsztyński, Lattices with real numbers as additive operators, Dissertationes Mathematicae LXII, Warszawa 1969, where the unbounded and non-distributive translation lattices are included (they all are called d-lattices there to stress the importance of the distance $d$).<|endoftext|>
TITLE: Borel functions on $\omega_1$
QUESTION [6 upvotes]: Endow $\omega_1$ with order topology. It is easy to show that each continuous function $f\colon \omega_1\to \mathbb{R}$ is eventually constant. Is the same true for Borel functions?

REPLY [8 votes]: It isn't true for Borel functions. To see this, observe that the set of successor ordinals (plus zero) is an open set in the order topology of $\omega_1$, and its complement is the set of limit ordinals, so these are both Borel sets in that topology. Thus, the characteristic function of the limit ordinals, with value $1$ on the limit ordinals and $0$ on successors and zero, is a Borel function, but it is not eventually constant.

REPLY [7 votes]: Assuming that by "Borel function" you mean (as is customary) a function such that the inverse image of each Borel set is Borel, the answer is no.  Consider the characteristic function of the set $L$ of countable limit ordinals.  The inverse image of any subset of $\mathbb R$ is $L$ (which is closed), the complement of $L$ (which is open), all of $\omega_1$, or empty.<|endoftext|>
TITLE: A geometric characterization for arithmetic genus
QUESTION [37 upvotes]: Let $X$ be a smooth projective variety over $\mathbb{C}$. The following information is all equivalent (any of these numbers can be computed by a linear equation from any of the others):

the arithmetic genus of $X$
the constant coefficient of the Hilbert polynomial of $X$
$\chi(X, \mathscr{O}_X)$
the "Todd genus" $\int_X \operatorname{td}(T_X)$, where $T_X$ is the tangent bundle of $X$ and $\operatorname{td}$ denotes the Todd class.


Is there a geometric characterization for any of these numbers?

If I understand correctly, characteristic classes (and in particular, Todd classes) can be defined entirely from the topology of $X$, or at least its structure as a smooth manifold. [Edit: This is not true--see the answer of "anonymous." If I understand correctly, the Todd class of a complex vector bundle is a smooth invariant. However, different complex structures on the same real manifold $X$ can give rise to non-isomorphic complex vector bundle structures on $T_X$; in fact, a complex vector bundle structure on $T_X$ is, by definition, an almost complex structure on a real manifold $X$.]  Thus, in some sense, item 4 provides a "geometric characterization" for the arithmetic genus of $X$ (and the other items on the list).  However, I personally find this description so far abstracted from actual geometric properties of $X$ as to be hardly geometric at all. (If anyone disagrees with me and can articulate a geometric intuition for the Todd genus, that would be a reasonable answer.) 
By comparison, I do consider the following characterizations of various properties "geometric":

The self-intersection number of the diagonal embedding of $X$ into $X \times X$. (the Euler characteristic)
The number of points in which a general linear space of complementary dimension meets $X \subset \mathbb P^n$. (the degree of $X \hookrightarrow \mathbb P^n$)
The genus of the curve $X \cap L$, where $L$ is a general linear space of dimension one greater than $\operatorname{codim} X$. (I don't know of a standard name for this, but in a particular sense, it is one of the coefficients of the Hilbert polynomial of $X$.)
The maximum number of copies of $S^1$ that can be removed from $X$ without disconnecting it. (the genus of $X$ if $X$ is a smooth curve, i.e., Riemann surface)

Note that either the first or the last point gives a geometric characterization for (information equivalent to) the genus of a curve. Without one of these, I would not consider the third bullet a "geometric characterization" of anything.  In a way, this provides part of my motivation for asking this question. Let $L_k$ be a general linear space of codimension $k$ in $\mathbb P^n$. Unless I am mistaken, knowing the Hilbert polynomial for $X$ is equivalent to knowing the arithmetic genus of $X \cap L_k$, for every $k \leq n$ such that this intersection is nonempty, via the formula
$$ \chi(\mathscr O_X(n)) = \sum_{k \ge 0} \chi(\mathscr O_{X \cap L_k}) \binom{n+k-1}{k}\;\text.$$
Thus, a geometric characterization for arithmetic genus would automatically give a geometric characterization for the Hilbert polynomial. (Again, in some sense, this is already provided by the Hirzebruch-Riemann-Roch Theorem; but I find this formula so abstracted as to be hardly geometric at all.)

REPLY [8 votes]: It is not true that quantities like the Todd genus (or the arithmetic genus) depend only on the underlying smooth manifold. There are examples of diffeomorphic Kaehler manifolds with different Todd genera.
Check the following papers for further information:
D. Kotschick, Topologically invariant Chern numbers of projective varieties, Adv. Math. 2012, doi 10.1016/j.aim.2011.10.020
D. Kotschick and S. Schreieder, The Hodge ring of Kaehler manifolds, Compos. Math. 2013, doi 10.1112/S0010437X12000759<|endoftext|>
TITLE: How do you tell if a system of linear inequalities has a solution?
QUESTION [9 upvotes]: A naive solution would be to optimize a dummy variable via linear programming and see if a result is returned.  I imagine there must be a more direct way.

REPLY [9 votes]: Most linear programming solvers check for feasibility first (usually this takes as much time as the second phase, which is actually finding the optimum, and uses the exact same algorithm). I would advise reading a standard textbook on the subject (Luenberger is good, Schrijver great if you are interested in integer/mixed problems).<|endoftext|>
TITLE: Why the letter "p" for genus?
QUESTION [8 upvotes]: Does anybody know why the genus (arithmetic or geometric) of a curve was historically denoted by $p$ ($p_a$ and $p_g$)? What does the letter "$p$" stand for? 
Any references would be greatly appreciated.

REPLY [2 votes]: By looking at Coolidge's "Algebraic Plane Curves" Ch. VII, one may guess that $p$ stands
for Plücker. You should have a look at the reference cited by Coolidge in his footnote to the first page of Ch. VII with title "Plücker's equations and Klein's equation" where the notion of genus is presented. The footnote says "For an historical account, see Berzolari, p. 343".
The citation is to:
Berzolari, `Allgemeine Theorie des höheren ebenen algebraischen Kurven', in
Enzyklopädie der Math. Wissenschaften, vol. iii, Part $2^1$, Leipzig, 1906, 99.<|endoftext|>
TITLE: Vector bundles vs principal $G$-bundles
QUESTION [32 upvotes]: It is well known that a (real) vector bundle $\pi : E\to B$ over a topological space (or manifold) $B$ is a fibre bundle whose fibres
$$F=\pi^{-1}(x), \ \ \ x\in B $$
over any $x\in B$,   are diffeomorphic to a vector space $V$.   On the other hand, a principal $G$-bundle is a fibre bundle $\pi : P\to B$ over $B$ with a right free action of a Lie group $G$ on $P$ such that for any open set $U\subset B$, the locally trivial fibrations defined by:
$$
\Phi_{U} : \pi^{-1}(U)\to U\times G,  \ \ \Phi_{U}(p)=(\pi(p), \varphi_{U}(p)).    
$$
Here
$$
\ \varphi_{U} :\pi^{-1}(U)\to G 
$$
 is a $G$-equivariant map, that is $\varphi_{U}(pg)=\varphi_{U}(p)g$, for all $p\in\pi^{-1}(U)$ and $g\in G$. In the last case the  fibers are submanifolds of $P$ which are always diffeomorphic with the structure group $G$. 
Although for any vector bundle $\pi : E\to B$, its  fibers  $F\cong V$   can be considered as Lie group with operation the vector addition, in general we do not include the vector bundles
as examples of principal $G$-bundles (Although, to every vector bundle we can associate  the frame bundle which is a ${\rm GL}_{n}\mathbb{R}$-principal bundle, but I don't speak here about associated bundles).
My question is about a good explanation about the fact that IN GENERAL vector bundles (themselves) do not give   examples of principal $G$-bundles. For example, the tangent bundle $TM$ of a smooth manifold is a prototype example of a vector bundle, but itself it cannot be considered as a principal bundle for a Lie group $G$, is this true? 
Thus I ask:
Which is the basic difference between a vector bundle an a $G$-bundle, which does not allows us (almost always??), to consider the vector bundles themselves as examples of $G$-bundles?
For example, a $G$-bundle is trivial (isomorphic to the product bundle), if and only if it admits a global section, but I think that this is not true  for vector bundles.
A second question is about  examples  of vector bundles which can be considered the same time as $G$-bundles for some Lie group (I think that such an example is a cylinder)

REPLY [15 votes]: I believe you have answered this yourself. Principal bundles are trivial iff they admit a global smooth section. Vector bundles always admit a global smooth section (zero section). 
Therefore vector principal G-bundles are always trivial. The only available examples of vector G-bundles are thus of the form M x G, where G is both a vector space and a Lie group. Any examples of such G except for Abelian?<|endoftext|>
TITLE: Which semigroups can be linearly ordered?
QUESTION [8 upvotes]: As usual I consider a semigroup to be a structure $(A, +)$ such that $+$ is an associative binary function over the set $A$. The notion of linearly-ordered semigroup corresponds to structures of the form $(A, + , \leq)$ such that $(A,+)$ is a semigroup, $\leq$ is a linear order on $A$, and this order $\leq$ is compatible with the binary operation $+$ (i.e., if $a \leq b$ and $a' \leq b'$, then $a + a' \leq b + b'$).
I am interested on known answers to the following questions (they follow the same pattern):

Is there some "useful" characterization of semigroups which can be linearly ordered? To be more precise, for which semigroups $(A, +)$ there is a linear order $\leq$ such that $(A, +, \leq)$ is a linearly ordered semigroup?
Is there some "useful" characterization of commutative semigroups which can be linearly ordered?

Perhaps it is worth pointing out that for the case of commutative groups it is well known that the criteria for admitting a linear order coincides with being torsion-free.

REPLY [3 votes]: I am going to show that any characterization of (linearly) orderable commutative semigroups should be as hard (or as easy, depending on your taste) as the characterization of orderable 3-nilpotent commutative semigroups and as hard as characterization of orderable commutative magmas.
A semigroup is called 3-nilpotent if it has 0 and $xyz=0$ for every $x,y,z$. Commutative 3-nilpotent semigroups have very explicit structure. Each such $S$ is a disjoint union of three subsets $A\sqcup B\sqcup \{0\}$ and the operation gives a symmetric function $A\times A\to B\cup \{0\}$. Conversely any pair of (non-empty) sets $A,B$ and any symmetric function $A\times A\to B\cup\{0\}$ defines a 3-nilpotent commutative semigroup (the product of elements from $A$ is defined using the function, all other products are 0; the associativity is automatic since $xyz=0$ for every $x,y,z$). 
I will use my old idea of $\Theta$-indicator functions (see, say,  Sapir, Mark V. Residually finite semigroups in varieties. Monash Conference on Semigroup Theory (Melbourne, 1990), 258–268, World Sci. Publ., River Edge, NJ, 1991.) If $\Theta$ is a property of countable semigroups and ${\mathcal A}, {\mathcal B}$ are two classes of semigroups (for simplicity we can assume that all countable semigroups have natural numbers as the underlying set). Then a $\Theta$-indicator is a map $\Phi$ from ${\mathcal A}$ to $\mathcal B$ such that 
(*) for every $S\in \mathcal A$, the operation in $\Phi(S)$ is computable given the oracle computing the operation in $S$, satisfying the following property 
(**) $S$ satisfies $\Theta$ iff $\Phi(S)$ satisfies $\Theta$.
The point is that if a $\Theta$-indicator exists, then the problem of describing semigroups from $\mathcal A$ satisfying $\Theta$ is as hard (or as easy) as the problem of describing semigroups from $\mathcal B$ with that property.
 Edit.  We can view the set $\mathcal F$ of all functions $\mathbb{N}\times\mathbb{N}\to \mathbb{N}$ as subsets of $\mathbb{N}^3$. They form a compact subset with the natural product topology. The set of all semigroup operations is a closed subset. So we can view $\Phi$ as a function from a closed subset of $\mathcal F$ to itself. As I learned from Simon Thomas, $\Phi$ satisfies (* ) if and only if it is continuous. See  Dave Marker's text, Lemma 3.11. Dave Marker told me that the result probably goes back to Kleene and Turing. So $\Phi$ is a $\Theta$-indicator if and only if it is a continuous map satisfying (**). I did not know it when I introduced these in 1974 (I was a second year undergraduate student then), in fact I did not know it till a few days ago. 
I used $\Theta$-indicators mostly to study  residually finite semigroups. But one can use it for other properties including the orderability. 
 Theorem. Let $\Theta$ be the property of being linearly orderable. Then there exists an (explicitly constructed) $\Theta$-indicator map from the class of commutative monoids to the class 
of 3-nilpotent commutative semigroups. 
 Proof.  Let $S$ be a commutative monoid. Consider the semigroup $T(S)$ which is a union of three sets $A\times \{1,2\}$ and $\{0\}$ with operation $(a,1)(b,1)=(ab,2)$, all other products are 0. 
It is easy to see that (*) is satisfied. To show (**), assume that $S$ is orderable, then order $T(S)$ by $(a,i)\le (b,i)$ iff $a\le b$, $(a,2) \lt (b,1)$, $0\le x$ for every $a,b\in S$ and any $x$. Clearly it is a linear order on $T(S)$. 
Conversely, if $T(S)$ is linearly orderable, define an order on $S$ by $a\le b$ iff $(a,2)\le (b,2)$. Note that $a\le b$ iff $(a,2)\le (b,2)$ iff $(a,1)(e,1)\le (b,1)(e,1)$ where $e$ is the identity element of $S$. Hence $(a,2)\le (b,2)$ iff $(a,1)\le (b,1)$. That implies $(ac,2)=(a,1)(c,1)\le (b,1)(c,1)=(bc,2)$ in $T(S)$ hence $ac\le bc$ in $S$. Q.E.D.
Thus one can say that orderable commutative monoids are described modulo 3-nilpotent commutative semigroups. 
On the other hand if $S$ is only a commutative magma (not necessarily associative groupoid) with unit, then $T(S)$ is still a 3-nilpotent commutative semigroup and $S$ is orderable iff $T(S)$ is orderable. Hence describing orderable commutative (3-nilpotent) semigroups is as hard as describing all orderable unitary commutative magmas.<|endoftext|>
TITLE: Sums of two same powers modulo $p$
QUESTION [6 upvotes]: It is related to this  question.  
 Question.  Suppose that $p$ is a prime, $p-1$ is divisible by $q^2$ for some $q$. Is it true that every number modulo $p$ is a sum of two $q$-th powers. 
If $q=2$, then (*) is true. Indeed in that case $p\equiv 1 \mod 4$. Take any number $t \mod p$. WLOG we can assume that $t$ is odd (the product of two sums of two squares is a sum of two squares). If $t\equiv 1 \mod 4$, then consider the arithmetic progression $4np+t, n\ge 0$. By Dirichlet, it contains a prime $p'\equiv 1\mod 4$. By Fermat, $p'$ is a sum of two squares, hence $t$ is a sum of two squares modulo $p$. If $p\equiv -1 \mod 4$, then consider the arithmetic progression $8np+t-2p, n\ge 0$. By Dirichlet, it contains a prime number $p'$ of the form $4k+1$ (since $p\equiv 1\mod 4$) and we are done. 
That question may be easier than the question cited above. Or it may be a known open problem. 
 Update  Several simplifications of the argument for $q=2$ were proposed. Although the statement is not true when $(p-1)/q^2$ is small, it is true when this quotient is large enough. I think this answers my question almost completely. Thanks to everybody who gave an answer or a comment.

REPLY [9 votes]: Suppose $p=q^2 + 1$. Then the set of $q$-th powers, including 0, has size $q+1$. The number of elements given by a sum of pairs of these is at most $(q+1) + \frac{(q+1)(q)}{2} = \frac{q^2 + 3q + 2}{2} < q^2 + 1$ whenever $q > 3$. So, for example, $6,7,10,11$ are not expressible as a sum of two fourth powers in $\mathbb{Z}/(17)$.<|endoftext|>
TITLE: Vertex transitive graphs
QUESTION [7 upvotes]: Does having vertex transitivity make the problem of calculating independence and chromatic numbers easier?

REPLY [11 votes]: Not particularly. There is a paper by Codenotti, Gerace, Vigna "Hardness results and spectral techniques for combinatorial problems on circulant graphs" Linear Algebra Appl. 285 (1998) 123-142 which shows that computing the chromatic number of a circulant graph is NP-hard.
(The pdf is available on Codenotti's web page.) Being vertex transitive guarantees that 
a $k$-regular graph has vertex connectivity at at least $2(k+1)/3$ and that its edge
connectivity is equal to $k$. Aside from this, it is not easy to identify useful consequences of vertex transitivity.<|endoftext|>
TITLE: Left ideals vs right ideals
QUESTION [6 upvotes]: By default, let all algebras be complex and unital. I am concerned with the non-commutative algebras. I am wondering if the following might be true (at least for some classes of algebras, like semi-simple algebras).
Suppose $A$ is a complex, non-commutative algebra with a maximal right ideal which is not finitely generated as a right ideal. Must $A$ contain a maximal left ideal which is not finitely generated as a left ideal? Is there any relation between generation of right/left ideals in general?

REPLY [8 votes]: The answer is No.  A ring is right (left) noetherian if all of its right (left) ideals are finitely generated. You can find an example of a ring R that is right noetherian but not left noetherian in:
Tsit-Yuen Lam, A first course in noncommutative rings.<|endoftext|>
TITLE: Is reflexivity an open condition?
QUESTION [9 upvotes]: Is the condition that a module is reflexive an open condition?  
That is, if $X$ is a smooth projective complex variety, $T$ a quasi-projective variety, and $F$ a finitely presented module on $X \times T$ that is $T$-flat, then we can form the locus $T' \subset T$ of points $t$ such that the restriction of $F$ to $X \times t$ is reflexive.
Is $T' \subset T$ open?
If not, is it locally closed?
Recall that a coherent sheaf $F$ is said to be reflexive if the natural map $F \to (F^{\vee})^{\vee}$ to the double dual is an isomorphism.

REPLY [9 votes]: This locus is indeed open. I will explain why using Kollar's "Hulls and Husks" (arXiv:0805.0576). More generally, this article studies in great detail when taking the double dual commutes with base change.
First, we may restrict to the open locus of $T$ where $F_t$ is torsion-free (because reflexive sheaves are torsion-free).
  Then, we choose an ample line bundle $H$ on $X$, and we compute the Hilbert polynomials relatively to $H$. The Hilbert polynomial $P(F_t)$ of $F_t$ is constant by flatness. From the exact sequence $0\to F_t\to F^{\vee\vee}_t\to F^{\vee\vee}_t/F_t\to 0$, we see that $F_t$ is reflexive exactly when $P(F^{\vee\vee}_t)=P(F_t)$, i.e. exactly when $P(F^{\vee\vee}_t)$ takes its minimal value. But the polynomial $P(F^{\vee\vee}_t)$ is constructible and upper semicontinuous by Proposition 28 (3) of Hulls ans Husks. This proves that this locus is open.<|endoftext|>
TITLE: A model category of spaces where strict commutative monoids are $E_\infty$-spaces
QUESTION [15 upvotes]: There are various strict monoidal model categories of spectra (e.g. symmetric spectra) where the honestly commutative monoid objects model the "coherently commutative" ring spectra (which might otherwise be expressed using, say, operads). Is there an analog for spaces? That is, there a monoidal model category, Quillen equivalent to spaces (in some monoidal sense), such that the category of commutative monoids in this category is (Quillen) equivalent to the category of algebras in spaces over some fixed and suitably free $E_\infty$-operad? In spaces, this is false; topological abelian groups are very far from modelling infinite loop spaces.

REPLY [5 votes]: You may also be interested in this paper.<|endoftext|>
TITLE: Small 4-chromatic coin graphs
QUESTION [11 upvotes]: A coin graph is a graph that can be represented by a set of disjoint, except possibly touching, unit disks in the plane (i.e. the disks are the vertices and the edges correspond to the pairs that touch each other). It's easy to show by induction that $\chi(G)\leq4$ for every coin graph $G$, as there's always a vertex of degree at most 3. 
My question is: what is the smallest order (i.e. the number of vertices) of a 4-chromatic coin graph?  
In this paper by Erdos http://www.renyi.hu/~p_erdos/1987-27.pdf there is a coin graph of order 19 that is 4-chromatic (see Figure 1) by I doubt it's the smallest one (it was constructed for a different purpose, having to do with the independence number). The question I asked was proposed for an IMO competition in 1979, see p. 138 question 73 in Djukic, Jankovic, Matic, Petrovic: the IMO Compendium (there is no solution there, however). 
Clearly, coin graphs are also unit distance graphs, for the definition see http://en.wikipedia.org/wiki/Unit_distance_graph. The smallest 4-chromatic unit-distance graph is probably the Moser spindle http://en.wikipedia.org/wiki/Moser_spindle that has 7 vertices.
There is a similar notion of matchstick graphs: those are unit distance graphs drawn in the plane with non-crossing straight-line segments, see http://en.wikipedia.org/wiki/Matchstick_graph Note that the Moser spindle is NOT a matchstick graph, although it's planar and unit-distance. 
The second (related) question is: what is the smallest order of a 4-chromatic matchstick graph?
I think the answer (to the second question) is 8.

REPLY [4 votes]: About matchstick graphs:
I think Paul's proof also shows that my construction on 9 vertices is uniquely optimal for matchstick graphs.
By the same argument, the cycle $C$ bounding the infinite face must be induced, and every vertex on it has a neighbor inside the cycle.
If there is only one vertex inside $C$, the graph is the wheel on $7$ vertices, but then it is $3$-colorable.
By planarity, notice that the neighborhoods of the internal vertices on $C$ are paths which can only intersect in the end points. Thus, $G$ consists of a number (one for each internal vertex) of partial wheels, where neighboring partial wheels either overlap in a point or are connected by an edge.
If there are at least three internal vertices,
as every internal vertex has degree at least $3$, this implies that $C$ has at least $6$ vertices. But in fact, $6$ is not possible, as three diamonds can not be arranged in a triangle.
So there are exactly $2$ internal vertices. Again, $C$ can not contain only $6$ vertices as otherwise the two internal vertices must be in the same place (and the resulting graph would be $3$-colorable anyways), so $C$ contains at least $7$ vertices. A short analysis of the possible sizes of the partial wheels shows that there are exactly two possible configurations ($2$ partial wheels with $5$ vertices each, or one with $4$ and one with $6$, overlapping in one vertex), and only the second one is not $3$-colorable.<|endoftext|>
TITLE: Left adjoint to the forgetful functor from finite product categories to symmetric monoidal categories
QUESTION [10 upvotes]: I recall reading that the forgetful functor $FinProdCat \to SymMonCat$ from categories with finite products and product preserving functors to symmetric monoidal categories and tensor preserving functors has a left adjoint. 
To make this precise one has to insert lax, weak or strict in several places -- I am interested in any combination of these (but most in a 2-adjunction between the categories with weakly product, resp. tensor, preserving functors).
Is something like this true at all? If so, can anyone give a true and precise statement and/or a reference? I wouldn't mind getting a concrete description of the left adjoint, but a confirmation of its existence would already be a treat.
Thanks!
(This is not a case of google laziness: I spent half a day looking for reference. I would imagine that the statement emerges after inserting the right things into long known results about enriched base change or 2-monads, but I wasn't able to find the right one) 

EDIT: What would be nice would be an argument along these lines: Both Symmetric monoidal categories and finite product categories are algebras for certain pseudomonads. Algebras for pseudomonads are are finite copower preserving functors from Cat-enriched Lawvere theories, see Power's Enriched Lawvere Theories, Thm 3.4. There should be a map (sort of an inclusion, since we demand less structure for a symmetric monoidal category) from the Lawvere theory for symmetric monoidal categories to that for finite product categories and the forgetful functor should be precomposition with it. Now the left adjoint could be obtained by taking Cat-enriched left Kan extensions along this map.
One problem is that I only know that left Kan extensions of product preserving functors along product preserving functors are product preserving again, but I don't know the corresponding statement for copowers. This could either be true or for our special Lawvere theories it could be enough to ask for product preserving functors, then the above might have a chance to work.

REPLY [6 votes]: The 2-categories of finite-product categories and symmetric-monoidal categories are both 2-monadic over $\mathrm{Cat}$, in all possible senses.  That is, there are strict 2-monads $P$ and $S$ on $\mathrm{Cat}$ such that $P$-algebras and strict, pseudo, lax, and colax $P$-morphisms coincide with finite-product categories and their morphisms, and likewise for $S$-algebras and symmetric-monoidal categories.  Moreover, both $P$ and $S$ are finitary, and we have a 2-monad morphism $S\to P$ which induces your forgetful functor(s) on categories of algebras.
First consider the strict case: (strict) limits of both $P$-algebras and $S$-algebras (and strict morphisms) are created in $\mathrm{Cat}$, so the forgetful functor preserves strict limits.  Thus, since the 2-categories of $P$- and $S$-algebras and strict morphisms are locally presentable, this functor has a (strict) left adjoint.
The pseudo case requires some more work.  One could in theory mimic the above argument entirely in the world of 2-categories, but I don't know if the machinery has been set up for that yet.  Alternatively, one can use pseudo-morphism classifiers to reduce the problem to the strict case.  This is done in the classic Blackwell-Kelly-Power paper "Two-dimensional monad theory", Theorem 5.12.  (They state the strict case as Theorem 3.9.)
I have no idea whether the lax or colax cases are true; offhand I would say it seems unlikely.<|endoftext|>
TITLE: Metric conditions on configurations of points with only finitely many solutions
QUESTION [11 upvotes]: There is an old puzzle, which I believe I learned from Nob Yoshigahara, that asks for all configurations of four (distinct) points in the plane such that the six pairwise distances assume only two distinct values.  (In other words, there exist two distinct positive reals $a$ and $b$ such that the distance between any two of the four points is either $a$ or $b$.)  Two configurations are considered equivalent if one can be obtained from the other by a dilation followed by a rigid motion.  What makes this a good puzzle is that there are only finitely many solutions.  MO readers may enjoy finding all solutions.
My question is, given $n$, $d$, and $k$, is there an efficient algorithm to determine whether there are only finitely many configurations of $n$ distinct points in $\mathbb{R}^d$ such that the $\binom{n}{2}$ distances between them assume only $k$ distinct values?  Of course, I'm using the same notion of equivalence as stated above.
EDIT: Peter Winkler's book Mathematical Mind-Benders provides the following information about the origin of the aforementioned puzzle: ‘[It] appeared as Problem 3a (submitted by S. J. Einhown and I. J. Schoenberg) in the “Puzzle Section” of the Pi Mu Epsilon Journal in 1985.  Later it showed up on page 1 of Nob Yoshigahara's Puzzles 101, where it was attributed to Dick Hess.’

REPLY [3 votes]: Not an answer but a reformulation: Such a configuration gives rise to a set of $n$ points
on a large $d-$dimensional sphere forming only $k$ different pairwise angles. Indeed,
add a very large last coordinate $\rho$ and make a suitable perturbation (always possible since
all non-zero determinants are roughly linear in $\rho$ and you need only a perturbation of 
order $1/\rho^2$). Considering the Gram matrix formed by scalar products of these $n$ points,
we get a symmetric positive definite matrix $G$ of rank $d+1$ with coefficients in a set
with $k$ elements. Subtracting $\lambda I_n$ from $G$ where $\lambda$ is 
the minimal eigenvalue of $G$, the problem is equivalent to the problem of determining
all positive definite matrices of rank $\leq d$ with constant diagonal
and off-diagonal coefficients in a finite set with $k$ elements.<|endoftext|>
TITLE: Random geometries
QUESTION [6 upvotes]: Let $M$ be a smooth $n$-dimensional manifold, and let $FM = GL(M)$ indicate its tangent frame bundle.  Let $G$ be a fixed linear subgroup of $GL(n)$, and consider the space $\mathcal S$ of all $G$-structures on $M$. Each element $Q \in \mathcal S$ is a $G$-subbundle of the frame bundle $FM$.  For example, if $G = O(n)$, then $\mathcal S$ gives a parametrization of Riemannian metrics on $M$.
Question 1) Does the space $\mathcal S$ of $G$-structures have a nice structure in its own right? e.g., can one express it in terms of bundles, quotients, etc?
Next, for each $G$-structure $Q \in \mathcal S$, let $\mathcal C_Q$ be a space of principal connections on $FM$ which are compatible with the structure on $Q$.
For example, in the Riemannian case $G = O(n)$, we might focus on torsion-free metric connections, in which case $\mathcal C_Q$ consists of a single point, the Levi-Civita connection for $Q$.  We could also focus on more general metric connections, in which case $\mathcal C_Q$ would be non-trivial.
In contexts I like to work in, a geometry consists of some sort of structure like a metric, represented here by the $G$-structure $Q$, and some notion of transport, represented by a choice of a connection $\Gamma \in \mathcal C_Q$.  Is there a more standard name for a geometry consisting of a $G$-structure and a connection?
Now, the space of connections is affine, so we may consider the bundle $\pi : \Omega \to \mathcal S$. A point in the bundle $\Omega$ consists of a $G$-structure and a compatible connection, so we may call $\Omega$ the space of geometries on $M$.  Again, I would be happy to use a more standard name for this space.
The space $\Omega$ is a fiber bundle, and need not globally decompose into a product $\mathcal S \times \mathcal C$ as is customary in probability theory.  Nonetheless, it locally looks a product which should be sufficient for most applications.
Question 2) As in question 1, does the space of geometries $\Omega$ have a nice structure in its own right?
Finally, let's get to probability.  Let $\mathcal F$ be the Borel $\sigma$-algebra of $\Omega$.  A probability measure $\mathbb P$ over the space $(\Omega, \mathcal F)$ is called the law for a random geometry, or simply a random geometry for short.   
My intuition is that a measure $\mathbb P$ is a deterministic object which represents a fuzzy geometry: the picture is that the fuzzy geometry is somehow a superposition of many deterministic realizations of geometries. This intuition can be made more precise. Let $f : \Omega \to \mathcal A$ be some observable of a geometry (measurable function), where $\mathcal A$ is some nice algebra (I'm thinking of the real numbers). Then we may want to take the expectation of $f$, which is simply the integral $\int_\Omega f(\omega) \mathbb P(\mathrm d \omega)$.  We don't evaluate $f$ for on any fixed geometry, but instead assign each $\omega$ an infinitesimal weight $\mathbb P(\mathrm d\omega)$ and add up the weighted contributions of $f(\omega)$.
Question 3) Does the space of random geometries $\mathcal P(\Omega)$ have nice geometric structure?
Of course, this space is quite large and there are many such probability measures. We must probably impose additional constraints to do some actual probability theory.
Suppose $M$ is a homogeneous space with symmetry group $\Sigma$ (i.e., $\Sigma$ is a Lie group with a transitive, faithful action on $M$).  Of course, when $M$ is equipped with a geometry $\omega \in \Omega$, the space $(M, \omega)$ will not be invariant under the action. This is not a problem, though, since we could naturally impose a symmetry constraint on the law of a random geometry, rather than the geometry itself.
The group $\Sigma$ naturally acts on $\Omega$ in the obvious way: instead of pushing forward a point through a symmetry transformation, we pull back a geometry.  This means that $\Sigma$ has a natural action on $\Omega$.  We now impose that the law $\mathbb P$ is invariant under the symmetries of the space $M$.  That is, we now impose the condition that $\mathbb P = \mathbb P \circ \varphi^{-1}$ for all transformations $\varphi \in \Sigma$.
Let $\mathcal P^\Sigma(M)$ denote the probability laws on $\Omega$ which are invariant under $\Sigma$, and call this the space of symmetric random geometries. Note that a specific realization of a symmetric random geometry need not be symmetric; rather, the law is smeared smoothly over the space $M$.
Question 4) Does the space of symmetric random geometries $\mathcal P^\Sigma(M)$ have nice structure?  Are there any simple, natural examples of any symmetric random geometries $\mathbb P \in \mathcal P^\Sigma(M)$?
Thanks for bearing with such a long question. Hopefully, it is accessible to both geometers and probabilists.  I thank Ben Bakker, Jarek Korbicz, and Kate Poirier for teaching me some real geometry over the past few weeks, and my apologies to them in advance for the errors which are probably lurking around this post.

REPLY [4 votes]: First of all, a canonical reference for special geometric structures is the book "Compact manifolds with special holonomy" by Dominic Joyce.
A1: As you observed, specifying an $O(n)$-structure is the same thing as picking a Riemannian metric, in other words a section of the bundle of positive symmetric 2-tensors. For other $G$'s you can also usually describe the space of $G$-structures as the space of sections of some bundle of tensors or forms (this way of thinking about them will also be quite useful, when you want to specify some probability measure) Here are a few more examples:
$G=Sp(n,\mathbb{R})$, nondegenerate 2-form $\omega$, almost symplectic structure
$G=GL(n,\mathbb{C})$, endomorphism $J$ of $TM$ with $J^2=-1$, almost complex structure
$G=G_2$, positive 3-form $\varphi$ (on a 7-manifold), almost $G_2$ structure
You might wish to focus on torsion-free $G$-structures. This amounts to imposing some integrability condition, in the above examples they are $d\omega=0,N_J=0$ and $d\varphi=0=d\star\varphi$ respectively. Also, in many cases the moduli-space of torsion-free $G$-structures is actually finite-dimensional, which might make life easier if you want to do analysis/probability. 
A2: When $G\subseteq O(n)$ then the Levi-Civita connection is the unique one to focus on. In general, I doubt that there is much extra geometric structure on the space of connections in addition to being an affine space.
A3: Maybe you can indeed try to put some nice extra structure on the space of probability measures $\mathcal{P}(\Omega)$. One idea would be to consider some Wasserstein-distance where the cost-function fits well to the geometric problem, e.g. for $G_2$ how much does it cost to transport $\varphi_1$ to $\varphi_2$....
Hope that helps at least somewhat... (the parts of your question that I did not address also sound very interesting!)<|endoftext|>
TITLE: Filling $\mathbb{R}^3$ with skew lines
QUESTION [7 upvotes]: I would like to know if it is possible to fill $\mathbb{R}^3$ with lines with the
following two properties:
(1) Every point $x \in \mathbb{R}^3$ is contained in precisely one line.
(2) Every neighborhood of every point is pierced by lines whose directions fill out the
sphere of possible line orientations, in this sense:
For every point $x$ and every $\epsilon > 0$, the lines that pass through a point
in the ball $B_\epsilon(x)$ of radius $\epsilon$ centered on $x$ have the property
that, were they all translated to pass through the origin, the closure of the set of points that
constitutes their intersection with an origin-centered sphere $S$, fills out $S$ completely.
This image below is meant to suggest the idea:
                


I am sure there is a more concise way to phrase the second condition; apologies for my
ungainly formulation.  I want to be able to find every line orientation within a neighborhood of
every point.
Perhaps condition (2) is not possible to achieve
in conjunction with (1). But I don't see an argument.
Any ideas/insights/pointers would be appreciated—Thanks!

REPLY [10 votes]: I have to give a lecture in a few minutes, so this will be just a quick sketch.  List, in a well-ordered sequence of length $\mathfrak c$ (the initial ordinal of cardinality continuum) the requirements that (1) some line passes through $x$ (one requirement for each $x\in\mathbb R^3$) and (2) some line passes through $B$ in direction $d$ (one requirement for each open ball $B$ and direction $d$).  Now go through the requirements, one at a time, and choose, for each one, a line fulfilling that requirement and disjoint from previously chosen lines.  (Exception: If you get to a requirement (1) and the relevant $x$ is on a previously chosen line, skip that requirement since it's already satisfied.)  I claim it's easy to check that you never get stuck, i.e., at any stage, the previously chosen, strictly fewer than $\mathfrak c$ lines, cannot block all the lines that would satisfy your current requirement.
Since this "construction" depends on well-ordering a set of the cardinality of the continuum, it will give a horrible decomposition of $\mathbb R^3$.  I don't see at the moment whether this can be done "nicely", for example with a Borel partition.<|endoftext|>
TITLE: Why does the Section Conjecture exclude curves of genus 1?
QUESTION [26 upvotes]: Let $X$ be an integral proper normal curve over a (perfect) field $F$, of genus $\geq 2$. One variant of Grothendieck's "section conjecture" states that the sections $G_F \rightarrow \pi_1(X)$ of the exact sequence
\begin{equation}
1 \rightarrow \pi_1(X_{\bar{F}}) \rightarrow \pi_1(X) \rightarrow G_F \rightarrow 1
\end{equation}
are, up to conjugation, in bijection with the $F$-rational points of $X$, where $G_F$ is the absolute Galois group of $F$ and $\pi_1$ is the algebraic fundamental group. 
Question: what is the reason for excluding genus 1 curves?
I understand why genus 0 curves must be excluded: if $F$ has characteristic zero, it is a general fact that the 'geometric' fundamental group $\pi_1(X_{\bar{F}})$ is just the profinite completion of the regular topological fundamental group of $X$, seen as a curve over $\mathbb{C}$. For genus 0, the topological fundamental group is trivial, and thus the above exact sequence induces an isomorphism $\pi_1(X) \rightarrow G_F$. Hence there is always at least one section even if $X$ has no rational points whatsoever. 
However, I don't know of a good reason why genus 1 curves should be excluded here. The above argument obviously won't do since the topological fundamental group is no longer trivial for genus 1. Are there even so known counter-examples for genus 1 curves? What goes wrong?
I know the philosophy is that one should expect 'anabelian behaviour' only when the fundamental group is 'far from being abelian', which excludes the genus 1 case. But I would be more satisfied with a more concrete, less philosophical, reason!

REPLY [28 votes]: I think that Grothendieck had already observed that the map from rational points to sections is injective (for curves of genus at least 2 over a number field)  and I believe that his proof works even for curves of genus $1$, so the thing that fails for curves of genus $1$ is surjectivity.
Consider any exact sequence of groups $1 \to A \to G \to H \to 1$ with $A$ abelian and assume that there is a section $\sigma:H \to G$. A simple calculation shows that if $\sigma$ is a section and $f: H \to A$ is any map, then the function $\tau: H \to G$ given by $\tau(h) = f(h)\sigma(h)$ is a section (i.e. also a homomorphism) iff $f$ is a $1$-cocycle. If the cocyle is not a coboundary then $\sigma$ and $\tau$ are not conjugate, so what we really care about is $H^1(H,A)$.
We now apply the foregoing in the situation of the question, so we are led to consider the group $C = H^1(Gal(\bar{F}/F), \pi_1(X_{\bar{F}}))$. Since $\pi_1(X_{\bar{F}})$ is a $\hat{\mathbb{Z}}$ module, so is $C$. Now suppose the curve $X$ has infinitely many rational points, so the group $C$ is also infinite (but finitely generated by the Mordell-Weil theorem). However, there are no fintely generated but infinite $\hat{\mathbb{Z}}$ modules. It follows that the map from rational points to sections modulo conjugacy cannot be surjective.<|endoftext|>
TITLE: Localization of a symmetric monoidal category at a single morphism
QUESTION [10 upvotes]: Let $C$ be a symmetric monoidal category, and $f : x \to y$ be a morphism in $C$. I would like to construct the localization $C_f$ explicitly, which solves the universal property
$$\mathrm{Hom}_{\otimes}(C_f,D) = \{F \in \mathrm{Hom}_{\otimes}(C,D) : F(f) \text{ iso}\}.$$
I am not interested in a general existence proof or alike; instead I would like to exhibit $C_f$ as an explicit full reflective subcategory of $C$, thereby also showing its existence. The reason is that I want to actually compute something in these localizations which a priori does not simply follow from the universal property.
There is a general construction of a localization of a plain category with respect to arbitrary sets of morphisms, which can be found on page 6 of Gabriel-Zisman's Calculus of fractions and homotopy theory. Thus, an object of the localization $C_f$ is an object of $C$, and a morphism is a class of a finite sequence of the form $f_1 s^{-1} f_2 \cdots s^{-1} f_n$ (perhaps without $f_1$ or $f_n$), where the sources and targets should fit, subject to the obvious cancellation rules. But this might not define a (small) set of morphisms, right?
Question 1. Which conditions have to be imposed on $C$ and $f$ so that $C_f$ exists (without leaving the universe)?
I know the basics about left/right multiplicative systems (as in Gabriel-Zisman, Weibel, Kashiwara-Schapira, etc.), saturations etc., but I could not find any answer to this question in the literature.
EDIT: As Theo points out, there is no set-theoretic problem if we localize a category at just one single morphism. But when $C$ is monoidal, there is no reason why the tensor product $C \times C \to C$ extends to a tensor product $C_f \times C_f \to C_f$, because this means that for every $x \in C$ the invertibility of $f$ forces the invertibility of $x \otimes f$ and $f \otimes x$ in the language of categories, which is unplausible. Instead we should better localize at all morphisms $x \otimes f$ and $f \otimes x$, where $x$ runs through all objects of $C$; this is a monoidal class in the language of Day's paper "A Note on Monoidal Localisation". But now there are set-theoretic problems in the description of $C_f$ above. On the other hand, we repair this easily if $C$ has a small colimit-dense subcategory, which happens to be the case when $C$ is presentable.
Question 2. Which conditions have to be imposed on $C$ and $f$ so that we can write down explicitly the localization $C_f$ in the $2$-category of symmetric monoidal categories? How does it look like?
Question 3. Actually I am interested in cocomplete symmetric monoidal categories and cocontinuous symmetric monoidal functors between them. How does the localization look like in this context?
Let me mention a special case where everything works out: Let $\mathcal{L} \in C$ be an object whose symmetry $\mathcal{L}^{\otimes 2} \to \mathcal{L}^{\otimes 2}$ is the identity and $f : 1_C \to \mathcal{L}$ a morphism (imagine a global section of a line bundle on a scheme). Let $C_f \subseteq C$ the full subcategory consisting of those $M \in C$ such that $M \otimes f : M \to M \otimes \mathcal{L}$ is an isomorphism. If $C$ is cocomplete, the inclusion $C_f \subseteq C$ has a left adjoint: It maps $M \in C$ to the colimit of $M \to M \otimes \mathcal{L} \to M \otimes \mathcal{L}^{\otimes 2} \to \dotsc$. Using this left adjoint, one can define tensor products and colimits in $C_f$ (one may cite Day's reflection theorem here) and verify easily that $C \leadsto C_f$ is the localization in the context of cocomplete symmetric monoidal categories.
One can also verify that if $\mathcal{L}$ is a line bundle on a scheme $X$ and $f \in \Gamma(X,\mathcal{L})$ is a global section, then we really have $\mathrm{Qcoh}(X)_f = \mathrm{Qcoh}(X_f)$, so this categorical localization is compatible with the scheme theoretic localization.
However, for other applications, I need more general morphisms $f$. This motivated my question. I am pretty sure that this should be standard in category theory, therefore the reference request tag.

REPLY [5 votes]: I'll try to bend David White's answer towards the actual situation of your question. The outcome is somewhat clumsy and it totally looks like model structures can be eliminated from it, but anyway:
Assume your category C is closed monoidal and locally presentable. Then it is a monoidal model category with cofibrations and fibrations all morphisms and weak equivalences the isomorphisms. This model category is cofibrantly generated: One can take the identity of the initial object as generating trivial cofibration and the set of all morphisms between the objects $G$ from some generating set as generating cofibrations.
This model category then satisfies the hypotheses of Barwick's Thm. 4.46 in this article for a Bousfield localization at the one element set containing $f$. The homotopy category for the localized model structure has the universal property you want and lives in the same universe. You have an adjunction between the homotopy category of the original model structure, which is the category itself, and the localization.
This adjunction is a reflection to an orthogonal subcategory as in Adamek/Rosicky, 1.35-1.38, namely to the full subcategory of all objects from whose point of view $f$ "was already an isomorphism" (i.e. $f$-orthogonal objects; precise definition via a unique-lifting-condition). This is much like in your example (but with the condition on the twist removed and without the domain of f having to be special). If you chase through the proofs, you also get an expression of the reflection functor as a colimit via the small object argument, resp. via Adamek/Rosicky's proof...
Barwick's Prop. 4.47 gives then a criterion for the homotopy category to be closed monoidal again: It suffices that any object $X$ which satisfies the unique right lifting condition with respect to $f$ also satisfies it with respect to $f \otimes G$ for every generating object $G$ (remember the category was locally presentable now) i.e. if $f$ induces an iso $Hom(f,X)$ then $Hom(f \otimes G,X)$ is an iso, too, for every generating object $G$.
edit: Sorry, I am no longer sure that the homotopy category of the Bousfield localization is in fact the localization along $f$ in the sense you asked for: When you localize with respect to an arrow you automatically invert together with it a bunch of other arrows. When you do plain category theory it is somewhat uncontrollable what those other arrows are, it seems to me. When you do Bousfield localization these other arrows are those having the left lifting property with respect to the $f$-local objects. Now I don't see a reason why the class of additionally inverted arrows should be the same in both cases. What Bousfield localization as sketched here probably yields, is the universal colimit preserving functor which inverts $f$.<|endoftext|>
TITLE: Can all convex optimization problems be solved in polynomial time using interior-point algorithms?
QUESTION [24 upvotes]: Just a new guy in optimization. Is it true that all convex optimization problems can be solved in polynomial time using interior-point algorithms?

REPLY [45 votes]: No, this is not true (unless P=NP). There are examples of convex optimization problems which are NP-hard.
Several NP-hard combinatorial optimization problems can be encoded as convex optimization problems over cones of co-positive (or completely positive) matrices. See e.g. "Approximation of the stability number of a graph via copositive programming", SIAM J. Opt. 12(2002) 875-892 (which I wrote jointly with Etienne de Klerk).
Moreover, even for semidefinite programming problems (SDP) in its general setting (without extra assumptions like strict complementarity) no polynomial-time algorithms are known, and there are examples of SDPs for which every solution needs exponential space. 
See Leonid Khachiyan, Lorant Porkolab. "Computing Integral Points in Convex Semi-algebraic Sets". FOCS 1997: 162-171 and Leonid Khachiyan, Lorant Porkolab "Integer Optimization on Convex Semialgebraic Sets". Discrete & Computational Geometry 23(2): 207-224 (2000).
M.Ramana in "An Exact duality Theory for Semidefinite Programming and its Complexity Implications" Mathematical Programming, 77(1995) shows that SDP lies either in the intersection of NP and co-NP, or outside the union of NP and coNP, and nothing better than this is known.
In "Semidefinite programming and arithmetic circuit evaluation" Discrete Applied Mathematics, 156(2008) Sergey P. Tarasov and Mikhail N. Vyalyi show that SDP can be used to  compare numbers represented by arithmetic circuits. (The latter is regarded as one of hard problems).<|endoftext|>
TITLE: families of genus four curves with only hyperelliptic reduction
QUESTION [8 upvotes]: Is it possible to construct a nonisotrivial family of genus four curves $X \rightarrow S$, with the following properties:
(1) $S$ is a complete curve;
(2) All the fibers are smooth;
(3) The generic fiber lies on a singular quadratic in $\mathbb{P}^3$.
It is not possible to construct such a family if we replace (3) by "all the fibers...". But the problem is that there could be smooth hyperelliptic reductions.

REPLY [2 votes]: No, there is no such $S$:
EDIT: (BIG) GAP BELOW I compute limits of certain linear series in the Hurwitz scheme, and then I make claims about limits of other (bigger) linear systems taken over curves in $\mathcal{M}_3$. However, the limit curves in the Hurwitz scheme are not stable (the first of them is analyzed in Donagi's paper quoted below in Example 2.10 (iii) ).
The trigonal construction (see Donagi's "the fibers of the prym map" - sadly the arxiv version is without the pretty pictures) gives you a 1-1 correspondence between:

Smooth genus $4$ curves $C$, a non trivial 2-torsion point $\alpha\in JC[2]$, and a base point free $g^1_3$ on $C$.

Smooth genus $3$ curve $X$ -- which is the Prym curve of the pair $(C,\alpha)$ -- with a $g^1_4$, such that no fiber of a the $g^1_4$ is of the form $2(p+q)$.


Assuming that this $g^1_4$ is not a canonical pencil, it is a complete linear system $|K_X+\beta|$. One then verifies that there is a natural isomorphism between the unramified double cover of $C$ associated with $\alpha$, and
$$\overline{X\times_{|K_X+\beta|} X\setminus\mathrm{diagonal}}/(x,y)\sim(y,x)\cong \Theta_X \cap (\Theta_X+\beta),$$
which means that the inverse of the trigonal construction on $|K_X-\beta|$ gives $C$ with a different $g^1_3$. Which means that under the 1-1 correspondence above, $C$s sitting on a singular quadric correspond to the case where the $g^1_4$ is canonical.
We now turn to the case where $C$ is hyperelliptic. One should work carefully on the Horwitz schemes on both sides of the correspondence (essentially using the bigonal construction dictionary from Donagi's paper, and using the fact that the intermediate Abelian varieties in the two sides of the bigonal construction are dual), but the upshot is that on the $X$ side we would have a $g^1_4$ on the prym of a hyperelliptic curve, which splits to two cases (denoting by $H_C$ the hyperelliptic class on $C$):

$\alpha+H_C$ is the sum of two Wierstrass points $w_1,w_2$, in which case $X$ is hyperelliptic. One may identify $|H_X|$ and $|H_C|$, such that the residual points to $w_1, w_2$ correspond to the Weierstrass points of $X$ (there is an exercise about it in ACGH, in the chapter on Prym varieties).

$\alpha+2H_C$ is the sum of four Weirstrass points $w_1,\ldots, w_4$, in which case $X$ is reducible: it is the "gluing" of genus $1$ and a genus $2$ curve. Similarly to above (but with the dualizing system on the $X$ side and the canonical system on the $C$ side), $w_1, \ldots w_4$ corresponds to the ramification points of the $g^1_2$ on the genus $1$ component, and the residual $6$ points correspond to the ramification points of the $g^1_2$ on the genus $2$ component.


Note that the first case here becomes closer to the second if we glue a rational tail to the genus $3$ curve. Indeed, the resulting curve is not longer semi stable, but we work with Hurwitz schemes, so by and large, not much harm is done.
Armed with all this, we now take the curve $S$, throw the hyperelliptic points, lift it from $\mathcal{M}_4$ to $\mathcal{R}_4$ (i.e. adding a choice of a non trivial 2-torsion point), apply the construction to get a family of genus $3$ curves together with canonical pencils, and lift the curve from $\mathcal{M}_3$ to $\mathcal{M}_3(2)$, (adding full level structure). We now have coordinates on the universal canonical system (e.g. by fixing the "first" four bitangents). I.e. we have a non projective curve $S'$, with

A surface $\tilde{X}\subset \mathbb{P}^2\times S'$: the universal $X$.

A curve $P\subset \mathbb{P}^2\times S'$: the universal point such that for every $s\in S'$ the projection of $\tilde{X}_s$ from $p_s$ to $\mathbb{P}^2$ is the fiber's $g^1_4$.


We can now close $S', \tilde{X}$ and $P$. If you believe the analysis of the hyperelliptic cases, then the limits in both cases are two double lines (representing the two components of $X$, which may be either genus $0$ and $3$, or genus $1$ and $2$), and where the limit of $P$ over this fiber does not hit any of the lines.
I.e. we now have a surface of bidegree $(4, d_1)$, and a curve $(1, d_2)$ inside $\mathbb{P}^2\times\overline{S'}$ which do not intersect; which is impossible.<|endoftext|>
TITLE: When quotient of a $k$-algebra by any maximal ideal is $k$?
QUESTION [5 upvotes]: Let $k$ be a valued field.
Is there a special term for a commutative (Banach) $k$-algebra $A$ such that for any maximal ideal $m$ we have $A/m=k$?
Is there an easy to check criterion that would imply this property?

REPLY [3 votes]: I don't think that there is a standard term for what you are looking for, but I would be inclined to call such an algebra "compact".  My reason is the following.  Suppose that $X$ is a manifold (resp. locally compact Hausdorff space).  Then the maximal spectrum of the ring $C^\infty(X)$ of smooth real-valued functions (resp. $C^0(X)$ of continuous real-valued functions) with the Zariski topology is homeomorphic to the Stone--Cech compactification $\beta X$ of $X$.  If $X$ is not astronomically large*, then the points $\mathfrak m \in \beta X$ for which the quotient field $C^\infty(X)/\mathfrak m$ (resp. $C^0(X)/\mathfrak m$) is isomorphic to $\mathbb R$ are precisely the points in $X \hookrightarrow \beta X$.  In particular, $X$ is compact iff $\beta X = X$ iff every maximal ideal comes from a point of $X$ iff every maximal ideal has quotient field $\mathbb R$.
I said this over $\mathbb R$, but really any infinite field $\mathbb K$ will do.  (Then I need to take "isomorphic" to mean "isomorphic as $\mathbb K$-algebras", so that if something is isomorphic to $\mathbb K$, then it is so uniquely.  Over $\mathbb R$, I can in fact take "isomorphic" to mean "isomorphic as rings", since $\mathbb R$ has no nontrivial ring automorphisms.)
*Footnote: The claim fails if there is a measurable cardinal $\lambda > |\mathbb K|$ with $|X| \geq \lambda$.  Note that in many universes there are no uncountable measurable cardinals at all, and if uncountable measurable cardinals exist, then they are strongly inaccessible, and hence very very large.<|endoftext|>
TITLE: A question about the Beurling-Selberg majorant
QUESTION [6 upvotes]: Beurling's majorant is defined as the unique entire function $B(z)$ such that the Fourier transform of $B(x)$ is compactly supported in the interval $[-1;1]$, $B(x) \geq \text{sgn}(x)$ and $B(x)$ minimizes the integral, 
$$
\int_{-\infty}^{\infty} B(x) - \text{sgn}(x) \text{d} x=1
$$
Explicitly,
$$
B(z) = \left ( \frac{\sin (\pi z)}{\pi} \right )^2 \cdot \left ( \frac{2}{z} + 
\sum_{n = 0}^{\infty} \frac{1}{(z-n)^2} - \sum_{n=1}^{\infty} \frac{1}{(z+n)^2} \right )
$$
Note that by Paley-Wiener the property that the Fourier transform of $B(x)$ is compactly supported in $[-1;1]$ is equivalent to the property that $B(z)$ is a function of exponential type, with $B(z) = O(e^{2 \pi |\Im z|})$. 
In general $B(x)$ is also a very good point-wise approximation to $\text{sgn}(x)$ and this makes it a very valuable function when one needs optimal numerical constants. The need for such arises sometimes in analytic number theory, for example. 
Seulberg modified Beurling's function, and considered, 
$$ S_{+}(z) = \frac{1}{2} B(\delta (z - a)) + \frac{1}{2} B(\delta ( b - z)) $$
This is a very good majorant for the characteristic function of the interval $[a;b]$. Here $\delta$ is a parameter that regulates the quality of the approximation. The price to pay for a larger $\delta$ is that $\hat{S_{+}}(x)$ vanishes when $x$ is larger, i.e when $|x| > \delta$.
Selberg's majorant has the properties that $S_{+}(x) \geq 1$ when $a \leq x \leq b$ and $S_{+}(x) \geq 0$ otherwise. Furthermore $\hat{S_{+}}(x) = 0$ when $|x| \geq \delta$ and finally 
$$
\int_{-\infty}^{\infty} S_{+}(x) \text{d} x = b - a + \frac{1}{\delta}
$$
When $\delta(a-b)$ is an integer this is in fact the optimal majorant satisfying these constraints, but in general it is quite good. 
After this long introduction, my question is the following: Is there a known construction of an optimal majorant $M_{+}(x)$ with the following properties,
1) $M_{+}(x) \geq 1$ when $a \leq x \leq b$, and $M_{+}(x) \geq 0$ otherwise
2) The fourier transform of $M_{+}(x)$ is compactly supported in $[-\delta;\delta]$
3) The difference
$$
\int_{-\infty}^{\infty} |M_{+}(x)|^2 \text{d} x - (b-a) 
$$
is as small as possible? How small can it be, in terms of $\delta$ ? (EDIT: I am particularly interested in upper bounds for the above quantity). 
I will be grateful for any insights, references concerning this question. I know that Valeer has done quite some work with Beurling and Selberg's majorants but I haven't been able to locate anything relevant to my question in his publications. 
You can view the Selberg majorant approximating the unit interval with $\delta = 3$
at http://www.freeimagehosting.net/xffsu . (If one of the mods could link the image into my question I will be grateful).

REPLY [3 votes]: Given such a $M$ note that $F(x)=M^2(x)$ will be (1) non-negative, (2) majorize $1_{[a,b]}$, and (3) has $\hat{F}$ supported in $[-2\delta,2\delta]$. Thus it will be a (possibly not optimal) solution to the standard $2\delta$ Beurling-Selberg problem. This implies that 
$$\int |M(x)|^2dx - (b-a) \geq \frac{1}{2\delta}$$
(at least in the cases when the standard Beurling-Selberg construction is known to be sharp).<|endoftext|>
TITLE: Can one find the hodge number by counting points over finite fields?
QUESTION [14 upvotes]: Given a proper smooth variety $X$ of dimension $n$ over $\mathbb{C}$, assume it has a model over a DVR of mixed characteristic $(0,p)$ with residue field $\mathbb{F}_q$, and assume the closed fiber $X_0$ is smooth.
By the Weil conjecture, one can find the Betti number of the complex manifold $X^{an}$ by counting $\mathbb{F}_{q^r}$-points of $X_0$. If by counting points we find $|X_0(\mathbb{F}_{q^r})|=\sum\pm u_j^r$ for all $r$ and $b_i$ of the $u_j$'s has absolute value (in $\mathbb{C}$) equal to $\sqrt{q}^{i}$, then the $i$-th Betti number is equal to $b_i$.
My question is, can one find the Hodge number of $X$, which is $h^{ij}=\dim H^i(X,\Omega^j)$ by counting points of the closed fiber $X_0$? (The reason I'm asking this is, I guess both should connect to the theory of weights on the motive. So even if one cannot find the Hodge number this way, the reason must be interesting.)

REPLY [15 votes]: It is worth mentioning the "Newton above Hodge" theorem. This is a way in which the point count can impose nontrivial conditions on the Hodge numbers, beyond knowing the Betti numbers. 
As I assume you know, the number of points of $X(\mathbb{F}_{q^k})$ is $\sum_{r=0}^{2n} (-1)^r \sum_{i=1}^{b_r} \alpha_{i,r}^{k}$ where $b_i$ is the $i$-th betti number and $\alpha$'s are algebraic integers obeying $|\alpha_{i,r}| = q^{r/2}$, where the left hand side is any archimedean absolute value. Also, we can number the $\alpha$'s such that $\alpha_{i,r} \alpha_{b_r+1-i, r} = q^r$. (This is a consequence of Hard Lefschetz.). In particular, $\prod_i \alpha_{i,r} = \pm q^{r b_r/2}$.
Fix $r$, so it will no longer appear in our notation. Let $F(x) = \prod_{i=1}^{b_r} (x-\alpha_{i,r})$; this is the characteristic polynomial of Frobenius on $H^r(X)$. Let $N$ be the $p$-adic Newton polytope of $F$. Since the constant term of $F$ is $\pm q^{r b_r/2}$, the endpoints of the Newton polytope are at $(0,0)$ and $(b_r, r b_r/2 \cdot v_p(q))$. The symmetry $\alpha_{i,r} \alpha_{b_r+1-i, r} = q^r$ means that the segments of slope $\mu$ and $r-\mu$ have the same length. In ulrich's example, all of the eigenvalues of Frob on $H^2$ have norm $q$, so $N$ is just a straight line from $(0,0)$ to $(22,22)$.
We now define the Hodge polygon $H$. This is also a piecewise linear convex curve joining $(0,0)$ to $(b_r, r b_r v_p(q)/2)$. All of the segments have slope of the form $k r v_p(q)/2$ for some integer $k$ between $0$ and $2r$, and the horizontal length of this segment is $h^{k,r-k}$. The symmetry of the Hodge diamond tells us that this polygon, also, has the property that the segments of slope $\mu$ and $r-\mu$ have the same length.
In Ulrich's example $X$, the Hodge polygon goes from $(0,0)$ to $(1,0)$, then to $(21,20)$, and then to $(22,22)$.

The Newton above Hodge theorem says
  that $N$ is above $H$.

I learned this from Kiran Kedlaya $p$-adic differential equations course, which he has now converted into a book, see Chapter 14. I am not familiar with the history of this result, but a quick scan of mathscinet suggests that the result was proved for hypersurfaces by Dwork, conjecturd in general by Katz (unpublished, I think), mostly proved by Mazur and the final details added by Ogus. Mazur's paper looks very readable -- that's where I would start.<|endoftext|>
TITLE: Homomorphic images of a Cartesian product of finite groups
QUESTION [10 upvotes]: What can be said about the class of groups which can be represented as a homomorphic image of an (infinite) Cartesian product (=unrestricted direct product) of finite groups? What would be simple groups in this class? Which "classical" groups, like various matrix groups, belong to this class?
The question can be (trivially) reformulated as a description of the smallest class of groups containing all finite groups and closed under taking Cartesian products and homomorphic images.
Remarks:

This class includes ultraproducts of finite groups, or, what is the same, groups having the same first-order theory as a set of finite groups (called pseudo-finite in the literature).
A countable Cartesian product of finite groups is a profinite group. I don't know how beneficial it would be to consider the question in the profinite context.
A finitely-generated group from this class is necessarily finite (Nikolov and Segal, see arXiv:1108.5130, Theorem 32).

REPLY [4 votes]: @Pasha, you might like to look at http://arxiv.org/pdf/1108.5130.pdf if my memory serves me right, you might find some answers to your questions.<|endoftext|>
TITLE: What is the Dunford Integral and why is it useful?
QUESTION [8 upvotes]: Wikipedia http://en.wikipedia.org/wiki/Pettis_integral defines the Pettis Integral for Banach space valued functions wrt to some measure space by duality. 
It calls the Pettis & Bochner integral the weak & strong integral respectively, which implies some kind of relationship; also, apparently there is a Dunford integral which specialises to the Pettis integral. 
My question is: Why are these weak integrals useful, what is the definition of the Dunford Integral and how does it specialise to the Pettis one, and what is the relationship between the Pettis Integral & the Bochner Integral?.
I've just noted: Weak and Strong Integration of vector-valued functions, which answers the part of my question about the Pettis Integral.

REPLY [12 votes]: Since you've answered part of the question, let me elaborate on the Dunford integral. If $f:\Omega\to E$ is weakly measurable and satisfies $\langle x^* ,f\rangle\in L^1(\Omega)$ for all $x^*\in E^*$ then it's possible to define $\int_X f\in E^{**}$. This acts on $E^{*}$ via the prescription
$$x^*\mapsto \int_X\langle x^{*},f(\omega)\rangle d\omega.$$
I first learnt this in Talagrand's AMS memoir:
http://books.google.co.uk/books?id=YqJ3t2oT5WgC&lpg=PP1&dq=pettis%20integral%20and%20measure%20theory&pg=PP1#v=onepage&q=pettis%20integral%20and%20measure%20theory&f=false
but a good text on infinite dimensional analysis will probably mention it. $f$ is Pettis integrable precisely when the Dunford integral $\int_Xf$ lies in the canonical image of $E$ in $E^{**}$ and then the two agree (modulo the spaces they live in).
Let's cook up an example. Take the function $f:[0,1]\to c_0$ defined by $f(x)=2^ne_n$ whenever $x\in [2^{-n},2^{-n+1})$. Define $f_n$ to be the obvious simple approximation: let it be $0$ on $[0,2^{-{n+1}})$ and agreeing with $f$ elsewhere. 
The problem with these step functions: we compute
$$\int_0^1{\| f(s)-f_n(s)\|ds}=\sum_{k=n}^\infty{2^k(2^{-k+1}-2^{-k})}=\infty$$
and note that our dumb approximation doesn't work in the strong sense.
But pick any $g\in\ell^1$ and note that
$$\int_0^1{\langle f(s)-f_n(s) , g\rangle ds} =\sum_{k=n}^\infty g(k) \to 0$$
as $n\to\infty$. So our function can be weakly approximated by step functions - in fact you can easily see $s\mapsto \langle f(s),g\rangle$ is $L^1$ with $\int\langle f(s),g\rangle=\lim_{n\to\infty}\int\langle f_n(s),g\rangle$. So $f$ is Dunford integrable.
Note that the integrals
$F_n=\int_0^1{f_n(s)ds}=\sum_{k=1}^{n-1}{e_k}$
do not converge to anything in $c_0$. We can think of them as members of $\ell^\infty$ and of course here they converge (in weak* topology) to the identity. In fact, $\int_0^1{\langle f(s), g\rangle ds}=\sum_{k=1}^\infty g(k)$ - so by definition $\int_0^1{f(s)ds}=1 \in \ell^\infty$. This is an example of the Dunford integral, when the Petis/Bochner integrals won't do.
Another example: modify the function $f$ in our example of Dunford integrability, so that $f(x)=2^ne_n/n$ instead of $2^ne_n$. The obvious step function approximation still can't be strong, as $\int_0^1{\|f(s)-f_n(s)\|}=\sum_{k=n}^\infty{1/k}$. Now the sequence of approximations $F_n$ converges in norm to $(1/k)_{k=1}^\infty$. Thought of as an element of $\ell^\infty$ this is the Dunford integral, but as an element of $c_0$ it's just the Pettis integral. (Note, I'm not even saying this function is not Bochner integrable, but the weak approximations make the Pettis integral very natural in this context).<|endoftext|>
TITLE: Lower bounds on dimensions of faithful representations of braid groups
QUESTION [12 upvotes]: Let $B_n$ be the braid group on $n$ strands.  It's a theorem of Daan Krammer and Stephen Bigelow that there is a a faithful representation
$$B_n \to GL_{n \choose 2} \mathbb Z[t^{\pm}, q^{\pm}] $$
i.e. the range is the group of invertible matrices of rank $n \choose 2$ whose entries are from a Laurent polynomial ring over the integers. 
My question for the community is, what kind of lower bounds are known on $k$ for there to be a faithful representation
$$B_n \to GL_k \mathbb C$$ 
Also, are there any analogous bounds for the same question but for mapping class groups? 
I have a vague recollection that there are some good answers to this question but I forget who they're due to.  I also strongly suspect if you restrict to unitary representations there are some stronger results, maybe due to Stoimenow or Marin?  
edit: Marin has a lower bound $k \geq n+1$ provided $n$ is sufficiently large. 
edit2: Korkmaz has a theorem that says no representation from the genus $g$ mapping class group to $GL_n \mathbb C$ can be faithful, provided $n \leq 3g-3$. http://arxiv.org/abs/1108.3241

REPLY [11 votes]: Slightly improving a theorem of Franks and Handel, Korkmaz proved that any representation from the genus $g$ mapping class group to $\text{GL}_n(\mathbb{C})$ is trivial if $n \leq 2g-1$.  This bound is sharp because of the standard symplectic representation.  See http://arxiv.org/abs/1104.4816.<|endoftext|>
TITLE: Efficient topological triangulations of non-convex polyhedra
QUESTION [8 upvotes]: Does every polyhedron in $\mathbb{R}^3$ with $n$ triangular facets have a topological triangulation with complexity $O(n)$?
Suppose $P$ is a non-convex polyhedron in $\mathbb{R}^3$ with $n$ triangular facets, possibly with positive genus.  A topological triangulation of $P$ is a simplicial complex whose underlying space is the closure of the interior of $P$, such that every facet of $P$ is a cell in the complex.  These boundary facets are true geometric triangles, but interior simplices may be arbitrarily bent and twisted.  In the more standard geometric triangulations, every simplex is the convex hull of its vertices.
Results of Chazelle and Shouraboura imply that every polyhedron has a geometric triangulation with complexity $O(n^2)$.  Moreover, a classical construction of Chazelle implies that the $O(n^2)$ bound is is optimal in the worst case, even when the genus is zero.
But we can get tighter bounds for topological triangulations, at least for genus-zero polyhedra.  If $P$ has genus zero, Steinitz's theorem implies that there is a convex polyhedron $Q$ that is combinatorially equivalent to $P$.  Alexander's extension of the Schönflies theorem implies that the interiors of $P$ and $Q$ are both homeomorphic to open balls.  Thus, applying a suitable homeomorphism to a minimal geometric triangulation of $Q$ gives us a topological triangulation of $P$ with complexity $O(n)$.  (Alternatively, we can triangulate $P$ by joining an arbitrary interior point to every facet.)
What makes the question tricky for higher-genus polyhedra is the possibility of knottedness; the topology of the interior of $P$ is not determined by its genus.  Intuitively, the question is how knotted the interior of a polyhedron can be, as a function of the number of facets.
The following question may be equivalent: Let $K$ be a closed polygonal chain (or "stick knot") in $S^3$ with $n$ edges.  Is there a topological triangulation of $S^3$ with complexity $O(n)$ that includes $K$ in its 1-skeleton?  Again, if we insist on geometric triangulations, $\Theta(n^2)$ tetrahedra are always sufficient and sometimes necessary, even if $K$ is unknotted.
Added for bounty (Apr 13): Partial results, subquadratic upper bounds, or references that imply this problem is open (or the crossing-number problem in my comment on the first answer) would be welcome.

REPLY [3 votes]: I've been thinking about the main question in the original post on and off for a few days.  All of my efforts have been in the direction of finding enough examples to prove a super-linear lower bound, following Misha's suggestion to use hyperbolic volume.  This hasn't worked yet - the problem appears to be tricky!  In any case, here is the most appealing of the constructions. 
Stick braids
Let $x, y, z$ be the usual coordinates on $\mathbb{R}^3$.  Let $D$ be the unit disk in the plane $z = 0$ and let $E$ be the unit disk in the plane $z = 1$.  Suppose that $\{a_i\}_1^n \subset D$ is a collection of points.  Let $b_i$ be the point in $E$ with the same $x$ and $y$ coordinates as $a_i$. Suppose that $\sigma \in \Sigma_n$ is a permutation.  Let $B = B(a, \sigma)$ be the collection of line segments where the $i$'th segment has endpoints $a_i$ and $b_{\sigma(i)}$. If the segments are pairwise disjoint then we call $B$ a stick braid.
It follows that the braid closure of $B$ is a link with stick number at most $5n$.  As a concrete example, the $(p,q)$-torus knot can be obtained by placing the points $a_i$ at the $p$'th roots of unity, taking $\sigma(i) = i + q$, modulo $p$, and taking a braid closure.   
Hyperbolic volume
Now we must use the Euclidean geometry of the braid $B$ to draw conclusions about the hyperbolic volume of the braid closure.  Consider the unit disk $D_t$ in the plane $z = t$.  As $t$ varies from $0$ to $1$, the points of intersection $D_t \cap B = \{a_i^t\}$ move along straight lines at speeds depending on the slope of the $i$'th strand.  Here is a lie: when two points $a_i^t$ and $a_j^t$ come much closer to each other than they are to any of the other points, then there is a definite contribution to hyperbolic volume.  Making this precise (ie, actually true) and then finding a braid $B$ that arranges superlinearly many such meetings would give the desired lower bound. 
One way to do this would be to take $n$ sufficently large, $\epsilon$ correspondingly small, and take the points $a_i$ to be a generic $\epsilon$--net in $D$.  Choose $\sigma$ to be a random permutation.  Let $B = B(a, \sigma)$.  Take the braid closure and plug everything into SnapPy.  I've not tried to do this yet, but it would at least give some data...
Edit
Before writing the above, I had the idea of generating a random stick knot using outer billiards -- namely, let $P$ and $Q$ be concentric spheres and build a knot by taking segments tangent to $Q$ with endpoints on $P$.  This has the virtue that when $Q$ has smallish radius, the expected crossing number will be quadratic.  But it seems easier to estimate volume using the braid construction, and it is volume that really matters to us. 
And then... after all this thinking and writing, I started poking randomly around the web and found O'Rourke's question on our very own MO.  O'Rourke gives a very simple model for random stick knots: just bounce around in a sphere.  The Thurstons suggest that the expected volume grows as $n^{3/2}$.<|endoftext|>
TITLE: Can Duflo type map be defined for invariant differential operators ? (In a way compatible with "Harish-Chandra" isomorphism defined by F. Knop)
QUESTION [5 upvotes]: Background 1) Knop's theorem
In fundamental paper in Annals 1994 F. Knop proved the following theorem.
Let G be a connected reductive group and X a smooth G-variety.
Theorem: Assume that X is either spherical or affine. Then the center Z(X) of the ring of G-invariant differential operators on X is a polynomial ring. More precisely, Z(X) is isomorphic to the ring of invariants of a finite reflection group.
Background 2) Duflo's map
M. Duflo defined fundamental construction: for any Lie algebra $g$ a map:
$DufloMap: S(g) \to U(g) $, such that it is 
1) Restricted to the $S(g)^g$ it will give isomorphism of commutative algebras $S(g)^g \to Z(U(G))$
2) It is isomorphism of graded vector spaces, moreover it is identity map on "principal symbolds"
3) Isomorphism of $g$ modules 
D. Calaque, C. Rossi "Lectures on Duflo isomorphisms in Lie algebras and complex geometry"
http://people.mpim-bonn.mpg.de/crossi/LectETHbook.pdf
See  Capelli determinant = Duflo ( determinant)  - was it known ?  , 
Is the Duflo map for Lie algs. unique ?
for some info on Duflo map.
Question
Let $G$ be a Lie group (reductive may be necessary). $M$ is manifold (may be affine).
Question
Is it possible to find such a map from Functions(T*M) to Differential operators on $M$,
such that  it satisfies requirements similar to the Duflo map,
where the role of $S(g)$ is played by $Fun(T^*M)$ and of $U(g)$ is played by $Dif(M)$ ?
Moreover we should require is is compatible with Knop-Harish-Chandra isomorphism.
If center of $Dif(M)^G$ is trivial, then there is no point to ask the question,
but if center is non-trivial the requirements seems quite non-trivial.

REPLY [4 votes]: Let $X=T^*M$, and apply Dolgushev's equivariant formality theorem (see http://arxiv.org/abs/math/0307212) to get a $G$-equivariant $L_\infty$-quasi-isomorphism
$$
\mathcal U:T_{poly}X \longrightarrow D_{poly}X
$$
Let $\pi$ be the Poisson structure on $X$ which comes from the canonical symplectic form on $T^*M$. Then one gets a $G$-equiavriant quasi-isomorphism of complexes
$\mathcal U_\pi^1$ from the Poisson cochain complex of $(X,\pi)$ to the Hochschild cochain complex of $(\mathcal O_X,\star_\pi)$, where $\star_\pi$ is the star-product the quantizes $\pi$ through $\mathcal U$.
Moreover, one has a homotopy betwen the cup-product one both sides, and it is $G$-equivariant (compatibility with cup-products is sketched in the original paper of Kontsevich http://arxiv.org/abs/q-alg/9709040, is rigorously proved in http://arxiv.org/abs/math/0106205 by Manchon and Torossian, and the globalization is adressed in http://arxiv.org/abs/0805.3444 - in this last reference you will easily see that if you start with a $G$-invariant connection to perform globalization then your homotopy will be $G$-invariant too).
Conclusion. taking $G$-invariants and restricting yourself to the degree zero part of cohomology (Poisson and Hochschild, respectively), you'll find that the Poisson center of $\mathcal O_X^G$ is isomorphic, as an algebra, to the center of the subalgebra of $G$-invariants elements in the quantization.
Things I have been hiding under the carpet. 1. issues involving $\hbar$. In this case you have to prove that the quantization is actually convergent and can be specialized to $\hbar=1$. 2. You have to prove that the quantization really gives back differential operators... this is probably the more tricky part.<|endoftext|>
TITLE: Fourier transform
QUESTION [8 upvotes]: Does anyone know what the Fourier transform (in the sense of distributions) of 
$$
f(x) = (x^2 - 1)^{1/2}x, \quad |x|\ge 1, 
$$
and $f(x) = 0$ otherwise, is?

REPLY [17 votes]: First of all observe that
$$ f(x)=\frac{1}{3}\frac{d}{dx} (x^2-1)^{\frac{3}{2}}_+, $$
where  for any real number $t$ we set $t_+=\max(t,0)$.   Thus it suffices  to compute the Fourier  transform of $(x^2-1)^{\frac{3}{2}}_+$.
In Section 2.5 Chapter 2  of   the book by Gelfand and Shilov, Generalized Functions, vol.1, Academic Press 1964,  the authors  compute the Fourier transform of $(ax^2+bx+c)^\lambda_+$. Your example corresponds to Case (3) discussed there.  More precisely  the Fourier transform of $(x^2-1)^\lambda_+$ is the function
$$\Gamma(\lambda+1)\sqrt{\pi}\left|\frac{\xi}{2}\right|^{-\lambda-\frac{1}{2}}\frac{\cos\pi(\lambda+\frac{1}{2}) J_{-\lambda-\frac{1}{2}}(|\xi|)-J_{\lambda+\frac{1}{2}}(|\xi|)}{\sin \pi(\lambda+\frac{1}{2})}, $$
where $J_\alpha$ denotes the Bessel function  of order $\alpha$.<|endoftext|>
TITLE: Finite sums of prime numbers $\geq x$
QUESTION [19 upvotes]: Let $S_x$ be the set of finite sums of prime numbers $\geq x$. In other words, let $S_x$ be the submonoid of $(\mathbf{Z}_{\geq 0},+)$ generated by the set $\mathcal{P}_{\geq x}$ of prime numbers $\geq x$.
It is easy to see that $S_x$ contains every sufficiently large integer. This follows from the classical fact that given two coprime integers $a$ and $b$, every sufficiently large integer, in fact every integer $\geq (a-1)(b-1)$, is of the form $ma+nb$ for some $m, n \in \mathbf{Z}_{\geq 0}$. See for example this page.
Let $N_x$ be the largest integer which is not in $S_x$.
Examples :
If $x=2$ then $S_2 = \{0\} \cup \mathbf{N}_{\geq 2}$ so that $N_2=1$.
If $x=3$ then $S_3 = \{0,3\} \cup \mathbf{N}_{\geq 5}$ so that $N_3=4$.
By definition, we have $N_x \geq x-1$ (in fact parity considerations imply that $N_x \geq 2x-2$ for $x \geq 3$).
On the other hand, given that there are at least two primes in the interval $[x,2x]$, the above classical fact implies that $N_x \ll x^2$.

What is the asymptotic behaviour of $N_x$ as $x \to \infty$ ?

REPLY [4 votes]: According to "The three primes theorem with almost equal summands" by Baker and Harman, every large odd $N$ is a sum of three primes each of size $\sim N/3$. (In fact, within $N^{4/7}$ of $N/3$-- this is much closer than we need, so we could use weaker results of earlier authors, if preferred.) 
For odd $N$, this gives $N$ as a sum of three primes, each at least $\sim N/3$. If $N$ is even, take a prime $p$ of size $\sim N/4$ (which exists by the prime number theorem), and write $N-p$ as a sum of three primes of size about $N/4$. So we get $N$ as a sum of four primes at least $\sim N/4$.<|endoftext|>
TITLE: The area of the intersection of convex sets with prescribed pairwise intersections
QUESTION [11 upvotes]: Consider two numbers $a>b>0$. Let $A_1,A_2,A_3$ be three convex sets in ${\mathbb R}^2$ such that $\mu(A_i)=a$, $\mu(A_i\cap A_j)=b$ ($\mu$ is the usual measure on ${\mathbb R}^2$). What is the minimal possible value of $\mu(A_1\cap A_2\cap A_3)$?
Surely, if we omit the convexity assumption, the answer is trivial. But it is not necessary realizable by convex sets. Consider, for instance, the case $a=2b$: the answer is nonzero!
It seems that the optimal construction is the following one. Take a triangle $ABC$, and cut three trapezoids $A_BA_CBC$, $B_AB_CCA$, $C_AC_BBA$ (here $A_BA_C\parallel BC$, $B_A,C_A\in BC$, similar relations for the others). E.g., for $a=2b$ the altitude of the trapezoid is $2/5$ of the altitude of triangle; hence the answer in this case seems to be $1/16$.
The generalizations are also interesting. E.g., what happens if we fix the areas but omit the relation that they are equal?

REPLY [2 votes]: Here is Ilya's construction:<|endoftext|>
TITLE: A quantum Grothendieck group?
QUESTION [11 upvotes]: Question 1: Given a co-commutative bialgebra, does there exist a sort of Grothendieck group type construction? Presumably this should take the form of a functor from co-commutative bialgebras to hopf algebras?
My motivation: the finite particle vectors in the symmetric Fock space $\mathbb{C}\Omega\oplus \bigoplus_{n=1}^\infty H^{\vee n}$ have the natural structure of a graded bialgebra. Just set

$m(f^{\otimes n},g^{\otimes m})=Sym(f^{\otimes n}\otimes g^{\otimes m})$,
$\eta(\lambda)=\lambda\Omega$,
$\Delta(f^{\otimes n})=\sum_{k=0}^n{ {n\choose k} f^{\otimes k}\otimes f^{\otimes n-k}}$, where $f^{\otimes 0}:=\Omega$, and
$\epsilon(\cdot)=\langle \Omega,\cdot\rangle$.

but it doesn't seem to have an antipode.
Question 2: Can we make a quantum group containing this bialgebra?
Actually it occurs to me that it has probably been looked at, since a finite dimensional $H$ gives us an algebra isomorphic to $\mathbb{C}[x_1,\ldots,x_{\mathrm{dim} H}]$? I should say also that I'm neither an algebraist, nor a quantum groupie, so I'd appreciate any references/constructions readable by a non-expert!

REPLY [5 votes]: The forgetful functor from the category of Hopf algebras to the category of bialgebras has a left adjoint.  This means that given a bialgebra $B$, there is a Hopf algebra $H(B)$ with a bialgebra morphism $\iota : B \to H(B)$ such that any bialgebra morphism from $B$ to a Hopf algebra $H$ factors through $\iota$ via a morphism of Hopf algebras.  
I do not know whether $H(B)$ is cocommutative if $B$ is, and I also do not know whether the morphism $\iota$ is always injective.  For the latter question I sort of suspect the answer to be negative, just as a semigroup does not always inject into its Grothendieck group.
For discussion of this and related issues, see http://arxiv.org/abs/0905.2613 (very short) and references therein, especially to the lecture notes of Bodo Pareigis available here: http://www.mathematik.uni-muenchen.de/~pareigis/pa_schft.html<|endoftext|>
TITLE: Frobenius splitting over non-algebraically closed fields
QUESTION [6 upvotes]: Let $X$ be a scheme over an algebraically closed field $k$ of positive characteristic $p$. Recall that the absolute Frobenius morphism $F : X \to X$ is the map which is the identity on points and the $p^{th}$ power morphism on functions. Recall also that we say that $X$ is Frobenius split if there is an $\mathcal O_X$-linear morphism splitting the $p^{th}$ power morphism $ \mathcal O_X \to F_* \mathcal O_X $.
Now, whenever one sees the definition of Frobenius splitting, it is always stated for an algebraically closed field $k$. However, the definitions above make perfectly good sense for any scheme over $\mathbb F_p$, and in fact many Frobenius-split schemes, eg flag varieties, are "split over $\mathbb F_p$" in the sense that the Frobenius splitting is the appropriate base-change of a morphism $F_* \mathcal O_X \to \mathcal O_X$ for a scheme $X$ over $\mathbb F_p$. (Although I defined the Frobenius morphism above for schemes over $k$ it also makes sense for any scheme over $\mathbb F_p$). My question is: Why is the definition of Frobenius splitting always stated for schemes over an algebraically closed field, when one can state it more generally for schemes over any field containing $\mathbb F_p$? Is this just convention or is there a deeper reason?

REPLY [12 votes]: There is of course the commutative algebra perspective on Frobenius splittings as well.  Indeed, the general case there is much more general than what I think you are even considering.  Let me make some comments on various generalizations of the notion of Frobenius splittings.
The perfect field case:  Essentially everything works without change as David Speyer already mentions.  Of course, there are the usual complications with working over non-algebraically closed field, but I am completely confident that there are no problems with the theory (many of my papers on this subject work in this generality, or the next one).  One problem of course is with the differentials, if you are working at non-$k$-points, things get slightly more complicated.
The $F$-finite field case:  A field $k$ is called $F$-finite if the $k$ is a finitely generated $k^p$ vector space.  This automatically holds for any residue field of a variety of finite type over a perfect field (in particular, the case that David Speyer mentioned above).  A variety over an $F$-finite field is still fairly well behaved.  In particular, such a variety $X$ is itself $F$-finite, meaning the Frobenius morphism:
$$F : X \to X$$
is a finite map.  I don't think the Cartier isomorphism works without doing some relative Frobenius stuff, but, things still mostly work. 
In particular, there is still a trace map:
$$T : F_* \omega_X \to \omega_X$$
Here's how you see it.  Fix $\omega_X$ on $X$ coming from the structural map to $k$ (in general, this is the first non-zero cohomology of $g^! k$ where $g : X \to k$ is the structural map).  There is a natural map $$\mathcal{H}om_X(F^e_* O_X, \omega_X) \to \omega_X.$$  This is the evaluation at $1$ map.  On the other hand, it is not difficult to show that in the context described, $\mathcal{H}om_X(F^e_* O_X, \omega_X) \cong F_* \omega_X$ (critically using the Frobenius is finite).  Thus we have obtained another trace map.
Notice, this map works for all $X$, not just normal $X$ (it even works for non-reduced schemes of finite type over $k$).  This map agrees with the map in the algebraically closed / perfect $k$ case, at least up to multiplication by a unit (in that isomorphism above, I have to make a choice).  
Standard arguments then imply, that for normal $X$, we have that Frobenius splittings of $X$ come from sections of $H^0(X, O_X((1-p)K_X)$.  Again, let me briefly describe this:  
Set $U = X_{\text{reg}} \subseteq X$, the regular locus.  Then we have that 
\begin{align} & & \mathcal{H}om_X(F_* O_X, O_X) \newline 
& \cong & i_* \mathcal{H}om_U(F_* O_U, O_U) \newline
& \cong & i_* \mathcal{H}om_U(F_* (\omega_U^p), \omega_U) \newline
& \cong & i_* \mathcal{H}om_{F_* O_U}(F_* \omega_U^p, \mathcal{H}om_U(F_* O_U, \omega_U))\newline
& \cong & i_* F_* \mathcal{H}om_U(\omega_U^p, \omega_{U})) \newline
& \cong & i_* F_* O_U( (1-p)K_U) \newline
& \cong & F_* O_X( (1-p)K_X).\end{align}
The map $i : U \to X$ is the inclusion and the isomorphsims involving adding or removing the $i_*$ just come because the sheaves are reflexive / S2.  
In particular, Frobenius splittings can still be viewed as divisors linearly equivalent to $(1-p)K_X$.  The only problem is that we don't quite have the same nice local differential form picture as far as I know (at least, it hasn't been written down).
The $F$-finite scheme case:  Now I want to consider schemes not of finite type over a base field, but which are still $F$-finite, in particular, $$F : X \to X$$ is a finite map.  Note by a result of Kunz, such schemes are automatically locally excellent, and by a result of Gabber, they all have dualizing complexes.  In particular, $\omega_X$ still exists and makes perfect sense.
We still have that $\mathcal{H}om_X(F_* O_X, \omega_X) \cong F_* \omega_X$ locally, but it is an open question whether this is a global isomorphism.  It is possible to deduce that there exists a line bundle $L$ such that 
$$\mathcal{H}om_X(F_* O_X, \omega_X) \cong F_* (\omega_X \otimes L).$$
For some discussion of this, see THIS QUESTION OF MINE
Regardless, in some applications, you can keep track of this line bundle.  In particular, Frobenius splittings correspond to sections of some fixed twist of $\omega_X^{(1-p)}$.  
The general scheme case:
The fact that $F_* O_X$ now need not be a finite $O_X$-module seems to make things much harder.  Indeed, in this case, requiring a Frobenius splitting doesn't even seem to be the right thing to do.
However, in the local setting, commutative algebra has a replacement.
Definition (Hochster-Roberts):  A local ring $R$ is called $F$-pure if for any module $M$, the map:
$$M = M \otimes R \to M \otimes F_* R$$
is injective.  This is a weaker condition than being $F$-split, and I think Mel Hochster told me an example at some point, but at this point I forget.  Regardless,we have the following.
Theorem:  An $F$-finite local ring is $F$-pure if and only if it is $F$-split.  
A good reference for this is the paper of R. Fedder, $F$-purity and rationality singularity
It seems that this condition is much better behaved in a number of ways.  If you search mathscinet for $F$-pure ring, you'll find many papers on this topic.  I should point out that there is even a replacement for an ``honest'' Frobenius splitting in this context.  
Set $E$ to be the injective hull of $R/\mathfrak{m}$.  It turns out that $R$ is $F$-pure if and only if there exists an injective map $$E \to F_* E$$ (this is a recent result of Rodney Sharp).  
Philosophical statement:  Maps $E \to F_* E$ should be viewed as replacements for maps $F_* R \to R$ for non-$F$-finite rings.
If $R$ is $F$-finite, such maps are Grothendieck-local-dual to maps $F_* R \to R$ (at least for $R$ complete), and injective ones dualize to surjective ones (which can easily be tweaked into splittings).  I think that even in the general context, such maps should still correspond to divisors, although I've never actually checked this.<|endoftext|>
TITLE: A counter example in obstruction theory
QUESTION [6 upvotes]: Let $K$ denote a simplicial complex and $Y$ a path-connected topological space. Let us also denote by $K^n$ the $n$-skeleton of $K$. I would like to have an example for the following situation or a proof of its impossibility: 

A map $f^1:K^1\to Y$ that can be extended to $f^2:K^2\to Y$ and yet no such extension can be further extended to $f^3:K^3\to Y$. 

The idea is that there is an obstruction to the existence of $f^3$ already on the one-dimensional level, but not by obstructing the existence of $f^2$. It is written in Hilton and Wylie's book that a bit more general phenomenon of this type is possible:

There is a complex $K$, a subcomplex $L\subseteq K$ and a map $f^0:L\cup K^0\to Y$, such that there is an extension $f^1:L\cup K^1\to Y$ which has an extension over $L\cup K^2$, but not over $L\cup K^3$ while $f^0$ has an extension over $L\cup K^3$.

In words, when trying to extend a given map $f:L\to Y$ over $K$, inductively through $f^n:L\cup K^n$ it is possible to get stuck with an $f^2$ that not only does not have an extension over $L\cup K^3$, it is even impossible to fix it by revising the last step and yet by revising the last two steps it is possible to extend the chosen $f^0$ over $K^3$. I could not find an example for this either so it would also be appreciated.

REPLY [15 votes]: Here is an example with CW complexes rather than simplicial complexes.  I doubt that there is an important difference, although the simplicial case will require more bookkeeping.
Take $K=\mathbb{R}P^3$ and $Y=\mathbb{R}P^2$.  We can give $K$ a CW structure with skeleta $\mathbb{R}P^k$ for $0\leq k\leq 3$.  Let $f^1:\mathbb{R}P^1\to Y$ be the evident inclusion.  Clearly this extends over $K^2$.  Now suppose we have an extension $f^3:K^3=K\to Y$ of $f^1$.   This will then give a graded ring homomorphism $(f^3)^*:H^*(Y;\mathbb{Z}/2)\to H^*(K;\mathbb{Z}/2)$, or in other words $(f^3)^*:(\mathbb{Z}/2)[y]/y^3\to (\mathbb{Z}/2)[x]/x^4$.  Because $f^3$ extends $f^1$ we must have $(f^3)^*(y)=x$.  This gives a contradiction because $y^3=0$ but $x^3\neq 0$.<|endoftext|>
TITLE: Area Enclosed by the Convex Hull of a Set of Random Points
QUESTION [12 upvotes]: Consider $n$ points generated randomly and uniformly on a unit square. What is the expected value of the area (as a function of $n$) enclosed by the convex hull of the set of points?

REPLY [21 votes]: Update: By a result of Buchta (Zufallspolygone in konvexen Vielecken, Crelle, 1984; available on digizeitschriften.de) there is a general formula for this expected value,  it is 
$$1 -\frac{8}{3(n+1)} \bigl( \sum_{k=1}^{n+1} \frac{1}{k} (1 - \frac{1}{2^k}) - \frac{1}{(n+1)2^{n+1}} \bigr) $$
yielding (starting with $n=3$): $11/144$, $11/72$, $79/360$, $199/720$, and so on.
The paper contains in fact a more general result, where the problem is solved for any convex $m$-gon; not just the square. 
For asymtotics see other answer(s).
--
Old version (highly incomplete and wrong guess)
For $n=3$ the expected value is $11/144$ and for $n=4$ it is $11/72$. 
This information is taken from a somewhat recent paper (2004) by Johan Philip where the respective distribution functions are studied in detail. I did not see any mention of exact values for other small values of $n$ there (the asymptocic result given already is mentioned though), so they might be unknown.

REPLY [12 votes]: For any convex set $K$ in dimension $d$ with volume $V(K)$, it is aymptotically $$V(K)-\frac{T(K)}{(d+1)^{d-1}(d-1)!}n^{-1}ln(n)^{d-1}+O(n^{-1}ln(n)^{d-2}ln(ln(n)))$$ (see {New perspectives in stochastic geometry} by W. Kendall and I. Molchanov, p. 49) where $T(K)$ is the number of "flags", i.e. of sequences $f_0\subset f_1 \subset ... \subset f_{d-1}$ where $f_i$ is an $i$-dimensional facet. There is an abundant literature for random convex hulls and if you're interested there might be an exact closed formula for the square.<|endoftext|>
TITLE: Prime divisors of the order of an automorphism group
QUESTION [7 upvotes]: Let $G$ be an unsolvable group of order $(p+1)t$, where $t\mid (p-1)/2$. Is it true that $p$ does not divide $|Aut(G)|$?

REPLY [5 votes]: I believe the answer is yes. Let $\alpha$ be an automorphism of $G$ of order $p$. By the Frattini argument, for any prime $q$ dividing $|G|$, a Sylow $q$-subgroup of $G$ is normalized by an element of order $p$ in the semidirect $S$ product of $G$ with $\langle \alpha \rangle$. Since all elements of order $p$ are conjugate in $S$, it follows that $\alpha$ normalizes some Sylow $q$-subgroup $Q$.
But, since the only prime that could divide both $p+1$ and $t$ is 2, we have $p > |Q|$ whenever $q>2$ and hence $\alpha$ centralizes $Q$. We also have $p > |Q|$ for $q=2$, and hence $\alpha=1$, except in the case when $p$ is a Mersenne prime.
So we have reduced to the case when $p$ is a Mersenne prime, and hence $t$ is odd. Now $\alpha$ centralizes a subgroup $T$ of $G$ of order $t$ and acts nontrivially on a Sylow 2-subgroup $Q$, which has order $p+1$ (and must be elementary abelian). In the semidirect product $S$, we have $[\langle \alpha \rangle, G] = [\langle \alpha \rangle, TQ] = [\langle \alpha \rangle, Q] = Q$, so $Q$ is normal in $G$. Hence, since the odd order subgroup $T$ is solvable, $G$ is solvable, contradiction.<|endoftext|>
TITLE: Automorphisms of $\mathbb{C}$
QUESTION [17 upvotes]: Is it true that $G_{\mathbb{Q}}$, the absolute Galois group of $\mathbb{Q}$, is a subgroup of $Aut(\mathbb{C})$ ?
Or a simpler question: can any automorphism of $\overline{\mathbb{Q}}$ be extended to an automorphism of $\mathbb{C}$?

REPLY [11 votes]: Your question depends on the axiom of choice.  As noted in other comments, if you assume AC, then the complex numbers have crazy automorphisms, and any automorphism of any subfield of $\mathbb{C}$ can be extended to an automorphism of all of $\mathbb{C}$.
However!  If you do not assume the axiom of choice, then it is consistent to say that the only automorphisms of the complex numbers are the identity and conjugation (this is consistent with ZF and implied by additional axioms such as the axiom of determinacy [which implies among other things that all sets are measurable]).  Note that this is inconsistent with AC, but it is not the negation of it.
In this case (without AC), the only automorphisms that can be extended to automorphisms of $\mathbb{C}$ are the identity and complex conjugation!
So with the axiom of choice, the automorphisms are really crazy and any automorphism of a subfield can be extended to an automorphism of all of $\mathbb{C}$.  But without the axiom of choice (and WITH some other consistent axioms), the automorphisms are all really boring!
Interesting stuff.
-Pat Devlin<|endoftext|>
TITLE: enumerative meaning of natural q-Catalan numbers
QUESTION [15 upvotes]: Define $[n]=(1-q^n)/(1-q)$ and $[n]!=[1][2][3] \cdots [n]$, so that $[2n]!/[n]![n+1]!$ is a polynomial in $q$ (the most algebraically natural $q$-analogue of the Catalan numbers); what enumerative interpretation(s) does it have, vis-a-vis the standard members of the "Catalan zoo"?  I would be especially interested in answers pertaining directly to triangulations of polygons, without any intervening bijections.

REPLY [3 votes]: NEWEST
Seem my random speculations below. It never hurts to look in the OEIS and it would appear that the coefficient of $q^k$ in $\frac{1}{[n+1]}\left[{2n\atop n}\right]_q$ is the number of Dyck words of length 2n having major index k . I have not had time to look into translating that into triangulations.
OLDER
We do know that the $q$-binomial coefficient $$\left[{m\atop k}\right]_q=\frac{[m]!}{[k]![m-k]!}$$ is a polynomial (with non-negative integer coefficients) equal to $\binom{m}{k}$ when $q=1$ and that the coeffcient of $q^d$ is the number of ordered lists of type $0^k1^{m-k}$ having $d$ inversions. Equivalently, the number of lattice paths from $(0,0)$ to $(k,m-k)$ which enclose an region of area $d$ (with the $x$-axis and $x=k$). 
Each of the lattice paths counted by $\binom{2n}{n}$ has an excedence $0 \le e \le n$ which is the number of horizontal edges above the diagonal. Each value of $e$ occurs $\frac{1}{n+1}\binom{2n}{n}$ times and the Catalan numbers count the paths of excedence $0.$
These $q$-Catalan polynomials $\frac{1}{[n+1]}\left[{2n\atop n}\right]_q$ again have non-negative integer coefficients and for $q=1$ equal the Catalan numbers which the excedence $0$ lattice paths from $(0,0)$ to $(2n,n)$. One could dream that there is an interpretation of the coefficient of $q^d$ but I certainly can't find one. A recurrence relation might be helpful in back constructing an interpretation.
The obvious greedy guess fails. For that one would have a recurrence $C_q(n+1)=\sum q^{(k+1)(n-k)}C_q(k)C_q(n-k).$ Those polynomials do not have a strong claim on being a natural q-analogue.<|endoftext|>
TITLE: Unitary representations of Quantum Groups
QUESTION [11 upvotes]: Let $\mathfrak{g}$ be a finite-dimensional complex simple Lie algebra and let $U_q(\mathfrak{g})$ be some incarnation of the quantized universal enveloping algebra of $\mathfrak{g}$; here I am assuming that $q \in \mathbb{C}^\times$ is not a root of unity.  I am interested in unitarizability of representations of $U_q(\mathfrak{g})$ with respect to various $*$-structures.  
The $*$-structures on $U_q(\mathfrak{g})$ have been classified; this can be found, for example, in section 9.4 of A Guide to Quantum Groups by Chari and Pressley, for example.  For the $*$-structure known as the compact real form of $U_q(\mathfrak{g})$, it is known that each finite-dimensional irreducible representation $V_\lambda$ admits an invariant inner product, i.e. a positive-definite Hermitian form such that
$$ \langle av, w \rangle = \langle v, a^* w \rangle    $$
for all $v,w \in V_\lambda$ and all $a \in U_q(\mathfrak{g})$.
Question: for an arbitrary $*$-structure on $U_q(\mathfrak{g})$, has anybody classified the set of dominant integral weights $\lambda$ for which $V_\lambda$ admits an invariant inner product?
Chari and Pressley mention in passing in their book that this is an open question, but that was in 1995 and I thought I'd see if anybody had resolved it in the meantime.  I would expect that for an arbitrary $*$-structure, most irreps do not have such an invariant inner product, since in the classical situation the corresponding real form of the group is not compact, so you can't just average over Haar measure as you do in the compact case.

REPLY [5 votes]: As far as I know, there is no such classification or even a good attempt. There are some results about unitary representations of quantum groups with the suitable choice of involution. L.Vaksman and his collaborators studied unitary representations for quantum analogs of $SU(n,n)$ (Most of these papers are in arxiv). Also there are papers of Klimyk, Pakulyak etc. But the classification is done only for the trivial case of quantum $SU(1,1)$.<|endoftext|>
TITLE: Jacobson's theorem and further
QUESTION [7 upvotes]: Jacobson's theorem states that 
If $R$ is a ring, and for every $x\in R$, there exists $n(x)\geq 2$ such that $x^{n(x)}=x$. Then 
$R$ is commutative. 
I wonder if the following stronger assertion(in case $R$ has unity) is true.
Let $R$ be a ring with unity. For every $x$ in $R$, there exists $n(x)\geq 2$ such that
$x^{n(x)} = x$.
Then, $R$ is embedded in a product(possibly infinite) of fields $F_i$, where
each $F_i$ is an algebraic extension of $F_{p_i}$ (prime field of $p_i$ elements).
If this is not true, then I am also interested in counterexample.
Thanks in advance.

REPLY [11 votes]: Yes, this is true. 
By Jacobson's theorem, $R$ is commutative. Now the radical of $R$ is the intersection of all primes $P$ of $R$. Hence we have an embedding 
$$\phi: R/rad(R) \to \prod_P R/P.$$
For each $x \in R/P, x \neq 0$ there is $n \ge 1$ such that $x(x^n-1)=0$ and since $R/P$ is a domain, $x^n =1$, i.e. $x$ is a unit. Thus $R/P$ is a field. Since elements $\neq \pm 1$ of $\mathbb Q$ aren't roots of unity, $R/P$ has prime characteristic and each element of $R/P$ is algebraic. Therefore $R/P$ is an algebraic extension of some $\mathbb{F}_p$. 
To finish the proof, we have to show $rad(R) = 0$. Let $x \in rad(R)$. Let $k>0$ be minimal with $x^k=0$ and let $n \ge 2$ with $x^n=x$. Suppose $k > n$. Write $k=qn+r$ with $q > 0$, $0 \le r < n$. Then $0 = x^k = (x^n)^q x^r = x^{q+r}$. Minimality of $k$ implies $k = q+r$, i.e. $n=1$, in contradiction to the assertion $n \ge 2$. Hence $k\le n$ and $x = x^n = x^k x^{n-k} = 0$.

REPLY [5 votes]: This is true. Let  $R$ be a ring satisfying your property. Then $R$ has no nilpotent element since if $x^a=0$, then for $b$ an integer such that $n(x)^b > a$ we have $x=x^{n(x)^b}=0$.
So the radical of $R$ is $0$, and since the radical is the intersection of all
prime ideals of $R$, we see that the natural map $R \mapsto \prod_{P} R/P$ is an injection.
It therefore suffices to prove the result for $R/P$, that is for a domain.
Assuming no that $R$ is a domain, the equation $x^{n(x)}-x=0$ factors as $x=0$ or $x^{n(x)-1}=0$. So, in $R$, every non-zero element is a root of unity. This is also true of the fraction field $K$ of $R$. No $K$ is of characteristic $p>0$, since otherwise it would contain
$\mathbb{Q}$ which contains many non-roots-of-unity such as $2$. And clearly, $K$ is algebraic
over the prime field. So we're done.<|endoftext|>
TITLE: Countable admissible ordinals
QUESTION [5 upvotes]: Jensen claimed that for any finite increasing sequence countable admissible  ordinals $\omega= \alpha_0<\alpha_1\cdots <\alpha_n$, there is a real $x$ so that, for each $m\leq n$, $\alpha_m$ is the $m+1$-th admissible ordinal relative to $x$.
Anybody knows the proof? Or where to find it?

REPLY [7 votes]: There is a model theoretic proof of the generalization to countable sequences which appears in a paper by Simpson and Weitkamp.
High and low Kleene degrees of coanalytic sets
Stephen G. Simpson & Galen Weitkamp
Journal of Symbolic Logic 48 (2):356-368 (1983) 
I believe that this proof is due to Harrington, based on Jensen's and Friedman's model theoretic proof of Sacks's theorem that every countable admissible ordinal is the least admissible relative to some real.<|endoftext|>
TITLE: Common extension of two sigma-additive measures
QUESTION [12 upvotes]: Let $\mathcal{A_1}$ and $\mathcal{A_2}$ be $\sigma$-algebras of subsets of some space X. Suppose $\mu_j$ is probabilistic measure on $\mathcal{A}_j$ for $j=1,2$. 
Question: What are the necessary and sufficient conditions for existence of common extension of these measures to some probabilistic measure on $\sigma(\mathcal{A}_1,\mathcal{A}_2)$?
The obvious necessary condition is as follows:
$\forall U_i\in \mathcal{A_i}$, $i=1,2$, if $U_1 \subset U_2$ then $\mu_1(U_1)\leq \mu_2(U_2)$ and vice versa. It is known that if we are interested in finitely-additive measures then this condition is sufficient. 
What about general $\sigma-$additive measures?

REPLY [4 votes]: The following example (457H, in Fremlin's Measure theory Vol. 4) shows that the obvious necessary condition isn't sufficient. 
Let $X = \{(x, y) \in [0, 1]^2: x < y\}$. Let $\pi_1, \pi_2 : X \to [0, 1]$ be projections on the $x$ and $y$ coordinates. Let $\Sigma_i = \{\pi_i^{-1}[E] : E \subseteq [0, 1] \text{ is Borel}\}$ and $\mu_i(\pi_i^{-1}[E]) = \mu(E)$ where $\mu$ is Lebesgue measure. Then it is easy to check that, although $\mu_1, \mu_2$ have a common finitely additive extension, no such extension can be countably additive.
To appreciate the difficulty in obtaining a simple necessary and sufficient criterion, Fremlin mentions the following concrete problem (457Z(a)): Characterize the Borel sets $X \subseteq [0, 1]^2$ for which the analogously defined measures in the above example admit a common countably additive extension.<|endoftext|>
TITLE: Integer polynomials mapping the unit disk into itself
QUESTION [5 upvotes]: Is there a complete characterization of those integer polynomials, that is $P\in{\mathbb Z}[X]$, such that $P(D)\subset D$, where $D$ is the unit disk ? At least, $P(X)=\pm X^k$ works, when $k\in\mathbb N$. But are there other ones, many other ones ?

REPLY [19 votes]: Since $P\in \mathbb{Z}[X]$, We have $P(0)\in \mathbb{Z}$. Suppose that $P(0)\neq 0$, then 
$|P(0)|\geq 1$. In that case $P(D)\subset D$ is not satisfied unless $P$ is constant by open mapping theorem. 
So, if $P$ is non-constant, then we must have $P(0)=0$. 
Write $P(X)=XQ(X)$. Then $Q(X)=P(X)/X$. On a disk $D_r$ of radius $0< r<1$ centered at $0$, 
We have by Maximum modulus theorem that
$$|Q(X)|\leq 1/r$$ 
since $|P(X)|\leq 1$.
Letting $r\rightarrow 1$, we have also that $|Q(X)|\leq 1$ whenever $|X|\leq 1$. 
By repeating the same argument for $Q(X)$, we obtain that 
$P(X)=\pm X^k$, when $k\in \mathbb{N}$ are the only polynomials in $\mathbb{Z}[X]$ satisfying the property.<|endoftext|>
TITLE: The prime divisors of a simple group
QUESTION [13 upvotes]: Let $G$ be a finite simple group such that $\pi (G)=\pi (A_{n})$, where $n\geq 23$ ($\pi (G)$ is the set of prime divisors of the order of $G$). Is it true that $G$ is isomorphic to an alternating group?

REPLY [6 votes]: I'm not sure what a definitive answer would look like, but maybe I can frame the question a little differently in a series of comments:
1) The underlying question really concerns an arbitrary pair of (not too small) simple groups $G, H$ for which $\pi(G) = \pi(H)$.  As observed by others, the classification of finite (nonabelian) simple groups should be in hand already.   Given only finitely many groups non-isomorphic to an alternating group or group of Lie type, the question basically involves $G, H$ of those two types.   (But there is no obvious reason to require that one of the groups be alternating.)
2) There is a long tradition, going back at least to Emil Artin in the 1950s, of sorting out which of the known finite simple groups having different origins are actually isomorphic.   So the question asked is not really about existence of isomorphisms: one knows already from the literature which groups of Lie type are isomorphic to others of Lie type or to alternating groups.   In practice all such accidental isomorphisms involve relatively small orders.   (On the other hand, simple groups coming from odd orthogonal groups and symplectic groups with the same rank and same finite field are well understood to be non-isomorphic in spite of having the same order.)
3) It's conceivable that some isolated group of Lie type will turn up whose order accidentally has the same prime divisors as some (not too small)    alternating group.   This would be a bit surprising, but probably not interesting unless something systematic is observed that involves more groups.
The key word here is accidental.   To prove rigorously that no accidents happen looks challenging, even if ultimately not too illuminating.
4) Why emphasize alternating groups?    There is perhaps one good reason: here the primes dividing the group order form a list (from 2 to the largest prime) having no gaps.   For groups of Lie type that are not very small, this behavior of $\pi(G)$ strikes me as purely accidental and somewhat unusual.    If that's not the case, it would be interesting and need an explanation.   But considering a family of groups such as $\rm{PSL}(2,p)$ (with $p>3$) suggests a skeptical attitude.<|endoftext|>
TITLE: What is the definition of exceptional divisor？
QUESTION [11 upvotes]: Does this concept is defined for every birational morphism？ What is the precise meaning？ 
Thank you for your comments.

REPLY [33 votes]: The exceptional set is defined for every birational morphism $\pi : Y \to X$.  This is defined as follows.  Set $\Sigma \subset X$ to be the smallest closed subset of $X$ outside of which $\pi : (Y \setminus \pi^{-1}(\Sigma)) \to (X \setminus \Sigma)$ is an isomorphism.
In this case the exceptional set is defined to be 
$$E = \pi^{-1}(\Sigma).$$
Now, $\Sigma$ itself is just a set, and not a scheme, so often people will give $E$ the reduced scheme structure and consider it as a closed subscheme of $Y$.
In many applications (perhaps even most), $E$ is actually a divisor, in other words, it is a union of subvarieties of codimension 1.  In this case, $E$ is called the exceptional divisor.  Indeed, if you obtained $\pi : Y \to X$ by blowing up $\Sigma$, and say $X$ is a normal variety with $\Sigma \subseteq X$ a codimension $\geq 2$ subset, then $E$ is indeed a divisor.  However:
Negative result:   There are many examples when $E$ is not a divisor.  This is particularly common for small resolutions (resolutions of singularities for which the exceptional set has no divisorial components), which are in many cases desirable.  For example, they are crepant resolutions.
The most common example of a small resolution is probably the following.
$$X = V(xy - uv) \subseteq \mathbb{A}^4.$$
Note $X$ has dimension 3.
In this case, consider blowing up the ideal $(x,u) \subseteq O_X$.  This is the ideal of a divisor on $X$.  I'm not going to do this for you, but it is a good exercise.  In this case, the blowup gives you a resolution of singularities of $X$, but the exceptional set is 1-dimensional.
Positive result:  If $X$ is smooth (or even factorial) and $\pi : Y \to X$ is birational with $Y$ quasi-projective, then the exceptional set is always a divisor.  Sándor explains this quite nicely in THIS QUESTION.
Positive result #2:  If you are willing to change your variety $Y$ by taking a further blow-up, then you can always turn the exceptional set into a divisor.  For example, simply take $\rho : Y' \to Y$ to be the (normalized) blow up of $E$.<|endoftext|>
TITLE: Proving that a variety is not (isomorphic to) a toric variety
QUESTION [11 upvotes]: Is there an algorithmic (or other) way to prove that a (projective)
variety is not isomorphic to a toric variety?
I'd be happy with an algebraic answer (for affine or projective varieties),
using the fact that toric ideals are binomial prime ideals.  There ne could
use that the coordinate rings are characterized as those admitting a fine
grading by an affine semigroup , i.e. presented by a binomial prime ideal (Prop. 1.11 in Eisenbud/Sturmfels "Binomial ideals").
This question resulted from an Example that I discussed with Mateusz
Michalek.  The example is: let $V$ be the Zariski closure of the image of
the parameterization: $$(p_1,p_2,a_1,a_2,a_3,b_1,b_2,b_3) \to 
\begin{pmatrix}
p_1a_1a_2a_3+p_2b_1b_2b_3 \\
p_1a_1a_2b_3+p_2b_1b_2a_3 \\
p_1a_1b_2a_3+p_2b_1a_2b_3 \\
p_1a_1b_2b_3+p_2b_1a_2a_3 \\
p_1b_1a_2a_3+p_2a_1b_2b_3 \\
p_1b_1a_2b_3+p_2a_1b_2a_3 \\
p_1b_1b_2a_3+p_2a_1a_2b_3 \\
p_1b_1b_2b_3+p_2a_1a_2a_3 \\
\end{pmatrix}$$
Implicitization using Macaulay2 is quick and yields a complete intersection:
$$\langle et-ry-qu+wo, wt-qy-ru+eo, we-qr-yu+to \rangle \subset k[q,w,e,r,t,y,u,o]$$
How to prove that $V$ is not toric?

REPLY [6 votes]: The question of algorithmically deciding if an ideal is binomial after a (suitable, e.g. linear) automorphism of affine space is decidable and various algorithms are discussed in "When is a polynomial ideal binomial after an ambient automorphism?" by Lukas Katthän, Mateusz Michałek, and Ezra Miller.
I discussed the methods in this paper with the authors and tried them on the example in the question, but the complexity was too high.  Together with Lukas Katthän, however, we found a different way to prove that this ideal is in fact toric.  It is a complete intersection and given some standard invariants and tables of toric complete intersections it was easy to identify a single candidate toric variety that would be isomorphic to this one.  
The ideal in question can be rewritten as: 
$$ I = \langle x_1x_2 + x_3x_4 - y_1y_2 - y_4y_3, x_1x_4 + x_2x_3 - y_1y_4 - y_2y_3, x_1x_3 + x_2x_4 - y_1y_3 - y_2y_4\rangle $$
This makes the structure a bit more visible.  The candidate isomorphic toric ideal is 
$$J = \langle x_1y_1 - x_2y_2, x_1y_1 - x_3y_3, x_1y_1 - x_4y_4\rangle$$
We examined the singular locus and in particular its minimal primes which are all linear. This revealed what the linear automorphism was: $$ \begin{align} x_1 & \mapsto x_1+x_2+x_3+x_4 + (y_1+y_2+y_3+y_4) \\ 
x_2 & \mapsto x_1+x_2-x_3-x_4 + (y_1+y_2-y_3-y_4)\\
x_3 & \mapsto x_1-x_2+x_3-x_4 + (y_1-y_2+y_3-y_4)\\
x_4 & \mapsto x_1-x_2-x_3+x_4 + (y_1-y_2-y_3+y_4)\\
y_1 & \mapsto x_1+x_2+x_3+x_4 - (y_1+y_2+y_3+y_4)\\
y_2 & \mapsto x_1+x_2-x_3-x_4 - (y_1+y_2-y_3-y_4)\\
y_3 & \mapsto x_1-x_2+x_3-x_4 - (y_1-y_2+y_3-y_4)\\
y_4 & \mapsto x_1-x_2-x_3+x_4 - (y_1-y_2-y_3+y_4).
\end{align}$$
The way it is written here it maps the toric ideal to the original ideal.<|endoftext|>
TITLE: Conservative differential equations "in the wild"
QUESTION [6 upvotes]: Dear MO world,
I'm teaching an undergraduate course on "fun with chaos". As part of a test (on bifurcations in differential equations), I asked students to sketch phase portraits for a family of (2d) differential equations. While preparing my solutions, I fed the differential equations (with one value of the parameter) into a phase portrait plotter on the web and found that there were families of closed curves as solutions, so that the system seemed to have a first integral. Changing the parameter, this persisted. I'm hoping someone can tell me why this should have been obvious to me!
The system was
$$
\eqalign{\dot x&=y-x+1\cr \dot y&=y-rx^2.}
$$
The phase portrait (with $r=0.15$) looks like this: 
Some computation with mathematica reveals that $2rx^3+3y^2+6y-6xy$ is a first integral. The question is why? Is this an outrageous coincidence? (I thought that your chance of bumping into a conservative differential equation by accident were nil unless the system was of the form $\ddot x=f(x)$).
So: should I go and buy a lottery ticket, or is there some reason that this I shouldn't be surprised by this?

REPLY [11 votes]: Divergence free vector fields in the plane are associated with Hamiltonian systems. Your vector field is easily seen to be divergence free.

REPLY [7 votes]: Let $a=y-x+1$, then $\dot{x}=a$ and $\dot{a}=-rx^2+x-1$, so $\ddot{x}=-rx^2+x-1$.
So perhaps it is possible to accidentally bump into a system of the form $\ddot{x}=f(x)$.
How to tell this from your integral: naively trying to put it into a simpler form by completing the square seemed to work pretty well.
How to tell this from your differential equation: Compute $\ddot{x}$ and $\ddot{y}$ and see if you can write them in terms of only $x$ or only $y$.<|endoftext|>
TITLE: What differential operators have Schur polynoms as eigenfunctions ? Can this be deformed to trig. Calogero and Jack polynoms ?
QUESTION [10 upvotes]: Consider Schur polynomials - symmetric polynoms of variables x_1, ... x_n, indexed by $d_1 \ge d_2 \ge ... \ge d_n$. 
I wonder what is known about the differential operators which have them as eigenfunctions ?
First, I am not sure such operators exist, but almost sure, by the reasons below.
Second, it seems to me (see reason below) that these operators have one parameter deformation which makes them trigonometric Calogero-Moser operators (may be modula some change of variables). Schur polynomials have also one parameter deformation - Jack polynomials, so it might be these deformations compatible, i.e. Jack polynoms - eigenfunctions of trigonometric Calogero-Moser.
Motivation 1
It is known (see Wikipedia) that Schur polynoms are specializations of Jack polynomials. Which are eigenfunctions of the so-called Sekeguchi operators (see e.g. first formula in 
Okounkov-Olshanski "SHIFTED JACK POLYNOMIALS, BINOMIAL FORMULA, AND APPLICATIONS" ).
So I guess specialization of these operators should give desired operators for Schur polynomials.
I cannot remember clearly, but some time ago Evgeny Sklyanin showed me something like how to make change of variables from Sekeguchi operators to Calogero-Moser operators, unfortunately I cannot find the notes.
Motivation 2
The followings steps seems can give the desired operators.
1) Schur polynomials can be obtained as characters of the represenations of GL_n,
restricted on diagonal matrices.
2) Characters are eigenfunctions for the center of universal enveloping acting by differential operators on functions on the group GL_n.
So we need to "restrict" these differential operators  on diagonal matrices - and that should be the answer to the question. It seems to me if it is correct that should be well-known.
Naive word "restrict" means "calculate the radial part", i.e. we should take into account that characters are functions invariant for conjugation action of GL_n on itself, center of U(gl_n) also invariant differential operator, so acting by inv.oper. on inv. func. we get inv. func. So we get desired operators.
Why I think this is related to the trigonometric Calogero-Moser.
By Etingof-Ginburg rational Calogero-Moser can be obtained by "deformation" of similar procedure applied to invariant differential operators with constant coefficients. 
Center of U(gl_n) I think plays the same role, but for trigonometric Calogero-Moser.
Probably this is too long to put details here... 
Some related questions:
representation theoretic interpretation of Jack polynomials
Jack polynomials as determinants
"The" Harish-Chandra homomorphism (see Etingof-Ginzburg) for invariant dif. opers. on gl_n - what are images of higher order Laplacians e.g. Tr(D^3) = d_ijd_jkd_ki ?

REPLY [10 votes]: Schur polynomials are defined to be the characters of irreducible representations of $G=GL_n$. By the Weyl character formula,
$$s_\lambda=\frac{1}{\delta}\sum_{w\in W}(-1)^{l(w)}e^{w(\lambda+\rho)},$$
where $\delta$ is the Vandermonde determinant.
Clearly, the numerator is an eigenfunction of the Laplace operator on the Cartan $\mathfrak{h}$:
$$L_1(\delta s_\lambda):=\Delta (\delta s_\lambda) = (\lambda+\rho, \lambda+\rho) \delta s_\lambda.$$
So, the Schur polynomials themselves are eigenfunctions of $M_1=\delta^{-1} L_1\delta$.
The operators $L_1$ and $M_1$ admit one-parameter deformations $L_k$ and $M_k=\delta^{-k} L_k\delta^k$ to the quadratic Sutherland (trigonometric Calogero-Moser) and the Sekiguchi operators. Eigenfunctions of $M_k$ are the Jack polynomials.
The degeneration Jack -> Schur corresponds to taking the coupling constant $k=1$ in which case the Sutherland Hamiltonians are all free.
You can see formulas for a general root system in hep-th/9403168: the conjugated quadratic Hamiltonian is given by (2.4).<|endoftext|>
TITLE: Number of relations and free subgroups
QUESTION [7 upvotes]: Is there a function $f$ such that for  any presentation $$G=\langle x_1,\ldots,x_n \mid r_1,\ldots,r_k\rangle\quad  \text{with}\quad |r_i|\leq 3$$ 
$k\leq f(n)$ implies that $G$ has non-abelian free subgroups.
Of course $f=0$ works trivially, I am asking for bigger functions.

REPLY [4 votes]: Another way to prove that $f(n)=n-2$ works is to recall Romanovskii's Freiheitssatz:

In any finitely presented group
  $$
>G=⟨x_1,…,x_n∣r_1,…,r_k⟩,
>$$
  some $n-k$ of {$x_i$} freely generate a free subgroup of $G$ (if $n\ge k$).

See
N.S. Romanovskii, "Free subgroups of finitely presented groups" Algebra i Logika , 16 (1977) pp. 88–97 (In Russian).<|endoftext|>
TITLE: Which colimits commute with which limits in the category of sets?
QUESTION [48 upvotes]: Given two categories $I$ and $J$ we say that colimits of shape $I$ commute with limits of shape $J$ in the category of sets, if for any functor $F : I \times J \to \text{Set}$ the canonical map $$\textrm{colim}_{i\in I} \text{lim}_{j\in J} F(i,j) \to \textrm{lim}_{j\in J} \text{colim}_{i\in I} F(i,j)$$ is an isomorphism.
The standard examples are a) filtered colimits commute with finite limits and b) sifted colimits commute with finite products. (Those statements can be regarded as definitions of which categories $I$ are filtered or sifted respectively, but both terms have independent definitions for which these commutation results are propositions.) A third, less known example is to take $I$ a finite group and $J$ a cofiltered category, in other words, if $G$ is a finite group and $X_j$ is an inverse system of $G$-sets, then the canonical map $$(\varprojlim_{j\in J} X_j)/G \to \varprojlim_{j \in J}(X_j/G)$$ is an isomorphism.
Now, all of these examples are easy to prove separately (here's a proof of the $G$-set result, for example) but I see no unifying pattern. Is there a simple criterion for when $I$-colimits and $J$-colimits commute in the category of sets?
[Note: It's true that $I$ is filtered (resp. sifted) if and only if for all finite (resp. finite discrete) $J$ the diagonal functor $I \to I^J$ is final; but I don't think that for arbitrary $I$ and $J$, if the diagonal $I \to I^J$ is final then $I$-colimits commute with $J$-limits. If I'm wrong and that condition on the diagonal actually is sufficient for commutation: why? and is it also necessary?]

REPLY [21 votes]: I'm sorry for having been so slow to react, I was only made aware of this discussion by a colleage of mine recently.. I have been working for several years now on almost exactly all of the above questions in the context of my thesis to be submitted in very near future. 
Concerning first question: That $I$-limits commute with $J$-colimits in set implies that the diagonal functor $J \to J^I$ is final, is a necessary fact wich is only/exactly sufficient when $I$ is a so-called sound class of "limit index", but not in general.
I suggest in my thesis that we define $J$ to be $I$-filtered when the diagonal functor $J \to J^I$ is final instead of when $I$-limits commute with $J$-colimits in set (to keep diagrammatic interpretations).
In fact though sound doctrines are easy to work with they are included in a (nearly just as nice) slightly larger "type of doctrines" (including "pullbacks+terminal objects") namely those classes of index "essentially closed" in the galois correspondence between the classes of "limit index" and classess of "colimit index" that commute in set. I give in my thesis a complete picture of these classes and by this we "decrypt" the abstract conditions of F. Foltz. I'm sorry that I don't yet have nice and short versions of my work to put down here, nor an article ready to refer to, but I'm happy to discuss or answer any related questions in detail either off-line or by email with anyone interested! While finishing written versions.. Cheers, Marie Bjerrum.<|endoftext|>
TITLE: Does such an infinite index subgroup exist?
QUESTION [13 upvotes]: Notation: If $G$ is a countable group and $H$ is a subgroup, for $g\in G$, let $|\mathcal{O}_{gH}|$ be the size of the $H$-orbit of $gH$ in the $H$-set $G/H$. 
Does there exist a countable group $G$ and a subgroup $H$ with $[G\colon H]=\infty$ such that:

There is a $g\in G$ with $|\mathcal{O} _{gH}|\neq |\mathcal{O} _{g^{-1}H}|$,
$|\mathcal{O} _{gH}|=\infty$ if and only if $|\mathcal{O} _{g^{-1}H}|=\infty$, and
There is a constant $M>0$ such that $\displaystyle \frac{|\mathcal{O} _{gH}|}{|\mathcal{O} _{g^{-1}H}|} \leq M$ for all $g\in G$ with $|\mathcal{O} _{gH}|,|\mathcal{O} _{g^{-1}H}|\neq \infty$.

For instance, 2 and 3 above would be satisfied if:

If $|\mathcal{O} _{gH}|\neq |\mathcal{O} _{g^{-1}H}|$, then $0< |\mathcal{O} _{gH}|,|\mathcal{O} _{g^{-1}H}|\leq M$ for some $M>1$ independent of $g$.

REPLY [7 votes]: I believe (1 and 2) and (3) are mutually exclusive. Here is a proof:
First, the commensurator 
$$
Comm_G(H) = \{g\in G : |\mathcal{O} _{gH}|, |\mathcal{O} _{g^{-1}H}|<\infty\}
$$ 
is a group. We will show:

Lemma: $\varphi\colon Comm_G(H)\to \mathbb{Q}_{>0}$ by 
  $g\mapsto \displaystyle\frac{[H\colon H\cap gHg^{-1}]}{[H\colon H\cap g^{-1}Hg]}$
  is a homomorphism.

From the lemma, if we assume there is a $g\in Comm_G(H)$ such that $\varphi(g)=x>1$ (criteria 1 and 2), then the order of $g$ must be infinite, since $x>1$ implies $x^n>1$ for all $n\geq 1$. Since the order of $g$ is infinite, criterion 3 cannot hold since eventually $\varphi(g^n)=\varphi(g)^n=x^n>M$ for any $M>0$.
Proof of the lemma:
We must show $\varphi(g_1g_2)=\varphi(g_1)\varphi(g_2)$. Define the following constants:

For $i=1,2$, $a_i = [H\colon H\cap g_iHg_i^{-1}]$ and $b_i = [H\colon H\cap g_i^{-1}Hg_i]=[g_iHg_i^{-1}\colon H\cap g_iHg_i^{-1}]$
$a=[H\colon H\cap (g_1g_2)H(g_1g_2)^{-1}]$ and $b=[(g_1g_2)H(g_1g_2)^{-1}\colon H\cap (g_1g_2)H(g_1g_2)^{-1}]$

Note that since $x\mapsto g_1xg_1^{-1}$ is an automorphism of $G$, we have:

$a_2=[g_1Hg_1^{-1}\colon g_1Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}]$ and $b_2=[(g_1g_2)H(g_1g_2)^{-1}\colon g_1Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}]$

Now look at the subgroup $K=H\cap g_1 Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}$, and define

$a_1'=[H\cap (g_1g_2)H(g_1g_2)^{-1}\colon K]$
$a_2'=[H\cap g_1Hg_1^{-1}\colon K]$
$b_1'=[g_1Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}\colon K]$

which are all finite, since if we have a quadrilateral of groups $L_1\cap L_2\subset L_1,L_2\subset G$, we must have $[L_1\colon L_1\cap L_2] \leq [G\colon L_2]$. Now since index is multiplicative, we have

$a a_1'=a_1a_2'$
$ba_1' = b_2 b_1'$
$a_2b_1'=b_1a_2'$

Solving for $a$ and $b$, we get
$$
\frac{a}{b} = \frac{a_1a_2'}{a_1'}\frac{a_1'}{b_2b_1'}=\frac{a_1a_2'}{b_2b_1'}.
$$
Now note that $\displaystyle \frac{a_2'}{b_1'}=\frac{a_2}{b_1}$, so we have
$$
\varphi(g_1g_2)=\frac{a}{b} = \frac{a_1}{b_2}\frac{a_2}{b_1}= \frac{a_1}{b_1}\frac{a_2}{b_2}=\varphi(g_1)\varphi(g_2).
$$<|endoftext|>
TITLE: Proving a messy inequality 
QUESTION [12 upvotes]: EDIT:
After much work I was able to reduce the inequality to a single variable function which I need to show is non-positive. That function is (for $0\leq p\leq\frac{1}{2}$)
$$\frac{p^2(\log(p))^2 - (1-p)^2(\log(1-p))^2}{1-2p} + \log(p)\log(1-p) + p\log(p)+(1-p)\log(1-p) \leq 0$$
This looks tractable, but is quite hard to prove since neither the function nor any of its derivatives are convex/concave/monotonic. Its graph looks like 
.     
Any inputs welcome!
END EDIT
I'm trying to prove this messy inequality I got while working on an information theoretic problem (more details below).    
Let $0 \leq p,q \leq \frac{1}{2}$. 
Let $H(x) = -x\log(x) - (1-x)\log(1-x)$ be the binary entropy function. 
Let $p\star q = p(1-q) +q(1-p)$ (think of this as addition of two Bernoulli random variables in $\mathbb{Z}_2$)  
The inequality is:
$$-\frac{(1-2q)}{(p \star q) (1- p \star q)\log\left(\frac{1-p}{p}\right)} + \frac{\log\left(\frac{1- p \star q}{p \star q}\right)}{p(1-p)\left(\log\left(\frac{1-p}{p}\right)\right)^2} - \frac{H(q)(1-2p)}{H(p)(p \star q) (1- p \star q)\log\left(\frac{1-q}{q}\right)} \leq 0$$
Some Remarks:  
If we write this as $-A+B-C \leq 0$ with $A,B,C$ being the corresponding quantities above. I can show that $-A +B \geq 0$, so the $-C$ term subtracts enough to make the expression negative. The plot of $-A+B-C$ as a function of $p,q$ looks like  
.  
It appears to decrease in $q$ for a fixed $p$, starting at $0$ and going to $-\infty$ as $q$ goes to $\frac{1}{2}$.  
Origin:
This inequality came from looking at a function $f(x,y) = H(H^{-1}(x)\star H^{-1}(y))$ where $f: [0,1]\times [0,1] \to [0,1]$. This function occasionally pops up in information theory, like when studying the binary symmetric channel or an inequality called  Mrs. Gerber's Lemma. The inequality I have implies that $f(x,y)$ is concave along lines through the origin (i.e $f(x,\theta x)$ is concave in $x$). What I've written is a simplified version $f''$ along a line through the origin.  
I'm at a complete loss as to how to attack this. All the plots I have show this is true, but a plot is not a proof. Any suggestions welcome!

REPLY [5 votes]: I think I managed to prove the entire inequality analytically. The whole proof is a bit long to post here (about 7 pages) and involves ugly looking expressions. I'll outline the general strategy I used:  

Start with the original inequality
$$-\frac{(1-2q)}{(p \star q) (1- p \star q)\log\left(\frac{1-p}{p}\right)} + \frac{\log\left(\frac{1- p \star q}{p \star q}\right)}{p(1-p)\left(\log\left(\frac{1-p}{p}\right)\right)^2} - \frac{H(q)(1-2p)}{H(p)(p \star q) (1- p \star q)\log\left(\frac{1-q}{q}\right)} \leq 0$$
and throw all the $p\star q$ terms on one side to get 
$$(p \star q)(1- p\star q)\log\left(\frac{1-p \star q}{p \star q}\right)  \leq(1-2q)p(1-p)\log\left(\frac{1-p}{p}\right) + \frac{p(1-p)(1-2p)\left(\log\left(\frac{1-p}{p}\right)\right)^2 H(q)}{H(p)\log\left(\frac{1-q}{q}\right)}$$
Now keep $p \star q$ fixed $ = k$ say, and try minimise the right hand side. Notice that when $p = k, q = 0$ we have equality. This might lead us to hope that the right hand side is a decreasing function of $p$, and when $p$ becomes $k$ we have equality.  
Now consider the partial derivative of the RHS in $p$, and try to show that this is negative. By some stroke of luck, it turns out that if we look at $\frac{\partial RHS}{\partial p}$ for a fixed $p$ and vary $k$, the value of $k$ which maximises $\frac{\partial RHS}{\partial p}$ is $k = p$. So its sufficient to prove $\frac{\partial RHS}{\partial p}|_{(p,p)} \leq 0$  
$\frac{\partial RHS}{\partial p}|_{(p,p)}$ when evaluated gives me the second inequality (the one I added in the EDIT). Thus it suffices to show
$$p^2(\log(p))^2 - (1-p)^2(\log(1-p))^2 + (1-2p)\left(\log(p)\log(1-p) + p\log(p)+(1-p)\log(1-p)\right) \leq 0$$  
Let me call the above expression $F(p)$. I showed that if $F(p) > 0$ somewhere, then $F'''(p)$ would have atleast $2$ zeros in $[0,0.5]$. (This is easy to show.) Explicitly computing $F'''(p)$ shows that it is $-\infty$ at $0$ and strictly  positive at $0.5$. So, if we happened to show that $F'''$ is concave,  it would have exactly one zero, and thus contradict the result that it should have $2$, and thus prove the result.  
Now I do something truly horrible. I explicitly compute the 5-th derivative and show that it is negative, thus $F'''$ is indeed concave.  
The inequality $F'''''(p) \leq 0$ can be somewhat simplified to get 
$$P_1(p)\log(p) + P_2(p)\log(1-p) + P_3(p) \leq 0$$
where $P_1, P_2,P_3$ are some $7$-th degree polynomials.  
I now use Sturm's theorem to show that over the range $[0,0.5]$ $P_1,P_2 \geq 0$. This, too, was quite horrible.  
Now I use a polynomial approximation to $\log$, and use the positivity of $P_1,P_2$ to conclude 
$$P_1(p)\log(p) + P_2(p)\log(1-p) + P_3(p)\leq P_1(p)\left( -(1-p) - \frac{(1-p)^2}{2}\right) + P_2(p)\left(-p - \frac{p^2}{2}\right) + P_3(p)$$  
By another stroke of luck, the RHS here factorises to give $$ -24(1-p)^2 (p-0.5)^2  p^2 (p^2-p+\frac{7}{3})
$$
which is clearly seen to be negative, and thus by step $9$ I can now say that $F'''''$ is negative, and $F'''$ is concave, and it has only one root in $[0,0.5]$ and thus $F$ has to be non-positive everywhere in $[0,0.5]$. 

In retrospect, I guess using numerical techniques (as advised in the comments section) would have been more prudent.<|endoftext|>
TITLE: A Game on Noetherian Rings
QUESTION [216 upvotes]: A friend suggested the following combinatorial game. At any time, the state of the game is a (commutative) Noetherian ring $\neq 0$. On a player's turn, that player chooses a nonzero non-unit element of the ring, and replaces the ring with its quotient by the ideal generated by that element. The player to make the last legal move wins, by passing the opponent a field.
So if the ring was $\mathbb C[x,y]/(x^2+y^2-1)$, and a player chooses $x$, the ring becomes $\mathbb C[y]/(y^2-1)$. This is a poor move, as his opponent can turn the ring into a field by choosing either $y+1$ or $y-1$, and win. A winning move would have been $x+iy+1$, which turns the ring into a field immediately and wins the game.
Problem: Given a ring, how can we tell if it is a winning position for the first player or for the second player?
(The game terminates since the original ring is Noetherian, and an unending game would be an infinite ascending chain in the original ring.)

REPLY [33 votes]: Let us call a Noetherian commutative ring $R$ an $\mathcal{N}$-position if the next player has a winning strategy; otherwise the previous player has a winning strategy and we will call it a $\mathcal{P}$-position. The zero ring is allowed and hence we will choose the misère play rule, i.e. the player with the last move loses (see Tom Goodwillie's comment). For example, it is clear that the zero ring is $\mathcal{N}$, that fields are $\mathcal{P}$, and that PIDs which are no fields are $\mathcal{N}$.
In general, $R$ is $\mathcal{P}$ iff $R/\langle x  \rangle$ is $\mathcal{N}$ for all $0 \neq x \in R$, and $ R$ is $\mathcal{N}$ iff either $R=0$ or $R/\langle x \rangle$ is $\mathcal{P}$ for some $0 \neq x \in R$. As a general theme, $\mathcal{P}$-positions are quite rare and hard to find, and every $\mathcal{P}$-position is responsible for many $\mathcal{N}$-positions which are  then more easy to find.
The following results are proven here:

If $R$ is $\mathcal{P}$, then $\mathrm{Spec}(R)$ is connected (Rem. 5.1).
Let $R$ be a Dedekind domain. If $R$ has some principal maximal ideal, then $R$ is $\mathcal{N}$; otherwise, $R$ is $\mathcal{P}$ (Prop. 5.6).
It follows that $K[X,Y]/\langle f \rangle $ is $\mathcal{P}$ when $f$ is a Weierstrass equation and $K$ is an algebraically closed field (Prop. 5.7).
One can also show that the coordinate ring of the cusp $K[X,Y]/\langle Y^2-X^3 \rangle$ is $\mathcal{P}$ (Prop. 5.13).
Hence, $K[X,Y]$ is $\mathcal{N}$ if $K$ is algebraically closed.
This actually holds for every field $K$, because $K[X,Y]/\langle X^2 \rangle$ is $\mathcal{P}$ (Prop. 5.16).
For this one needs that $K[X,Y]/\langle X^2,f^{n+1},f^n X \rangle$ is $\mathcal{P}$ for every irreducible $f \in K[Y]$ (special case of Prop. 5.15).

I conjecture that $K[X_1,\dotsc,X_n]$ is $\mathcal{N}$ for all $n \geq 1$.

We can also play such a game with groups. One starts with a group $G$. A move consists in replacing $G$ by $G/\langle\langle a \rangle\rangle$, i.e. we quotient out the smallest normal subgroup containing some element $a \neq 1$. The ending condition holds iff the ascending chain condition on normal subgroups holds (do these groups have a name?). When $G$ is abelian, this means that $G$ is finitely generated.
Actually we can play this game for every algebraic structure: We start with an algebra $A$ of a given signature. A move consists in replacing $A$ by $A/(a \sim b)$, where $a,b \in A$ with $a \neq b$.
I have tried to analyze this game for abelian groups, non-abelian groups, and rings in the mentioned article. There are lots of scattered examples, but for abelian groups the structure theorem makes it possible to give a general answer which ones are $\mathcal{P}$ under both play rules:
If $A$ is a finitely generated abelian group, then

$A$ is a normal $\mathcal{P}$-position if and only if $A$ is a square, i.e. $A \cong B^2$ for some abelian group $B$.
$A$ is a misère  $\mathcal{P}$-position if and only if $A$ is either a square, but not elementary abelian of even dimension, or elementary abelian of odd dimension.<|endoftext|>
TITLE: Homotopy Groups of Connected Sums
QUESTION [14 upvotes]: This was sparked because I wanted to compute $\pi_2(Sym^2(\Sigma_2))$ via $Sym^2(\Sigma_2)\approx \mathbb{T}^4$# $\bar{\mathbb{C}P}^2$.
We know how to compute $\pi_1$ of $M$ # $N$ via van-Kampen's theorem. But what about higher homotopy groups? I looked in the literature and google without luck, and so I am wondering if no such procedure exists.  Are there any results for calculating $\pi_n$ of connected sums?
There was mention of "higher van Kampen theorem"... has this actually been used to do such computations? I'd be interested in references if not just examples.

REPLY [5 votes]: Just a comment on the role of the Higher Homotopy Seifert-van Kampen Theorems:
they should be regarded as an extra tool in algebraic topology. There are quite severe conditions on their applicability but when they apply they compute quite a lot. Just as the 1-dimensional theorem, in its groupoid formulation, is about calculating 1-types, so the 2-dim theorem is about computing 2-types, in the form of crossed modules (over groupoids). However computing the second homotopy group from this 2-type may not be straightforward. But then the situation is the same for the 1-dim theorem, as is evidenced by the complications of theorems such as the Kurosh subgroup theorem, which can be seen to be  about the fundamental group(s) of a cover of a wedge of $K(G_i,1)$'s. 
As a taster, based on the 2-d theorem, work with Chris Wensley enabled the computation of the crossed module representing the 2-type of the mapping cone of a map $Bf: BG \to BH$ induced by a morphism $f: G \to H$. Of course. the second homotopy group, even as a module over the fundamental group, is but a pale shadow of the 2-type. You can see some of this in our book (pdf available from my web page on the book). 
R. Brown, P.J. Higgins, R. Sivera,  Nonabelian algebraic
topology: filtered spaces, crossed complexes, cubical homotopy
groupoids, EMS Tracts in Mathematics Vol. 15, 703 pages. (August
2011).
November 8, 2013: As a taster, let $X$ be the homotopy pushout  of the classifying spaces of the two maps of groups $P \to P/M, P \to P/N$ where $M,N$ are normal subgroups of the group $P$. The the homotopy 2-type of $X$ is determined by the crossed module $M \circ N \to P$, the coproduct of the two crossed $P$-modules, which is given by the pushout of crossed modules 
$$\begin{matrix} (1 \to P) & \to & (N \to P) \cr
\downarrow && \downarrow \cr 
(M \to P)& \to  & (M \circ N \to P)
\end{matrix} $$. It follows that 
$$ \pi_2(X) \cong (M \cap N)/ [M,N]. $$
(Of course we know $\pi_1 X$ by the 1-dimensional van Kampen Theorem.) This result is applied in Bardakov, Valery G; Mikhailov, Roman; Vershinin, Vladimir V.; Wu, Jie, 
"Brunnian braids on surfaces". Algebr. Geom. Topol. 12 (2012), no. 3, 1607–1648.<|endoftext|>
TITLE: Separating vectors for C$^*$-algebras
QUESTION [9 upvotes]: (I asked this on math.stackexchange, without response).
Let $A$ be a C$^*$-algebra, concretely acting on a Hilbert space $H$.  Suppose that $\xi_0\in H$ is cyclic and separating for $A$ (that is, the map $A\rightarrow H, a\mapsto a(\xi_0)$ is injective with dense range).  Let $M=A''$ the von Neumann algebra generated by $A$.

Need $\xi_0$ still be separating for $M$?  That is, $x\in M, x(\xi_0)=0 \implies x=0$?

It is standard (and easy to prove) that this is equivalent to $\xi_0$ be cyclic for $M'$.  However, the usual proof breaks down, and does not show this to be equivalent to $\xi_0$ being separating for $A$.
I think I can prove this using left Hilbert algebras.  We turn $\mathfrak A = \{ a(\xi_0) : a\in A \}$ into a left Hilbert algebra algebra in the obvious way.  Then run the Tomita-Takesaki machinery (actually not needed in full generality as we start with a state, not a weight).  Then the von Neumann algebra generated by $\mathfrak A$ is nothing but $M$, and so the general theory tells us that $\varphi(x) = \|x\xi_0\|$ will be a faithful weight on $M$, which is what we need.  Actually, it's not at all clear to me that this is correct-- I don't see why the map $S:\mathfrak A \rightarrow \mathfrak A; a\xi_0 \mapsto a^*\xi_0$ is preclosed.  So now I suspect there might be a counter-example...

REPLY [11 votes]: The answer is yes for trace vectors, but no in general. Take a closed nowhere dense subset $C\subset[0,1]$ with positive measure and consider the state $\phi$ on $C([0,1],M_2)$ defined by $$\phi(f)=\int_C f(x)_{11}\, dx + \int_{[0,1]\setminus C}\mathrm{tr}f(x)\,dx.$$ Here $f(x)_{11}$ is the $(1,1)$-entry of $f(x) \in M_2$. Then, $\phi$ is a faithful state, and the GNS vector is a cyclic separating vector. However it is not separating for the double commutant (which is $L^\infty([0,1],M_2)$).<|endoftext|>
TITLE: Is there a general result that theorems about finite structures proved in ZFC can be proved in ZF?
QUESTION [8 upvotes]: The title question is too vague so let me be specific.
Much of modern finite semigroup theory uses profinite semigroups and properties of profinite semigroups that depend on the existence of prime ideals in Boolean algebras, which is a choice principle weaker than AC. Often results are proved first via profinite methods and then explicit proofs using finite means are found later.  
To narrow my question down, let me consider the following situation.  Let V be what is sometimes called a variety of finite semigroups, that is, a class of finite semigroups closed under finite products, subsemigroups and homomorphic images.  A pro-V semigroup is an inverse limit of semigroups in V. It is known, using Tychnoff's theorem for products of finite spaces, that every continuous finite quotient of a pro-V semigroup belongs to V. So if you want to show a variety W of finite semigroups is contained in a variety V it suffices to show each element of W is a continuous quotient of a free pro-V semigroup. Sometimes one has good information on the structure of free pro-V semigroups and can exploit this. 
For example, Almeida proved that the smallest variety of finite semigroups containing all finite commutative semigroups and all finite groups is the variety of finite semigroups with central idempotents using the approach sketched above.  Later Auinger gave a proof using only finite semigroups. 


Question. Is there some general result in logic or set theory that would imply that  the existence of a proof in ZF+Boolean prime ideal theoerem that a variety W of finite semigroups is contained in a variety V of finite semigroups implies the existence of a proof in ZF that W is contained in V?


Many results of this sort in finite semigroup theory were motivated by questions in automata theory and in principle one would like to avoid choice in this context. 
Caveat: I know very little set theory or model theory so please take that into account in answer.

REPLY [8 votes]: As long as your varieties have reasonable definitions, the axiom of choice will be eliminable from proofs like these.  The point is that any finite semigroup has an isomorphic copy whose underlying set consists of natural numbers, and that copy will be in Gödel's constructible universe $L$, where the axiom of choice holds.  So if there were a semigroup in $W$ that isn't in $V$, then the same would be true in $L$.
Now about that proviso "reasonable definitions": You can sneak a lot of set theory into the definition of a variety.  Consider the variety that consists of all groups if the continuum hypothesis holds and consists of all commutative semigroups if the continuum hypothesis fails.  With such "cheating" you can surely arrange for $W$ to be included in $V$ iff the axiom of choice holds.  My argument in the preceding paragraph tacitly assumed that the definitions of the varieties were absolute, in the sense that the constructible universe $L$ and the whole universe agree as to whether any particular finite semigroup is in the variety.  By working harder, one can get by with weaker absoluteness requirements, but one can't get rid of them completely.

REPLY [8 votes]: There is a standard method to handle this kind of situation which uses the absoluteness of certain statements over models of set theory.
The basic idea is this. If you're investigating the properties of a certain "well-behaved" set $A$, then instead of working in the full set-theoretic universe $V$, you can work in the smallest universe $L[A]$ that contains $A$ and all the ordinal numbers. By classical results of Gödel and others, this smaller universe $L[A]$ satisfies some very strong forms of choice and many other nice properties. You can then freely use these facts to show that "$A$ has property $X$" holds in $L[A]$. As is this doesn't really mean much, but if "property $X$" is upward absolute, then you can conclude that "$A$ has property $X$" holds in the full set-theoretic universe $V$.
There are several tricks to analyze when a property is absolute or not. Many of them are syntactic in nature, for example Shoenfield's absoluteness theorem states that $\Sigma^1_3$ properties are always upward absolute. Assuming the existence of some large cardinals, a whole lot of statements can be shown to be absolute.
For the situation you describe, I believe everything is absolute for even simpler reasons. The constructible universe $L$ already contains isomorphic copies of all finite objects in the real universe $V$. Since the definition of a variety is finitary, varieties of finite semigroups are essentially the same in $L$ as they are in $V$. Therefore, inclusions of varieties of finite semigroups are absolute between $L$ and $V$. So you can freely use the axiom of choice in $L$ and the conclusions drawn automatically transfer from $L$ to $V$.<|endoftext|>
TITLE: Geodesic path on the unit sphere with the sup norm
QUESTION [10 upvotes]: Let $X$ be the unit sphere of $\mathbb{R}^n$ with the sup norm, i.e. $X=\{x\in\mathbb{R}^n: \|x\|_{\infty}=1\}$.  Let the metric $d$ on $X$ be the geodesic metric induced by the sup norm, i.e for any two points $x,y\in X$, $$d(x,y)=\inf \{\ell(\alpha): \alpha \mbox{ a continuous curve on }X \mbox{ joining }x,y \},$$ where $\ell(\alpha)$ denotes the arc length of $\alpha$ measured with the sup norm. It is obvious that one only has to consider piecewise linear curves.
My question is: Has anyone obtained formulae for $d(x,y)$ and the geodesic paths (curves of minimum length) joining $x,y$?

REPLY [7 votes]: Dear TCL,
If one does not confuse geodesics with minimal geodesics, then one can say that if you develop the cube onto the plane, then straight lines are geodesics. Of course, not all geodesics arise in this way and I've never considered the question of giving an explicit formula for the distance function.
If $P$ is an $(n-1)$-dimensional polyhedron in an $n$-dimensonal normed space, which we consider as a piecewise flat Finsler space, then a large class of geodesics that do not pass through the $(n-3)$-skeleton of $P$ can be easily described as follows:
(1) On each facet they are straight lines.
(2) They cross the $(n-2)$-dimensional faces transversely and at each crossing they satisfy the generalized Snell-Descartes law (of refraction).
This statement almost obvious because straight lines are geodesics in normed spaces (and hence on the facets, which I consider as pieces of normed spaces with the induce norm) and because the generalized Snell-Descartes law follows from Fermat's principle. 
In the case of the cube with the piecewise flat $\ell_\infty$ metric, this gives that straight lines in the developed cube are geodesics on the cube.
The generalized Snell-Descartes law
This is a good time to say that what I'm writing now is basically an annoucement of an essay I'm writing for a book with A.C. Thompson ("An invitation to Minkowski geometry").
The presentation here is a bit sketchy at some points (otherwise it would be too long), but
if you reconstruct the pictures, I think everything will be clear.
A well-known secret in physics is that everything becomes simpler when you use momentum (a truly physical notion) rather than velocity (which is mere kinematics) and so it is with the laws of reflection and refraction. Moreover, once formulated in terms of momenta, the generalization to the Finsler (or normed-space) setting is obvious.
Correspondence between velocity and momentum or one minute with the Legendre transform
Let $v$ be a unit vector on a normed space $(X,\|\cdot\|)$, the momenta associated to $v$ is the set of unit covectors 
$\xi \in X^*$, the dual normed space, such that the hyperplane $\xi = 1$ supports the unit ball of $X$ at the point $v$. We extend the definition to all non-zero vectors by homogeneity. You can also write this in terms of the subdifferential of the square of the norm.
(Here I really should have inserted a nice picture!) 
The same procedure allows us to associate to every non-zero momentum a set of velocities. When the unit spheres of $X$ and $X^*$ are strictly convex, the correspondence between velocities and momenta is a bijection. 
The Snell-Descartes law
Consider an $(n-1)$-dimensional cooriented subspace $W \subset {\mathbb R}^n$ (the wall) and consider a norm on each half-space that describes the propagation properties of light in this anisotropic medium. The norms do not have to agree on the wall, although they do when we look at piecewise flat Finsler metrics on polyhedra and so for simplicity (I do not want to deal here with "critical angles" at which refaction becomes reflection) I will assume both norms agree on the subspace $W$.
Law of refraction: If the light ray hits the wall $W$ transversally at the origin with incoming unit momentum $\xi$ (i.e. $\|\xi\|^*_1 = 1$), then the outgoing momenta $\eta$ (there may be an infinity of light rays refracting from a single ray!) are characterized by the conditions:
(a) $\|\eta\|^*_2 = 1$, 
(b) The restrictions of the covectors (= linear forms) $\xi$ and $\eta$ to the subspace $W$ agree,
(the third condition is easier to draw than to state and it distinguishes refraction from reflection)
(c) The points at which the hyperplane $\eta = 1$ supports the unit ball 
$B_2 = \{v : \|v\|_2 \leq 1\}$ lie on the same side of $W$ as the points at which the 
hyperplane $\xi = 1$ supports the unit ball $B_1 = \{v : \|v\|_2 \leq 1\}$.
Indeed if we change (c) by 
(c') The points at which the hyperplane $\eta = 1$ supports the unit ball 
$B_2 = \{ v : \|v\|_2 \leq 1 \}$ and the points at which the 
hyperplane $\xi = 1$ supports the unit ball $B_1 = \{ v : \|v\|_2 \leq 1 \}$ lie on different sides of $W$.
we obtain a slight generalization of the Law of reflection for Minkowski spaces given in
Corollary 3.2 of the paper by Gutkin and Tabachnikov).
Remarks.
1. The laws of refraction and reflection admit a very pretty synthetic interpretation in terms of pencils of hyperplanes. Have fun!
2. When condition (b) cannot be met is exactly the case where we have a "critical angle" and light is reflected instead of refracted. A good exercise is to check these constructions in the standard case and rediscover the usual laws of reflection and refraction (and the condition for critical angles) that you find in any physics textbook.
3. When the normed spaces are such that their unit balls and their duals are strictly convex, then to each incoming light ray corresponds one and only and outgoing light ray. 
4. On the upper half-space take the usual Euclidean norm, on the lower half-space take the norm whose unit ball is a double cone with base equal to the unit disc on the wall and apexes $(0,0,\dots,0,1)$ and $(0,0,\dots,0,-1)$. You can verify that all light rays coming from the upper-half plane get refracted into the sheaf of vertical rays.<|endoftext|>
TITLE: Source for NBG+Equipollence conservative over ZFC?
QUESTION [9 upvotes]: The "conservative" class theory, NBG, proves no new theorems about sets (with respect to ZFC).  The choice function used here is set choice, and it's not too hard to prove (if M is a ctm for ZFC, then D(M) is a ctm for NBG and has the same set universe).
However, if we add the axiom that there is a bijection (in the class universe) between $\mathbf{V}$ and $\mathbf{ON}$, the classes of sets and ordinals, respectively, this is apparently still conservative over ZFC (with a much stronger class choice axiom).  However, I can't find a reference for this.
Apparently this fact is credited to Easton and Solovay, published by Easton in 1964, and apparently it uses class forcing, but I can't find any more specific information on this topic, or the paper itself.  Does anyone have more specific information on this, or better search skills than I do?

REPLY [11 votes]: Yes, NBG + Equipollence, which is equivalent to NBG + Global Choice, is a conservative extension of ZFC; this was independently discovered by many people (at least Cohen, Felgner, Grishin, Jensen, Kripke, and Solovay). A detailed proof can be found in

Ulrich Felgner: Comparison of the axioms of local and universal choice, Fundamenta Mathematicae 71 (1971), pp. 43–62 (EUDML, PLDML).

The basic idea of the proof is pretty straightforward: you take the class of injective (set) functions whose domain is an initial segment of On as your forcing conditions, and then a generic filter gives you a bijection of On and V. (Actually, Felgner uses choice functions as forcing conditions, but that does not make much of a difference.)
The result is also included in the follow-up surveyish paper

Ulrich Felgner: Choice functions on sets and classes, in Sets and classes: On the work of Paul Bernays (Gert H. Müller, ed.), North-Holland, Amsterdam, 1976, pp. 217–255, doi 10.1016/S0049-237X(08)70895-4.

This paper also discusses a generalization of the result to set theory without the foundation axiom ($\mathrm{ZFC}_-$, $\mathrm{NBG}_-$). It turns out that in absence of foundation, equipollence has nontrivial consequences even for sets: $\mathrm{NBG}_- + \mathrm{Equipollence}$ is a conservative extension of $\mathrm{ZF}_-$ extended by the schema
$$\forall x\:\exists y\:\phi(x,y)\to\forall\alpha\in\mathrm{On}\:\exists f\colon\alpha\to\mathrm{V}\:\forall\beta<\alpha\:\phi(f\restriction\beta,f(\beta))$$
(a sort of class version of $\mathrm{DC}_\alpha$; an equivalent formulation: any class tree whose height is bounded by an ordinal has a maximal path). This schema implies AC and collection (assuming replacement), but one can show that it is strictly stronger than that (i.e., independent of $\mathrm{ZFC}_- + \mathrm{Collection}$).
Note that without foundation, equipollence still implies global choice, but not vice versa. It is apparently an open problem if $\mathrm{NBG}_- + \mathrm{Global\ Choice}$ is conservative over $\mathrm{ZFC}_-$.<|endoftext|>
TITLE: Why is the fundamental group of a compact Riemann surface not free ?
QUESTION [148 upvotes]: Consider a compact Riemann surface $X$ of genus $g$.
It is well-known that its fundamental group $\pi_1(X)$ is the free group  on the generators $a_1,b_1,...,a_g,b_g$ divided out by the normal subgroup generated by  the single relator $[a_1,b_1]\cdot \ldots\cdot [a_g,b_g]$.
(This has of course nothing to do with the complex structure of $X$, but may be  computed by considering the underlying topological manifold as a cell complex.)
This group is trivial for $g=0$ and free abelian on two generators for $g=1$. 
For $g\geq 2$, however, I had always taken for granted  that it is not free  but I have just realized that I cannot prove that. 
So, although I guess   the answer is no, I'll ask my official question in an open way : Is   $\pi_1(X)$ free for $g\geq 2$ ? 
Edit: Users have now brilliantly solved the problem in multiple ways.
Non-freeness is definitely established, with 12 proofs plus sketches of proofs in the comments!
It is clearly impossible to select in a reasonable way an answer for "acceptance" among all these great answers .
Since the software forces me to make only one choice, I have chosen Daniel's answer because of its merit, but also because it acknowledges Vitali's  contribution: Vitali was  the first to sketch a solution (in the comments) .

REPLY [3 votes]: Free groups are residually finite.
Finitely generated and residually finite implies Hopfian.

So $F\{a_1,b_1,...,a_g,b_g\}$ is Hopfian so it is not isomorphic to the quotient group by any nontrivial normal subgroup, hence
$$F\{a_1,b_1,...,a_g,b_g\}\not\cong \pi_1(X)=\dfrac{F\{a_1,b_1,...,a_g,b_g\}}{\langle[a_1,b_1]\cdot...\cdot[a_g,b_g]\rangle}\quad (1)$$ 
But taking the epimorphism $\pi_1(X)\to\pi_1(X)_{ab}\cong \mathbb{Z}^{2g}$ we see that $\pi_1(X)$ has least set of generators of size $2g$ and from $(1)$ it isn't isomorphic to the only free group in $2g$ generators.<|endoftext|>
TITLE: How to assess research "impact" for tenure/promotion committees 
QUESTION [23 upvotes]: Over the last several years, the college-level promotion & tenure committee at my university has increasingly been seeking to apply "objective" criteria for assessing the impact of candidates' research.  Journal impact factors have been a favored metric, and we have tried to argue that these are not a reliable measure, for all the usual well-documented reasons.  (The AMS statements on "The Culture of Research and Scholarship in Mathematics" have been helpful, if not entirely convincing.)  Now they want to use more individual metrics, such as h-indices and g-indices.  We would like to discourage this, but it's hard to argue convincingly to non-mathematicians about the flaws of such measures, and at the end of the day, they're demanding SOME numerical measure that they can use to compare candidates, both internally and to faculty at peer institutions.  
So, I'd like to hear how other math departments have dealt with such pressures.  In particular, how can you articulate standards in such a way as to maintain high expectations, but also to minimize the damage to candidates who might be doing well-respected research, but for whatever reasons this results in relatively few papers, or papers with relatively few citations?  And if we do end up having to use something like an h-index, is there any way to collect data from comparable math departments so we can at least say something about what is a "good" value for a particular index for mathematicians?

REPLY [2 votes]: This is not a direct answer to your question, but I think it is related. To get a perspective on why such things have been happening, I recommend the paper Neo-liberalism and Marketisation: the implications for higher education especially the section The Implications of Marketisation, starting on page 6.<|endoftext|>
TITLE: Does every polycube tiling imply a regular polycube tiling?
QUESTION [8 upvotes]: Let's define d-polycubes to be a union of unit hypercubes from the $\mathbb Z^d$ tiling of d-dimensional Euclidean space which has connected interior. Given a tiling of $\mathbb R^d$ by identical copies of a d-polycube, we call this tiling regular if all the centers of the hypercubes have integer coordinates (i.e. the hypercubes fit together in a $\mathbb Z^d$ lattice).
I haven't been able to make much progress on the following seemingly simple and natural question:

If a given d-polycube tiles $\mathbb R^d$, must it also tile this space regularly?

I believe I have the $d=2$ case by a simple argument. As long as one has a connected tiled region with a concave boundary one can show that the adjacent 2-polycube (a.k.a. polyomino) at that corner must be placed to fit next to the squares of the already placed polyominoes so that edges meet edges and vertices meet vertices. We continue this until we tile the entire plane or we reach a convex region. Since the allowed angles are $\pi/2, \pi, 3\pi/2$ then we must have a rectangle. Now we can tile the space with translations of this rectangle.
However this doesn't work in $d\geq 3$ and I'm having trouble coming up with an argument in that case.

REPLY [7 votes]: I am almost sure that the answer is no. Your question is strongly related to Keller's conjecture which turned out to be false. There a tiling of $\mathbb Z^{10}$ was given by translates of the unit cube such that the center of each translate is a halfinteger. Moreover, the tiling has a very strong lattice-like structure, 1024 cubes whose center is in $[0,2]^{10}$ are translated by every vector from $2\mathbb Z^{10}$.
Of course, the cube won't give a counterexample to your question. But you can divide each cube from this construction to like $3^{99}$ little cubes and then "dig little tunnels" among adjacent cubes to make sure the resulting polyominos only fit in the given way. This leaves many details to work out and I am not 100% sure it can be done but looks like a promising approach.<|endoftext|>
TITLE: Fastest decay of Fourier Transform for Generalized Functions of compact support
QUESTION [8 upvotes]: What is the fastest decay possible for the Fourier transform of a generalized function with compact support and finite value at the origin? I know that regular functions cannot attain exponential decay, due to the Paley-Wiener Theorem and that a short paper by Ingham (1934) suggests that the fastest decay in that case is of the order $e^{-|t| \epsilon(|t|)}$, where $\epsilon(t)$ satisfies $\int_0^\infty \frac{\epsilon(t)}{t} dt < \infty.$ 
Optimally, it seems we may take $\epsilon(t) = (\lg t \cdot \lg\lg t\cdot\lg\lg\lg t\ldots)^{-1}$, but not $\epsilon(t) = 1/\lg t$. My question is two-fold:

Are there (regular, as in non-generalized) functions with compactly supported and bounded Fourier transform whose asymptotic decay is $e^{-|t|/\lg|t|}$, or faster? In other words, how tight is Ingham's construction based on products of sinc functions?
Is it possible to beat the Paley-Wiener theorem if we move to the space of generalized functions (such as Gelfand-Shilov spaces), while keeping the Fourier Transform compactly supported and bounded at the origin?

A positive answer to the above questions would have great implications for stability results in Quantum Mechanics, such as the stability of topological quantum order under weak perturbations. In particular, it would greatly improve bounds on the strength of arbitrary quasi-local perturbations to topological quantum systems, which preserve the topological nature of the system.

REPLY [6 votes]: I think this question (at least for usual functions) is related to what is known as the "Beurling-Malliavin multiplier theorem".  A recent survey is "Beurling-Malliavin multiplier theorem: the seventh proof", by Mashreghi, Nazarov and Havin, St Petersburg Math J. 17 (2006), 699-744 (see http://www.ams.org/journals/spmj/2006-17-05/S1061-0022-06-00926-5/home.html ).
If I read this right, decay $\exp(-|x|/\log (2+|x|))$ is impossible for $L^2$ functions at least (see the Theorem in section 1.1 of the paper, which says that one can not even find an $L^2$ function with such decay and Fourier transform supported on a half-line; indeed, there is a necessary and sufficient condition for the latter).<|endoftext|>
TITLE: Reference on Geometric Topology
QUESTION [7 upvotes]: Geometric topology is more motivated by objects it wants to prove theorems about. Geometric topology is very much motivated by low-dimensional phenomena -- and the very notion of low-dimensional phenomena being special is due to the existence of a big tool called the Whitney Trick, which allows one to readily convert certain problems in manifold theory into (sometimes quite complicated) algebraic problems. The thing is the Whitney trick fails in dimensions 4 and lower.
As to my background, I've learnt Boothby's book "An Introduction to Diffential Manifolds ...". I recently want to dive in some depth into Geometric Topology. But I found the literature is quite a mess. Could anyone suggest a textbook or at least a sequence of books and papers(but not too many) that leads to the frontier of this field?

REPLY [16 votes]: Geometric topology is not really a unique field in the way that algebraic topology and general topology are. Its various subareas may share something of a common feel (and indeed an arxiv category), but are often too diverse to have any common techniques. Those areas include, for instance:
Low-dimensional topology (classical knots, 3-manifolds, 4-manifolds, etc.)
Morse theory, simple homotopy theory and algebraic K-theory of spaces
Dimension theory (of separable metrizable spaces)
Topology of manifolds (surgery theory, codimension two knots, etc.)
Singularity theory (of smooth maps), geometric immersion theory (dealing e.g. with 4-tuple points of sphere eversions)
PL topology (block bundles, collapsing, bistellar moves, etc.)
Generalized manifolds, wild knots, etc.
Group actions on manifolds
Manifold structures (smoothing/trangulability and the Hauptvermutung; also Lipschitz structures, CD-manifolds, ... )
Embedding theory (smooth embeddings of projective spaces, PL embeddings of polyhedra,
etc.)
I surely forgot to mention many important subjects here; even the grouping of items in this list is rather arbitrary (and the order is random). The point is, you will probably not get far with diving in some depth into geometric topology unless you're more specific on what you're interested in. If unsure, try some knot theory or low-dimensional manifolds. These now cover more than half of all geometric topology by any count. Ryan and Jim gave some good suggestions of starting points in their math.SE answers, such as Rolfsen's 'Knots and links'. There are also other flavors of low-dimensional topology.
There exist some books and courses mentioning 'geometric topology' in the title, but they are often specialized and/or advanced. For instance, the 'geometric topology' notes by Sullivan and Lurie are mostly focused on manifold structures, and are firmly grounded in methods which are very clever and useful, but kind of external to geometric topology (localization, Galois theory and simplicial sets). Likewise, Bing's 'Geometric topology of 3-manifolds' and Moise's 'Geometric topology in dimension 2 and 3' are mostly about wild things. (There's definitely a trend in the literature that if geometric topology gets explicitly mentioned, things are likely not all smooth or PL.) 
Arguably, closer to the point are Fenn's 'Techniques of geometric topology' and Ferry's 'Geometric topology notes'. Even these two virtually don't overlap with each other, so they are certainly not equivalents of some canonical algebraic topology text such as Spanier's or Hatcher's. But perhaps closer to such an equivalent than anything else that I can think of.<|endoftext|>
TITLE: Graph containing every n-vertex graph as an induced subgraph
QUESTION [12 upvotes]: Let $f(n)$ denote the minimum number of vertices in a graph $G$ which contains every graph on $n$ vertices (up to isomorphism) as an induced subgraph. I want to estimate $f(n)$. A simple counting argument gives a lower bound : there are essentially $2^{C(n,2)}/n!$ isomorphism classes of graphs on $n$ vertices, while a graph with $N$ vertices contains $C(N,n) \approx N^{n}/n!$ induced subgraphs of size $n$. From this we can deduce that, for example, 
\begin{eqnarray*}
\liminf_{n\rightarrow \infty} f(n) ^{1/n} \geq \sqrt{2}.
\end{eqnarray*}
Can we replace the inequality by an equality here ? I assume the answer is known, though I didn't find it after some basic googling. I know that the liminf above is $O(1)$, for example this follows from the main result of [1]. Indeed, it follows from [1] and the standard probabilistic argument for lower-bounding Ramsey numbers that there is some $c > 0$ for which the random graph $G(c^n, 1/2)$ almost surely, as $n \rightarrow \infty$, contains every graph on $n$ vertices as an induced subgraph. I assume the best-possible $c$ here is not known, since any $c < 4$ would yield an improvement on the best-known upper bound for Ramsey numbers. 
[1] H.I. Prömel and V. Rödl, Non-Ramsey graphs are $c \log n$-universal, J. Combin. Theory Series A 88 (1999), 379-384.

REPLY [4 votes]: @Tony and Chris : Thanks for your very helpful comments. I had a look at Vu's paper and it references an earlier paper 
B. Bollobas and A. Thomason, Graphs which contain all small graphs, Europ. J. Combinatorics 2 (1981), 13-15.
There it is proven (according to the review on MathSciNet, I could not find the original paper online) that almost every graph on approximately $n^2 2^{n/2}$ vertices is what they call $n$-full, i.e.: contains every graph on $n$ vertices as an induced subgraph. This answers all my questions : my function $f(n)^{1/n}$ does indeed tend to $\sqrt{2}$, and $G(c^n,1/2)$ is almost surely $n$-full for any $c > \sqrt{2}$. It was incorrect of me to suggest, when I posed the question, that the latter fact would have any consequences for estimating Ramsey numbers.<|endoftext|>
TITLE: What is the physical difference between states and unital completely positive maps?
QUESTION [9 upvotes]: Mathematically, completely positive maps on C*-algebras generalize positive linear functionals in that every positive linear functional on a C*-algebra $A$ is a completely positive map of $A$ into $\mathbb{C}$. Furthermore, we have the Stinespring construction as a powerful generalization of the GNS construction. 
Certainly, the relationship between completely positive maps and positive linear functionals can only go so far. I am curious about what physics has to say about this analogy/generalization. It seems that completely positive maps should serve as generalized states of a quantum system, but I've mostly seen cp maps arise in the discussion of quantum channels and quantum operations. I'd like to know precisely in what sense a completely positive map can be viewed as a generalized physical state.

Question: What is a completely positive map, physically? Particularly, in what precise sense can a completely positive map be regarded as a generalized (physical) state?

If there are nice survey papers discussing the above relationship, such a reference may serve as an answer to my question.

REPLY [9 votes]: The main interpretation, which is fundamental in quantum information theory, is that the transpose of a UCP map $E$ is a linear map on quantum states that represents a realistic information channel.  This is the correct generalization of a Markov map or a stochastic map in classical probability theory.  Such a map $E^T$ has to be linear, or otherwise it violates superposition of classical probabilities.  $E$ has to be unital, or otherwise $E^T$ does not conserve total probability.  It has to be positive, or otherwise negative probabilities can be created.  It has to be completely positive, or otherwise $E$ tensored with doing nothing on a companion system is not positive.  Then Stinespring's theorem says that the necessary conditions can be interpreted as sufficient for $E^T$ to be realistic.
There is also the Choi isomorphism theorem that identifies CP (but not UCP) maps between two systems with positive states on one system.  This is also important, but much more secondary.  The category of $C^*$-algebras with UCP maps as the morphisms, and with the arrows reversed to express the transpose, is a realistic category of quantum probability.<|endoftext|>
TITLE: When is the connected sum of manifolds orientation-independent?
QUESTION [10 upvotes]: Given $M$ and $N$, two connected orientable manifolds of the same dimension, when is $M$ # $N$ diffeomorphic to $M$ # $\overline{N}$, where $\overline{N}$ is $N$ with the orientation reversed? 
If $N$ has an orientation-reversing automorphism, is this a necessary or sufficient condition for $M$#$N$ to be diffeomorphic to $M$#$\overline{N}$? If it isn't a necessary condition, what invariants can be used to distinguish $M$ # $N$ from $M$ # $\overline{N}$? 
(As a baby example, how does one show that $CP^2$ # $CP^2\ncong CP^2 $#$ \overline{CP ^2}$ ?)
Zygund

REPLY [9 votes]: In the oriented category it's not true that if $M\#N$ is oriented diffeomorphic to $M\#(-N)$ then $N$ admits an orientation reversing diffeomorphism. I suspect there are easier counterexamples but the following works. Take $M^{4k+3}=S^3\times \mathbb{CP}^{2k}$ with $k\ge 1$. Then there exists an exotic sphere $\Sigma^{4k+3}$ such that $\Sigma$ does not admit an orientation reversing diffeomorphism but $M\#\Sigma$ is oriented diffeomorphic to $M$ (and hence also to $M\#(-\Sigma)$). Note that $M$ obviously admits an orientation reversing diffeomorphism because $S^3$ does.
I read about this fact in a paper by Belegradek, Kwasik and Schultz "Codimension two souls and cancellation phenomena". Not sure if this is the earliest reference, perhaps Igor can clarify this - he visits MO regularly.
More specifically they show that if $I(M)$ is the inertia group of $M$ (the group of oriented exotic $4k+3$-spheres $\Sigma$ such that the standard homeomorphis $M\#\Sigma\to M$ is homotopic to a  diffeomorphism)  then $I(S^3\times \mathbb{CP}^{2k})\cap bP_{4(k+1)}$ has index 2 in  $bP_{4(k+1)}$ where $bP_{4(k+1)}$ is the group of exotic $(4k+3)$-spheres bounding parallelizable manifolds. It's known that $bP_{4(k+1)}$ is cyclic or order exponentially growing in $k$ so any nontrivial element $\Sigma$ of $I(S^3\times \mathbb{CP}^{2k})\cap bP_{4(k+1)}$ of order different from 2 works.<|endoftext|>
TITLE: Math blog directory
QUESTION [7 upvotes]: Does anyone have a list of high quality mathematics (or related) blogs. I am of course aware of Terry Tao's most excellent blog, and also of ldtopology.wordpress.com, but I am sure the complete list is far longer.
EDIT As I say in my comment, the key point is that I am looking for high quality blogs. nLab and mathblogging both give a VERY long list, and while they are both useful resources (neither of which I was aware of before asking the question) neither is sufficiently selective to be really useful.

REPLY [2 votes]: John Roe had a nice blog http://www.math.psu.edu/roe/ for things in non commutative geometry and else too ; but unfortunately, it's not up to date (you can still probably find good stuffs)<|endoftext|>
TITLE: Liftings of Nichols algebras over racks via Doi twist
QUESTION [6 upvotes]: As a more nontrivial example for my Dissertation thesis, I'd require some example of the following type (of course I'll "cite" ;-) ), so thanx in advance:
Andruskiewitsch/Grana have by a new construction given very interesting new liftings of finite dimensional Nichols algebras e.g. over $S_4$ ("Examples of liftings of Nichols algebras over racks", 2004). 
On the other hand Masuoka has shown, that all the "well-known" liftings of Nichols algebras over abelian groups (finite dimensional, all prime divisors>7, by Andruskiewitsch/Schneider) are Doi/Cocycle-twists of the unlifted/graded one's ("Abelian and non-abelian second cohomologies of quantum enveloping algebras", 2008)
....is there an easy way to see that this is also true for the "new" liftings over nonabelian groups?...and how could I easily read off the precise 2-cocycle?

REPLY [6 votes]: At the moment it is not easy to see that this will also hold for the non-abelian case, or even for the abelian cases yet to be computed, although there are no counter examples. It holds in most known examples as shown in 
A. G. I. and Mombelli, M. Representations of the category of modules over pointed Hopf algebras over S_3 and S_4 (joint work with Martín Mombelli). Pacific Journal of Mathematics, 252 (2) (2011), pp. 343–378. Available at  arXiv:1006.1857v5. 
for liftings over S4 and S3, and in 
G. A. García and M. Mastnak. Deformation by cocycles of pointed Hopf algebras over non-abelian groups. Preprint: arXiv:1203.0957v1
for liftings over D4. In this last paper, moreover, the cocycles are explicitly computed.<|endoftext|>
TITLE: Generalizing a square wheel to a body rolling on a surface
QUESTION [9 upvotes]: A square wheel rolling on a catenary road maintains the wheel center at a fixed
height, a well-known construction previously discussed on MO
(e.g.,
"Generalizing square wheels rolling on inverted catenaries").
I wondered if this fundamentally one-dimensional example could be generalized to
two dimensions, in the following sense:

Is there a solid body $B$ and a non-flat surface $S$ which together have the property that,
  from some one, special fixed position of $B$ resting on $S$, $B$ can roll on $S$ in any 
  horizontal direction $v$ so that some point in $B$ (its center) remains at a fixed height?

Of course if $B$ is a sphere and $S$ is a plane, then the constant-height property
holds.  Note I am asking that this only hold for some special initial position
of $B$, but 
demand that rolling in any direction of the full $360^\circ$ retains constant height along
that ray.
My guess is that the requirement that this hold for every position of $B$ on $S$
forces a sphere on a plane. 
The following is meant to be suggestive only!
           


What brought this to mind is the traditional Easter Egg Roll. :-)

Addendum. Here is another suggestive image, not metrically accurate, of a revolved 
square diamond that can roll on a revolved catenary, as per Anton's answer.

REPLY [8 votes]: It happens if and only if
in the initial position it is the surfaces or revolution around $z$-axis and the intersection of $B$ with say $(yz)$-plane is rollable plane figure.
Note that  $B$ and its initial position overdetermines the surface;
it gives the surface as the foot points of rolling in all directions plus it gives its gradients at this foot points.
Both of these data are periodic along each ray.
For close rays they the foot-point-trajectories spread apart, but the gradients stays perodic.  Therefore the data can feet together only if the gradient is parallel to the ray. 
The later implies that we have a surface of revolution.<|endoftext|>
TITLE: (weak?) BN-Pair / Tits System for Sporadic Groups
QUESTION [18 upvotes]: The structure of finite simple groups of Lie type of arbitrary rank can be described well via BN-pairs. BN-pairs basically generalize the Bruhat decomposition of matrices into monomial $N$ and triangular $B$ matrices and come with a "Weyl group" $N/(B\cap N)$, that has to be a Coxeter group.
Conversely, Tits showed in 1974 that a group with a spherical BN-pair of rank at least 3 is of "Lie type". (Ironically, the general group identification needed in the classification for lowest rank 3 was the "quasi-thin" case solved last by Aschbacher). 
Now the existence of a rank 1 BN-pair for a group $G$ is equivalent to the existence of a doubly-transitive action of $G$ on a set $X$ (which can be taken to be $G/B$). This implies that also the sporadic simple Mathieu groups have a BN-pair (of rank 1).
Now my question: The construction of the Monster group uses also a rank 1 "BN-construction" that is not proper: The "triality" element in the Weyl group $S_3$ takes the (non-normalizing) role the transpositions ought to have. Sometimes the term "weak BN-pair" is mentioned. However, I could not find a proper definition. Can one briefly explain this concept and how it is related to the usual BN-pairs? Is it as generic or rather ad-hoc? Can one suggest good introductory literature? 
EDIT: Found e.g. in appendix F of Aschbachers "Classification of Quasithin groups"

REPLY [10 votes]: There has been a lot of work done on various generalizations of the concept of the building, to apply them to sporadic groups. These generalizations are variously known as diagram geometries, chamber systems, etc.
Names like G.Stroth, S.Smith, M.Ronan, A. Delgado, D. Goldschmidt, B. Stellmacher, etc. spring to mind. There is an "elementary" book on diagram geometries by A.Pasini (a review of the latter is 
 here.)
There is a series of books by A.A.Ivanov (some of them are jointly with S.Shpectorov) developing a theory of this sort to deal with a majority of sporadics. 
Indeed, one needs a weakening of the classical buildings to cover sporadics.  Instead of starting from a weak BN-pair, one can weaken Tits' axioms from his "Local approach to buildings" to develop a theory dealing with sporadics. E.g. Witt designs for Mathieu groups (already from 1938) are extensions (in certain well-defined way) of the affine plane of order 3 and of the projective plane of order 4. Similar things can be done with $HS$, $Suz$, Fischer's sporadic groups, $He$, $McL$, $Co_3$, $Co_2$, and $BM$. (E.g. --- cannot resist citing myself here: the 3-transposition graph for $Fi_{22}$ can be characterized as the extension of the polar space for $U_6(2)$.) This appears to work when the underlying combinatorics is not too complicated (and the corresponding permutation representation has low rank). 
Regarding the $BN$-pairs approach, I must say I don't recall details, having done very little work on these things in past 15 years. In a nutshell, one cannot hope for "real" apartments, etc., so one instead looks at amalgams of parabolic subgroups. Instead of a definition, let me giev you a toy example,  $GL_4(2)$ and its Borel subgroup $B$ (taken to be the upper-traingular matrices, say). Then you have "minimal parabolics" $P_i$, i.e. subgroups generated by $B$ and $e_{i+1,i}$, for $i=1,2,3$ (here $e_{ij}$ denotes the matrix with 1 at position $ij$ and on the main diagonals, and 0s elsewhere). Then, you get maximal parabolics, $P_{ij}$, generated by $P_i\cup P_j$. This is what is called a rank 3 amalgam (as you have 3 minimal parabolics).
Your geometry then consists of cosets of $B$, $P_i$'s, $P_{ij}$'s in the whole group and in each other. The amalgam is now the set-theoretic union of  $B$, $P_i$'s, $P_{ij}$'s, and you can study its universal completion, i.e. the biggest group where is can be embedded into. By tweaking the groups which can arise as  $B$, $P_i$'s, $P_{ij}$'s, one covers more cases than buildings, and tries to stay away from infinite universal completions for ranks at least 3.
PS. IMHO, Aschbacher sometimes tends to ignore prior work, re-inventing the wheel in different terminology.<|endoftext|>
TITLE: Definable map from all the ordinals to the surreal numbers with a dense image?
QUESTION [5 upvotes]: I'm trying to understand analogies and disanalogies between ${\Bbb R}$, the reals numbers, and ${\bf No}$, the surreal numbers.
${\Bbb R}$ admits countable dense sets such as the rationals.  This leads me to wonder whether ${\bf No}$ might contain a dense subclass somehow smaller than the whole.  
On one hand, any  sub-set of ${\bf No}$ must be closed in the interval "topology" since every element of ${\bf No}$ lies in a nest of shrinking intervals ordered like ${\bf On}$, the class of ordinals.  So that makes any dense subclass a proper class.  
On the other hand, ${\bf ZFC}$ doesn't entail a well-ordering of the whole universe (right?) so that means one might (consistently) regard the proper class ${\bf On}$ as small compared with the class of all sets.  
${\bf Question}$: Can we also regard ${\bf No}$ as a larger proper class than ${\bf On}$, in the sense that there does not exist any surjection from ${\bf On}$ to ${\bf No}$?  
In other words, might ${\bf ZFC}$ not guarantee any well-ordering of all of ${\bf No}$.  (This despite a well-ordered class of birthdays and well-orderings for all elements with any set of birthdays). 
It nevertheless seems reasonable to wonder whether ${\bf ZFC}$, or even ${\bf ZF}$, entails a dense subclass of ${\bf No}$ that admits a well-ordering hence my main 
${\bf Question}$: Does there exists a (definable) map from all the ordinals to the surreal numbers that has a dense image?

As with my last question, I confess uneasiness about working with proper classes, so please feel free to correct any misconception that underlies the formulation of my question.  In particular, please tell me if I need to change set theories to get a better question.

REPLY [7 votes]: The answer is that the existence of a definable class embedding
like that is independent of ZFC. In fact, it is equivalent to the
axiom V=HOD.
Theorem. The following are equivalent. 

There is a definable bijection from Ord to No. 
There is a definable surjection from Ord to No. 
There is a definable map from Ord to No with dense image.
V=HOD. 

Proof. If V=HOD holds, then there is definable bijection between
the class of ordinals and the entire universe $V$, and from such an
embedding one can construct a definable bijection between Ord and
No. So 4 implies 1, which implies 2, which implies 3. 
Convesely, suppose that V=HOD fails. Since in ZFC every set is
coded by a set of ordinals, it follows that there must be some
$\alpha$-length binary sequence $s$ that is not in HOD, which means
that $s$ is not definable from ordinals, and neither therefore is
any longer sequence than $s$. But the binary sequence $s$
determines a certain interval in No, by following the digits of $s$
left-and-right through the tree representation of the surreal
numbers. If there were a definable map from Ord to No with dense
image, then there would be OD element of No in this interval.
Suppose that the $\beta^{\rm th}$ element was in that interval. It
follows that $s$ would be definable from $\beta$ and $\alpha$ as
the $\alpha$ length sequence describing the interval at that level
of the tree inside of which the $\beta$-th surreal was to be found.
So $s$ would be ordinal definable, a contradiction. So 3 implies 4. QED
The standard way to force $V\neq \text{HOD}$ is to add a Cohen real $V[c]$. This forcing is almost homogeneous, which implies that every ordinal definable object in the extension $V[c]$ is already in the ground model. In particular, in the extension $c$ itself is not ordinal definable, but it still determines an interval in the surreals, which no ordinal definable element can fill. 
A generalization of the argument will handle the case that one might allow parameters in the definition. This is because by class forcing one can ensure that $V\neq HOD(A)$ for any fixed
set $A$. In this case, there wouldn't even be a map from Ord to No
with dense image, which was definable from parameters.
Meanwhile, I paste below the answer I had posted on math.SE to a similar question, concerning surjections from Ord to No.


From my answer to Willem
    Norduin's
    question on math.SE
The existence of a bijection between the class of ordinals $On$ and
    the class of surreal numbers $No$ is independent of the axioms of
    set theory. There are several interesting possibilities:

If ZFC is consistent, then there is a model of ZFC in which there is a definable such bijection. This is true in Goedel's constructible universe $L$, for example, for in $L$ there is a definable well-ordering of the universe, and we can use this well-ordering to well-order the surreals, which provides the desired bijection.
More generally, there is a first-order definable bijection between $On$ and $No$ if and only if the axiom known as $V=HOD$ holds. For the one direction, if $V=HOD$ holds, then there is a definable well-ordering of the universe and hence in particular a definable well-ordering of the surreals. Conversely, under ZFC if there is a definable bijection between $On$ and $No$, then there is a definable well-ordering of $No$. This allows us to construct a definable well-ordering of the class of sets of ordinals, since any set of ordinals determines a transfinite binary sequence of some ordinal length, and we can interpret this sequence as a $\pm 1$ sequence, which determines a unique surreal number by climbing through the tree of left-right cuts. Thus, we can well-order the class of sets of ordinals. But in ZFC every set is coded by a set of ordinals, and so we can construct a well-ordering of the entire universe, by looking for the least ordinal mapping to a surreal whose $\pm 1$ representation codes that set. So in this case, V=HOD holds.
Another way to summarize this argument is to say that if you can well-order $No$---and this is what your bijection to $On$ amounts to---then you can well-order every class.
If you drop the requirement that the bijection be definable, then we should move to the Goedel-Bernays context, in order to treat classes. The assertion that there is a bijection between $On$ and $No$ is equivalent over ZFC+GB to the axiom of Global Choice, which asserts that there is a well-ordering of the universe. This is by the same argument as above. (Note, we need AC for sets in order to make the last step of the argument; the class bijection in effect allows us to sew the set sized well-orderings together into a class well-ordering.) Thus, the theory ZFC+GB+(your bijection) is equivalent to GBC.
Because of this, if ZFC is consistent, then there are models of ZFC that have no bijection between $On$ and $No$, either definable or definable-from-parameters or otherwise. This is because it is known that ZFC does not imply global choice. One can construct such models by performing a class forcing iteration, adding a Cohen subset to every regular cardinal.
Meanwhile, every model of ZFC has a class forcing extension in which there is a class well-ordering of the universe, simply by forcing to add a global well-ordering, and this forcing extension adds no new sets, only classes. In this sense, it is compatible with every model of ZFC set theory to have the desired bijection as a class, without adding any new sets.
Further, every model of ZFC has a class forcing extension in which there is a definable bijection between $On$ and $No$, since we can force $V=HOD$. (This forcing, however, does add new sets.)

Lastly, upon reading your question again, I see that you asked for
    a surjection from $On$ onto $No$, rather than a bijection. But
    these are equivalent, since if there is a surjection, then we can
    remove the redundant ordinals from the domain by only using the
    least ordinal that maps to a given surreal, and this gives a
    bijection from a proper class of ordinals to $No$. But every proper
    class of ordinals is bijection with $On$ simply by collapsing to
    the order type of the predecessors.<|endoftext|>
TITLE: More general form of inequality?
QUESTION [7 upvotes]: I have proved a simple Lemma that I need for a larger result, and I was wondering whether it is actually another more famous result in disguise.
The lemma says that for any set of vectors in $\mathbb{R}^n$, it is possible to choose a set of coefficients in ${-1,+1}$ such that the weighted sum of the vectors is close to the origin. The interesting thing (to me, non-mathematician) is that how close depends on the dimensionality of the space and not on the number of vectors.
Lemma Consider a set of $m\geq1$ vectors  $\{v_{i}\}_{i=1}^{m}$ in $\mathbb{R}^{n}$, $n\geq1$, such that  $||v_{i}||_{\infty}\leq c$. Then there exists a vector of coefficients  $\{x_{i}\}_{i=1}^{m}$ ,  $x_{i}\in\{-1,+1\}$ , such that  $\left\Vert \sum_{i=1}^{m}x_{i}v_{i}\right\Vert \leq nc$.
(Note that the bound depends on the dimension $n$, but not the number of vectors $m$.)

REPLY [10 votes]: I don't have access right now to search tools, but the result is is due to Steinitz and might be described in 
Kadets, Mikhail I.; Kadets, Vladimir M. (1997). "Chapter 2.1 Steinitz's theorem on the sum range of a series, Chapter 7 The Steinitz theorem and B-convexity". Series in Banach spaces: Conditional and unconditional convergence. Operator Theory: Advances and Applications. 94.
Anyway, the right book by V. Kadets contains the result and best known estimates for the constant in terms of the dimension.
EDIT: I asked Gideon Schechtman for a good reference, and he provided
this<|endoftext|>
TITLE: Smooth representatives for elements of $\pi_7(\text{exotic $S^7$})$
QUESTION [12 upvotes]: Let $M$ be $S^7$ with an exotic smooth structure. Since one can smoothen maps, there exist smooth maps $f:S^7\to M$ which are homotopic to the identity (relative to a base point, if you want). 

Can one make explicit one such map? Can such a map be an homeomorphism?

Little addendum.  The smooth homeomorphism constructed in Ryan's answer below is of course not a diffeomorphism. 

Does one have some control on the non-smooth locus of the inverse of smooth homeomorphims, or on the type of their non-smoothness there?

The inverse of Ryan's map is non-smooth only at the bad pole and I guess the initial map $h$ arises as the "conical differential" of the map there, so the singularity there is pretty bad. Maybe one can find other smooth homeomorphisms whose inverse has a larger non-smooth locus but with tamer non-smoothness there?

REPLY [10 votes]: Duràn wrote down an explicit formula for such map in "Pointed Wiedersehen Metrics on Exotic Spheres and Diffeomorphisms of $S^6$".  That is he wrote an explicit formula for an exotic diffeomorphism from $S^6$ to $S^6$ which is homotopic but not isotopic to the identity. This the produces an explicit homeomorphism from $S^7$ to an exotic sphere by glueing as described by Ryan in his answer.
Geometric properties of that particular map were later studied by various people.
For example, it's written down explicitly on page 1 in "Bootstrapping $Ad$-Equivariant Maps, Diffeomorphisms and Involutions"  by Duràn and Rigas and Sperança (this link is freely accessible unlike the first one).<|endoftext|>
TITLE: sums modulo p^2
QUESTION [5 upvotes]: Let p>3 be a prime. For every a=1,2,...,p-1 let x(a) be the number in {1,...,p-1} such that ax(a) is congruent to 1 mod p (its inverse mod p). Let S be the sum of ax(a)^2 and let T be the sum of x(a) (in both cases sum over all a=1,2,...,p-1). Is it true that S-2T is always divisible by p^2? I noted this by computing for small values of p and I wonder if this is a known theorem or if there is a simple proof.

REPLY [11 votes]: This follows from well-known results (although perhaps not in a totally straightforward way). 
Let us write $x(a) = a^{-1} + p y(a)$, where $a^{-1}$ denotes the inverse of $a$ modulo $p^2$, and $y(a) \in \{0, \dots, p-1\}$. Then we find that
$$ a x(a)^2 = a^{-1} + 2p y(a) \pmod{p^2} $$
so
$$ S = \sum_{a = 1}^{p-1} a^{-1} + 2p \sum_{a = 1}^{p-1} y(a) \pmod{p^2} $$
while
$$ T = \sum_{a = 1}^{p-1} a^{-1} + p \sum_{a = 1}^{p-1} y(a) \pmod{p^2}. $$
So $p^2 \mid S - 2T$ if and only if $\sum_{a = 1}^{p-1} a^{-1} = 0 \pmod {p^2}$, and this latter statement is a consequence of Wolstenholme's theorem.<|endoftext|>
TITLE: Lie algebra "generated" by matrix-valued curve?
QUESTION [7 upvotes]: Let $A(t)$ be a $n\times n$-matrix-valued continuous (plus possibly other niceness conditions; see below) curve, with the matrix entries being complex in general. If I am not mistaken, $A(t)$ generates a minimal Lie algebra $\mathfrak{g}$ of matrices, in the sense that $\mathfrak{g}$ is the intersection of all $n\times n$ matrix Lie algebras containing $A(t)$ for all $t$. 
Now consider the equation $$\frac{d U(t,t_0)}{d t}=A(t) U(t,t_0)$$ with initial condition $$U(t_0,t_0)=\mathbf{1}_n$$ where $\mathbf{1}_n$ is the $n \times n$ unit matrix. 
I am interested in the matrix logarithms of $U(t,t_0)$ for arbitrarily large $t$. For $t$ close to $t_0$ at least one of them lies in $\mathfrak{g}$ because the condition $  \int_{t_0}^t \|A(s)\| ds < \pi$ is satisfied -as long as $A(t)$ is nice- and thus the Magnus series for the logarithm converges. For large $t$ I have no reason to expect that the Magnus series will continue to converge; however $U(t,t_0)$ has at least one logarithm, because the matrix exponential is surjective when considering matrices with complex-valued entries. My question is this:
Where do the matrix logarithms of $U(t,t_0)$ lie? Is it possible that they all lie outside $\mathfrak{g}$?
I would also appreciate it if anyone knows any good references dealing with these equations. Thank you!
EDIT: Robert Bryant gave an example of a $U$ whose logarithm lies outside $\mathfrak{g}$. However one may extend $\mathfrak{g}$ by adding the unit matrix, so that that logarithm is now included. Would this $\mathfrak{g'}$ include at least one of the logarithms of $U(t,t_0)$ in general?

REPLY [8 votes]: It can't always be done.  Take $n=2$, and start with a smooth curve $V:[0,1]\to SL(2,\mathbb{C})$ such that $V(0) = \mathbb{1}_2$ and 
$$
V(1) = \pmatrix{-1&1\\\\ 0&-1}.
$$
Set $A(t) = V'(t)V(t)^{-1}$.  Then $A:[0,1]\to{\frak{sl}}(2,\mathbb{C})$, and it satisfies $V'(t) = A(t)V(t)$ with $V(0) = \mathbb{1}_2$, but $V(1)$ is not the square of any element of $SL(2,\mathbb{C})$ and hence is not the exponential of any element of ${\frak{g}} = {\frak{sl}}(2,\mathbb{C})$.
Added after the question was edited:  Giving yourself just the extra 'room' of adding the center of ${\frak{gl}}(n,\mathbb{C})$ to your algebra $\frak{g}$ won't do it.  Here's a counterexample:  Consider $SL(2,\mathbb{C})$ as a subgroup of $GL(3,\mathbb{C})$ in the obvious way, and let $W:[0,1]\to GL(3,\mathbb{C})$ be defined by 
$$
W(t) = \pmatrix{V(t)&0\\\\ 0&1},
$$
where $V$ is as above.  Set $A(t) = W'(t)W(t)^{-1}$.  Then $A:[0,1]\to {\frak{g}} =  {\frak{sl}}(2,\mathbb{C})\subset {\frak{gl}}(3,\mathbb{C})$, and it satisfies $W'(t) = A(t)W(t)$ with $W(0) = \mathbb{1}_3$.  But it is easy to see that $W(1)$ is not the exponential of any element of ${\frak{g}'} = \mathbb{C}\cdot\mathbb{1}_3\oplus {\frak{sl}}(2,\mathbb{C}) = \mathbb{C}\cdot\mathbb{1}_3\oplus {\frak{g}}$.<|endoftext|>
TITLE: Cesaro means and Banach limits
QUESTION [10 upvotes]: Consider the class of bounded sequences to which every Banach limit (non-negative shift-invariant continuous functional on $l^\infty$ taking convergent sequences in the usual sense to their limits) assigns the same limit value.
 Does a sequence belong to this class if its Cesaro means have a limit? 
Also, is the converse true?

REPLY [4 votes]: If $(x_n) \in \ell^\infty$. According to Lorenz the Banch limit is unique (also known as almost convergent) iff
$$\lim_{p\mapsto\infty} \frac{ x_n + x_{n+1} + \cdots + x_{n+p}}{p} = L  \quad (*) $$
uniformly in $n$. Setting $n=0$ yields Cesaro summability.
As Aaron says, the converse is false. If each $x_n$ is chosen uniformly at random from $\{0,1\}$ then this sequence almost never has property $(*)$ (see Connor's appropriately named article Almost none of the sequences of 0's and 1's are almost convergent)
However the Cesaro limit of this random sequence $(x_n)$ is almost always $1/2$ by the law of large numbers.<|endoftext|>
TITLE: Two approaches to compute the signature of a Kaehler manifold
QUESTION [11 upvotes]: Given a compact Kaehler manifold $M$ of complex dimension $2n$, there are essentially two ways to compute its signature $\sigma(M)$, i.e. the index of the intersection form on $H_{2n}(M,\mathbb{R})$:
1.by Hodge index theorem $\sigma(M)=\sum_{p,q}(-1)^p h^{p,q}$, here $h^{p,q}$ stands for the Hodge numbers.
2.by Hirzebruch signature theorem $\sigma(M)=L[M]$, here $L[M]$ stands for the $L$-genus, i.e. the characteristic number of the top $L$-class.This approach is more general since it works on any $4k$ dimensional real manifolds.
My questions are
1.Since these two approaches rest on different levels of cohomology theory, how are they interrelated? 
2.Of course, one possible way to answer Question 1 is to generalize both by the Hirzebruch-Riemann-Roch on Kaehler manifolds, a point already mentioned in Hirzebruch's Neue topologische Methoden. However, I am wondering if someone could relate these two approaches on a more fundamental level. 
To be precise,
Is there a formula to express the Chern numbers/Pontryagin numbers out of the Hodge numbers on a compact Kaehler manifold $M$ of complex dimension $n$? Surely it is the case for $c_n[M]$ interpreted as the Euler characteristic number. 
Or, does anyone know such counterexamples that two Kaehler manifolds(notably, Kaehler surfaces, I guess)  have the same Hodge numbers but different Chern numbers?
Many thanks!

REPLY [5 votes]: "counterexamples that two Kaehler manifolds have the same Hodge numbers but different Chern numbers?"
As you explained above, Chern numbers of surfaces can be expressed in terms of the Euler number and the first Pontrjagin number, so you need dimension at least 3 for a counterexample.
In dimension 3, consider a projective space and a smooth quadric threefold. These two have the same Hodge numbers, same $c_3 = 4$ (Euler number), same $c_1 c_2 = 24$ (by Todd's theorem), but distinct degrees $c_1^3$: for the projective space it equals $64$ and for quadric it is $54$.<|endoftext|>
TITLE: Geometric interpretation of Hida isomorphism
QUESTION [16 upvotes]: [EDIT]: After getting a very nice answer by Kevin Buzzard I realize that my question was a little bit too vague and I try to restate it more precisely.
As the title says, I would like to understand an isomorphism of Hida from a more geometric perspective than what I normally read. What bothers me is that there are two construction of the universal (ordinary) Hida-Hecke algebra and they turn out to give isomorphic objects: fix a prime $p\geq 5$, and a tame level $N$ prime to $p$.

Take the projective limit over the level $r$ of the Hecke algebra acting on $S_k(\Gamma_1(Np^r),\mathbb{Z}_p)$ where $k$ is any weight. By applying the usual idempotent, one gets the Hida-Hecke ordinary algebra $h_k^0(Np^\infty;\mathbb{Z}_p)$, where I adopt notations as in Hida's paper in Inventiones, 1986, "Galois representations into $\mathrm{GL}(2,\mathbb{Z}_p[[X]])$...".  
Consider now the injective limit over the weight of the spaces of cusp forms $S_k(Np;\mathbb{Z}_p)$. By taking a suitable completion of this injective limit, one sees that the projective limit (over the weight, now) of Hecke algebras acts on the above completion. Applying again the idempotent, we get the Hida-Hecke ordinary algebra $h^0(N,\mathbb{Z}_p)$.

Theorem 1.1 in the quoted paper by Hida shows that these two algebras are isomorphic (in the most compatible way one can dream of, in particular inducing the same Hecke action on spaces of cusp forms) but his proof is entirely algebraic.
My question is: is there a reasonable way to prove this isomorphism geometrically?As Kevin Buzzard suggested, several papers of Katz (and successive work by Coleman-Mazur, Buzzard himself et al.) discuss geometric interpretation of $p$-adic modular forms and $p$-adic families of modular forms. Still, I do not understand how Hida's isomorphism comparing the Hecke algebra as acting on the projective limit over the level (so ''at the top of the modular tower'') or on the inductive limit over the weight (so, ''over the first curve $X_1(Np)$'') can be given a geometric interpretation.

REPLY [6 votes]: I am not sure what your criteria would be for a proof to be given a geometric interpretation, but the reason why weights "disappear" when we take the inverse limit on the level stems from the contraction property of Hecke operators (at $p$), or informally from the fact that Hecke operators at $p$ diminish the level.
As you know, the proof of the isomorphism between the two different Hecke algebras requires the definition of a map between Hecke algebras acting on forms of weight 2 and forms of weight $k$. Because theses two algebras are sub-algebras of endomorphisms generated by the same abstract elements but acting on different objects, this amounts to constructing a map between the cohomology of (one of the level of) the modular tower with coefficients in the constant sheaf and (one of the level of) the modular tower with coefficients in a sheaf of weight $k$ (or the same thing with the modular tower replaced by the Igusa tower, as in Kevin's answer). This last map is really no big deal: if memory serves, on the sheaves it is just projection on the last component. The remarkable fact is that the map on cohomology then is surjective with a finite kernel (and is an isomorphism in the ordinary case); the proof of this assertion being exactly the contraction property. Note that the proof necessarily requires the choice of a level at some point; how else would you even state the result?
Note for instance that for a tower of more general Shimura varieties, it is not at all obvious how the contraction property will play out: group-theoretic properties of $\operatorname{GL}_{2}(\mathbb Q_{p})$ really do play an important role in the proof. See the reference below though, for an answer to these questions.
So in the end, the isomorphism between the two Hecke algebras seems to me to come from the interplay between the cohomology of modular varieties and group-theoretic properties of the Hecke algebra. A very general formulation of this fact can be found in D.Mauger Algèbres de Hecke quasi-ordinaires universelles.  Ann. Sci. École Norm. Sup. (4) 37 (2004) (section 2.4 to be precise)<|endoftext|>
TITLE: Solubility of the quintic?
QUESTION [26 upvotes]: Over the p-adics, every Galois group is solvable.  Does this imply that the quintic (and higher-order polynomials for that matter) can be solved by radicals over $\mathbb{Q}_p$?
EDIT: The original place I learned that the p-adic galois groups were solvable was in Milne's Algebraic Number Theory text (Chapter 7, Cor 7.59).  
As was pointed out the comments, I should clarify that I meant to ask 2 questions.  Namely, whether the general quintic can be solved by radicals in this context (still no) and whether any given one can be (which I now believe is yes).

REPLY [4 votes]: Try Lazard (Daniel),
Solving quintics by radicals, in The legacy of Niels Henrik Abel, 207–225, Springer, Berlin, 2004.
MR2077574 (2005g:12002) says : 
Let $F$ be a field of characteristic different from 2 and 5. Let $f$ be a univariate irreducible polynomial of degree 5 over $F$. The polynomial $f$ is said to be solvable by radicals if the Galois group over $F$ of the field generated by all the roots of $f$ is solvable. In the paper the author gives a formula for solving by radicals any polynomial $f$ of degree 5 which is solvable by radicals. The field extension which is generated by the radicals which appear in the result is always minimal, when only one root is produced, as well as when all roots are given. This formula has been implemented in Maple.
Reviewed by Jerzy Urbanowicz.<|endoftext|>
TITLE: Finding topological obstructions for a complex manifold to be Kaehler
QUESTION [12 upvotes]: Well, it is of the "straightforward" questions one may ask. I propose it here to see if someone could tell me more on the recent status of this quite long-standing problem.
To initiate, let me give a brief description of the "classical" invariants. Essentially they are provided by either the symplectic structure or the Hodge structure, most of which relating to vanishing theorems and integral theorems. As far as a compact manifold $M$ of complex dimension $n$ is concerned, we have the following:
$b_2 \ne 0$ for the symplectic structure and many more;
$ b_{2k+1}$ are even and $b_{k-2} \leq b_k$ for $k \leq n$ for the Hodge structure, which date back to Lefschetz;
Various integral results on Chern numbers by Hirzebruch-Riemann-Roch on Kaehler manifolds;
and etc.
To make the question precise, I would like to ask:
Has there been any "higher" or essentially new invariants discovered so far? Particularly, one may observe that the above invariants are all torsion-free(as Yau did in his problem list), so the torsion invariants would be rather interesting, suppose they do exist. 
Any comments are welcomed and thanks a lot!

REPLY [6 votes]: You might want to consult the really excellent book by Amoros,Burger, Corlete, Kotschick, Toledo called "Fundamental groups of compact Kahler manifolds".<|endoftext|>
TITLE: Choice of base point in a Waldhausen category
QUESTION [7 upvotes]: Waldhausen's definition of a category with cofibrations includes the choice of a distinguished zero object. Probably this also means that an exact functor should preserve zero objects on the nose (Waldhausen writes "takes $\*$ to $\*$"). Isn't this rather evil from the general perspective of category theory? Even if we had fixed zero objects, we should only require that an exact functor preserves them up to isomorphism (which is unique anyway).
Remark that Weibel's K-Book doesn't demand this choice; there it is only required that some zero object exists. Isn't this more natural? On the other hand, the theories look quite the same.
So a question might be why Waldhausen has chosen his definition and if one definition of the two definitions has any real advantage in practice against the other.
This question has a topological analogue. If $(X,x)$ and $(Y,y)$ are pointed spaces, then why not defining a map $(X,x) \to (Y,y)$ to be a map $f : X \to Y$ together with a path $f(x) \to y$ (oplax) or $y \to f(x)$ (lax)? This has the advantage that we are more flexible as concerning the base points, we still have a well-defined induced map $\pi_n(X,x) \to \pi_n(Y,y)$ on homotopy groups, but on the other hand this will be more complicated than the usual notion of a pointed map. Any references are welcome.

REPLY [9 votes]: When Waldhausen defines $K$-theory, he considers sequences of the form
$* \rightarrow A_1 \rightarrow A_2 \rightarrow \cdots \rightarrow A_n$
for varying $n$, with all maps cofibrations. For any fixed $n$, these sequences together with the obvious morphisms again form a category, and these categories nearly assemble to a simplicial object in categories. Degeneracy maps are defined by inserting identities and the face maps except $d_0$ are defined by composition. $d_0$ of the above sequence is obtained by ignoring the $*$ and dividing out $A_1$ from the sequence to obtain
$A_1/A_1 \rightarrow A_2/A_1 \rightarrow \cdots $
However, this does not really make sense since choices of quotients are involved; in particular a choice of zero object $A_1/A_1$. You may make arbitrary choices, but then you only obtain a simplicial object up to equivalence: For example, $d_0 s_0$ is not the identity functor, but merely isomorphic to it. 
This problem is solved by considering a different category: We plug more information into our objects by declaring an object to be a sequence
$* \rightarrow A_1 \rightarrow A_2 \rightarrow \cdots \rightarrow A_n$
together with choices of quotients $A_i/A_j$, $i \geq j$. This yields an equivalent category, but now we have a shot to get an honest simplicial object in categories.
But then a problem in the degeneracy maps arises: In a degenerate sequence of the form
$* \rightarrow \cdots A_k = A_k \rightarrow \cdots$
we now also have to specify the quotient $A_k/A_k$, and again making arbitrary choices will not cut it. Letting this quotient be the chosen zero object works when one also restricts to sequences starting with this chosen zero object. I think this is the main technical reason to pick a zero object once and for all.
Such evil concepts are actually quite common in $K$-theory: Often, it is not enough to have diagrams of categories commuting up to natural isomorphism, so one has to wriggle the definitions of the involved categories a bit to obtain different, equivalent categories with a strictly commuting diagram of functors.<|endoftext|>
TITLE: Applications of Descent?
QUESTION [5 upvotes]: The technique of faithfully flat descent, and, in the case of vector spaces, Galois descent has been used quite a bit in Algebraic Geometry. However, the question of whether, say, a given $k$-vector space $V$ arises from some $L$-vector space $W$ seems like it could be asked in a wide variety of settings. I'm wondering, in particular, if anyone has seen descent in modular representation theory.

REPLY [5 votes]: Descent does appear in modular representation theory (of finite groups) in a slightly surprising form. Indeed, descent techniques allow us to discuss extension from $Stab(kH)$ to $Stab(kG)$, where $H$ is a subgroup of $G$, of index prime to the characteristic of the field $k$, and where $Stab(-)$ denotes the stable category of modules modulo projectives. 
To understand how an extension problem turns out to be a descent problem, one should start by the following "monadicity" observation: There exists a commutative separable ring object (i.e. monoid) $A$ in the symmetric monoidal category $C:=Stab(kG)$ such that the category $A-Mod_C$ of $A$-modules in that category $C$ is equivalent to the stable category $Stab(kH)$, and this in such a way that restriction from $G$ to $H$ coincides with extension-of-scalars with respect to $A$.
So restriction to a subgroup is in fact a extension-of-scalars!
The condition that the index of $H$ in $G$ is prime to the characteristic exactly expresses the faithfulness of the ring object $A$. (Flatness is sort of built-in when you deal with tensor triangulated categories.) Hence extension from $H$ to $G$ is really faithful(ly-flat) descent!
One can then reason rather geometrically, pretty much as with (finite) etale extensions. There is even a Grothendieck topology and a stack hiding in the bushes... This is all explained in the preprint
"Stacks of group representations"
available on my publication page.<|endoftext|>
TITLE: Lie algebras and non-smoothness of centralisers in bad characteristic
QUESTION [6 upvotes]: Let $G$ be a simple algebraic group over an algebraically closed
field $k$ of characteristic $p>0$. For $x\in G$, let $C_{G}(x)$
denote the centraliser, considered as a group scheme over $k$. If
$p$ is a very good prime for $G$ it is known that $C_{G}(x)$ is
smooth (over $ $$k$). Assume that $p$ is not very good for $G$.

Is there a general description of the elements $x\in G$ for which
  the centraliser $C_{G}(x)$ is not smooth? 

It does not seem too difficult to do this via a direct case by case check, so I'm mainly asking whether there is a uniform argument and whether there are any relevant references.
Let $\mathfrak{c}_{\mathfrak{g}}(x)$ denote the centraliser in the
Lie algebra $\mathfrak{g}$ of $G$. We have $\dim C_{G}(x)=\dim\mathfrak{c}_{\mathfrak{g}}(x)$ if $C_{G}(x)$ is smooth. 

In the cases where $C_{G}(x)$ is not smooth, are the values $\dim\mathfrak{c}_{\mathfrak{g}}(x)-\dim C_{G}(x)$
  known?

The second question is related to Problem 12 of Steinberg's 1966 ICM
contribution, which asks for a proof of the statement that $\mathfrak{c}_{\mathfrak{g}}(x)$ 
is always the sum of the Lie algebra of $C_{G}(x)$ and the centre of
$\mathfrak{g}$. Steinberg's problem is mentioned in Humphreys's book
Conjugacy classes in semisimple algebraic groups (1.10) but there is
no indication as to whether Steinberg's problem has been solved or
not.

REPLY [5 votes]: For all groups of exceptional types the answer can be deduced from the paper "Jordan block sizes of unipotent elements in exceptional algebraic groups" by Ross Lawther, published in
Comm. Algebra, 23, Issue 11, 1995, 4125-4156. In order to determine the Jordan block sizes of ${\rm Ad}\ u$, where $u\in G$ is unipotent, Lawther used comuter-aided computations. The case of a general $x\in G$ reduces quickly to the case where $x$ is unipotent by using the Jordan-Chevalley decomposition in $G$.
Of course, the number of Jordan blocks is just $\dim\ \mathfrak{c}_{\mathfrak g}(u)$, or rather $\dim\ \mathfrak{g}^{{\rm Ad}\ u}$,  which is one of the numbers you are interested in.
This number could then be compared with $\dim\ C_G(u)$ which is much easier to find in the literature (Spaltenstein's book on unipotent classes might have an answer). 
Regarding Steinberg's Problem 12, when $p=2$ the Lie lagebra of type $E_8$ is simple, hence has zero centre, but there are about 20 instances when the value of $\dim\ \mathfrak{g}^{{\rm Ad}\ u}$ jumps and becomes bigger than that for the counterpart of $u$ in good characteistic (these cases are starred in Lawther's paper). Since the number of "new" unipotent classes in characteristic $2$ is far less than 20, I think, there will be instances when  $\dim\ \mathfrak{g}^{{\rm Ad}\ u}>\dim\ C_G(u)$. The orbit labelled $E_7$ could do the trick, but I didn't check.
There is also a version of Lawther's results for nilpotent elements in exceptional Lie lagebras in the literature; see "Varieties of nilpotent elements
for simple Lie algebras II: Bad primes" by University of Georgia VIGRE Algebra Group, published in J. Algebra, 292, 2005, 65–99. This paper also relies on extensive computer-aided computations.
As far as I'm aware there is no computer-free proof of the above results at the present time, hence there is no uniform proof either. One could even say, on a more philosophical note, that nothing is uniform when $p$ is bad.<|endoftext|>
TITLE: Why the underlying function of a monomorphism may not be an injection
QUESTION [12 upvotes]: In category theory, a monomorphism (also called a monic morphism or a mono) is defined to be a left-cancellative morphism. It seems that this definition generalizes the definition of injections. However, even in a concrete category, a monomorphism may not be an injection. Why could this happen? I know examples of monomorphisms which are not injections but what is the reason behind the existence of such examples? Isn't monomorphism a generalised concept of injection? Similar questions can be asked about epimorphisms and surjections.

REPLY [5 votes]: I would like to add some remarks (although there is already an accepted answer).
A morphism $f : X \to Y$ is a monomorphism precisely iff $f_\*(T) : \mathrm{Hom}(T,X) \to \mathrm{Hom}(T,Y)$ is injective for all $T \in C$, iff $f_\* : \mathrm{Hom}(-,X) \to \mathrm{Hom}(-,Y)$ is a monomorphism in the category of presheaves $\widehat{C}$. This leads to the following intuition: The notion of monomorphism is only well-behaved if there are enough presheaves.
If $U : C \to D$ is any functor, of course it may fail to preserve or reflect monomorphisms. However, if $U$ has a left adjoint, then it is easy to verify that $U$ preserves monomorphisms. Moreover, when $U$ is faithful, then it is easy to check that $U$ reflects monomorphisms. Thus, if $(C,U)$ is a concrete category over a category $X$ (a category $C$ equipped with a faithful functor $U : C \to X$; see Section 5 in The Joy of Cats), such that $U$ has a left adjoint, then a morphism $f$ in $C$ is a monomorphism iff $U(f)$ is monomorphism in $X$. This applies in particular to the well-behaved case that $C$ is algebraic over $X$.

REPLY [2 votes]: Note that monomorphism is not the only way to generalise an injective map.  There are stronger notions, like a section.  A morphism $f:A\rightarrow B$ is a section if $\exists g:B\rightarrow A$ such that $f\circ g = 1_B$, where $1_B:B\rightarrow B$ is the identity morphism.  Moreover, there also are different kinds of monomorphisms (regular monomorphisms, extremal monomorphisms), most of which are generally stronger notions than a monomorphism, but weaker than a section.  The situation with epimorphisms is dual.<|endoftext|>
TITLE: Detecting tilings by toric geometry
QUESTION [10 upvotes]: This is probably a silly question, but I figured that if there is a good answer, this would be a good place to ask.
Ever since I got my hands on the book "Toric Varieties" by Cox, Little and Schenck, I've been excited to learn about the different combinatorial properties of polytopes that one can deduce from the corresponding toric varieties. In fact, toric varieties can prove combinatorial theorems not only about polytopes but also about many other objects living in $\mathbb Z^n$. One such thing would be tilings of $\mathbb R^d$ by integral polytopes.
I believe the following comes as a natural question:

Can one tell if a convex polytope $P$ tiles Euclidean space by looking at its corresponding projective toric variety? Can one deduce properties of the tiling this way?

In case there is no simple answer, do tilings by polytopes correspond to algebraic gadgets in the same spirit that polytopes are in bijection with projective toric varieties with a specified ample line bundle?

REPLY [11 votes]: A related question (but not exactly the one you asked) is: 

Can one tell if a convex polytope $P$ and its translations by $\mathbb Z^n$ tile $\mathbb R^n$? Which polytopes $P$ have this property?

Fix some positive quadratic form $q$ on $\mathbb R^n$ and the corresponding distance function. Let $P^0$ be the set of points in $\mathbb R^n$ which are closer to $0$ than to any other integral (i.e. in $\mathbb Z^n$) point. The closure $P$ of $P^0$ is called the Voronoi polytope w.r.t. $q$. Then $P$ obviously has the above property.
Voronoi's conjectured circa 1907 that the opposite is true, i.e. any such $P$ is a Voronoi polytope w.r.t. some $q$. 
This conjecture is known for $n\le 4$ due to Delaunay and for zonotopes by Erdahl "Zonotopes, Dicings, and Voronoi Conjecture on Parallelohedra". It is still open in general, I believe.
So what is special about the toric variety $X_P$ corresponding to $P$? I am not sure. If you look at the Delaunay tiling which is dual to the Voronoi tiling $P+\mathbb Z^n$, then the polytopes in that tiling and the corresponding toric varieties have a clear geometric meaning: they describe degenerations of principally polarized abelian varieties. But this is a dual picture.
Note by the way that Delaunay polytopes have vertices in $\mathbb Z^n$, so they indeed correspond to projective polarized toric varieties. In contrast, the Voronoi polytope for a generic $q$ will have irrational vertices. Also, when you vary $q$ continuously, the Voronoi polytope will vary continuously. But the Delaunay polytopes will jump discretely, and there are only finitely many Delaunay polytopes modulo $GL(n,\mathbb Z)$.
One place where the Voronoi tilings appear is tropical geometry. Indeed, a principally polarized tropical abelian variety $A$ is just the real torus $\mathbb R^n / \mathbb Z^n$ together with the positive definite form $q$. Then the $(n-1)$-skeleton of the Voronoi tiling modulo $\mathbb Z^n$ is the theta divisor on $A$. See Mikhalkin-Zharkov http://arxiv.org/abs/math/0612267 for more details.<|endoftext|>
TITLE: Concentration of Gaussian vectors
QUESTION [5 upvotes]: If $f: \mathbb{R}^n \to \mathbb{R}$ is a Lipschitz function and $X$ is a standard $n$-dimensional Gaussian vector with $\mathbb{E} f(X) = 0$, then $f(X)$ is subgaussian (in a way that does not depend on $n$). If $f$ is $\mathcal{C}^1$, this is equivalent to saying that $|\nabla f|$ bounded implies $f(X)$ is subgaussian.
There seem to be two natural generalizations of this. The first is to ask for weaker bounds on $|\nabla f|$. For example, if $|\nabla f|$ is subgaussian, then $f$ should be subexponential. The second generalization concerns functions $f: \mathbb{R}^n \to \mathbb{R}^k$. If I want to control $|f|$ independently of $k$, it is no longer enough to assume that $f$ is Lipschitz, since for the function $f(x) = (x_1, \dots, x_k)$, $|f|$ concentrates around $\sqrt k$. The natural condition seems to be a bound on the Frobenius norm of $D f$ (the matrix of partial derivatives).
The following statement contains both generalizations simultaneously (and is not hard to prove): If $f: \mathbb{R}^n \to \mathbb{R}^k$ is continuously differentiable and $\mathbb{E} f(X) = 0$ then
$$
\big(\mathbb{E} |f(X)|^p\big)^{1/p} \le c \sqrt p \big(\mathbb{E} \|Df\|_F^p\big)^{1/p}.
$$
My question is whether a statement like this is known and (if so) where I can find a reference.

REPLY [3 votes]: To answer my own question, this follows from a more general result that is mentioned in "On measure concentration of vector valued maps" by Ledoux and Oleszkiewicz, Theorem 4: for any convex function $\Psi: \mathbb{R}^k \to \mathbb{R}$,
$$
\mathbb{E} \Psi(f(X)) \le \mathbb{E} \Psi(\frac{\pi}{2} Y \cdot Df(X))
$$
where $X$ and $Y$ are independent standard Gaussians. If you condition the right hand side on $X$ and integrate $Y$, a standard result on the moments of order-2 Gaussian chaos gives
$$
\mathbb{E} (\frac{\pi}{2} Y \cdot Df(X))^p \le (cp)^{p/2} \mathbb{E} \|Df\|_F^p
$$
which is what I claimed above. (By following the references a little more carefully, you can even get the sharp constant.)<|endoftext|>
TITLE: Triply graded spectral sequence?
QUESTION [12 upvotes]: As we know, most of the spectral sequences are doubly graded. However, this "doubly graded" condition is not a part of the formal definition of spectral sequence. Is there any useful triply (quadruply, quintuply, etc.) graded spectral sequences? If not, is there a hope that some meaningful work can be done with this topic?

REPLY [3 votes]: J. L. Verdier dissertation (written in the 60s), pre-Ravenel's book,  covers  multiple-graded  complexes.  It was reprinted "recently"

Des catégories dérivées des catégories abéliennes. (French. French summary) [On derived categories of abelian categories] 
  With a preface by Luc Illusie. Edited and with a note by Georges Maltsiniotis. 
  Astérisque No. 239 (1996), xii+253 pp. (1997).<|endoftext|>
TITLE: Dual of the space of Hölder continuous functions?
QUESTION [23 upvotes]: Let $X=C^{\alpha}(\Omega,\mathbb{R})$ be the space of Hölder continuous functions. What is its dual?

REPLY [18 votes]: Your question can be interpreted in several ways, but I guess you are asking "is it a known space, or just some weird new Banach space?"
If $\Omega=R^n$ and $s$ is noninteger, the dual of $C^s$ is known in the above sense (almost). Indeed, one can identify $C^s$ with the Besov space $B^s_{\infty,\infty}$, and duals of Besov spaces are well studied. The precise result is: denote with $\dot C^s$ the closure of the Schwartz space of rapidly decreasing functions in $C^s$, then $(\dot C^s)'=B_{1,1}^{-s}$. I bet similar results should be true also on more general open sets $\Omega$ but I'm not sure. A good starting point are Triebel's books (Theory of Function Spaces I, II and III).<|endoftext|>
TITLE: On the multidimensional generalisation of Gamma function
QUESTION [5 upvotes]: Gamma function is defined as
$$
  \Gamma(z) = \int\limits_{0}^{+\infty} x^{z-1} e^{-x} \; dx
$$
I'm looking for multidimensional generalisation of this definition. I consider the class $Q$ of positive, concave and positively homogeneous of order one functions from $\mathbb{R}^n_+$ to $\mathbb{R}_+$. Examples of such functions are linear function $q(x) = \langle p, x \rangle$ for $p > 0$ and CES function
$$
   q(x) = \left( \alpha_1 x_1^{-\rho} + \ldots + \alpha_n x_n^{-\rho} \right) ^{-\frac{1}{\rho}}, \;\;\; \sum\limits_{i=1}^{n}\alpha_{i} = 1, \;\;\; \alpha_{i}, \rho > 0 \; (i = 1,\ldots, n)
$$
I try to define
$$
  \Gamma_{q}(z) = \int\limits_{\mathbb{R}^n_+} x^{z-1}e^{-q(x)} \; dx \equiv \int\limits_{\mathbb{R}^n_+} x_1^{z_1-1}\ldots x_n^{z_n-1} e^{-q(x_1,\ldots,x_n)} \; dx_1 \ldots dx_n 
$$
If $n = 1$ then $\Gamma_{q}(z) = \Gamma(z)$ for any $q \in Q$. 

The main question is if there is some literature on the similar generalisation?
If there is some other multidimensional generalisation of Gamma function? 
If there are some special types of $q(x)$ for which we can represent $\Gamma_{q}(z)$ in terms of well-known functions?

REPLY [2 votes]: I can suggest an answer to Question 2. E W Barnes studied multiple gamma functions early last century and there has been some sporadic recent interest in those functions. His generalization seems to be rather different from what you had in mind, but you should refer to his works, and that of recent authors, to be sure. These can be found on this Wikipedia page:
http://en.wikipedia.org/wiki/Multiple_gamma_function
The relevant references are:


Barnes, E. W. "The Theory of the Double Gamma Function", Philos. Trans. of the Royal Society of London. Series A, 196 (1901), 265–387.  
Barnes, E. W. "On the theory of the multiple gamma function", Trans. Cambridge Philos. Soc. 19 (1904), 374–425. 
Friedman, Eduardo; Ruijsenaars, Simon. "Shintani–Barnes zeta and gamma functions", Adv. in Math. 187 (2004), no. 2, 362–395. 
Ruijsenaars, S. N. M. "On Barnes' multiple zeta and gamma functions", Adv. in Math. 156 (2000), no. 1, 107–132.<|endoftext|>
TITLE: Generalization of finitely generated, finitely presented modules?
QUESTION [10 upvotes]: Let $R$ be a commutative ring and $M$ an $R$-module.
The module $M$ is finitely generated iff there is an exact sequence $R^{k_0} \to M \to 0$.
Similarly, $M$ is finitely presented iff there is an exact sequence $R^{k_1} \to R^{k_0} \to M \to 0$.
It seems we could generalize this as follows:
for $n \in \mathbb{Z}_{\ge 0}$ let us call $M$ a finitely $n$-presented module,
if there is an exact sequence $R^{k_n} \to \dotsm \to R^{k_0} \to M \to 0$.
So finitely generated = finitely $0$-presented, and finitely presented = finitely $1$-presented. You could also define finitely $\infty$-presented modules, which have a resolution with finite free modules. I would even try defining $M$ to be finitely $\omega$-presented, if it has a finite free resolution $0 \to R^{k_m} \to \dotsm \to R^{k_0} \to M \to 0$ (I think these have been studied a lot).
I haven't seen these notions defined before (except for the last one). Couldn't they be useful, or have they been used? I think both finitely generated and finitely presented modules are important, although if you are only interested in noetherian rings, there is no difference.
For example, there is a result saying that if $0 \to M' \to M \to M'' \to 0$ is exact, $M''$ is finitely presented and $M$ is finitely generated, then $M'$ is also finitely generated. You could generalize this: if $M$ is finitely $n$-presented and $M''$ is finitely $(n+1)$-presented, then $M'$ is finitely $n$-presented.
Also we could look at submodules: $M$ is noetherian iff every submodule of $M$ is a finitely generated. $M$ is coherent iff it is finitely generated, and every finitely generated submodule of $M$ is finitely presented. We could generalize this as follows: $M$ is $n$-coherent ($n \in \mathbb{Z}_{\ge 0}$) iff it is $(n-1)$-presented, and every finitely $(n-1)$-presented submodule of $M$ is finitely $n$-presented. So noetherian = $0$-coherent, and coherent = $1$-coherent. You could also define $R$ to be $n$-coherent iff it is an $n$-coherent $R$-module. The category of noetherian/coherent $R$-modules is abelian, so I guess the same should hold for the category of $n$-coherent $R$-modules.

REPLY [10 votes]: All these notions have been defined and studied long time ago. Serre called a module type $FL_n$ if it is finitely $n$-presented in your terminology. Type $FL_\infty$ and type $FL$ is used for finitely $\infty$-presented and finitely $\omega$-presented. They are studied a lot for group rings ($R$ does not need to be commutative) and show up in the definition of $G$-theory for general rings. Modules of type $FP_\infty$ are sometimes also called pseudo-coherent, a name/definition that goes back to SGA 6, I.2.9, see for Example 7.1.4 in Chuck Weibel's book on Algebraic K-theory.
A good starting point might be K.S. Brown's book "Cohomology of groups", Chapter VIII is about finiteness conditions.

REPLY [6 votes]: There is a big literature about this at least for modules over group rings $R=\mathbb{Z}[G]$.  The standard terminology is to say that $M$ has type $FP_n$ if it is finitely $n$-presented according to your definition.  (Or maybe the indexing is shifted by one, I don't remember.)  Peter Kropholler is one of the main authors so if you look for his papers you will get lots of references.<|endoftext|>
TITLE: Rock-paper-scissors...
QUESTION [23 upvotes]: A directed graph whose underlying undirected graph is complete is called a tournament. Let us call a (finite) directed graph balanced if every vertex has as many incoming as outgoing edges. The question is: Have balanced tournaments been classified? (The weakest form of "classify" might be: given $n$, determine the number of balanced tournaments on $n$ vertices up to isomorphism.)
Here are some elementary observations:

for even $n$ there is no balanced tournament on $n$ vertices.
for odd $n$ there is a standard balanced tournament on $n$ vertices: take as vertex set $\{0,\ldots,n-1\}$ and include an arrow from $i$ to $i+j \mod n$ for $1 \le j \le (n-1)/2$. (The automorphism group is cyclic of order $n$.)
for $n=1$, $n=3$, and $n=5$ the only balanced tournament is the standard one.
for $n=7$ there is a non-standard balanced tournament: to construct it invert an appropriate $3$-cycle in the standard b.t., to see that it is non-standard look at the "out-link" of an appropriate vertex. (The automorphism group is trivial.)
One can take a sort of wreath product to construct examples whose automorphism groups are abelian, non-cyclic. There are other constructions to produce examples.

The motivation for this question is simply the following: the b.t. on $3$ vertices encodes the game rock-paper-scissors. The one on $5$ vertices encodes the game rock-paper-scissors-lizard-Spock (if you don't know it, you can figure it out once you know that "lizard poisons Spock" [edit, thanks to Ramiro de la Vega:] and that "paper disproves Spock"). From $7$ on there is some choice, how much and which?
Also: does someone know the proper expression for "balanced"? [edit: according to David Speyer the term in this context is "regular tournament"]

REPLY [24 votes]: Your sequence is Sloane A096368. Sloane links to this page, which has files of all examples up to $13$ vertices. MathSciNet has 30 papers with "regular tournament" in the title, none of which seem to know much about enumeration up to isomorphism. A quick scan of the papers with "balanced tournament" in the title suggests that that term means something else, so I would search on "regular tournament".
McKay proves an asymptotic formula for the number of LABELED regular tournaments of the form $2^{\binom{n}{2}} e^{-O(n \log n)}$. (See his paper for a much more precise statement.) Since the size of $n!$ is "only" $e^{O(n \log n)}$, we can deduce that the number of isomorphism classes is also $2^{\binom{n}{2}} e^{-O(n \log n)}$, and in particular goes to $\infty$ as $n \to \infty$.

I can say a bit more about the labeled problem, where we don't quotient by automorphism. This is Sloane A007079.
The number we want is the coefficient of $\prod_{i=1}^{n} x_i^{(n-1)/2}$ in $\prod_{1 \leq i < j \leq n} (x_i+x_j)$; each factor corresponds to an edge and choosing $x_i$ or $x_j$ corresponds to orienting this edge. This polynomial is the Schur polynomial $s_{(n-1)(n-2)\cdots 321}(x_1, \ldots, x_n)$; this follows from the ratio of alternants formula. So you are looking for the Kotska number
$$K_{(n-1)(n-2)\cdots 321,\ mmm \cdots m} \ \mbox{where} \ m=(n-1)/2.$$
I don't think there is a closed formula for this Kotska number.

Here's something interesting that I couldn't get to work, but maybe will help someone else. Essentially by definition, this Kostka number is the dimension of the space of diagonal invariants in the representation of $SL_n$ associated to the partition $(n-1, n-2, \cdots, 2,1)$. And this vector space comes with a natural $S_n$-action, from the embedding of $S_n$ into $SL_n$. 
Unfortunately, they don't match! When $n=3$, there are two labeled tournaments. (Orient the triangle clockwise or counterclockwise.) So a permutation $\sigma$ in $S_3$ either preserves or switches the tournaments based on the sign of $\sigma$. The corresponding permutation representation of $S_3$ is the direct sum of the trivial and the sign rep. By way of contrast, the representation of $S_3$ on the diagonal invariants of $V_{21}$ is the $2$-dimensional irrep of $S_3$. 
Still, it would be really cool if we could find a deeper relation between these two representations of $S_n$ than simply the fact that they have the same dimension. In particular, remember that the number of orbits in a permutation representation is always the multiplicity of the trivial rep, so one could imagine counting isomorphism classes by representation theory if we can find a good alternate description of the tournament representation.<|endoftext|>
TITLE: Amenable groups of deficiency $1$
QUESTION [20 upvotes]: Let $G=\langle X;R\rangle$ be a finitely presented group. The rank of $G$ is defined to be the size of smallest generating set of $G$. The deficiency ${\rm def}(G)$ of $G$ is defined to be the maximum of $|X| - |R|$ over all finite presentations $G = \langle X;R \rangle$.
The deficiency of an amenable group can be at most $1$. One (maybe the only?) way to see this is to note that there is a Morse inequality for $\ell^2$-homology
$$1-{\rm def}(G) \geq b_0^{(2)}(G) - b_1^{(2)}(G) + b_2^{(2)}(G),$$
where $b_i^{(2)}(G)$ denotes the $i$-th $\ell^2$-Betti number of $G$. Cheeger and Gromov showed that an amenable group satisfies $b_i^{(2)}(G)=0$ for $i \geq 1$. This implies in particular that ${\rm def}(G) \leq 1$ for $G$ amenable.
Now, apart from $\mathbb Z$ and Baumslag-Solitar groups $BS(1,n) = \langle a,b; a^nba^{-1}b^{-1} \rangle$, I do not know of any amenable groups which realize ${\rm def}(G) =1$. In particular, I do not know any examples of rank $\geq 3$.

Question: Does every amenable group of deficiency $1$ have rank $\leq 2$.

It can be shown that any amenable group with ${\rm def}(G)=1$ must have cohomological dimension $\leq 2$, which puts severe restrictions on $G$. What else is known about amenable groups with deficiency $1$?

REPLY [7 votes]: Inspired by John Wilson's result which was mentioned by Mark in his answer (which he has deleted by now), I can answer my own question now for elementary amenable groups.
Let us first argue that any amenable $G$ with ${\rm def}(G)=1$ has cohomological dimension $\leq 2$. This follows from an argument that I learned from [1]. Consider the presentation $2$-complex $X$ (with one 0-cell) of a presentation which realizes the deficiency. The cellular chain complex of the universal covering looks like
$$0 \to {\mathbb Z}G^n \stackrel{d}\to {\mathbb Z}G^{n+1} \to {\mathbb Z}G \to 0.$$
The kernel of $d$ is $\pi_2(X)$ and the only thing to show is that $\pi_2(X)=0$. Then $X$ is aspherical and ${\rm cd}(G)\leq 2$. Now, the $\ell^2$-homology of $X$ is computed by 
$$0 \to ({\ell^2}G)^n \stackrel{d}\to ({\ell^2}G)^{n+1} \to {\ell^2}G \to 0.$$
Now, the zeroeth and first $\ell^2$-Betti number of $X$ agrees with the $\ell^2$-Betti number of $G$ and has to vanish by Cheeger-Gromov since $G$ is amenable. This implies that $d$ must be injective on $(\ell^2 G)^n$. Hence, it is injective on $(\mathbb Z G)^n$. We conclude that $\pi_2(X)=0$.
Let us come to the main part of the argument. Jonathan Hillman has defined the Hirsch length $h(G)$ for any elementary amenable group $G$ (Theorem 1 in [3]) and shown that it is bounded above by the cohomological dimension of $G$, see Lemma 2 in [3]. Now, Theorem 2 of [3] implies $G/T$ is solvable where $T$ is the maximal locally finite normal subgroup of $G$. However, since the cohomological dimension is finite, $T$ is trivial and we conclude that $G$ itself is solvable. Now, Theorem 5 from [2] says that every solvable group of cohomological dimension $\leq 2$ must be solvable Baumslag-Solitar, $\mathbb Z^2$ or $\mathbb Z$.
[1] Jon Berrick and Jonathan Hillman "The Whitehead conjecture and $L^2$-Betti numbers", Guido's Book of Conjectures, Monographies de L'Enseignement Mathématique, L'Enseignement Mathé matique, (2008), 35–37. 
[2] Dion Gildenhuys, "Classification of soluble groups of cohomological dimension two" Math. Z. 166, 1  (1979), 21-25.
[3] Jonathan Hillman, "Elementary amenable groups and 4-manifolds with Euler characteristic 0" J. Austral. Math. Soc. (Series A) 50 (1991), 160-170.<|endoftext|>
TITLE: Condition to ensure that the product of closed maps be closed
QUESTION [5 upvotes]: If $f_i : X_i \to Y_i$ with $i=1,2,\ldots,n$ are closed maps between topological space it is known that their product map 
$$f : X_1 \times \cdots \times X_n \to Y_1 \times \cdots \times Y_n : (x_1, \ldots, x_n) \mapsto (f_1(x_1), \ldots, f_n(x_n))$$
doesn't need to be closed. However the question is: are there some nice conditions on $X_i$, $Y_i$ (like compactness, connection, Hausdorff...) such that $f$ will be closed? I have looked in "Bourbaki - General Topology Part 1,2" and I have found nothing about it. By the way I'm more interested to the case $X_1 = \cdots = X_n$, $Y_1 = \cdots = Y_n$. 
Thank you for your help!

REPLY [2 votes]: If each $X_i$ is compact and each $Y_i$ is Hausdorff then $X:=X_1 \times \cdots \times X_n$ is compact and $Y:=Y_1 \times \cdots \times Y_n$ is Hausdorff. So any map $f:X \to Y$ is closed, whether it is a product of (closed) maps or not.<|endoftext|>
TITLE: Existence of "good" concretizations
QUESTION [13 upvotes]: In this question it was asked what relation, if any, exists between the notions of a monomorphism in a general category $\mathcal C$ and an injective map.  Of course the question only makes sense if $\mathcal C$ is a concretizable category.  If $\pi:\mathcal C\to Set$ is a faithful functor, we may regard a morphism $f:X\to Y$ of objects of $\mathcal C$ as an injection (relative to $\pi$) if $\pi(f)$ is injective in $Set$.
One then notices that this notion depends very much on the choice of concretization, and thus that this notion is not an intrinsic property of the category itself.
However, Emil Jerabek noticed that it is natural to ask the following:  given a concretizable category, does it admit a concretization which preserves monomorphisms?
Apologies if this question is trivial, hopelessly too general, or of no interest to category theorists.

REPLY [9 votes]: An easy sufficient condition for monomorphism-preserving concretizability is the existence of a generating set (as opposed to a proper class).  A generating set is a set $\mathcal S$ of objects such that, for any distinct parallel morphisms $f,g:A\to B$, there is a morphism $h:S\to A$ with $S\in\mathcal S$ and $fh\neq gh$.  This says that the coproduct of the representable functors Hom$(S,-)$, over all $S\in\mathcal S$, is a faithful set-valued functor, i.e., a concretization.  As a coproduct of representable functors, it preserves monomorphisms.
I would not expect this sufficient condition to be necessary.  So I don't claim that this is  a complete answer --- it just takes care of an easy situation, which happens to cover a lot of categories.<|endoftext|>
TITLE: Extremals versus minima for variational problems
QUESTION [8 upvotes]: A geodesic on a Riemannian manifold is generally not the shortest among nearby curves with the same endoints.  But it can always be divided into parts each the shortest among nearby curves between its endpoints.  
How much does that generalize?  Is there some natural description of Lagrangians with the following property:  for each solution $f$ to the Euler-Lagrange equation on a given domain, that domain can be covered by subdomains such that each restriction of $f$ to one of those subdomains actually minimizes (or maximizes) the Lagrangian integral on that subdomain?

REPLY [6 votes]: You are asking a very classical question, which is the question of sufficient conditions for a minimum in the calculus of variations.  Quite a lot is known about conditions on Lagrangians that ensure minimality for local solutions.  
For example, it is known that the Lagrangian for $k$-dimensional area in a Riemannian manifold has this property for all $k$.  (The case $k=1$ is the case of geodesics.)  The technique that works in this case is the technique of calibrations, developed to a very high degree in the fundamental paper Calibrated Geometries by Harvey and Lawson.
For the more general case, you should take a look at Volume II of Giaquinta and Hildebrandt's Calculus of Variations.  There you will find methods due to Weyl, de Donder, Caratheodory, etc.<|endoftext|>
TITLE: De Finetti's theorem, the pointwise ergodic theorem, and reverse martingales
QUESTION [10 upvotes]: De Finetti's theorem says that an exchangeable sequence of random variables $X_i$ is a mixture of i.i.d. random variables.  In other words, if $\mu$ is a measure on $\mathbb{R}^\infty$ that is invariant under exchanging finitely many coordinates (a symmetric measure), then there is some probability measure $\eta$ on probability measures such that $\mu = \int \nu^\infty \, d \eta(\nu)$.
Further, I know the following.

The product measures of the form $\nu^{\infty}$ are the extreme points for the convex set of symmetric measures.  They are also ergodic with respect to the group of transformations which exchange finitely many coordinates.  So $\mu = \int \nu^\infty \, d \eta(\nu)$ is an ergodic decomposition.
For $\mu$-a.e. $x=\{x_i\}_{i\in\mathbb{N}}\in \mathbb{R}^\infty$, there is some probability measure $\nu_x$ on $\mathbb{R}$ such that for all measurable sets open balls $A \subseteq \mathbb{R}$, 
(A) $\quad$ ${\displaystyle \lim_{k\rightarrow\infty} \frac{1}{k} \sum_{i<k} \mathbf{1}_A(x_i) = \nu_x(A) }$.
Moreover, if $P^n_k$ is the set of all injective functions $\pi \colon [n] \rightarrow [k]$, then for all bounded continuous functions $f\colon \mathbb{R}^n \rightarrow \mathbb{R}$,
(B) $\quad$ ${\displaystyle  \lim_{k\rightarrow\infty} \frac{1}{|P^n_k|} \sum_{\pi \in P^n_k} f(x_{\pi(0)},\ldots ,x_{\pi(n-1)}) = \int_{\mathbb{R}^n} f\, d \nu_x^n}$.
Equations (A) and (B) and de Finetti's theorem can all be proved using reverse martingales.  Indeed, $M_{-k}(x) = \frac{1}{|P^n_k|} \sum_{\pi \in P^n_k} f(x_{\pi(0)},\ldots ,x_{\pi(n-1)})$ is a reverse martingale.

My questions are as follows.


To what extent are equations (A) and (B)
  instances of some variant of the pointwise
  ergodic theorem?  (I guess (A) is
  just Birkoff's pointwise ergodic
  theorem with the shift map---although I am not sure why the shift map comes in.  But (B) is not so clear to me.)
When may an ergodic average be represented as a reverse martingale?
Similarly, for which types of pointwise ergodic theorems and ergodic decompositions is there a proof using reverse martingales?


Pointers to any relevant references would be helpful.

REPLY [3 votes]: (My understanding of this material has significantly gone up in the months since I asked it, and I will attempt to answer my own question.)
In general, if $(\Omega,\mathcal{B},\mathbb{P},\{T_g\})$ is a measure preserving system where $\{T_g\}$ is a commuting (semi-)group action, then one can consider the limits of averages over any chain of finite subsets $H_1 \subseteq H_2 \subseteq … \subseteq G$, that is averages of the form
$$A_n (f) = \frac{1}{|H_n|} \sum_{g \in H_n} f \circ T_g.$$
In this particular case of de Finetti's theorem, we can take $G$ to be finite permutations of $\mathbb{N}$, and $T_g$ to be the transformations on $\Omega = \mathbb{R}^\infty$ associated with that permutation (by permuting the coordinates of $x=(x_n) \in R^\infty$.  Then $H_n$ is the set of permutations which only permute the first $n$ coordinates.
Notice, in this particular case, that each $H_n$ is a subgroup and therefore
$$\frac{1}{|H_n|} \sum_{g \in H_n} f \circ T_g = \mathbb{E}[f \mid \mathcal{I}nv(H_n) ]$$ where $\mathcal{I}nv(H_n)$ is the sigma-algebra of sets which are  $T_g$-invariant for all $g \in H_n$.
Now one has a reverse martingale $\mathbb{E}[f \mid \mathcal{I}nv(H_n) ]$, which converges to $\mathbb{E}[f \mid \mathcal{I}nv(G) ]$, since $\mathcal{I}nv(G) = \bigcap_n \mathcal{I}nv(H_n)$. Hence, the ergodic decomposition $\mu_x$ comes from the conditional probability $\mu_x = \mathbb{P}[ \cdot \mid \mathcal{I}nv(G)] (x)$.  
In the de Finetti case, since each $\mu_x$ must be invariant under the permutations of coordinates $\{T_g\}$, it is a product measure of the form $\mu_x = \nu^\infty_x$ for some $\nu_x$. 

To answer the first question is seems that yes, equations (A) and (B) are examples of a pointwise ergodic theorem.
To answer the second question, it seems that such ergodic averages can be represented as reverse martingales when the $H_n$ are finite subgroups of $G$.  (This leads to an observation that even when $H_n$ are not finite, if they are subgroups, we could define an ergodic average of $f$ over $H_n$ as $\mathbb{E}[ f \mid \mathcal{I}nv(H_n)]$.
As for the third question, many pointwise ergodic theorems can be proved using reverse martingale techniques.  The idea is to first work on the group $G$ itself rather than a probability space.  In this setting, the ergodic averages become
$$A_n (f) (x) = \frac{1}{|H_n|} \sum_{g \in H_n} f(gx)$$
for $x \in G$ and $f \in \ell_1 (G)$.  This is reminiscent of the averages in the Lebesgue differentiation theorem---think of $\{gx: g \in H_n\}$ as a ball around $x$. 
These averages can be approximated by reverse martingales on $\ell_1 (G)$: rather than average over a ball around $x$, partition $G$ and average over the part that $x$ is in (if the $H_n$ are subgroups then the parts can be the cosets and the two types of averages are the same!). 
Then on can use the geometry of $G$ to turn a theorem about reverse martingales (for example a maximal theorem) into a theorem about ergodic averages. (This is similar to how one can prove the Lebesgue differentiation theorem using (forward) martingales, with some additional lemmas to handle the geometry.) Finally, one can use the Calderon transfer principle to transfer the result on $\ell_1(G)$ to a result on any measure preserving system with a $G$-action.<|endoftext|>
TITLE: Checking if one polytope is contained in another
QUESTION [6 upvotes]: I have two sets of inequalities, say, $Ax \leq 0$ and $Bx \leq 0$. I would like to know if they both define the same polytope. Or, even, whether one is contained in the other.
At the moment I am checking, for each row $a \in A$, whether there exists some $x$ satisfying $ax > 0$ but $Bx \leq 0$. Is there a more efficient approach?
Thanks

REPLY [10 votes]: I would like to draw your attention to the 2002 survey by Volker Kaibel and Marc Pfetsch,
"Some Algorithmic Problems in Polytope Theory," arXiv:math/0202204v1, which contains this on p.6:
 


As you probably know, an $\cal{H}$-description is by halfspaces, whereas a $\cal{V}$-description is by vertices.
Reference [20] is: R. M. Freund and J. B. Orlin, "On the complexity of four polyhedral set
containment problems," Math. Program., 33 (1985), pp. 139–145.
Reference [17] is B. C. Eaves and R. M. Freund, "Optimal scaling of balls and polyhedra," Math. Program., 23 (1982), pp. 138–147.<|endoftext|>
TITLE: Is Stone-Čech compactification of 0-dimensional space also 0-dimensional?
QUESTION [14 upvotes]: What is an example of a 0-dimensional locally compact Hausdorff space $X$ for which the Stone-Čech compactification $\beta(X)$ is not 0-dimensional?
It is known that if $X$ is a 0-dimensional locally compact Hausdorff space which is also paracompact, then $\beta(X)$ is 0-dimensional. (Engelking 1989, Th. 6.2.9). I would expect a counterexample in the non-paracompact case.
Another way of asking the question is to look at a Boolean ring L (without assuming a unit).  If X is the Stone space of L then X is a 0-dimensional locally compact Hausdorff space and L is isomorphic with the ring of compact open sets in X.  The Boolean algebra $\mathrm{Cl}(X)$ of closed-open subsets is clearly a Boolean algebra extension of L, and the Stone space of $\mathrm{Cl}(X)$ is a certain compactification of $X$.  I believe it is easy to see that this compactification of $X$ is just $\beta(X)$ if and only if $\beta(X)$ is 0-dimensional.
Thus the question can be expressed in the algebraic side of the duality.  Find conditions on the Boolean ring $L$ so that $\beta(X)$ is 0-dimensional, where $X$ is the Stone space of $L$.
I think this may be true if, for example, $L$ is a Boolean $\sigma$-ring.  My original question asks for an example showing this is not true without some additional conditions on $L$.

REPLY [3 votes]: Thanks to KP.  This is the class $\Psi$ of Gillman-Jerison 5I, but apparently Terasawa explained more some 16 years later. 
There is a little history here. As pointed out by Figa-Talamanca and Franklin, "Multipliers of Distributive Lattices", Indian J. Math. 12 (1970) p. 159, John Kelley asserted in his topology book (p. 169) that the Cech-Stone compactification is always zero-dimensional.  They attempt to correct this error by citing an unspecified example in the 1955 paper of Dowker, "Local Dimension of Normal Spaces", Quart. J. Math. 2 (1955) 101-120, and they say this example was pointed out to them by J. Isbell and P. Dwinger.  However, I have not yet been able to verify that any of the non-strongly zero-dimensional examples in Dowker's paper are locally compact.  I wonder which one they had in mind?  If they were right, this would be a much earlier example than Terasawa's $\Psi$.  Could all four of them been mis-reading Dowker?
Here is a link to my copy of the Dowker paper. Does it contain a counterexample?<|endoftext|>
TITLE: What are the generalizations of the 27 lines on a cubic surface?
QUESTION [20 upvotes]: The following doubtlessly naive heuristic suggests to me that there might be some generalizations.  I don't know whether, at one extreme, the story is classical, or at the other extreme, the heuristic just fails.
Consider a generic hypersurface $S$ of degree $d$ in ${\Bbb P}^n$.
The intersection of $S$ with a generic plane $P$ should form a curve of degree $d$, hence a curve of genus $g=(d-1)(d-2)/2$.
The possible planes $P$ range over a Grassmannian ${\rm Gr}(n+1,3)$ of dimension $3(n-2)$.
One thus gets a morphism from ${\rm Gr}(n+1,3)$ to the moduli space of curves of genus $g$, which has dimension $3(d-1)(d-2)/2 - 3$ (unless $d=3$).
If the numbers work out right, one can try to make $n$ large enough, but not too large, so that one gets a 0-dimensional set of planes where the intersection gives rise to curves with some desired amount of degeneration.  With enough degeneration perhaps, the original curves of genus $g$ will acquire components of smaller genus.  So one might get interesting configurations of comparatively low genus curves (not necessarily all of the same genus).
For example, with $S$ of degree $4$ one might look for a configuration of genus 1 and 2 curves.  (Personally, I don't know enough about compactifying moduli spaces even to guess the details at this point.)
In any case, with the 27-lines on the cubic surface, elliptic curves would seem to degenerate into a finite set of lines sharing common points, making this classic object an example of the heuristic above.
All that said, I'll make my question the broad one in the title.

REPLY [2 votes]: I agree with @Scarnahan that you do not make clear what you want to generalize. Perhaps I am too naive but e.g. you can also make the question not from the point of view of the surface with the lines but rather you could make the question: Is there an analogue of $P^2$  namely the well known construction of $ r \le 9 $ points in the projective plane with a special geometric configuration then by blowing up the  plane in these points with the given configuration you(e.g. Manin's book on Cubic forms, Elsevier. ) blow them up and obtain the cubic surface and the exceptional divisors turn out to be the lines.
So you could make  the question if there is an analogon say for $P^3$ and instead of points take codimension one hypersurfaces  with special configuration of lines, etc; You can play the same game for $P^n$ but now you have  $r $ hyperplanes with a special  geometrical configuration and the blow up with center at these hyperplanes? Depending on r you have to study the linear system defined by the blown up $P^n$. This should be found in the classical literature as e.g. Baker's principles of geometry. It is also well known that  there are no surfaces of degree $n$ that are normal in $P^n$ for $ n \ge 9$  and for $ n \leq  9  $  the surfaces with such property arise as projections from a point ending with the cubic surface with the 27 lines. (See e.g. Semple and Roth's Introduction to algebraic geometry, Oxford University Press, spec. Chap. 7 ).<|endoftext|>
TITLE: Are algebraic groups defined by their invariants in tensor spaces?
QUESTION [9 upvotes]: Let $K$ be a field of characteristic zero, and let $G \subseteq \mathrm{GL}_V$ be an algebraic group over $K$, acting faithfully on a finite dimensional vector space $V$. Let $H \subseteq \mathrm{GL}_V$ be the largest algebraic subgroup with the following propertes:
(1) If a subspace $V_1 \subseteq V$ is invariant under $G$, then it is also invariant under $H$.
(2) Given $G$-invariant subspaces $V_1$ and $V_2$ of $V$, and integers $a,b\geq 0$, the equality
$$\mathrm{Hom}_G(V_1^{\otimes a}, (V/V_2)^{\otimes b}) = \mathrm{Hom}_H(V_1^{\otimes a}, (V/V_2)^{\otimes b})$$
holds.
The second condition means that $G$ and $H$ have the same fixed points in any tensor space that can be formed out of subquotients of $V$. The inclusion $G\subseteq H$ is tautological, and my question is:

do we have $G=H$?

If $G$ is reductive, then the answer iy yes, because in that case $V$ and all its tensor powers are semisimple, but the equality $G=H$ also holds for example if $G$ is the group of upper triangular matrices.

REPLY [6 votes]: Yes, see Prop. 3.1 and Remark 3.2 here.<|endoftext|>
TITLE: Estimating a partial sum of weighted binomial coefficients
QUESTION [9 upvotes]: There is a well-known estimate for the sum of all binomial coefficients $\binom{n}{k}$ satisfying $k \leq \alpha n$ for some $\alpha$ satisfying $0 < \alpha \leq 1/2$:
$$ \sum_{k=0}^{\alpha n}\binom{n}{k} = 2^{(H(\alpha) + o(1))n}$$
where $H(\alpha) = -\alpha\log_2(\alpha) - (1-\alpha)\log_2(1-\alpha)$ is the binary entropy function. 
My question is whether there exists a similar estimate when we weight the $k$-th binomial coefficient by $\lambda^k$ for some $\lambda > 0$. That is, I would like to estimate the following sum:
$$ \sum_{k=0}^{\alpha n} \binom{n}{k} \lambda^k $$

REPLY [8 votes]: Expanding on the previous answers. I'm taking $\lambda$ and $\alpha$ to be constants which do not vary as $n\to\infty$. 
If $α<λ/(λ+1)$ then the sum is within a constant of the last term. In fact the largest terms are approximately in geometric progression so you can get it quite accurately by computing the ratio.
If $α>λ/(λ+1)$, almost all of the complete binomial expansion is present, so the sum equals $(1+o(1))(1+λ)^n$.
If $α≈p$, then Russell's normal approximation will be good. (This needs some work to clarify whether the geometric approximation of the lower tail is good right up to the point where the normal approximation begins to be good. I think it is.)<|endoftext|>
TITLE: James-Stein phenomenon: What does it mean that a James-Stein estimator beats least squares estimator?
QUESTION [19 upvotes]: Background James-Stein estimator  and Stein's phenomenon, as described in Wikipedia are rather counterintuitive and amazing. 
It is claimed that if one wants to estimate the mean $\Theta$ of
Gaussian distributed vector $ y$~$ N(\Theta, \sigma^2 Id)$, then the naive estimation
 - (i.e. just take $y$ as an estimation) is not good for size of vector greater or equal 3.
"Not good" means that (quote Wikipedia) "James–Stein estimator always achieves lower (Mean squared error (MSE) than the least squares estimator".
Question please clarify the sentence above.
I wonder the following - usually when we calculate MSE we need some distribution
on the estimated  parameter $\Theta$ and averaging in MSE is taken over this distribution also.
So what distribution is assumed ? Or may be for ANY distribution it holds true ?
PS 
Stein's example is more general (quote Wikipedia):
Stein's example (or phenomenon or paradox), in decision theory and estimation theory, is the phenomenon that when three or more parameters are estimated simultaneously, there exist combined estimators more accurate on average (that is, having lower expected mean-squared error) than any method that handles the parameters separately. This is surprising since the parameters and the measurements might be totally unrelated.
A Paradox?
Popular articles have appeared hailing the James-Stein estimator a paradox; one should use the price of tea in China to obtain a better estimate of the chance of rain in Melbourne!
(Quote from Deane Yang's suggested page http://jmanton.wordpress.com/2010/06/05/comments-on-james-stein-estimation-theory/ )

REPLY [8 votes]: There is an excellent issue of Statistical Science that address the James-Stein phenomena from various aspects.
https://www.jstor.org/stable/i23208816
Question What does it mean that a James-Stein estimator beats least squares estimator?

It means that the JS estimator has a smaller risk than LSE w.r.t. a
  prescribed risk function $R(\delta)=E_{\theta}L(\delta,\theta)$; which
  is equivalent to say that if we choose $L^2$ loss function, then when
  the dimension is higher than 3, the LSE for the mean is no longer
  admissible. The only admissible estimator should be JS estimator.

The original JS estimator assumes multi-dimensional Gaussian distribution. But later with consideration of admissibility, there are various cases where shirkage estimator(JS estimator) actually beats the LSE(also MLE) and other frequentist estimators. And "beating" also depends on the $L^2$ loss function you choose, and hence the risk function $R$ you choose for this decision problem. To illustrate this point, we consider Gaussian case with $L^2$ loss function in the explanation below.
There are two approaches that seem very intuitive to me.
(1)Linkage between JS estimator and the Diffusion process.
Brown [1] pointed out that under the framework of decision theory, it could be shown that the admissibility and the recurrence of Brownian motion is equivalent. Moreover, Brown shown in his major theorem that via a variational minimization problem, the admissibility of an estimator can be discussed using recurrence of corresponding Brownian motion. Such a variational approach can be extended to various other situations.

Theorem 5.1 in [1], modified. A necessary condition for $\delta$ to be
  admissible with given risk $R$ is that there exists a non-negative measure $F$ s.t. its
  density $f^{*}<\infty$ and $\delta(x)=\delta_{F}(x)$(generalized Bayes
  rule w.r.t. F) for almost all $x\in E^{m}$ w.r.t. Lebesgue measure.
  Furthermore
(A) If $\{Z_{t}\}$ is transient then $\delta$ is inadmissible.
(B) If $\{Z_{t}\}$ is recurrent and the risk set w.r.t. $R$ is bounded uniformly
  then $\delta$ is inadmissible.
where $\{Z_{t}\}$ is the diffusion process in $E^{m}$ along with its
  infinitesimal generator with local mean
  $\nabla(logf^{*})=\delta_{F}(x)-x$ and a local variance covariance
  matrix $2I$.

Therefore since 1d and 2d Brownian motions are recurrent, so is the mean estimator; but when it goes to 3d, Brownian motions are transient and hence mean estimators are no longer admissible and beaten by JS estimator.
This is probably one of the most celebrated results derived from Bayesian statistics and it integrated the stochastic process so seamlessly that I believe it even lead to later MCMC simulation developments. 
(2)Geometric interpretation of JS estimator.(taught by Prof.M.P. :)
Zhao and Brown [3] pointed out that if we restricted ourselves to spherically symmetric estimators, then the naive geometric optimal estimator derived from simple Euclidean geometry $$\delta_{NGO}(Z)=\left[1-\frac{p-1}{\left\Vert Z\right\Vert ^{2}}\right]Z$$ is providing exact amount of shrinkage as JS estimator did. This approach is intuitive in sense that spherical symmetric estimators of the form $\delta(X)=\tau(\|X\|)X$ with a scalar function $\tau$, the key of using this class of estimators is that (i) By admissibility consideration, estimators that are not spherically symmetric are inadmissible (ii) By structure of the spherical estimator, it provides a natural geometry that is isometric to Euclidean geometry. By (i)(ii), the problem of choosing an admissible estimator becomes a "compass-and-ruler" problem of geometry. Amazing!
Reference
[1]Brown, Lawrence D. "Admissible estimators, recurrent diffusions, and insoluble boundary value problems." The Annals of Mathematical Statistics 42.3 (1971): 855-903. http://projecteuclid.org/euclid.aoms/1177693318
[2]https://stats.stackexchange.com/questions/13494/intuition-behind-why-steins-paradox-only-applies-in-dimensions-ge-3
[3]Brown, Lawrence D., and Linda H. Zhao. "A Geometrical Explanation of Stein Shrinkage." Statistical Science (2012): 24-30.<|endoftext|>
TITLE: Optimal wireframe sphere
QUESTION [14 upvotes]: Suppose you have a length $L$ of metal pipe at your disposal,
and you would like to build a wireframe unit-radius sphere,
by bending, cutting, and welding the pipe into a connected structure $F$.
Your goal is to minimize the height difference between
where the center of the unit sphere would be ($1$) and where the center
of your structure $F$ is, in any position resting on flat ground.
To be more precise, let $S$ be a unit radius sphere with center $o$,
and $F$ a connected structure,
built from pieces that are topological arcs, inscribed in contained within $S$.
Let $H$ be the convex hull of $F$, required to enclose $o$, and
define $\delta$ to be $1$ minus
the minimum distance from $o$ to any point on $H$.
So $\delta$ is the difference between the spheres centered on $o$
circumscribed about and inscribed in $H$.

Q1.
For a given $L$, what is $\delta_{\min}(L)$, the minimum
value of $\delta$ for any connected structure $F$?
Equivalently, given $\delta$, what is $L_{\min}(\delta)$,
the minimum length of all pieces in $F$ together that achieve $\delta$?

For example, given $L=4 \pi$, one might construct two orthogonal hoops:

 
 
 
 
 


This achieves $\delta = 1 - \sqrt{2}/2 = 0.29$.
But surely this is not optimal.  For example, one could remove
some pipe near the poles and add it to the equator to lower $\delta$.
This suggests this question:

Q2.
Is an optimal frame structure $F$ always composed of straight segments?
I.e., does it ever help to bend the pipe?

I think not.
The question can be generalized to any dimension.
Even in $\mathbb{R}^2$ it seems not uninteresting:

 
 
 
 
 



Q3.  What are the optimal structures $F$ inscribed
in a unit circle that minimize $\delta$ for a given $L$?
[Added:]
Could it be that a regular $n$-gon minus one edge is optimal for that $L$,
as suggested by the hexagon example above?

This two-dimensional version especially feels like it should have been investigated
previously, but I am not finding any literature on it.
Ideas and/or pointers welcomed—Thanks!
Addendum.
Thanks to Gerhard Paseman and Aaron Meyerowitz, it is now clear that for the 2D
question Q3, the Steiner tree spanning the vertices of a regular $n$-gon
is a strong candidate for optimality, and certainly improves upon the $n$-gon minus
an edge (except for $n=6$, as per Aaron's remark).

REPLY [3 votes]: For Question 3, in the case that $L=3\sqrt{2} \approx 4.24$ one can do better than 3 sides of a square. The two diagonals have length 4 with the extra left for improving thigs a bit. Probably a Steiner tree would be even more efficient (I think that your 5/6 of a hexagon example with all the 120 degree angles lucks into being a Steiner tree itself.)
LATER
Here is a structure F with $L=5$ and $\delta=0.08897$ I think this construction gives the minimum $\delta$ at least for certain $L.$ It certainly is an improvement over 5/6 of a hexagon and works for any $L.$

The illustration shows the unit circle and a concentric circle whose radius happens to be $r=0.9110275245$ along with three tangents. The straight segments run from $[2r^2-1,\pm2r\sqrt{1-r^2}]$ to $[r,\pm\sqrt{1-r^2}]$.
For $L \le \frac{\pi+2}{\sqrt{2}} =3.635655$ We achieve $r=\frac{L}{\pi+2} $ using the (left) half of a circle of radius $r$ along with two horizontal segments. The Steiner tree competition would presumably achieve $r=\frac{L}{6}$ using three segments of length$\frac{L}3$ to get convex hull an equilateral triangle.<|endoftext|>
TITLE: Correlation-Function for Random Graph Ising Model
QUESTION [9 upvotes]: For non-Ising'ers: Given a graph, we study the probability-distribution on the set of colorings ("Spin-up" and "-down") generated by a given correlation ("force to equality") between adjacient nodes (Bolzmann-Term depending exponentially on a "Temperature Parameter"). The limit of high Temperature is no-correlation whatsoever, the limit of low Temerature is all colors coincide for sure (are "parallel").
In this paper (Dommers, Giardinà, van der Hofstad, 2010) the partition function of the Ising model on a sparse random graph was calculated (sparse = almost surely no triangle, quadrangle etc )...well, I must add "calculated" refers to a random-variable fixpoint equation - I've put some effort into deriving an explicit expression thereof, but did not suceed ... a different story and maybe later a different question ;-)
The proof techniques very basically relay on the transfer matrix method for a local tree inside the random graph (which is defined by a given branching distribution, e.g. Poisson); this ultimately leads to a recursion formula for the up/down-random-variable of a single site and the authors proof the existence of a unique solution (the mentionen fix-point). Partition function etc. are derived as various expectation vals.
For use in an own work I would like to get ahold of the correlators, i.e. the probability of two sites of given distance $a$ to coincide. Though I do not expect this to be easier than explicitely calculate the partition function, a similar fix-point equation as in the original work would totally suffice! I tried the trick one uses in 1D, but I didn't get through...although I would naively expect, that it's easier than 2D, as you have better control over the walks...?
Moreover, it would be interesting to have multi-correlators in the sense of knowing the probability of given up-down configurations e.g. in an arbitrary $abc$-triangle depending on the edge-length's $a,b,c$. $\;\;$ I don't even recall such from classical (2D) Ising model....any sources or ideas? Is that in "physically-sound" language the n-point-functions in this case?

REPLY [4 votes]: There is a method to study these physical systems called belief propagation (BP), which yields exact methods for interaction graphs that are trees, and empirically works pretty fine when you have physical systems with loopy interaction graphs that locally-look-like trees. Randomly-generated sparse interaction graphs are of this form (its explicitly mentioned in the paper), and you could apply Belief Propagation to your model to obtain good estimates of the following quantities.

Marginal probabilities $\sigma_i$.
Joint probabilities $p(\sigma_i, \sigma_j)$, in particular, the correlators you mention.
Conditional probabilities $p(\sigma_i | \sigma_j)$
Partition functions, energies, free energies, entropies.

And, in addition, BP can be adapted to sample from the probability distribution $p(\sigma)$. In physics, systems like these are sometimes called 'diluted spin-glasses'.
I summarise the mean features of Belief propagation below. Mézard, Montanari's book contains a very pedagogical exposition of the method, also rather up-to-date and complete; you can read the draft of the book online without buying it. I personally used this book to prepare a seminar and I can recommend it. The connections between Belief Propagation and the ferromagnetic spin Isin chain are explained in Part D, section 14.1 of the online version.
Overview
Belief propagation is an iterative "message-passing" algorithm. For each site/spin/particle (vertexes of the graph) of your system, the belief propagation methods assigns incoming and leaving "messages" to its neighbouring edges (which represent interactions) in the interaction graph. These messages can be understood as "probability weights" that contribute to the marginal probability that the particle say on site $i$ is on a given local configuration $\sigma_i$, which I denote by $p(\sigma_i)$. The messages are updated through local computations (thus, not costly) for each particle, following a set of equations known as the belief-propagation update-rules. I will not write the update-rules here, for they are lengthy.
What it is important is the following: if on site $i$, with neighbours indexed by $a$, we denote the leaving/incoming messages on time $t$ by $p_{i\rightarrow a}^{t}(\sigma_i)$ and $p_{a\rightarrow i}^{t}(\sigma_i)$, then BP method gives a way to estimate the marginal probability $p(\sigma_i)$ as follows:
$$p(\sigma_i)\simeq p^{t}(\sigma_i)=\prod_{a\in\partial_i}p_{a\rightarrow i}^{t-1}(\sigma_i)$$
For trees, this iteration converges to fix point in finite time $t_{*}$, and this bound has a known value. Moreover, the method provably computes the marginal probabilities exactly. In fact, for trees this method and transfer-matrix methods used e.g. for the classical Ising spin chain turn out to be equivalent.
If your system it is not a tree, you do not get exact results and the message-update iteration is not promised to converge. However, the method has been used with wide experimental success on graphs that locally-look-like-trees (this seems to set an upper bound to the correlation-distance that you can have) to provide good estimations of these marginal probabilities.
Finally, the method can be adapted to estimate all the other quantities I mentioned.
Drawbacks
The main disadvantage of this method is that it is rather complicated to find analytical bounds for errors or analytical conditions that guarantee convergence; both are fields of current research. For a recent study regarding convergence you can check [1]; regarding errors, I found this ones [2].
And, of course, another disadvantage is that if you add a lot of random interactions to your graph, at some point your interaction-graph stops being tree-like and the method would behave poorly. In fact, there is a threshold were a gigantic connected component appears in the interaction graph [3].<|endoftext|>
TITLE: Triangulations of polytopes and tilings of zonotopes
QUESTION [9 upvotes]: Consider a set $A = \{ a_1,a_2,\ldots, a_n \} $ of vectors in $\mathbb{R}^d$, which lie in a common affine hyperplane. Two convex polytopes may be obtained from $A$, namely the convex hull of the vectors in $A$, $conv(A)$ and the zonotope generated by vectors in $A$, $Z(A)$.
Both polytopes can be viewed as the projection of a polytope $P$ which sends the unit vectors in $\mathbb{R}^n$ to the columns of $A$, where $P$ is the standard $n-1$-simplex in the first case and the unit $n$-cube in the second. The general study of affine projections of convex polytopes is developed in this paper of Billera and Sturmfels entitled "Fiber Polytopes" 
The first question may be common knowledge, but is there some relation between the triangulations of $conv(A)$ and cubical tilings of $Z(A)$ which arise from this projection? I haven't seen this explicitly written down, but in the light of fiber polytopes both should be related concepts.

Question: Is there a reason why the triangulations of $conv(A)$ should be considered in some sense "more natural" than the cubical tilings of $Z(A)$ (both subdivisions seen as arising from the projection)?.

The reason I ask this is because I have seen many instances in the literature of toric ideals where triangulations of a convex polytope are used to characterize some algebraic construction but, as far as I remember, none performing similar characterizations in terms of tilings of zonotopes. As an example, a theorem of Sturmfels here characterizes the radicals of the initial ideals of a toric variety (associated to a matrix $A\in \mathbb{Z}^{d\times n}$) as the radical of the Stanley-Reisner ideals of regular triangulations of the convex hull of the columns of $A$.
However, when it comes to the combinatorial information of a vector configuration, the zonotope associated to it seems to relate more directly to the oriented matroid of the vector configuration. Recall, for instance the Bohne-Dress theorem relating the set of zonotopal tilings of the zonotope generated by the vector configuration and the one-element liftings of its oriented matroid.
I would be very satisfied with answers which, say,

give pointers to characterization of algebraic objects (e.g. Groebner bases, minimal free resolutions) from the theory of toric ideals in terms of some associated zonotopes.
give pointers to a general relation between triangulations of $conv(A)$ and cubical tilings of $Z(A)$,
indicate why, and in which context, one of both concepts should be more natural than the other.

p.s. may be someone reputed enough would like to create the tag oriented-matroids?

REPLY [4 votes]: Triangulations of polytopes are "more fundamental" than cubical tilings of zonotopes. By the Cayley trick, every cubical tiling of a zonotope can be seen as a triangulation of the Cayley lifting of the segments defining it. The latter is equal to their Lawrence lifting. 
This can be seen as an addition to the Bohne-Dress theorem, relating tilings of a zonotope to liftings of the associated oriented matroid.<|endoftext|>
TITLE: Does there exist any "quantum Lie algebra" embeded into the quantum enveloping algebra U_q(g)?
QUESTION [11 upvotes]: We have known that any finite dim Lie algebra can be embeded into it's enveloping algebra $U(\mathfrak{g})$, my question is: is there any "quantum Lie algebra" embeded into the quantum enveloping algebra $U_q(\mathfrak{g})$?
The related question is, take $sl(2)$ generated by $\{X,Y,H|[XY]=H, [HX]=2X, [HY]=-2Y\}$ for example, consider the representation on polynomial $K[x,y]$, $K[x,y]$ is in fact a module-algebra over $U(sl(2))$, the  elment of $sl(2)$ can be represented by $X=x\frac{\partial}{\partial y}, Y=y\frac{\partial}{\partial x}, H=x\frac{\partial_q}{\partial x}-y\frac{\partial_q}{\partial y}$ (see  Kassel "Quantum groups" (GTM155), pp. 109). In fact, $\{x\frac{\partial}{\partial y}, y\frac{\partial}{\partial x}, x\frac{\partial_q}{\partial x}-y\frac{\partial_q}{\partial y}\}$ generated a three dim Lie subalgbebra (isomorphic to $sl(2)$ under the above correspondence) of derivation algebra of $K[x,y]$.
Similariy, Is there quantum Lie algebra contained in $U_q(sl(2))$? In fact, by Kassel "Quantum groups" (GTM155), pp. 146-149, there is an action of $U_q(sl(2))$ on quantum plane $K_q[x,y],  E=x\frac{\partial_q}{\partial y}, E=y\frac{\partial_q}{\partial x}, K=\sigma_x\sigma_y^{-1}, K^{-1}=\sigma_y\sigma_x^{-1}$ , so is there any finite dim quantum Lie algebra generated by $E,F,K,K^{-1}$, or does the operators  $x\frac{\partial_q}{\partial y}, y\frac{\partial_q}{\partial x}, \sigma_x, \sigma_y^{-1}, \sigma_y, \sigma_x^{-1}$  generate a Lie subalgebra of  of derivation algebra of $K_q[x,y]$?

REPLY [4 votes]: I hope no one gets offended if I summarize a couple of comments adding few details to it. Truly $U_h(\mathfrak g)$ as an associative algebra is not different from $U(\mathfrak g)[[h]]$ (here $\mathfrak g$ is semisimple, everything is char=0). 
This results is just a rigidity result on the associative algebra $U(\mathfrak g)$ (which reflects rigidity of the Lie algebra $\mathfrak g$). What I wrote inside the bracket seems innocent but it is not: one may say it depends on the fact that the Hochschild cohomology of $U(\mathfrak g)$ is isomorphic to the Chevalley-Eilenberg cohomology of $\mathfrak g$. I find this is nicely explained in http://people.mpim-bonn.mpg.de/crossi/LectETHbook.pdf .
When I first saw this result (which is already in the by now classical Chari-Pressley's book) my first impression was "so what's all the fuzz about quantum groups?". The point is that: 

They are non trivial deformation of the universal enveloping algebra as a Hopf algebra.
The isomorphism as associative algebras is neither explicit not canonical. We know it exists from purely cohomological arguments...

How does this connect to the embedding $\mathfrak g\hookrightarrow U(\mathfrak g)$? The canonical way to reconstruct $\mathfrak g$ inside $U(\mathfrak g)$ is to identify it with the set of primitive elements (primitive means $\Delta X=X\otimes 1+1\otimes X$). Therefore this embedding depends on the whole Hopf algebra structure (coproduct to determine primitive elements and product to show that they form a Lie algebra and generate a PBW basis). 
The set of primitive elements in $U_h(\mathfrak g)$ is trivial and certainly does not allow to reconstruct a PBW basis.
One may look for twisted primitive elements with respect to some group-like element different from 1. This is an interesting object, it contains the analogue of simple root vectors, but its algebraic properties are rather weak. Still: starting from twisted primtive elements and performing $q$-commutators, in the context of global quantization $U_q(\mathfrak g)$ it is possible to reconstruct all "root vectors" giving a PBW basis. But there is no obvious algebraic structure even on this set of $q$-root vectors.
Of course one may try to understand some kind of embedding $\mathfrak g_h\hookrightarrow U_h(\mathfrak g)$ as a deformation of $\mathfrak g\hookrightarrow U(\mathfrak g)$ as was done in some of the mentioned reference but everything is non canonical and, in my opinion, in the long run it just turns out to be a way of building up an explicit algebra isomorphism; which is known to be technically very complicated.
(this comment does not touch on the "braided" side of the story; that, I do not understand)<|endoftext|>
TITLE: do you know this determinant (basic commutative algebra)?
QUESTION [6 upvotes]: Let $\ell_1,\dots,\ell_n$  be $d+1$-variate linear forms over complex numbers in variables $X=(X_0,\dots,X_d)$. Consider the $(n-d)$-fold products 
$$\ell_{i_1}(X)\ell_{i_2}(X)\dots\ell_{i_{n-d}}(X)=\sum_{|I|=n-d}a_{I,J}X^I,\ J=(i_1,\dots,i_{n-d}),\\ 1\leq i_1<i_2<\dots  < i_{n-d}\leq n.$$ 
Define $\binom{n}{d}\times\binom{n}{d}$-matrix $A$ with entries $A_{IJ}=a_{I,J}$. Then it appears that 
$$ \det A=C\prod_{K}L_K,\quad K=(k_1,\dots,k_{d+1}),\ 1\leq k_1< k_2<\dots<k_{d+1},$$
where each $L_K$ is a $(d+1)\times (d+1)$-minor of the $(d+1)\times n$-matrix $L$ of the coefficients of $\ell_1,\dots,\ell_n$, and $C\neq 0$.
For instance, let $n=4$, $d=2$, and 
$$L=\left(\begin{array}{rrrr}
a_{0} & a_{1} & a_{2} & a_{3} \\
b_{0} & b_{1} & b_{2} & b_{3} \\
c_{0} & c_{1} & c_{2} & c_{3}
\end{array}\right)$$
Then $$ A = {\scriptsize \left(\begin{array}{rrrrrr}
a_{0} a_{1} & a_{0} a_{2} & a_{0} a_{3} & a_{1} a_{2} & a_{1} a_{3} & a_{2} a_{3} \\
a_{1} b_{0} + a_{0} b_{1} & a_{2} b_{0} + a_{0} b_{2} & a_{3} b_{0} + a_{0} b_{3} & a_{2} b_{1} + a_{1} b_{2} & a_{3} b_{1} + a_{1} b_{3} & a_{3} b_{2} + a_{2} b_{3} \\
a_{1} c_{0} + a_{0} c_{1} & a_{2} c_{0} + a_{0} c_{2} & a_{3} c_{0} + a_{0} c_{3} & a_{2} c_{1} + a_{1} c_{2} & a_{3} c_{1} + a_{1} c_{3} & a_{3} c_{2} + a_{2} c_{3} \\
b_{0} b_{1} & b_{0} b_{2} & b_{0} b_{3} & b_{1} b_{2} & b_{1} b_{3} & b_{2} b_{3} \\
b_{1} c_{0} + b_{0} c_{1} & b_{2} c_{0} + b_{0} c_{2} & b_{3} c_{0} + b_{0} c_{3} & b_{2} c_{1} + b_{1} c_{2} & b_{3} c_{1} + b_{1} c_{3} & b_{3} c_{2} + b_{2} c_{3} \\
c_{0} c_{1} & c_{0} c_{2} & c_{0} c_{3} & c_{1} c_{2} & c_{1} c_{3} & c_{2} c_{3}
\end{array}\right)}$$
and $$ \det A=(-a_{3}b_{2}c_{1}+a_{2}b_{3}c_{1}+a_{3}b_{1}c_{2}-a_{1}b_{3}c_{2}-a_{2}b_{1}c_{3}+ a_{1}b_{2}c_{3}) \\ \times (a_{3} b_{2}c_{0} -  a_{2} b_{3} c_{0} -  a_{3} b_{0} c_{2} + a_{0} b_{3} c_{2} +
a_{2} b_{0} c_{3} -  a_{0} b_{2} c_{3})\\ \times (- a_{3} b_{1} c_{0}+a_{1} b_{3} c_{0} + a_{3} b_{0}c_{1}-a_{0} b_{3} c_{1}-a_{1}b_{0} c_{3} + a_{0} b_{1} c_{3})\\ \times (a_{2} b_{1} c_{0} -  a_{1} b_{2}c_{0} -  a_{2} b_{0} c_{1} + a_{0} b_{2} c_{1} + a_{1} b_{0} c_{2} - 
a_{0} b_{1} c_{2})\\ 
=L_{(234)}L_{(123)}L_{(124)}L_{(134)}. $$
This (and more - namely we would like to know how $A^{-1}$ looks like) must be well-known, but we cannot find relevant references. 
Update II. If one instead takes $n-d-1$-fold products of $\ell_i$, then one gets, in the same way, a  $\binom{n-1}{d}\times\binom{n}{d+1}$-matrix, with determinants of $\binom{n-1}{d}\times\binom{n-1}{d}$-minors factoring into products of $L_K$ as above. More precisely, if a $\binom{n-1}{d}\times\binom{n-1}{d}$-minor $M$ misses one $\ell_i$ then one arrives to the situation outlined above, which we know how to deal with, thanks to David's answer. Otherwise, we still can see that $\det M$ is divisible by $L_K$, where $K$ is one of $d+1$-subsets of $(1,\dots,n)$ distinct from the complement of $J=(i_1,\dots,i_{n-d-1})$ in $(1,\dots,n)$, where is $J$ corresponding to a column of $M$; there are $\binom{n-1}{d+1}=\binom{n}{d+1}-\binom{n-1}{d}$ such $K$. If $L_K$ vanishes then the forms $\ell_t$ comprising its columns have a common zero $z$, and as $K\cap J\neq\emptyset$, the vector $(z^I)$ is in the left kernel of $M$. Degree count now shows that $\det M$ factors into the product of $L_K$.
(Some $\det M$ vanish identically, and this apparently has to do with the homology of a simplicial complex related to the index sets $J$ of its columns).

REPLY [14 votes]: Here's a proof; I don't know any references. I will show that, if $L_K=0$, then $\det A=0$. Since the $L_K$ are distinct irreducible polynomials, this shows that $\prod L_K$ divides $\det A$, and the two sides have the same degree.
Suppose that $L_K=0$. Then the linear forms $\ell_{k_1}$, ..., $\ell_{k_{d+1}}$ have a common zero; call it $(z_0, z_1, \ldots, z_n)$. Every $I$ has nonempty intersection with $K$, since $|I|+|K|=(n-d)+(d+1) = n+1$. So every product $\ell_{i_1} \cdots \ell_{i_{n-d}}$ vanishes at $(z_0, z_1, \ldots, z_d)$.
That means that the vector of monomials $(z^J)$ in in the left kernel of $A$, so $\det A=0$.<|endoftext|>
TITLE: Alexandrov geometry techniques  for Finsler manifolds.
QUESTION [7 upvotes]: Hi, first I would like to apologize for my English. It's not my native language and I feel my grasp of it is limited.
I've been reading Burago's book on metric geometry  and  I've that it mentions Finsler manifolds as an example of Alexandrov space. However it restricts to reversible Finsler metrics $F(x, \lambda y) = |\lambda|F(x,y).$
In general Finsler metrics only hold this:
$$F(x, \lambda y) = \lambda F(x,y) \quad \lambda > 0. $$
The problem is that since the goal of book is to study a metric induced like this:
$$d_{F}(x,y) = \inf_{\gamma}\int_{a}^{b}F(\gamma,\dot{\gamma})dt.$$
The resulting distance in the general case is not symmetric.My question is:
Does anyone know any approach that might be pursued to use similar techniques as those presented in the book to study Finsler manifolds in the general case?

REPLY [4 votes]: There is a generalization of the Alexandrov curvature conditions in the dissertation `Construction of compatible Finsler structures on locally compact length spaces' of M.G. Knecht.  He shows that a length space with this curvature condition has a differentiable structure and a Finsler function that generates the original lengths. The differentiable structure and the Finsler function are not $C^\infty$ though.<|endoftext|>
TITLE: What is a good reference that compact resolvent implies Fredholm operator?
QUESTION [13 upvotes]: Suppose $A \in \mathcal{L}(E_1, E_0)$ is a bounded linear operator between Banach spaces $E_1$ and $E_0$, and we also have that $E_1$ is densely, continuously embedded in $E_0$ (i.e. $A$ can be regarded as a closed, unbounded operator on $E_0$). It is well known that, if the resolvent $R(\lambda, A) := (\lambda I - A)^{-1}$ is a compact operator in $\mathcal{L}(E_0)$ for $\lambda \in \rho(A)$ (which is equivalent to $R(\lambda_0, A)$ is compact for one particular $\lambda_0 \in \rho(A)$) then the operator $\lambda I - A$ is a Fredholm operator for every $\lambda \in \mathbb{C}$, however I cannot seem to find a reference which states this result. It seems that most references regarding unbounded operators with compact resolvents conclude their investigation with a proof that the spectrum is composed of isolated eigenvalues with finite multiplicity and regard Fredholm operators only long enough to discuss the essential spectrum of an operator. I have looked through texts by Dunford and Schwartz, Kato, Engel and Nagel, and Hormander (among others...), without finding the reference which I am hoping to find. 
$\bullet$ To be clear, I am looking for a reference which proves that if $R(\lambda, A)$ is compact for $\lambda \in \rho(A)$ then $\lambda I - A$ is Fredholm for $\lambda \in \mathbb{C}$.
A proof might go as follows: In the case that $\lambda \in \rho(A)$, the conditions of a Fredholm operator are trivial. Meanwhile when $\lambda \in \sigma(A)$ I can prove the result using a spectral projection $P_{\lambda}$ and the fact that $E_0$ decomposes into the direct sum of a finite dimensional space $P_{\lambda}E_0$ and a residual space $(1 - P_{\lambda})E_0$ on which $\lambda I - A$ is bijective. Although this proof is not too complicated, it seems unnecessary that I should have to include it, as the result should show up in previous references. This is my last ditch effort before I break down and either include the proof for myself or else pass it off as a "it is well-known" without reference, so any suggestions or opinions would be helpful.
Thank you.

REPLY [3 votes]: See Theorem 3.4.3, page 93, of these notes for a detailed  proof of the fact that $T: H\to H$ is Fredholm if and only if  there exists  $Q:H\to H$  such that $QT-1$ is compact.  If $Q=(T-\lambda)^{-1}$ is compact then
$$Q T= Q(T-\lambda)+\lambda Q=1+\lambda Q$$
so that
$$  QT-1=\lambda Q =\mbox{compact}. $$<|endoftext|>
TITLE: Explicit Bijection between Central Simple Algebras and twists of $\mathbb P^n$
QUESTION [13 upvotes]: The automorphism group of the algebra of $n$-dimensional matrices over a field $K$ is $PGL_n(K)$. The automorphism group of $n-1$-dimensional projective space over $K$ is also $PGL_n(K)$. Therefore, twists of them over $\bar{K}$ are both classified by the same Galois cohomology group, $H^1(PGL_n(K))$. 
Twists of $M_n(K)$ are central simple algebras of rank $n^2$. Twists of $K\mathbb P^{n-1}$ are projective varieties geometrically isomorphic to $\mathbb P^{n-1}$. 
There should be a bijection between these two sets.
How explicit and geometrically/algebraically nice can we make it? 
What properties of one object correspond to properties of the other?

REPLY [13 votes]: References:
Artin, M.
Brauer-Severi varieties. Brauer groups in ring theory and algebraic geometry (Wilrijk, 1981), pp. 194–210, 
Lecture Notes in Math., 917, Springer, Berlin-New York, 1982. 
Knus, Max-Albert; Merkurjev, Alexander; Rost, Markus; Tignol, Jean-Pierre
The Book of Involutions. (English summary) 
With a preface in French by J. Tits. American Mathematical Society Colloquium Publications, 44. American Mathematical Society, Providence, RI, 1998
Let $A$ be a central simple $K$-algebra of degree $n$ (i.e., of rank $n^2$). Let $Gr(n, A)$ be the subspace of the grassmannian of rank $n$ dimensional subspaces. Let $X \subset Gr(n, A)$ be the subvariety of subspaces which are also right ideals of $A$. Then $X$ is the corresponding space, called the Severi-Brauer, Brauer-Severi, or Chatelet variety of $A$, and is denoted here by $SB(A)$.
When $A = End_K(V)$ is trivial, every right ideal of $A$ of dimension n (as a $K$-vector space) is of the form $\operatorname{Hom}_K(V, U)$, where $U$ is a 1-dimensional vector space. So the space is $\mathbb{P}_K(V)$.
I am not sure of an explanation in the other direction as nice. It is the case that for a Severi-Brauer variety $SB(A)$, you can consider the following subbundle $\mathcal{I}$ of the constant bundle $SB(A) \times A$. The bundle $\mathcal{I}$ consists of pairs $(I, a)$, where $a \in I$. Then $\mathcal{I}$ is a locally free of rank $n$ and the global endomorphism ring $\operatorname{End}_{SB(A)}(\mathcal{I})$ is isomorphic to $A$ (or the opposite algebra of $A$, I can't remember which). 
However, this doesn't answer your question of how to start with a twist of $\mathbb{P}^{n-1}$ and nicely produce a central simple algebra of rank $n^2$. I think for that you may have to argue by galois descent.<|endoftext|>
TITLE: Can Morley's theorem be generalized?
QUESTION [26 upvotes]: Morley's theorem states that in any triangle, the three points of intersection of the adjacent angle trisectors form an equilateral triangle.
In a talk some years ago, David Rusin made the provocative claim that Morley's theorem is a rare example of a striking theorem that defies generalization.  The first ideas that come to everyone's mind—passing to higher dimensions or hyperbolic geometry for example—don't work.
The proof by Alain Connes yields a mild generalization of sorts, but not a very satisfying one in my opinion.  Wikipedia claims that there are "various generalizations" of Morley's theorem, but by this it seems to mean extensions of Morley's theorem, i.e., further equilateral triangles that one can construct.  This is not what I would, strictly speaking, call a "generalization."
So is David Rusin correct?

Are there no satisfactory generalizations of Morley's theorem?

REPLY [2 votes]: A recent "generalization":

Tran, Q. H. "Morley’s trisector Theorem for isosceles tetrahedron." Acta Mathematica Hungarica (2021): 1-8.
DOI.

Abstract.
We extend Morley’s trisector theorem in the plane to an isosceles tetrahedron in three-dimensional space. We will show that the Morley tetrahedron of an isosceles tetrahedron is also isosceles tetrahedron. Furthermore, by the formula for distance in barycentric coordinate, we introduce and prove a general theorem on an isosceles tetrahedron.
A "Morley tetrahedron" is determined by planes trisecting each
dihedral angle.
Tran, Q. H. is likely MO user TranQuangHung.<|endoftext|>
TITLE: how to construct a $C^\infty$ stack from a holomorphic stack
QUESTION [6 upvotes]: Given a complex manifold, you can `weaken' its structure to give a smooth manifold. Is there an analogous construction that constructs a stack over the category of smooth manifolds from a stack over the category of complex manifolds? 
Obviously, this is possible for a stack represented by the quotient of a complex manifold by a group action, so I'd imagine that this should  at least be possible for DM stacks, but I can't think of any general construction that doesn't involve some kind of atlas.

REPLY [2 votes]: Denote by $$u:CxMfd \to Mfd$$ the forgetful functor from complex manifolds to smooth manifolds. Let $$u_!:St\left(CxMfd\right) \to St\left(Mfd\right)$$ denote its 2-categorical prolongation. Explicitly, this is given by the bicategorical Kan extension of $y_{Mfd} \circ u$ along the Yoneda embedding $$y_{CxMfd}:CxMfd \to St\left(CxMfd\right),$$ where $y_{Mfd}$ is similarly defined. $u_!$ is the unique weak colimit preserving functor which agrees with $y_{Mfd} \circ u$ on representables.
I claim that $u_!$ sends holomorphic stacks (stacks coming from groupoid objects in complex manifolds) to differentiable stacks.
Indeed, let $\mathcal{X}$ be a holomorphic stack coming from a groupoid object $X_1 \rightrightarrows X_0.$ Then, $\mathcal{X}$ is the weak colimit of the truncated semi-simplicial diagram $$X_2\mspace{5mu} \{(3\mspace{5mu} parallel \mspace{5mu} arrows)\}\mspace{5mu} X_1 \rightrightarrows X_0,$$ viewing each $X_i$ as a representable presheaf on $CxMfd$. Applying $u_!$ to this diagram, yields that $u_!\left(\mathcal{X}\right)$ is the weak colimit of the same diagram, now viewing each $X_i$ as a representable presheaf in $Mfd$. This in turn implies that $u_!\left(\mathcal{X}\right)$ is the stackification of the weak presheaf of groupoids arising canonically from $X$ viewed as a Lie groupoid. In particular, this implies that $u_!$, when restricted to holomorphic stacks agrees with the answer of David Roberts, only, it makes no explicit reference to atlases.<|endoftext|>
TITLE: How large is TREE(3)?
QUESTION [38 upvotes]: Friedman, in _Lectures notes on enormous integers shows that TREE(3) is much larger than n(4), itself bounded below by    $A^{A(187195)}(3)$ (where $A$ is the Ackerman function and exponentiation denotes iteration). But actually, using the fast-growing hierarchy, $n(p)$ is smaller than $f_{\omega^{\omega^\omega}}(p)$, shown by Friedman in

Long finite sequences, Journal of Combinatorial Theory, Series A 95 Issue 1 (2001) pp. 102-144, doi:10.1006/jcta.2000.3154, author pdf

while it seems that TREE grows faster than $f_{\Gamma_0}$ (${\Gamma_0}$ being the Feferman-Schütte ordinal). So it could well be that in fact TREE(3) is larger than, say, n(n(4)), or even any number expressible by iterations of n. What is known on this question?
For reference, I should have added that TREE(3) is the incredibly (at first, or even second look) large answer to the question : "which is the length of the  longest sequence $(T_2,T_3,T_4,\dots,T_n)$ of labeled trees such that $T_k$ has at most $k$ nodes labeled $a$ or $b$,   and $T_i$ is not a subtree of $T_j$ for $i < j$ ?".
Here, trees are rooted trees, and are treated as poset on their sets of vertices. A tree $T$ is called a subtree of $T'$ if there is an inf-preserving embedding from $T$ into $T'$, (that is, an injective map $h:Vertices(T) \to Vertices(T')$ such that $h(\inf(x,y)) = \inf((h(x), h(y))$) that respects the labeling by $a$ or $b$.

REPLY [2 votes]: I'd like to post an improvement of what originally posted here, by Deedlit and myself. Another related post is here, which shows an even better bound, though there are questions how the order in the last post exaclty works, i.e. it does not contain evidence why it should be like this. Anyway, we start with these trees:
T1  {}
T2  [[]]
T3  [()()]
T4  [(()())]
T5  [(((())))]
T6  ([((()))][])
T7  ([((()))]()())
T8  ([((()))]()())
T9  ([((()))]((())))
T10 ([((()))](()()))
T11 ([((()))](((()))))
T12 ([((()))]((())))
T13 ([((()))](()))
T14 ([((()))]())
T15 ([((()))])
T16 ([(())][(())][(())][(())][(())])
T17 ([(())][(())][(())][(())]([()][]))
T18 ([(())][(())][(())][(())]([()][]))
T19 ([(())][(())][(())][(())]([()]()()))
T20 ([(())][(())][(())][(())]([()]())[()])
T21 ([(())][(())][(())][(())]([()]())(([])))
T22 ([(())][(())][(())][(())]([()]())([][][]))
T23 ([(())][(())][(())][(())]([()]())([][])([]))
T24 ([(())][(())][(())][(())]([()]())([][])[][][])
T25 ([(())][(())][(())][(())]([()]())([][])[][]()())
T26 ([(())][(())][(())][(())]([()]())([][])[][]((())))
T27 ([(())][(())][(())][(())]([()]())([][])[][](()))
T28 ([(())][(())][(())][(())]([()]())([][])[][]())
T29 ([(())][(())][(())][(())]([()]())([][])[][])
T30 ([(())][(())][(())][(())]([()]())([][])[]#)

where # is the first tree of $\mathrm{tree}(8)$. 
This is very important. We can now reduce # for $\mathrm{tree}(8)$ steps. In general, we can reduce [] for as many steps as it allows. Generally this will be comparable to the number of brackets allowed in total. 
This means that we can preform basic recursion around tree already. We can spawn a ([]) to $n$ []s, so this is already at $f_{1}$ where $f_0 = \mathrm{tree}$.  We can spawn a () to $n$ ([])s so it is at at $f_2$. Similiarly, we can take ([]#) for any tree # and reduce it just like as we would as in tree. We can reach $f_{\vartheta(\Omega^{\omega})}(n)$ in the modified hierarchy this way. $f$ is a certain variant of the fast growing hierarchy. 
But $f_{\vartheta(\Omega^{\omega})}(n)$ in the modified hierarchy is just $F_{\vartheta(\Omega^{\omega},0)2}(n)$, where $F$ is a certain variant of the normal fast growing hierarchy (with $F_0=n+1$.)
Similiarly, we have ([][]#), reaching $F_{\vartheta(\Omega^{\omega})3}(n)$ at ([][][]) and $F_{\vartheta(\Omega^{\omega})\omega}(n)$ at (([])).  $F_{\vartheta(\Omega^{\omega})(\omega+1)}(n)$ is reached by  (([])[]), $F_{\vartheta(\Omega^{\omega})(\omega2)}(n)$ by  (([])([])), and $F_{\vartheta(\Omega^{\omega})\omega^2}(n)$ at (()).  With # we work in the product, reaching $F_{\vartheta(\Omega^{\omega})^2}(n)$ at (([][])) and $F_{\vartheta(\Omega^{\omega})^\omega}(n)$ at ((([]))).
A pattern emerges. We can in fact reach $F_{\varepsilon_{\vartheta(\Omega^{\omega})+1}}$ at [()] which would evaluate to (((...((([])))...))) with as many pairs as possible. 
We see that TREE(3) is already far beyond the tree function itself.<|endoftext|>
TITLE: Local splitting of the tangent bundle with interesting properties
QUESTION [5 upvotes]: Let $(M,g)$ be a Riemannian manifold and let $U\subset M$ be an open subset. Suppose that the tangent bundle over $U$ splits into two orthogonal distributions $TU=\mathcal{E}\oplus \mathcal{F}$.
Is it possible that the two $C^{\infty}(U)$-bilinear maps 
\begin{align*}I:\mathcal{E}\times\mathcal{E}&\to \mathcal{F}\\
&(X,Y)\mapsto pr_{\mathcal{F}}(\nabla_X Y)
\end{align*}
and
\begin{align*}I:\mathcal{F}\times\mathcal{F}&\to \mathcal{E}\\
&(X,Y)\mapsto pr_{\mathcal{E}}(\nabla_X Y)
\end{align*}
are both antisymmetric in $X$ and $Y$ without vanishing?
If $\mathcal{E}$ and $\mathcal{F}$ both were integrable, both maps would be symmetric. So is this in some sense the most non-integrable way, distributions can be?

REPLY [5 votes]: Yes, this can happen.  A little experimentation with the structure equations shows that there is a $3$-parameter family of homogeneous examples in dimension $4$:  Let $c_1,\ldots,c_4$ be nonzero constants satisfying $c_1c_2=c_3c_4$, and consider the simply-connected $4$-dimensional Lie group $G$ that has a basis of left-invariant $1$-forms $\omega_1,\ldots,\omega_4$ that satisfy the structure equations
\begin{aligned}
d\omega_1 &= 2c_1\ \omega_2\wedge\omega_3 + 2c_3\ \omega_4\wedge\omega_3\ ,\\\\
d\omega_2 &= 2c_1\ \omega_3\wedge\omega_1 \ ,\\\\
d\omega_3 &= 2c_2\ \omega_4\wedge\omega_1 + 2c_4\ \omega_2\wedge\omega_1\ ,\\\\
d\omega_4 &= 2c_2\ \omega_1\wedge\omega_3 \ .
\end{aligned}
Now endow $G$ with the Riemannian metric $g$ for which the $\omega_i$ define an orthonormal coframing, let $e_1,\ldots,e_4$ be the dual ($g$-orthonormal) vector fields, and let $\mathcal{E}$ be the $2$-plane field spanned by $e_1$ and $e_2$ while $\mathcal{F}$ is the $2$-plane field defined by $e_3$ and $e_4$.  
One easily checks that this is an example of the desired type:  If $\nabla$ is the Levi-Civita connection of this metric, then
$$
\nabla_{e_1}e_1\equiv0\ ,\quad\nabla_{e_1}e_2\equiv c_4e_3, \quad
\nabla_{e_2}e_1\equiv-c_4e_3\ ,\quad\nabla_{e_2}e_2\equiv 0 \mod \mathcal{E}
$$
and
$$
\nabla_{e_3}e_3\equiv0\ ,\quad\nabla_{e_3}e_4\equiv c_3e_1, \quad
\nabla_{e_4}e_3\equiv-c_3e_1\ ,\quad\nabla_{e_4}e_4\equiv 0 \mod \mathcal{F},
$$
as desired.
There are non-homgeneous examples in dimension $4$ as well.  A little more work with the structure equations shows that there exists a $4$-parameter family of examples of cohomogeneity $2$.  (I don't know how many of these are complete.)  If there is interest, I can give the structure equations of these examples as well.<|endoftext|>
TITLE: Abelian category which is not well-powered
QUESTION [10 upvotes]: Can you give an example of an abelian category which is not well-powered? If not, maybe you can give any reason why there are such abelian categories?

REPLY [7 votes]: Here's a simpler, but less consequential, example.
Take the category of "eventually constant" functors from ordinals (considered as a category with a single morphism $\alpha\to\beta$ when $\alpha\leq\beta$) to abelian groups, meaning functors $F$ for which there is some ordinal $\alpha$ such that $F(\beta)\to F(\gamma)$ is an isomorphism for all $\alpha\leq\beta\leq\gamma$. 
The "eventually constant" condition ensures that this is a locally small category. It is not well-powered, since for any ordinal $\alpha$, the constant functor taking value $\mathbb{Z}$ has a subfunctor $F_\alpha$ with
$$F_\alpha(\beta)=\begin{cases}0&\text{if $\beta<\alpha$}\\
\mathbb{Z}&\text{if $\beta\geq\alpha$.}
\end{cases}$$<|endoftext|>
TITLE: j-invariant fixed point?
QUESTION [8 upvotes]: If we view the j-invariant of a lattice as a map from the upper-half plane to the complexes by $\tau\mapsto j([1,\tau])$, then it is surjective, holomorphic, and has quite a number of other wonderful properties (see the third part of Cox's Primes of the Form for a great introductory reference).  
My question is: does $j$ have any fixed points?  If so, do we know what any/all of them are?  
I'm in particular curious what goes into the proof.  Specifically, whether the answer is immediate from some complex analysis, or whether you need to have a good handle on $j$ itself (or both!). A professor I asked suggested thinking about $j(\tau)-\tau$ on the compactification of the fundamental domain of $SL(2,\mathbb{Z})$, but we weren't able to clean it up.

REPLY [3 votes]: As already pointed out by Alexandre Eremenko, the $j$ function is a finite type map  in the sense of my thesis. It follows that any open set intersecting the domain boundary (here the extended real line) contains infinitely many fixed points which are repelling in the sense that the derivative there has modulus greater than 1. 
The normalization of $j$ is irrelevant: for any Mobius transformation $M$, the composition $M\circ j$ is also a finite type map.<|endoftext|>
TITLE: strong contactomorphism group inside contactomorphism group
QUESTION [8 upvotes]: Let $(M, \xi)$ be a closed contact manifold with co-oriented contact structure $\xi = \ker \alpha$.  Let $\mathrm{Cont}(M, \alpha)$ be the group of diffeomorphisms that preserve the contact form $\alpha$, and let $\mathrm{Cont}^+(M, \xi)$ be the diffeomorphisms that preserve $\xi$ with its co-orientation (i.e. that pull $\alpha$ back to a positive multiple of $\alpha$). 

Does the inclusion $\mathrm{Cont}(M, \alpha) \hookrightarrow \mathrm{Cont}^+(M, \xi)$ induce a weak homotopy equivalence?

My guess is that this is too much to hope for, but I don't have any candidate for a counter-example.
Motivation: This question came up in understanding the question of when a contact fibre bundle admits a global contact form with diffeomorphic Reeb dynamics on every fibre.

REPLY [9 votes]: One of the problems with this question is that $\text{Cont}(M,\alpha)$ depends heavily on the choice of contact form $\alpha$, so while the previous answer shows that there is a choice of contact form for which $\text{Cont}(M,\alpha)$ and $\text{Cont}^+(M,\xi)$ can't be homotopy equivalent, it's not immediately clear if this is true for all choices of $\alpha$.
That said, here's a connected example for which the inclusion also fails to be surjective on $\pi_0$. (Attribution: this emerged out of discussions with Hansjoerg Geiges.)
Let $(M,\xi)$ be $T^3$ with its standard contact structure $\xi_0$, and using coordinates $(x,y,\theta)$ on $T^3$, write the standard contact form as
$\alpha_0 = \cos(2\pi\theta) dx + \sin(2\pi\theta) dy$.
Now for any $A \in \text{SL}(2,\mathbb{Z})$, the direct sum of $A$ with the identity defines a linear map on $\mathbb{R}^3$ which descends to $T^3$ as a diffeomorphism $f : T^3 \to T^3$.  It is easy to show that for any such map, $f^*\alpha_0$ can be deformed through contact forms to $\alpha_0$, hence by Gray's stability theorem, $f$ is isotopic to a contactomorphism $f_0$.
However, if $f_0$ is a strict contactomorphism with respect to $\alpha_0$, then it must preserve the corresponding Reeb vector field, and this is a very strong restriction.  It means for instance that the Morse-Bott torus of Reeb orbits at $\{\theta=0\}$ is mapped to another Morse-Bott torus of orbits with the same period, and the only such orbits that exist for $\alpha_0$ point in either the same, opposite or an orthogonal direction. With arguments like this one can show that $f_0$ cannot be a strict contactomorphism unless the matrix $A$ is orthogonal... in fact, I believe it must be a fourth root of the identity.
With a little more work one can say something similar for a much larger set of contact forms, using the fact that a strict contactomorphism must always map Reeb orbits to Reeb orbits of the same period.  (I'm fairly sure that for a generic choice of contact form, not only are all Reeb orbits nondegenerate but no two of them have the same period.)  Unfortunately, I still don't know how to turn this into any statement for all contact forms, but the evidence is certainly against $\text{Cont}(M,\alpha) \hookrightarrow \text{Cont}^+(M,\xi)$ ever being bijective on $\pi_0$.<|endoftext|>
TITLE: Riemann zeta at even integers
QUESTION [9 upvotes]: I am talking about this in a course I am teaching, and hence am wondering: what are the various derivations of the values of Riemann zeta function at even integers? There are two incredibly cool proofs in Don Zagier's paper (section 1), but there must several other proofs floating around. Also, I recall reading that Euler originally proved the formula for $\zeta(2)$ by thinking of $\sin(x)$ as a polynomial -- has this argument been made rigorous since?
EDIT I did not realize that this was known as the "Basel Problem", so did not find @Yemon's answer myself. I conjecture, however, that the Robin Chapman list is incomplete, since I have found yet another proof, not contained in Robin's list, so maybe there are more yet out there...

REPLY [4 votes]: If we evaluate the Fourier trigonometric series expansion for the function defined in $\left[ -\pi ,\pi \right] $ by $f(x)=x^{2p}$ and extended to all of ${\mathbb R}$ periodically with period $2\pi,$ we get
$$\begin{equation*}x^{2p}=\frac{\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos nx\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x.\tag{1}\end{equation*}$$
So, for $f(\pi )=\pi ^{2p}$ we have
$$\begin{equation*}\pi ^{2p}=\frac{\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x,\tag{2}\end{equation*}$$
where the integral
$$\begin{equation*}I_{2p}:=\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x\tag{3}\end{equation*}$$
satisfies the following recurrence, as can be shown by integration by parts
$$\begin{equation*}I_{2p}=\frac{2p}{n^{2}}\pi ^{2p-1}\cos n\pi -\frac{2p(2p-1)}{n^{2}}I_{2\left( p-1\right) },\qquad I_{0}=0.\tag{4}\end{equation*}$$

For $p=1$, we obtain $$\begin{equation*}I_{2}=\frac{2}{n^{2}}\pi \cos n\pi.\end{equation*}\tag{5}$$ and

$$\pi ^{2}=\frac{\pi ^{2}}{3}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi\left(\frac{2}{n^{2}}\pi \cos n\pi \right)=\frac{\pi ^{2}}{3}+4\sum_{n=1}^{\infty }\frac{1}{n^{2}}.\tag{6}$$ Consequently, $$\zeta (2)=\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi ^{2}}{6}\tag{7}$$
By using $(1)$ to $(4)$ we can generate recursively the sequence $(\zeta(2p))_{p\ge 1}$. For instance, I evaluated $\zeta(4)$ and $\zeta(6)$ in this math.SE answer.<|endoftext|>
TITLE: Representation of μ-recursive functions
QUESTION [5 upvotes]: Can every μ-recursive function be defined using a single instance of the μ operator applied to a primitive recursive function?
According to Wikipedia, any μ-recursive function can be expressed as the μ-operator over a primitive recursive function (source):

A consequence of this result is that
  any μ-recursive function can be
  defined using a single instance of the
  μ operator applied to a (total)
  primitive recursive function.

From this I conclude that given a μ-recursive function $f(x_1,\ldots,x_n)$, I can write it as $\mu y.R(x_1,\ldots,x_n,y)$ with $R$ being a primitive recursive function.
If so, let's say there's a function $h(x,y)$ which is μ-recursive but not primitive recursive and its range is $0,1$. As $h$ is μ-recursive, then I can write it as $\mu z.R(x,y,z)$ for some $R$ primitive recursive. However, as $h$'s range is bounded, the 1-bounded μ-recursive operator (which is primitive recursive) over $R$ should give us $h$ as primitive recursive. Absurd! Regarding the existence of $h$, in this review they show how to construct one of these if I understand correctly.
The question, then, is: what am I misunderstanding in Wikipedia's remark?
Final note: this seems to be an example in Kleene's book Introduction to metamathematics (§58), but I don't see how it is solved.

REPLY [9 votes]: The answer is that you have to apply another primitive recursive function after the $\mu$ operator. Specifically, the Kleene normal form is that every recursive function $f$ has the form $f(n)=U(\mu x T(e,n,x))$, where both $U$ and $T$ are primitive recursive. The predicate $T(e,n,x)$ asserts that $x$ is the code of a halting computation of program $e$ on input $n$, and the function $U$ extracts the output value from this code. 
It is the step involving $U$ on which your proposed argument flounders.

REPLY [7 votes]: Every $\mu$-recursive function can be expressed as a primitive recursive function (usually called $U$) applied to the result of applying the $\mu$ operator to another primitive recursive predicate (usually called $T$).  Wikipedia is right that you only need one $\mu$ and it's applied to a primitive recursive $T$, but, as you noted (and as Wikipedia doesn't actually deny) you need additional work ($U$) after you've applied $\mu$.
The idea is that the value of a recursive function $f$ at input $x$ can be described as "the output of the first terminating computation for $f$ on input $x$."  The predicate $T$ of being a (Gödel number of a) terminating computation of a particular machine on a particular input is primitive recursive, and so is the function $U$ that extracts the final answer from a computation.<|endoftext|>
TITLE: topology of $\mathbb{P}_2 (\mathbb{C}) \setminus \mathbb{P}_2 (\mathbb{R})$?
QUESTION [8 upvotes]: What is the topology of $\mathbb{P}_2 (\mathbb{C}) \setminus \mathbb{P}_2 (\mathbb{R})$? For example what is the homology of this manifold with coefficients in $\mathbb{Z}$. I know that this is known but I can't find a good reference for it. Can anyone give me a reference?

REPLY [4 votes]: There is a beautiful (and elementary) paper by V. I. Arnold, which discusses this and generalizations.<|endoftext|>
TITLE: Minimum distance between two arbitrary circles in space?
QUESTION [5 upvotes]: What is the minimum distance between two arbitrary circles in space?
I am working out the problem with Maxima, but I am surprised by how complicated this rapidly turns out to unfold for such a "simple" question.
Monstrous equations, maybe someone has a bright idea?

REPLY [4 votes]: I didn't manage to solve the problem (edit: in the meantime an answer was posted which says a precise formula using radicals cannot be found), but I can post a proof that the line joining the points where the minimal/maximal distance is achieved is perpendicular to the tangent line at the circles in those contact points. (inspired by the comment of Gerhard Paseman)
To do this, choose $\vec{a}$ and $\vec{d}$ the position vectors of the centers and $\vec{b},\vec{c}$, respectively $\vec{e},\vec{f}$ be two pairs of orthogonal unit vectors which span the planes of the first and respectively the second circle. Denote by $r,s$ the radii of the two circles. Consider the circles parametrized as (in fact, the argument works for any parametrization)
 $$ p(\theta)=\vec{a}+r\cos\theta\ \vec{b}+r\sin\theta\ \vec{c}, \  \theta \in [0,2\pi] $$
 $$ q(\tau)=\vec{d}+s\cos\tau\ \vec{e}+s\sin\tau\ \vec{f},  \tau \in [0,2\pi]$$
 and denote $F(\theta,\tau)=|p(\theta)-q(\tau)|^2$. Then the pair of points which realize the minimal/maximal distance must satisfy 
 $$ \frac{\partial F}{\partial \theta}=\frac{\partial F}{\partial \tau}=0. $$
We have
$$ \frac{\partial F}{\partial \theta}=2\sum_{i=1}^3 [p_i(\theta)-q_i(\tau)]p_i'(\theta)=2 (p(\theta)-q(\tau))\cdot p'(\theta) $$
$$ \frac{\partial F}{\partial \tau}=-2\sum_{i=1}^3 [ p_i(\theta)-q_i(\tau) ] q_i'(\tau)=-2 (p(\theta)-q(\tau))\cdot q'(\tau) $$
where "$\cdot$" is the usual dot product. Therefore when $\theta,\tau$ correspond to the minimum/maximum value, the partial derivatives vanish and $p'(\theta)\perp (p(\theta)-q(\tau))$ and $q(\tau)'\perp (p(\theta)-q(\tau))$ where $p'(\theta),q'(\tau)$ are the tangent vectors in the contact points and $p(\theta)-q(\tau)$ is the vector connecting the points where minimal/maximal distance is achieved.<|endoftext|>
TITLE: Mysterious property of $\mathbb{Q}$
QUESTION [6 upvotes]: Hi,
I am currently working through the paper by Bousfield and Gugenheim on rational homotopy theory, and have come to a point where they show why it is important to work over $\mathbb{Q}$, and not just any field (remark 9.7).  They assert that if $Hom_k(Hom_{\mathbb{Z}}(G, k), k)=G$ for some non-trivial abelian group $G$, and some field $k$ [edit: of characteristic $0$] (thanks to Fernando Muro for pointing out that we need characteristic $0$), then we must have $k=\mathbb{Q}$ (and also that $G$ is finite dimensional as a $\mathbb{Q}$-vector space, which is clear enough).  I can't quite see what property of $\mathbb{Q}$ this reduces to, and it really makes my head spin trying to think through it.  The paper is now available online, just google "on pl de rham theory and rational homotopy type".
Any suggestions would be great.
Thanks
Brian

REPLY [16 votes]: First, $k$ needs to be assumed of characteristic $0$. The mysterious property is that $\mathbb{Q}$ is a prime field:
If $Hom_k(Hom_Z(G, k), k)=G$, $G$ has to be a $k$-vector space, say $G=kT$ for a basis $T$.
$Hom_Z (kT;k)$ is a $k$-vector space. 
Let me first argue that $T$ has to be finite. Since $Hom_k (kT;k) \to Hom_Z (kT;k)$ is injective, the dimension of $Hom_Z (kT;k)$ is at least the dimension of $(kT)^{\ast}$. Thus it follows that the dimension of $kT$ is at least the dimension of the bidual of $kT$. Thus can only be true if $T$ is finite. We can assume that $T$ has one element; therefore
$$dim_k (Hom_k (Hom_Z (k,k);k))=1.$$
If $k$ is not a prime field, then $dim (Hom_Z(k,k))>1$, which contradicts the above formula.
The same argument works in characteristic $p$ and shows that $k$ has to be $\mathbb{F}_p$.<|endoftext|>
TITLE: Integrability of the Cohen map
QUESTION [22 upvotes]: In the 1990's, Henri Cohen asked whether the map $(x,y) \mapsto (\sqrt{1+x^2}-y,x)$ from $\mathbb{R}^2$ to itself is integrable.  In other words, are the orbits confined to the level curves of some nice function?  It certainly looks like they are if you make plots (the postscript file at http://www.math.washington.edu/~cohn/cohen.ps shows some orbits and is easy to modify), [Image (rotated) added by O'Rourke:]
     

but no algebraic function could play this role (a result of Rychlik and Torgerson; http://nyjm.albany.edu/j/1998/4-5.html).  Years ago, someone told me it was unlikely that it was really integrable, because careful investigation identified hyperbolic periodic orbits, but I never learned more and I didn't see this in my crude plots.  Unfortunately, I don't remember who told me.
Is the map actually integrable?  This may well be an open problem: I looked at the papers that cite the Rychlik and Torgerson paper, except for one I couldn't access, but I found no evidence that this question has been resolved.  On the other hand, there don't seem to be many papers on this topic, and perhaps that's because the conjecture turned out to be false, with a disproof that was never published.
If it isn't integrable, then what could explain the near integrability?  From my naive perspective, a simple system like this that looks integrable but isn't would be really amazing.  Are such things more common than I realize?

REPLY [26 votes]: I have run some simple simulations of this system in Python, and it looks nothing like a nice integrable system once you zoom in on the orbits. Certainly, the orbits do not look like closed curves about the the fixed point, as suggested by the plot in the question. I'll give some numerical evidence of the chaotic behaviour of the map here, rather than actual proofs. However, I think that much of this behaviour could be demonstrated rigorously with nothing much more complicated than the the intermediate value theorem and some (very tedious) calculations with careful error bounds. Also, I think the kind of behaviour demonstrated by this map is reasonably common for area-preserving maps which are close to being integrable, although I'm not expert on this (the question just seems very interesting). When you perturb an integrable map, while retaining the area-preserving property, the orbits with irrational rotation numbers are relatively stable whereas the orbits at rational rotation numbers break up.
First, the function $f\colon\mathbb{R}^2\to\mathbb{R}^2$, $f(x,y)=\left(\sqrt{1+x^2}-y,x\right)$ is easily seen to be an invertible map with a single fixed point at $p_0=\left(1/\sqrt{3},1/\sqrt{3}\right)$. It has Jacobian matrix $\left(x/\sqrt{1+x^2},-1;1,0\right)$ which has unit determinant, so $f$ is area-preserving. At the fixed point, the Jacobian has eigenvalues $(1\pm i\sqrt{15})/4$ which is not a root of unity so, near the fixed point, $f$ is approximately a rotation (after a linear change of variables) by the irrational number $\theta_0=(2\pi)^{-1}\cos^{-1}(1/4)\approx0.2098$ of turns. On the other hand, far from the fixed point, $f(x,y)=(\vert x\vert-y,x)+O(1)$, so $f$ is approximated to leading order by $(x,y)\mapsto(\vert x\vert-y,x)$. This is integrable, with polygonal orbits (see the file linked by Sylvain Bonnot in the comments, and also the paper linked in the question, which mentions that this map has rotation number $2/9\approx0.222$). I think integrability of $f$ in the limit as you go very far or very close to the fixed point is guaranteed by the fact that it is area preserving and linear on radial lines, hence reduces to a homeomorphism of the circle.
Plotting the orbits of a set of initial points chosen along the x-axis demonstrates this behaviour, and the map looks very well-behaved so far.
     


Next, consider the line $\lbrace p_0+(x,0)\colon x\in\mathbb{R}^+\rbrace$. This should cut through the orbits, and we can plot the rotation number of $f$ on the orbit as a function of $x$. This can be done by appying $f$ some large number $n$ times, counting how many times the iterates rotate about the fixed point, and dividing by $n$. It appears to be an increasing function of $x$ going from about 0.210 to about 0.222, in agreement with the explanation above.
     


The rational number in this range with the smallest denominator is $3/14\approx0.2143$, which occurs at $x\approx1.119$. We can zoom into a small range about this value and calculate the rotation numbers again. I also multiply them by 14, so that it is clear where they pass through the integer value 3.
     


There is clear mode locking as the rotation number passes through 3/14. If true, this is inconsistent with the orbits being confined to closed curves about the fixed point! To show this, consider a point within the mode-locked region which is not a fixed point of $f^{14}$. There must be such points otherwise, by the fact that $f$ is analytic, $f^{14}$ would be constant everywhere, which is not the case (near the fixed point, for one thing). Then, take a closed ball which does not contain fixed points of $f^{14}$. Iteratively applying $f^{14}$ will coverge to a finite set of fixed points of $f^{14}$ on each orbit, which has measure zero, contradicting the area conservation property of $f$.
So, what does happen within the mode-locked regions? I chose a number of initial points in and around the mode-locked region and plotted a graph of the iterates of $f$ (rotated and rescaled to fit the graph).
     


The mode-locked region corresponding to rotation number 3/14 actually consists of 14 separate small regions (homeomorphic to closed balls) joined together, each of which contains a point about which iterates of $f^{14}$ rotate. These points form an orbit of period 14 for $f$, as do the points at which the small regions are joined (which look like hyperbolic periodic points).
It's not too difficult to see why this should be so. If you plot the angle through which $f^{14}$ rotates points on an orbit with rotation number less than 3/14, it will be less than 3 everywhere. As the rotation number of the orbit increases to 3/14, then the peaks of this graph increase to 3, representing the hyperbolic fixed points. Similarly, you can approach the rotation number of 3/14 from above, with the troughs the graph decreasing to the hyperbolic fixed points. These two limits give the two extreme (inner and outer) orbits of rotation number 3/14 joined at the hyperbolic fixed points. As they must be distinct orbits, they enclose the "mode-locked" region.
The region with rotation number 3/14 was easiest to check, as it has the lowest denominator, but I would expect that there is similar behaviour at other rational numbers. Furthermore, within the small mode-locked regions, where we saw that $f^{14}$ has orbits rotating about a central point, I would hazard a guess that we have similar behaviour again, giving a fractal structure to the orbits of $f$.

Let me now look at this from a theoretical point of view. It can be proven that it is not possible for the orbits of $f$ to all lie on closed curves about the fixed point. I'll first show that there are only finitely many periodic points of a given order (which is not a multiple of 3, although this constraint is maybe not necessary). This is enough to severely restrict the possible behaviour of the orbits of $f$ (the argument below should be perfectly rigorous once you fill in the bits I glossed over quickly).

Theorem: For any positive integer $n$, not a multiple of 3, there are only finitely solutions to $f^n(x)=x$.

Proof: First, let us upgrade $f$ to a 2-valued function by taking both signs for the square root, $f((x,y))=(\pm\sqrt{1+x^2}-y,x)$. Then, $f^n(x)$ takes up to $2^n$ values for any $x\in\mathbb{R}^2$, and we can rewrite $f^n(x)=x$ as $x\in f^n(x)$. I'll show that this actually has finitely many solutions in $\mathbb{C}^2$. Note that the set $S=\{x\in\mathbb{C}^2\colon x\in f^n(x)\}$ is algebraic (i.e., the zero set of a set of polynomials in x). As an affine variety, this is either of dimension 0 (finitely many values) or of positive dimension (uncountable and, in fact, unbounded). We just have to rule out the possibility of $S$ being unbounded. Defining the 2-valued function $g(x,y)=(\pm x-y,x)$, then $f(x)=g(x)+O(1)$. So, if there was a sequence $x_k\in S$ with $\Vert x_k\Vert\to\infty$, then $x_k/\Vert x_k\Vert\in g^n(x_k/\Vert x_k\Vert)$ up to an $O(1/\Vert x_k\Vert)$ term. Taking the limit $k\to\infty$, we have a nonzero solution to $x\in g^n(x)$. However, $g^n$ corresponds to multiplying by a matrix from the set $M_{\pm}=(\pm1,-1;1,0)$ n times. So, for $N$ equal to one of the n-fold products $N=M_\pm M_\pm\cdots M_\pm$, we would have ${\rm det}(N-I)=0$. As these are integer matrices, we can reduce mod 2. Note that, $M_+$ and $M_-$ both reduce to $M=(1,1;1,0)$ mod 2 and that $M^3=I$ (mod 2). Therefore, $N=M^n$ which is equal to one of $M$ or $M^2$ (mod 2), as $n$ is not a multiple of 3. This implies that ${\rm det}(M-I)=0$ or ${\rm det}(M^2-I)=0$ (mod 2), which you can check is not the case. QED
The previous theorem is enough to show that many of the curves in your plot must break up when you zoom in. Recall that $p_0\in\mathbb{R}^2$ denotes the fixed points of $f$.

Corollary: The set of closed curves about $p_0$ preserved by $f$ cannot cover all of $\mathbb{R}^2\setminus\{p_0\}$.

Proof: We can compute the rotation number of $f\vert_C$ for any closed curve $C$ surrounding $p_0$. For any $x\in C$, this can be calclated by counting the number of times that $f^n$ rotates $x$ about $p_0$, dividing by $n$, and taking the limit $n\to\infty$. This is a continuous function of $x$, which I denote by $R(x)$. As $x\to p_0$ we have $R(x)\to\cos^{-1}(1/4)/(2\pi)\approx0.2098$, and $R(x)\to5/9\approx0.222$ as $\Vert x\Vert\to\infty$. Choosing any rational number $p/q$ (with $q$ not a multiple of 3) between these limits then, by continuity of rotation numbers, $R(x)=p/q$ for some $x$. As we move out radially along a line from $p_0$ to infinity, one of two things can happen. (i) $R(x)=q$ on a non-trivial interval. As explained above, this contradicts the area conservation property of $f$. Or, (ii) $R(x)=q$ on a nowhere dense set. But, as the rotation number passes through $q$, you always get mode-locking unless $f$ is conjuate to a rotation of angle $p/q$ -- i.e., $f^q(y)=y$ for all $y$ on the closed curve passing through $x$ and preserved by $f$. However, this implies infinitely many solutions to $f^q(y)=y$. So, we can rule out both possibilities. QED
By a modification of this argument, you can show that the small orbits of $f^{14}$ in my plot must also break up. So, it is almost definite that some of the orbits of $f$ are in fact chaotic, and don't lie on any finite union of curves (I think you can prove this using the ideas above, but would take a lot more work to make it rigorous).<|endoftext|>
TITLE: Sum of $\sum_{k=1}^nd(k^2)$
QUESTION [15 upvotes]: There is a literature dealing with
$$
\sum_{k\le x}d(f(k))
$$
where $f$ is an irreducible polynomial and $d(n)$ is the number of divisors of $n$. Erdos 1952 shows that the sum $\asymp x\log x,$ which was improved to $Ax\log x+O(x\log\log x)$ by Bellman-Shapiro (cited in Scourfield). But these results only apply to irreducible polynomials.

What asymptotics are known for $\sum_{k\le x}d(k^2)$?
Are there good methods for calculating this sum quickly?

The literature includes: Dirichlet 1850, Voronoi 1903 and van der Corput 1922, Kolesnik 1969, Huxley 1993, Nowak 2001 (linear); Scourfield 1961, Hooley 1963, McKee 1995, McKee 1997, McKee 1999, Broughan 2002 (quadratic). The sequence is in the OEIS as A061503 but there is no real information there.

REPLY [3 votes]: It is interesting that $d(n^2)$ problem admits square root cancelation in the error term:
$$\sum_{n\leq x} d(n^2) = x P_2(\log x) +O\left(x^{1/2}\exp\left(-c\frac{\log^{3/5}x}{\log^{1/5}\log
x}\right)\right).$$
(See section 4.12 from Postnikov A. G. Introduction to analytic number theory. He reissues result from the article Stronina M. I.
Integral points on circular cones. (Russian)
Izv. Vysš. Učebn. Zaved. Matematika 1969.)<|endoftext|>
TITLE: Submersions of closed manifolds
QUESTION [7 upvotes]: Let $f\colon\thinspace M\to N$ be a map of closed smooth manifolds, with $\dim M > \dim N$. Recall that a submersion is a smooth map whose differential is surjective at every point in the domain.

Can one give conditions which guarantee that $f$ is homotopic to an submersion?

These conditions would necessarily have to be homotopy invariants. I am thinking there may be something in terms of relations amongst characteristic classes.
I am aware of the theorem of Phillips
MR0208611 (34 #8420) Phillips, Anthony Submersions of open manifolds. Topology 6 1967 171–206.
which says roughly that, when $M$ is open, $f\colon\thinspace M\to N$ is homotopic to an immersion if and only if the differential $df\colon\thinspace TM\to TN$ is homotopic to a bundle epimorphism. I'm also aware of the subsequent work of Thomas
MR0225332 (37 #926) Thomas, Emery On the existence of immersions and submersions. Trans. Amer. Math. Soc. 132 1968 387–394. 
giving applications of Phillips' theorem. However, I couldn't find any more modern references dealing with the case $M$ closed. Are there any, or is there a good heuristic reason why such conditions cannot be given?
Remark: By Ehresmann's Theorem, and since $M$ is compact, it is equivalent to ask whether $f$ is homotopic to the projection of a locally trivial fibration.

REPLY [6 votes]: Let $F$ be the homotopy fibre of $f$ (ie the space of pairs $(x,u)$, where $x\in M$ and $u$ is a path from $f(x)$ to a specified baspoint in $N$).  If $f$ is homotopic to a submersion $f'$, then (using Ehresmann) $F$ will be homotopy equivalent to a closed manifold $(f')^{-1}\{\text{point}\}$, and in particular, it will be equivalent to a finite CW complex.  I do not know whether the converse is also true, but I suspect that any counterexamples would be quite exotic.  So the first thing to do is to try to estimate the size of $F$.
Let $k$ be a field, and take cohomology with coefficients in $k$.  At least is $N$ is simply connected, there is an Eilenberg-Moore spectral sequence $\text{Tor}_{H^*N}(H^*M,k)\Rightarrow H^*F$, where $F$ is the homotopy fibre.  If $H^*M$ is a free module over $H^*N$ then this collapses to an isomorphism $H^*F=H^*M\otimes_{H^*N}k$, and this will be a finite-dimensional $k$-algebra.  I think it will also automatically have Poincare duality (provided that $M$ and $N$ are oriented).  If $H^*M$ is not free over $H^*N$ then the $E_2$ page will typically be very large, and in some cases it may be possible to show that no possible pattern of differentials will leave a finite-dimensional $E_\infty$ page.  If so, we can conclude that $f$ is not homotopic to a submersion.
Alternatively, if $f$ is homotopic to a submersion then the tangent bundle $\tau_M$ maps surjectively to $f^*(\tau_N)$, which implies that the Stiefel-Whitney polynomial $w(\tau_M,t)=\sum_iw_i(\tau_M)t^{\dim(M)-i}$ is divisible by $f^*w(\tau_N,t)$ in $H^*(M;\mathbb{Z}/2)[t]$.  This should generally be easy to check.  There is a similar criterion with Chern polynomials of the complexified tangent bundles in integral cohomology.<|endoftext|>
TITLE: Why S^3-K and SL(2,R)/SL(2,Z) are diffeomorphic? Here K is a trefoil in S^3.
QUESTION [11 upvotes]: I've heard this result from my differential manifold class, and I don't know how to prove it.
Does anyone know how to construct such diffeomorphism? Please tell me, thanks a lot.
Any comments are welcome.

REPLY [5 votes]: There is a proof without words in Etienne Ghys' ICM address. and a short proof with words by Rubinstein and Gardiner. (Compositio, 1979)<|endoftext|>
TITLE: Advice on doing physics under the umbrella of mathematics and the converse
QUESTION [5 upvotes]: Note: This is a question directly copied from Theoretical Physics SE primarily to get the advice of people indulged in mathematics.
In the current scenario of research in QFT and string theory (and related mathematical topics), which of the following would an undergraduate student, like me, be advised to do and why if s/he is interested in both the foundations and pushing the frontiers of these subjects and doesn't differentiate between them or thinks that different set of skills and attitudes are required (plus the student is not so strict about mathematical rigor and even precise, logical, hand waving arguments suffice to convince him/her)?
Get into a maths department for graduate studies and work under people having similar interests and probably also take advice from people in physics department.
Get into a physics department for graduate studies and work under people having similar interests and probably pick up the mathematics one needs along the way.
Moreover, what would such a student majoring both in mathematics and physics advised to concentrate on during his/her undergraduate education?
It seems to me after reading this post and general experiences of mine that often mathematicians are more willing to accept physical ideas than physicists willing to accept mathematical ideas. Also, the number of people willing also seem to be more in mathematics, at least to me. Also, it seems that some physicists often develop a kind of hatred for mathematics and always remain skeptical that any mathematics can ever do good to physics. On the basis of this, I am inclined towards option 1. Please feel free to correct me.

REPLY [9 votes]: As someone working in the field of mathematical physics, I believe that for a person with your interests it is somewhat better to be a graduate student in the math department. The reason is while there are quite a few mathematicians which regard physics as an important source of problems and insights, there are far fewer physicists who think that mathematical physics is worth working on. In particular, foundational issues in QFT and strings are not very popular with most physicists.<|endoftext|>
TITLE: Link between internal groupoids and stacks on a topos ?
QUESTION [12 upvotes]: Hello !
If I a have a grothendieck Site (C,J), I can consider :

The Stacks on (C,J) : category fibered in groupoid over C which statisfy suitable descent condition with respect to the covering sieve of J...
Internal Groupoid in the topos Sh(C,J).

Clearly those two notions are very close. but not exactly equivalent. I found a lot of book/article/thesis who define stacks but none of them were in term of the internal logic of the topos and it appears to me that some properties would be a lot more natural if they were stated in terms of an internal groupoid and the internal logic.
My feeling is that stacks corresponds to internal groupoids "up to weak categorical equivalences" (we invert the functor between groupoid which are internally fully faithfull and essentially surjective). But that this equivalence rely on the (external) axiom of choice. Am I right ? (and if I am, how can we make this more explicit ? )
Oh, and What about Higher Stacks ?
Thank you !

REPLY [12 votes]: The $2$-category of internal groupoids in $Sh(C,J)$ is equivalent to the $2$-category of sheaves of groupoids on $(C,J)$. To see this, given a groupoid $G$ in sheaves,the strict $2$-functor $$C \mapsto \left(G_0\left(C\right) \rightrightarrows G_1\left(C\right)\right)$$ is a sheaf of groupoids. So, there is a $2$-functor
$$F:Gpd(Sh(C,J)) \to Sh_J(C,Gpd) \to St(C,J)$$ which sends $$G \mapsto a(\left(G_0\left(\mspace{3mu}\cdot \mspace{3mu} \right) \rightrightarrows G_1\left(\mspace{3mu}\cdot \mspace{3mu}\right)\right),$$ where $a$ is the stackification $2$-functor.
I claim that $a$ is essentially surjective. Indeed, by the Grothendieck construction, one can see easily that every stack is equivalent to a strict $2$-functor from $C^{op}$ into the $2$-category $Gpd$ of groupoids. Such a strict $2$-functor, is the same as a functor into the $1$-category of groupoids, which is the same as a groupoid object in presheaves (by the same argument). One can sheafify this into a sheaf of groupoids, and then apply $F$ to it, and the result is equivalent to the starting stack.
The fact fact that stacks are equivalent to groupoids in sheaves up to "weak categorical equivalence" can be found in (I believe) in Bunge's Stack completions and Morita equivalence for categories in a topos, but can also be proven directly by examining $F$. As far as the internal logic and the axiom of choice, you are on to something, but I am not a logician. But, what I can say, is that the axiom of choice fails, in the sense that not every epimorphism splits in a topos, and if every epi DID split, then the notion of Morita equivalence would agree with the notion of categorical equivalence of internal groupoid objects.
What about higher stacks? Well, one can see this by the fact that given a site $(C,J),$ there is a model structure on simplicial $J$-sheaves whose associated $(\infty,1)$-category is equivalent to the $\infty$-topos of $\infty$-sheaves. Ok, so how does this answer your question? Well, for simplicity, lets work with the projective model structure. There is the global projective model structure on $Sh_J(C)^{\Delta^{op}}$, which does not yet  model $\infty$-sheaves and its associated infinity category corresponds to its fibrant objects which are those simplicial sheaves which are object-wise Kan-complexes, i.e., infinity groupoids. It follows that its associated $(\infty,1)$-category is the $(\infty,1)$-category of $\infty$-groupoid objects in sheaves. One can then left-Bousfield localize this model structure with respect to $J$-covering sieves, and one arrives as the local model strucrure on simplicial sheaves, and its associated $\infty$-category is the $\infty$-topos of $\infty$-sheaves. At the level of $\infty$-categories, this localization corresponds to a left-exact localization $$Sh_\infty(C,J) \leftrightarrows\infty-Gpd\left(Sh(C,J)\right).$$ The left-adjoint corresponds to fibrant replacement at the level of model categories, and $\infty$-stackfiication at the level of $\infty$-categories. So, every higher stack arises from an internal higher groupoid in $Sh(C,J)$.<|endoftext|>
TITLE: how to make the category of chain complexes into an $\infty$-category
QUESTION [22 upvotes]: I'd like to have some simple examples of quasi-categories to understand better some concepts and one of the most basic (for me) should be the category of chain complexes.
Has anyone ever written down (more or less explicitly) what the simplicial set corresponding to the quasi-category associated with the category of (say unbounded) chain complexes on an abelian category looks like?
I am not looking for an enhancement of the derived category or anything like this, I'm thinking of the much simpler infinity category where higher morphisms correspond to homotopies between complexes. My understanding is that the derived category should then be constructed as a localization of this $\infty$-category.
I am guessing my problem lies with the coherent nerve for simplicial categories.

REPLY [15 votes]: As everybody's said, there's an obvious thing to do. As Yosemite Sam cites, it's done in Section 13 of the ArXiv version of DAG I -- you think of chain complexes as enriched over simplicial sets via Dold-Kan, and then apply the nerve construction.
But there's an explicit thing you can do for any dg category, and I find it useful because it's given in terms of formulas. Moreover there's an obvious (if tedious) way to generalize this formula for any $A_\infty$-category so it's a cool thing to know. It's in the latest (February 2012) version of Higher Algebra. Since chain complexes obviously form a dg-category, this explicit method might be what you're looking for in case you want to produce some simplices in your quasi-category.
Specifically, Construction 1.3.1.6 tells you how to get a quasi-category from any dg category. Then Construction 1.3.13 and Remark 1.3.1.12 should convince you that it's equivalent to the "Dold-Kan + Simplicial nerve" construction cited by everybody else. (Lurie summarizes this equivalence in Proposition 1.3.1.17.) I would write out the formulas here but I don't want to re-TeX the long discussions. So here's at least a link to the latest Higher Algebra.<|endoftext|>
TITLE: Equivariant colimits and homotopy colimits
QUESTION [6 upvotes]: Suppose we are given a diagram of topological spaces. We can restrict ourselves to the diagrams over finite partially ordered sets and let all spaces be good enough (e.g. CW-complexes). One can take the colimit and homotopy colimit of this diagram and there is a set of rules to manipulate these objects, I mean first of all:
1) If $D_1\to D_2$ is coherent map of diagrams and $D_1(p)\to D_2(p)$ is homotopy equivalence for each $p$, then there is a homotopy equivalence $hocolim D_1\to hocolim D_2$.
2) If diagram $D$ is cofibrant in the Reedy category of diagrams (or some other equivalent conditions on $D$), then the canonical map $hocolim D\to colim D$ is a homotopy equivalence.
The question is about the equivariant version of these statements. Suppose topological group G acts (e.g. at the left) on each space of the diagram and all the maps in the diagram are equivariant. Then colim D and hocolim D seems to carry the induced G-action. 

1) Is it true that: If $D_1\to D_2$ is coherent G-equivariant map of diagrams of G-spaces and $D_1(p)\to D_2(p)$ is equivariant homotopy equivalence for each $p$, then there is an equivariant homotopy equivalence $hocolim D_1\to hocolim D_2$?
2) What conditions should be imposed on the diagram $D$ to guarantee the equivariant homotopy equivalence $hocolim D\to colim D$?

To be strict, here by hocolim I mean the specific topological space (not the homotopy type) given by the bar-construction.
What are good references for this subject? I had found some references on hocolims and colims in the categories of $G$-spaces. But I still don't understand, what is the connection between these objects and usual colims and hocolims with induced $G$-action.

REPLY [12 votes]: Shulman's paper ``Homotopy limits and colimits and enriched homotopy theory''
(on arXiv) gives a thorough study of homotopy colimits in enriched contexts,
and his enriching category $V$ can be taken to be the category of $G$-spaces
for a topological group $G$.  It specializes to answer your questions, in greater
generality than you ask.  The domain category for your diagrams, which I will 
denote $D$ as in Shulman, is implicitly a classical category (discrete 
hom sets), whereas Shulman allows hom objects in $V$.  He defines "corrected'' 
weighted homotopy colimits in terms of two-sided bar construction in sections 13 and
20 and shows in his Theorem 13.7 and Corollary 13.12 that this gives the appropriate 
left derived functors of weighted colimits.  (This becomes more general and explicit 
in his Theorems 20.4 and 21.1)  The upshot is that anything you can prove for spaces 
will generalize appropriately to G-spaces, with the usual kind of CW replacements. Making $D$ explicit helps answer your last quandary.  If your $D$ is just a plain 
category, a functor from $D$ to $G$-spaces is just a diagram of $G$-spaces and you just 
see things in spaces but with induced $G$-actions.  The story is more interesting when 
$D$ has $G$-spaces of morphisms between objects and you consider enriched diagrams 
($G$-maps of $G$-spaces on Hom's).<|endoftext|>
TITLE: Mathematical foundations of Quantum Field Theory 
QUESTION [29 upvotes]: Is there any reasonable approach, essentially different from Wightman's axioms and Algebraic Quantum Field Theory, aimed at obtaining rigorous models for realistic Quantum Field Theories? (such as Quantum Electrodynamics).
EDIT: the reason for asking "essentially different" is the following. It is possible to intuitively think "states" as solutions of the equations of motion (in some sense, in a "multiparticle space"). In realistic interacting QFT, the equations of motion are nonlinear. So, according to my chosen intuitive concept, a reasonable state space should be nonlinear (something like a Hilbert manifold). Meanwhile, in Wightman or AQFT frameworks, state spaces are Hilbert spaces. This seems to correspond with the fact that it is very very difficult to construct interacting QFT's in these frameworks. So, as a desire... there should be a different, more interaction-friendly framework where realistic models arise in a more natural way.
Does something in this direction already exist?

REPLY [6 votes]: Essentially nothing like the what you describe in the more detailed form of your question exists for the important reason that your "intuitive conception" of a hypothetical non-linear state space is incorrect.  
The linear structure in a quantum theory HAS NOTHING TO DO with whether or not the equations of motion are linear.  It is an exact feature that persists independent of any interactions. Indeed the Hilbert space arises from the fact that a configuration of a quantum system is given as a complex valued linear functional from the vector space $V$ which has a basis in bijective correspondence with the possible classical states of the system.  (Typically these functionals are required to be square normalizable in a suitable sense but this issue is secondary to the present discussion.) Linear functionals of course always have the structure of a vector space and as a result so too does the state space of a quantum mechanical theory including in particular quantum field theory.   
As a specific example consider quantum mechanics of a single particle moving on a line $\mathbb{R}$. The space of classical configurations is of course the real line, meaning exactly that to specify the all classical information means to specify a function $x(t)$ which tells you at which point $x$ the particle can be found at time $t$.  Quantum mechanically this is modified as follows.  We introduce a vector space $V$ with a basis of states in bijective correspondence with the set of all allowed classical configurations.  Thus in this example $V$ has a basis for each point in $\mathbb{R}$.  The Hilbert space of the theory is then the dual space (over $\mathbb{C}$) to $V$.  This is true no matter what non-linear interactions may be occurring, and a wide variety of exactly solvable non-linear models exist.  In this example the dual space is spanned by Dirac delta functions, typically denoted by $|x\rangle$ which indicates the functional that takes a non-zero value on the basis element corresponding to point $x$ and zero at all other basis elements associated to points $y\neq x$.  As a result the Hilbert space can be described as (suitably normalizable) distributions from $\mathbb{R}$ to $\mathbb{C}$. The interpretation of an element of this Hilbert space $\Psi(x): \mathbb{R}\longrightarrow \mathbb{C}$ is that the square of its norm specifies the probability distribution for the quantum mechanical particle to be observed at the position $x$.  The time evolution of the wave function $\Psi(x)$ is then always governed by the linear Schrodinger equation.
In a quantum field theory the abstract structure is the same, however $V$ is typically a much larger vector space since it has a basis in correspondence with fields, i.e. functions from space to say $\mathbb{R}$.<|endoftext|>
TITLE: Why is the classifying space of the natural numbers homotopy equivalent to the circle?
QUESTION [21 upvotes]: Is there a direct way to seeing that $B{\mathbb{N}}\simeq S^1$, i.e. the classifying space of the monoid of natural numbers is homotopy equivalent to the circle? 
Here, since the natural numbers ${\mathbb{N}}$ is not a group, some care is needed to define the classifying space $B{\mathbb{N}}$ properly. One way to do this is to consider ${\mathbb{N}}$ as a discrete simplicial monoid, then set $B{\mathbb{N}}:=|N{\mathbb{N}}|$ to be the geometric realization of its nerve.
This fact is a special case of (yet surprisingly, equivalent to) a larger theorem of James, namely that James' construction $J[X]$ on a pointed simplicial set $X$ is weakly equivalent to $\Omega\Sigma |X|$. Here $J[X]$ is the free simplicial monoid on $X$ modulo the basepoint $*$. Take $X=S^0$ gives $|N{\mathbb{N}}|\cong |NJ[S^0]|\simeq B\Omega\Sigma S^\simeq |\Sigma S^0|\simeq S^1$.

REPLY [10 votes]: Applying the functor "free $\mathbb Z$-module", the abelian monoid becomes a commutative ring $\mathbb Z[T]$, a polynomial ring in one variable. The bar construction, a simplicial set, becomes a simplicial abelian group: the bar construction for the augmented $\mathbb Z$-algebra. It follows that the homology of the classifying space is $Tor^{\mathbb Z[T]}(\mathbb Z,\mathbb Z)$, which is the homology of the circle. That, plus the fact the fundamental group is what it should be, plus the fact that the classifying space inherits its own commutative monoid structure, gives the result.<|endoftext|>
TITLE: Hausdorff and Naive Set Theory 
QUESTION [13 upvotes]: Erhard Scholz, in his article "Felix Hausdorff and the Hausdorff edition" writes the following:
"Hausdorff considered the contemporary attempts to secure axiomatic foundations for set theory as premature.  Working on the basis of a 'naive' set theory (expressedly understood as a semiotic tool of thought), he nevertheless achieved an exceptionally high degree of argumentation."
I have some questions regarding the above quote:

Is there anything to Hausdorff's view that the attempts to secure axiomatic foundations for set theory were premature (of course modern set theorists have, by their actions, essentially said 'no', but it still might be interesting to reconsider the question)?   I take the term "premature" to mean that axiomatizing a certain area of mathematics 'freezes' mathematical practice in that area for critical analysis, such as producing independence results and that Hausdorff's view (possibly) was that the attempts of his contemporaries to secure axiomatic foundations for set theory were premature in that current mathematical practice regarding set theory was not mature enough to provide adequate axiomatizations (of course Godel's results show that any attempt to axiomatize set theory might arguably be 'premature').
If Hausdorff worked in a 'naive' set theory "expressedly (Scholz's term) understood as a semiotic tool of thought" (whatever that means), how did he resolve the paradoxes in his 'naive' set theory?
Can one adequately resolve the paradoxes of naive set theory (Russell's, Burali-Forti, and Cantor's, etc.) via semiotics and if so, why is no one seeming to work in this 'reformed' naive set theory?

REPLY [9 votes]: I'll attempt an answer to question 1. Hausdorff was entitled to
think that set theory was not yet mature, because his own 1914
book made considerable advances on what had been done previously
(notably by Cantor and Zermelo). It is worth reading the glowing
review in the 1920 Bulletin of the AMS to see how his book
changed the perception of set theory by mathematicians. Just to
mention two of his contributions: definition of a topological
space, and the paradoxical decomposition of the sphere that paved
the way for the Banach-Tarski paradox.<|endoftext|>
TITLE: $(\infty, 1)$-Yoneda embedding via the Grothendieck construction
QUESTION [6 upvotes]: Let $C$ be a quasi-category. Then there is an imbedding
$$ C^{op} \times C \to \mathrm{Kan}$$
where $\mathrm{Kan}$ is the quasi-category of Kan complexes. This is essentially constructed in Lurie's book by choosing a model for $C$ as the nerve of a fibrant simplicial category $\mathfrak{C}$ and then taking the nerve of the ordinary Yoneda imbedding $\mathfrak{C} \times \mathfrak{C}^{op} \to \mathrm{Kan}$. 
However, by the Grothendieck construction, the Yoneda imbedding should correspond to a left fibration (i.e., a quasi-category cofibered in Kan complexes) over $C \times C^{op}$. Is there a direct way to construct such a left fibration? I'm wondering if there is a way to do this without appealing to the theory of simplicial categories. I'd even be happy with something weakly equivalent to this left fibration in the covariant model structure. 
In Lurie's book, a notion of bifibration (as Mike Shulman observes, this is elsewhere called a two-sided fibration) over a product $S \times T$ of simplicial sets is introduced, to correspond to the idea of a Kan complex functorial in two variables, but covariantly in one and contravariantly in another. I'm not very familiar with this theory, but a more general question would be how to turn a bifibration into a right or left fibration.

REPLY [11 votes]: Let $\mathcal{M}$ be the simplicial set defined by the formula $Hom( \Delta^{J}, \mathcal{M} ) =Hom( \Delta^{J^{op} } \star \Delta^{J}, \mathcal{C} )$, so that an $n$-simplex of $\mathcal{M}$ is a $(2n+1)$-simplex of
$\mathcal{C}$. The inclusions of $\Delta^{J^{op} }$ and $\Delta^{J}$ into $\Delta^{J^{op} } \star \Delta^{J}$ induce a left fibration $\mathcal{M} \rightarrow \mathcal{C}^{op} \times \mathcal{C}$, which is the left fibration you are looking for. For details see section 4.2 of my paper "Derived Algebraic Geometry X", entitled "Twisted Arrow $\infty$-Categories".
Your more general question can be phrased as follows: given a coCartesian fibration
$q: X \rightarrow S$ classified by a functor from $S$ into $\infty$-categories, how can you explicitly construct a Cartesian fibration classified by the same functor? First, construct a simplicial set $Y$ with a map $Y \rightarrow S$, such that maps $T \rightarrow Y$ classify maps $T \rightarrow S$ together with maps from $X \times_{S} T$ to the $\infty$-category of spaces. Then $Y \rightarrow S$ is a coCartesian fibration, whose fibers over a vertex $s \in S$ is the $\infty$-category of presheaves on $X_{s}^{op}$. Restricting to representable presheaves determines a full simplicial subset of $Y_0 \subseteq Y$, and the map
$Y_0^{op} \rightarrow S^{op}$ is the Cartesian fibration you're looking for.<|endoftext|>
TITLE: is f a polynomial provided that it is "partially" smooth?
QUESTION [30 upvotes]: Let $f$ be a $C^\infty$ function on $(c,d)$ ,and
let $O=\cup_{n\in \mathbb{Z}^+} (a_n,b_n)$ where $(a_n,b_n)$  are disjoint open interval in $(c,d)$ and $O$ is dense in $(c,d)$.
Suppose for each $n\in \mathbb{Z}^+$ ,$f$ coincides with a polynomial on $(a_n,b_n)$.
Is it necessary that $f$ coincide with a polynomial on $(c,d)$?

REPLY [22 votes]: The existence of such functions being established by now, there remains a curiosity
for the concrete situations that may generate them. Here is one, which is also
interesting as an example of a $C^\infty$, non-analytic solution, of a simple  first order linear delay differential equation.
Let's consider the  problem
$$f'(x)=\lambda f(3x),\qquad x < \frac{1}{2}  $$
where we seek for a function $f\in C^1(\mathbb{R})$ with support in
$I:=[0,1]$, and with the condition $f(x)=f(1-x)$.
I will show that it has a simple eigenvalue $\lambda= \frac{9}{2}$,
corresponding to an
eigenfunction with $f(\frac{1}{2})=1$ (see the details below). Then, it is
very easy
to check that $f$ is locally polynomial on the complement of the Cantor
set; and it
is, of course, $C^\infty$.
Indeed, since $f (3x)=0$ for $  x > \frac{1}{3}$, from the equation we
have that
$f'(x)$ vanishes identically on the interval $(\frac{1}{3},\frac{2}{3})$,
so that
$f$ is the constant $1$ therein. By induction, it is easy to see that $f$ is
polynomial of degree $k-1$ on each component of the complement of the
closed set
$F_k$ in $I$, inductively defined by $F_0:=I$ and $F_{k+1}:= \frac{1}{3}F_k
\cup\left(\frac{1}{3}F_k + \frac{2}{3}\right)$. Note that  $\{F_k\} _
{k\in\mathbb{N}}$ is the well-known nested sequence of closed sets whose
intersection is the Cantor set.
$$*$$
Construction. Consider the linear   operator $T$ on
$L^\infty(0,1)$ (thought as subspace of $L^\infty(\mathbb{R})$ via extension by zero),
defined by
$$Tu(x)= \cases{  \int_0^{3x}u(t)\, dt\qquad \qquad 0 < x < \frac{1}{3}\\\ 0
\qquad \qquad \qquad \qquad \frac{1}{3} < x < \frac{2}{3}\\\
\int_0^{3(1-x)}u(t)\, dt  \qquad\frac{2}{3}< x < 1.} $$
Since $T$ and $T ^ 2$ are integral operators with non-negative kernels,
their norms are attained on the constant function $1$ on $[0,1]$. We have, by simple
computations $T1(x)=3\min(x,1-x)\chi _ { F _ 1 }(x) $,
$\|T\| = \|T1\| _ \infty=1$ and $\|T^2\|=\|T(T1)\|_\infty=\frac{1}{3}$.
Therefore $\rho(T)\le\frac{1}{\sqrt 3} < \frac{2}{3}$ and $I-\frac{3}{2}T$ is
invertible. The function
$$f:= \left(I-\frac{3}{2}T\right) ^ {-1}\chi _
{[\frac{1}{3},\frac{2}{3}]} = \sum _ {k=0} ^ \infty  \left( \frac{3}{2}\right) ^ k T
^ k \chi _
 { [\frac{1}{3}, \frac{2}{3}] } $$
verifies
$$f(x)=\frac{3}{2} \int_0^{3x}f(t)\; dt\qquad \mathrm{for\; all}\quad 0 < x <
\frac{1}{3}\, ,$$
$$f(x)=1\qquad \mathrm{for\, all}\quad \frac{1}{3} < x < \frac{2}{3}\, ,$$
and $$f(x)=f(1-x)\qquad \mathrm{for\, all}\quad x\in\mathbb{R} \, .$$
The only points in $\mathbb{R}$ where the continuity of $f$ is not immediate and has
to be checked, are  $\frac{1}{3}$ and  $\frac{2}{3}$. We have
$$\int_ 0 ^ 1 f(x)dx=\int _ 0 ^ \frac{1}{3} f(x)dx+\int _ \frac{1}{3} ^ \frac{2}{3}
f(x)dx+\int _ \frac{2}{3} ^ 1 f(x)dx= 2\int _ 0 ^ \frac{1}{3} f(x)dx+ \frac{1}{3}\,
,$$
and
$$ 2  \int  _ 0  ^ \frac{1}{3}    f(x)dx 
=  3\int _ 0 ^ \frac{1}{3}  \int _ 0 ^ {3x} f(t) dt dx= \int _ 0 ^ 1 \int _ 0 ^ x
f(t) dt dx=$$
$$= \int _ 0 ^ 1 (1-t) f(t) dt = \int _ 0 ^ 1 tf(1-t)dt=   \int _ 0 ^ 1 tf(t)dt = $$
$$ =\frac{1}{2}\left(\int _ 0 ^ 1 (1-t)f(t)+ tf(t)dt\right) =\frac{1}{2} \int _ 0 ^
1  f(t)dt\, . $$
So
$$  \int _ 0 ^ 1f(t)dt = \frac{2}{3}\, ,$$
which implies the continuity of $f$ at $  \frac{1}{3}$ (and by symmetry at $
\frac{2}{3}$ ), since
$$f(\frac{1}{3}-)=\frac{3}{2}\int_ 0 ^ 1 f(x)dx =1= f(\frac{1}{3}+) \, .   $$
As a consequence, $f$ actually satisfies
$$f(x)=\frac{3}{2} \int_0^{3x}f(t)\, dt  $$
for all $  x < \frac{2}{3}\, ,$
and we conclude that $f$ is in $ C ^ 1(\mathbb{R})$,   and has the stated properties.

Here is  a Pudding Function $f$, and its integral function, $F(x):=\int_0^x f(t)dt$. Note how the latter coincides with the first third of the former, up to rescaling. Simple considerations based on odd and even symmetry show that $F(1)=\frac{2}{3}$.
Edit 2021 I learned from this answer by Gerald Edgar that  a function quite similar (but not easily related) to this is known since 1966 as the Fabius function. It verifies $F'(x)=2F(2x)$ instead of $f'(x)=\frac{9}{2}f(3x)$.<|endoftext|>
TITLE: Where else do the (topology) separation axioms turn up?
QUESTION [16 upvotes]: As an undergraduate I learned point-set topology from Munkres's book, as did many others.
One topic that gets a lot of attention is the separation axioms. For example, a space $X$ is normal if any two closed, disjoint subsets of $X$ can be separated by open neighborhoods.
Some of the axioms (e.g. Hausdorff) turn up a lot, but I feel like I virtually never hear topological spaces described as regular or normal (and there are a host of other properties too!) However, there are some fields I don't interact with too much.
In what contexts outside of general topology and set theory do these axioms, and the theorems you prove from them (Tietze extension, Urysohn's lemma, etc.), play an important role?
Thank you! -Frank

REPLY [5 votes]: I highly recommend Chapter 16 of Eric Schechter's Handbook of Analysis and its Foundations (AP, 1997), which deals with separation axioms and their individual importance in quite a bit of detail. One reason that separation axioms are implicit (rather than explicit) in many mathematical discussions outside general topology is that basic topological spaces (such are $\mathbb{R}^n$) that we use to construct other spaces already satisfy all most of the separation axioms, and these properties survive these constructions so do not need to be checked separately (key words here are hereditary, productive, initial).
If you squint at some of the separation axioms, you'll notice they mostly have to do with pairs of points and pairs of open/closed sets containing them. Thus, separation properties of a topological space $X$ can be naturally expressed as topological properties of the Cartesian product space $X\times X$ and its diagonal subset, $X\to X\times X$. Recall that relations on $X$ are subsets of $X\times X$, so one might expect some non-trivial interplay between relations on topological spaces (like orders or equivalence relations) and separation axioms. For instance, some separation properties fail to survive the quotient of a topological space by an equivalence relation. Then they have to be discussed explicitly. For example, famously the space of leaves of a smooth foliation of a manifold may fail to be Hausdorff.<|endoftext|>
TITLE: fundamental groups of smooth projective variety.
QUESTION [10 upvotes]: Is there a discrete group G which is the fundamental group of a compact Kahler
manifold but which is not the fundamental group of any smooth projective complex algebraic variety?
It is known that there are cohomology rings of compact Kahler manifolds not realisable by smooth projective complex algebraic varieties (some of the recent results are due to Voisin).

REPLY [8 votes]: As far as I know, still open. In the 2011 paper HOMOMORPHISMS BETWEEN FUNDAMENTAL GROUPS OF KA ̈HLER MANIFOLDS, one Botong Wang shows that there is a homomorphism between Kahler groups not realized by a holomorphic map between projective varieties, but this seems to be closest to a counterexample.<|endoftext|>
TITLE: Yang-Mills and Chern-Simons functionals as Morse functions
QUESTION [13 upvotes]: Can the Yang-Mills or Chern-Simons action functionals be considered as [possibly perfect] Morse functions? I assume we would be in an equivariant scenario due to considering the configuration spaces with gauge-groups/transformations. Or at least how far away are they from a Morse-Bott function (and from being perfect)?

REPLY [11 votes]: There's a generalization of Morse-Bott called Morse-Bott-Kirwan 
that you can read about in Kirwan's book. Basically this condition
guarantees that the unstable sets are manifolds, but not the stable sets, 
so the negative of a function that's Morse-Bott-Kirwan may not be.
If one defines a "Yang-Mills functional" very generally to be the
norm-square of a moment map, Kirwan proves that for $M$ finite-dimensional,
this norm-square is a perfect Morse-Bott-Kirwan function. (Of course the
original example of such is on $M$ a space of connections, where
Atiyah-Bott did the same, as I recall.)

REPLY [8 votes]: This got too long for a comment.
Atiyah and Bott showed that the Yang-Mills functional on a Riemann surface is equivariantly perfect, i.e. it's perfect for gauge-equivariant (integral) cohomology.  To be a little more precise, they showed that a certain stratification (the Harder-Narasimhan stratification) of the space of connections is perfect in this sense, and Daskalopoulos showed (using Uhlenbeck compactness among other things) that this stratification does in fact agree with stable manifolds of the Yang-Mills functional.  (Atiyah-Bott had conjectured this, but did not prove it in their paper.  Note that Uhlenbeck's compactness theorem came just after Atiyah-Bott.)
For non-orientable surfaces, the situation is different: in some cases the YM functional is "anti-perfect" in a certain sense, and in some cases it's neither perfect nor anti-perfect.  These ideas are discussed in recent work of Melissa Liu and Nan-Kuo Ho, and also in recent work of Tom Baird.<|endoftext|>
TITLE: Triangle theorems and cubic curves
QUESTION [5 upvotes]: One may think of a plane triangle as a kind of degenerate cubic curve.  Many Euclidean geometry theorems posit a map from plane triangles to plane triangles (e.g., take all the angle bisectors) and then predict that one gets an even more degenerate output.
Can one lift these theorems (systematically?) to general (or at least less degenerate) cubic curves?

REPLY [11 votes]: Any triangle is equivalent under projective transformations to the standard triangle whose sides are on the lines $x=0,y=0,z=0$ of the projective plane. If you select a point $(0:y_1:z_1),(x_2:y_2:0),(x_3:0:z_3)$ on each of these lines not at the vertices, then they are collinear if and only if $(y_1/z_1)(z_2/x_2)(x_3/y_3)=-1$ as you can check. This is Menelaus's theorem of classical plane geometry. You can think of the lines as copies of $\mathbb{G}_m$ and make a group out of $\mathbb{G}_m \times \mathbb{Z}/3$ in a suitable way and interpret Menelaus's theorem as the three points are collinear if and only if they add to zero in this group. Once you do that, it is easy to view this statement as a degenerate case of the fact that three points on a cubic are collinear if and only if they add to zero in the group law.
Ceva's theorem, which guarantees that certain lines from the vertices of a triangle (e.g. bisectors) are collinear, is the dual of Menelaus and thus admits a similar, but less obvious, interpretation.<|endoftext|>
TITLE: How to balance writing professionally and being able to have a conversation with your reader?
QUESTION [27 upvotes]: Background.
I am someone at the very very beginning stages of my career as a mathematician, and am, for example, almost ready to (re)submit my first paper. I have already read a lot of essays on how to write well (i.e. http://terrytao.wordpress.com/advice-on-writing-papers/ is golden), but I still sometimes have doubts on what the best course of action is in certain situations. So that's the reason of this question. If it gets closed, because the only reasonable answer is 'it depends', I completely understand. But it would be great if you could bear with me for a moment.
Problem. To me, an article is (or at least feels like) a written monologue. But, on the other hand, the purpose of a paper is to explain new ideas to your reader, which generally is much more easy in a dialogue. Because in a dialogue you can interact with someone, ask questions to make sure he or she understands and re-explain things in a different way if he/she so desires. So if I am writing mathematics, I try to remind the reader where we are in the argument, why we are there, where we are going next, what the main ideas are behind all the calculations, etc. Let me give you an example: when I read something mathematical that Timothy Gowers wrote, it is immediately clear to me that his goal is to make me understand his ideas. If I remember correctly, his article on his proof of Szemeredi's Theorem by Fourier-analytic methods is over 100 pages long, but his writing style is very leisurely and he first goes through the case of arithmetic progressions of length 4, to make some important ideas clear, to show what else is needed to complete the proof, etc. To me, this is a delight. So I'd like to model my writing style after his. But, unfortunately, I'm not Timothy Gowers. So I'm not sure if I can get away with writing like he does, or if being that leisurely and involved with the reader actually makes me look unprofessional. And thus my question is:

When in doubt, should I acknowledge the fact that I am a human being, writing and explaining concepts to another human being? Or should I try not to break 'the fourth wall'?

REPLY [15 votes]: I can't help but reiterate here Serre's great maxim: "Precise yet informal". Striving for this goal in mathematical writing will avoid your looking unprofessional, but will make you look more human at the same time. It seems that mathematics that does this best is very professional (I can't help but think of Milnor's style, here).
Regarding writing that appears purely to guide the reader between theorems: as far as the guidance ALSO is precise yet informal, there should be no foul. It is when these comments are not really true...but close to true...that they should be omitted or left to the talk. Too many "sort-ofs" should be avoided in the body of a mathematical paper. Serre's maxim seems a good guidepost for this, too.

Comments to supplement original answer:
Also, as long as mathematical papers are valued for being interesting it seems that there will always be a bit of 'ought' along with the 'is'. Otherwise, one could just rattle along formally proving theorems without regard to how they will be received. This is evidence of the human part of mathematics that stands in opposition to the dispassionate precision of formal proof. Elegance is far from formal perfection, and is perhaps closer to human utility or aesthetic sense. Somehow, Serre's maxim expresses something important about mathematical life and work.<|endoftext|>
TITLE: Groups where every two generator subgroup is free
QUESTION [6 upvotes]: Is there a name for the class of groups in the title, and any sort of characterization? Free groups and surface groups are in the class, but presumably there are many more...

REPLY [3 votes]: There is a widely used term binary finite (бинарно конечная, in Russian) group, see papers of Shunkov, Chernikov and their school. This means a group where all 2-generated subgruops are finite.
For instance, the Kourovka Notebook contains the following question (still open, as far as I know).
Question 4.74(b) (V.P.Shunkov, 1973). Does there exist a simple infinite binary finite 2-group?
Also, I stumbled across binary solvable and binary nilpotent groups... As you can guess, I am hinting that
you may call your groups 
binary free
if you like this terminology.<|endoftext|>
TITLE: A characterization of Lagrange multiplier. Where to find a proof?
QUESTION [7 upvotes]: Let $F,G\in C^1(\mathbb{R}^n,\mathbb{R})$. Assume for 
$s\in(s_0-\varepsilon,s_0+\varepsilon)$, 
\begin{align}
E(s) = \min F\quad\mbox{subject to}\quad G=s
\end{align}
is achieved at some $x(s)\in\mathbb{R}^n$. If $\nabla G(x(s))\ne 0$, then
\begin{align}
\nabla F(x(s)) = \lambda(s)\nabla G(x(s))
\end{align}
for some $\lambda(s)\in\mathbb{R}$. $\lambda(s)$ is called the Lagrange multiplier. This is well-known and is taught in a standard Calculus course. In contrast, it's until recently that I know there is a rule says 
\begin{align}
E'(s_0)=\lambda(s_0),\tag{1}
\end{align}
and I'm already a PhD student (in math). I suspect it's my problem, and asked some of my friends (also PhD students in math) whether they ever saw this rule, and the answers are totally no. So far, I still cannot find a clearly written down theorem and proof concerning this fact. Note that if $x(s)$ is itself differentiable, the proof is very simple:
Since $s=G(x(s))$, $1 = \nabla G(x(s))\cdot x'(s)$. And hence
\begin{align}
E'(s)=\nabla F(x(s))\cdot x'(s)=\lambda(s)\nabla G(x(s))\cdot x'(s)=\lambda(s).
\end{align}
However, it's also not hard to give counterexamples. 
I saw this rule in a paper, where the domain of $F,G$ is a Banach space and the authors justify it directly, instead of applying some general theorem, and their proof uses some convexity property of the functional they studied. Suspecting it should be true in some general setting and ought not to be justified case by case, I finally found that indeed if the minimizers $x(s)$ can be chosen so that 
$x(s)\to x(s_0)$ as $s\to s_0$, then (1) holds. And it seems OK too when the domain of $F,G$ is a Banach space.
The problem is: I do need this fact in my current paper, and I think it's so ``classical'' that it needs not to be proved there. Does anyone know a reference for it?

REPLY [3 votes]: I am sure you noticed this, but here is a simple counterexample which shows that you need to assume the continuity of the minimizer $x(s)$ at $s_0$: 
$$F(x_1,x_2) = x_2 \cos x_1, \qquad G(x_1,x_2) = x_2.$$
Then $E(s)=-|s|$ and, for $s\neq 0$, $\lambda(s)= - \mathrm{sgn} \; s$. For $s=0$, $\lambda(0)$ can be any number bewteen -1 and 1, depending on which minimizer you choose for the constant function $F(x_1,0) \equiv 0$.<|endoftext|>
TITLE: Is there an explicit formula for $\sum x^n/(n!)^2$?  
QUESTION [5 upvotes]: Let $f(x)=\sum_{n\geq 0}\frac{1}{n!n!}x^{n}$, is there an explicit formula for $f(x)$?

REPLY [15 votes]: This is, for $x \ge 0$, a special case of the modified Bessel function of the first kind. Have a look here: http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html
$$
\sum_{n = 0}^{\infty} \frac{x^{n}}{(n!)^{2}} = \mathrm{I}_0 \left(2 \sqrt{x}\right)
$$<|endoftext|>
TITLE: Using Polya's enumeration theorem for explicit generation of instances
QUESTION [5 upvotes]: I have a simple application of Polya's enumeration theorem to counting nonisomorphic pentago boards with a given number of stones.  In terms of the theorem, the color generating function is $f(x) = 1+x+y$ corresponding to empty, black, or white, there are $6 \times 6 = 36$ beads, and the group is $Z_4^4 \rtimes D_4$ of size 2048.  This all works well.
Next I want to explicit generate a set of representative instances, one from each isomorphism class, without generating all instances and reducing.  Is there an analogue of Polya's theorem that would assist in explicit enumeration of instances?
If necessary I can do this by manually unwrapping the structure of my particular group, but am curious if there is a more abstract technique applicable to arbitrary groups.

REPLY [3 votes]: There are two very general methods for this type of problem, called orderly generation, and generation by canonical construction path.  One place that describes these is the book Classification Algorithms for Codes and Designs (Springer 2006) by Kaski and Östergård (where they call the second method "canonical augmentation"). The second method would certainly work for this, and maybe the first as well.
However, in this case you have quite a small group and a more brutal approach should be faster if coded carefully.  Define some (lexicographic-type) order on all boards.  Then find those boards which are mimimal in their equivalence class according to that order. The crudest way is to generate all boards and reject those which are mapped by a group element to a smaller board; this is easily feasible for your problem.  However, by defining the order carefully, you should be able to do much better than that.  For example, define the order so that each of the four sub-boards of a minimal board are also in some minimal form (according to some ordering of the set of sub-boards), then make all the minimal sub-boards in advance. Then for each set of four minimal sub-boards, throw it out if it isn't minimal overall.
[Added:] Thinking about this a bit more: There are about 5000 equivalence classes of sub-board; make them all anyhow.  Then you just need to select four of these in inequivalent ways according to the action of $D_8$.  The inequivalent boards with four inequivalent sub-boards are the tuples $(A,B,C,D)$, $(A,C,D,B)$ and $(A,D,B,C)$ with $A\lt B\lt C\lt D$; i.e., coset representatives of $D_8$ in $S_4$. Then there more types but much smaller counts where some of the sub-boards are equivalent.<|endoftext|>
TITLE: Pólya's conjecture on the spectra of the Laplacians
QUESTION [7 upvotes]: Recently I've learned something about the spectra of the Laplacians. Given a bounded domain $\Omega \subset \mathbb{R}^n$ with $\partial \Omega$ smooth, we can consider eigenfunctions of Dirichlet type, i.e. $u \in C^2(\Omega)\cap C(\partial \Omega)$ s.t. $-\triangle u=\lambda u$ and $u|_{\partial \Omega}=0$. By standard results in functional analysis, $-\triangle$ has a discrete spectrum $0<\lambda_1 \leq \lambda_2 \leq \cdots$ with $\lambda_k \to +\infty$.
A well-known asymptotic formula by Weyl says $\displaystyle \lambda_k \sim W_n (\frac{k}{V(\Omega)})^{2/n}$. We refer $W_n$ as the Weyl constant. And Pólya conjectured that $\displaystyle \lambda_k \geq W_n (\frac{k}{V(\Omega)})^{2/n}$ holds.
As far as I know, the best known result is due to Li and Yau. They proved the conjecture in the sense of "average": $\displaystyle \sum_{j=1}^k \lambda_j \geq \frac{nW_n}{n+2}{k}^{(n+2)/n}{V(\Omega)}^{-n/2}$. 
I find their argument is elementary, only employing some standard Fourier tricks. And the big picture is quite clear if put into the quantum framework. My question is in some sense "soft", but it does make me feel absurd: what makes it so difficult to estimate the eigenvalues one by one while their average is so well understood? Does anyone work on this problem by carrying further Li & Yau's analysis? I do know one instance: Kröger has transplanted their proof to Neumann settings, but how about the original Dirichlet problem?

REPLY [4 votes]: Since your question is "soft", I think it is okay if I give a soft answer in the case of a hyperbolic compact Riemann surface.
The Selberg trace formula describes the spectrum of the Beltrami-Laplace operator pretty well. You get an identity
$$ \sum\limits_{\lambda} f(\lambda) = \sum\limits_{\gamma} \widehat{f( \log \gamma)} + \dots ,$$
where $\lambda$ is asymptotic to the square of an eigenvalue and $\gamma$ to the lengths of closed geodesics. 
The best known error term for the Weyl law her is $O(\sqrt{T} / \log T)$, so approxiamtely the square root of the main term.
Laplace-Beltrami Operator on Surfaces
Eigenvalues of Laplacian-Beltrami operator
Now why can we not do better? One of the main problems, is that you are only allowed to plugin holomorphic $f$, so it will be hard to estimate single objects. On the other hand, if you would be allowed to pluggin something compactly supported, then $\widehat{f}$ becomes entire, and is not compactly supported.
The fact that the support of $f$ and $\widehat{f}$ can not be simultaneously small, is a first instance of the uncertainity principle of Fourier analysis. The name is certainly derived from the Heisenberg uncertainity prinicple, eigenvalues are here "waves" and length of closed geodesics here "particles".
Similar things are happening for prime numbers and zeros of the Riemann zeta function (Weil's excplicit formula) or in quantum chaos (Gutzwiller trace formula). Already in finite group theory, if you try to compare traces of irreducible representations with conjugacy classes.
Perhaps Paley-Wiener theorem's give you a good flavor for first instances of Fourier uncertainity. Stein-Shakarchi "Complex Analysis" has a good treatment, chapter 4, I guess.
Best, Marc.<|endoftext|>
TITLE: Is Zeta function discrete-analytic?
QUESTION [5 upvotes]: Let's define discrete-analytic functions as functions that are equal to their Newton series expansion:
$$f(x) = \sum_{k=0}^\infty \binom{x-a}k \Delta^k f(a)$$
My question is whether $\zeta(s,q)$ ($q$=const) is discrete-analytic against $s$?
That is whether its Newton series converges and is equal to the function itself.
For comparison, in the following graphic there are four functions:

red is the function $\zeta(x,3)$
blue is $\frac{\cos (\pi x)\psi_b^{(x+1)}(3)}{\Gamma(x+2)}$ where $\psi_b$ is the balanced polygamma
yellow is $\frac{\cos (\pi x)\psi^{(x+1)}(3)}{\Gamma(x+2)}$ where $\psi$ is the polygamma as implemented in Mathematica
green is the partial Newton expansion of the above functions taken at first 20 terms.

The three first functions and the Newton expansion, if it converges, have the same values at non-negative integer arguments.
notation
$\zeta(x,q)$ is the Hurwitz zeta function, LINK 
Balanced polygamma LINK

REPLY [7 votes]: I hope this is OK.
$$
\zeta(s,q) = \sum_{n=0}^\infty \frac{1}{(n+q)^s}
\tag{1}$$
Let's first consider just one term, $f(s) = 1/(n+q)^{s}$.  Then
$$
\Delta f(s) = f(s+1) - f(s) = \frac{1}{(n+q)^{s+1}} - \frac{1}{(n+q)^s}
=\left(\frac{1}{n+q}-1\right)\frac{1}{(n+q)^s}
$$
$$
\Delta^k f(s) = \left(\frac{1}{n+q}-1\right)^k\frac{1}{(n+q)^s}
$$
$$\begin{align}
\sum_{k=0}^\infty\binom{x-a}{k} \Delta^k f(a) &=
\frac{1}{(n+q)^a}\sum_{k=0}^\infty\binom{x-a}{k}\left(\frac{1}{n+q}-1\right)^k
\cr &=\frac{1}{(n+q)^a}\left(1+\frac{1}{n+q}-1\right)^{x-a}
\cr &= \frac{1}{(n+q)^x}=f(x)
\end{align}$$
We applied the binomial theorem, which requires
$$
\left|\frac{1}{n+q}-1\right| < 1
$$
so this works for $q>1/2$.  
Thus  The question is whether the convergence in (1) is good enough that we can interchange two sums and get our conclusion...  
But in fact for $q \gt 1/2, a \gt 1, x \gt 1$ we can interchange the sums in
$$
\sum_{k=0}^\infty\sum_{n=0}^\infty\frac{1}{(n+q)^a}\binom{x-a}{k}\left(\frac{1}{n+q}-1\right)^k
$$
because, for $k \gt x-a, n \gt 1$, all terms have the same sign.<|endoftext|>
TITLE: Analogues of the Riemann-Roch Theorem
QUESTION [15 upvotes]: In his thesis, Tate derives a Poisson formula on the adeles and a theorem which he calls the "Riemann-Roch Theorem". More specifically, given a continuous, $L^1$ function $f$ on the adeles such that certain sums converge uniformly, then for all ideles $a$, we have
$\frac{1}{|a|}\displaystyle\sum_{\xi\in k}\hat{f}(\xi/a)=\displaystyle\sum_{\xi\in k}f(a\xi)$.
Tate further refers to this theorem as the "number theoretic analogue of Riemann-Roch". My question is how this relates to the geometric Riemann-Roch theorems and why this deserves to be called an analogue of these theorems.

REPLY [6 votes]: Perhaps you might want to look in the book project of Frankenhuijsen http://research.uvu.edu/Math/machiel/RH.pdf and search for Riemann Roch theorem.

Quote from page 3: The function $ζ_C$ (zeta function of a curve) satisfies the functional equation $ζ_C (1 − s) = ζ_C (s)$. This functional equation can be proved using the Riemann–Roch theorem
  $$l(D) = deg D + 1 − g + l(K − D),$$
  which is the analogue of (1) above.<|endoftext|>
TITLE: Intermediate extension of a Prikry-Silver extension?
QUESTION [9 upvotes]: Prikry-Silver forcing $\mathbb{V}$ (sometimes just Silver forcing) is the forcing notion consisting of all partial functions $p:\omega\rightarrow 2$ with co-infinite domain. In "Combinatorics on ideals and forcing with trees" Marcia Groszek mentions (without proof) that a Prikry-Silver real has minimal real degree, but not minimal degree.
That the Prikry-Silver real $r$ has minimal real degree means that whenever $s$ is a real in $V[r]$ that doesn't belong to $V$ we have $V[r]=V[s]$. A proof of this can be extracted from some more general results in the seminally named "Combinatorics on ideals and forcing" by Serge Grigorieff.
That $r$ doesn't have minimal degree means that there is some object $A\in V[r]$ for which $V[A]$ is different from both $V$ and $V[r]$. Can anyone point me to a proof of this fact?

REPLY [11 votes]: I don't know a reference for this, but the $A$ that you are looking for is the collection of
all domains of conditions in the Silver generic filter.
Equivalently, you can consider the set of all complements of domains of conditions in the filter.  This is a non-principal ultrafilter on $\omega$.
Silver forcing can be considered as an iteration of the following two forcing notions:
First you force with $\mathcal P(\omega)/fin$, which gives you a nonprincipal ultrafilter on $\omega$, and then you force with Grigorieff forcing  with respect to that ultrafilter.  (Grigorieff forcing is like Silver forcing, only that the complements of the domains are in the filter, not just infinite.)
The first forcing, $\mathcal P(\omega)/fin$, is $\sigma$-closed and therefore doesn't add any reals at all.  The second forcing does all the adding of reals.
The extension is not minimal since you force with an iteration.  I am leaving out the details, 
but it shouldn't be hard to show that the iteration that I am talking about is equvalent to Silver forcing.  (Consider instead of $\mathcal P(\omega)/fin$ the equivalent p.o. of infinite subsets of $\omega$. Map each Silver condition $p$ to the pair 
$(\omega\setminus\mbox{dom}(p),p)$.  This should be a dense embedding of Silver forcing into the iteration.)
I don't have access to the Grigorieff paper or to MathSciNet right now, but possibly it says something about this iteration.<|endoftext|>
TITLE: Trivial cobordism group in dimensions 1, 3, 7 related to H-space structures on the spheres in these dimensions?
QUESTION [7 upvotes]: Is there a connection between the existence of H-space structures on $S^1$, $S^3$ and $S^7$ and the fact that every (closed) 1-manifold, 3-manifold and oriented 7-manifold is a boundary, or is this a pure numerical coincidence? 
[It's possible that my question amounts to groundless mysticism, but I had to wonder.]
On a related note: is it known which (oriented or not) cobordism groups are zero?

REPLY [3 votes]: How about $n\ne 0,2,6$ instead of $n=1,3,7$? The Adams theorem on the vanishing of the stable Hopf invariant, which implies the non-existence of an H-structure on $S^n$ for $n\ne 1,3,7$, also implies the theorem of R. L. W. Brown and Liulevicius (and Mahowald) that if $M^n$ immerses in $\Bbb R^{n+1}$, then $M^n$ is a boundary or is cobordant to $\Bbb RP^0$, $\Bbb RP^2$ or $\Bbb RP^6$ (and these three do immerse in $\Bbb R^{n+1}$).<|endoftext|>
TITLE: Does your dissertation matter for industry research jobs?
QUESTION [23 upvotes]: Suppose Mr A. is a graduate student who, for some reason, does not want to go into teaching or academic research.  Mr A. did a Doctoral dissertation, but it is in a narrow area.  This area is not relevant to several industries which hire mathematicians. Is it prudent for Mr. A to apply to these jobs anyway?
In other words:  if one wants to work in industry does the dissertation topic matter?

REPLY [3 votes]: I am a software engineer, working in an industry (approximation algorithms, computational lithography, semiconductor manufacturing) completely unrelated to my Math PhD (low-dimensional topology, moduli spaces). Sometimes I wish I majored in Applied Math instead. But then I remember why I majored in Pure Math to begin with and regret nothing.
I don't know about others, but for me going into Pure Math was not about developing a career but about following my intellectual curiosity. That doesn't make much practical sense, but neither does education in Music or Art History. People do that because the subjects are interesting to them, not because their education will be useful to somebody else. Similarly, the topic one one's dissertation should be something of interest to himself, regardless of practical applicability.
That said, education in Mathematics has this odd quality of being useful. Most of the things I do as an engineer involve Calculus, Statistics, Linear Algebra, and a lot of Numerical Analysis and Algorithms, and a little bit of Functional Analysis. Most of the above is a part of any graduate education in Mathematics. The rest could be picked up by an average mathematician in a relatively short time. Computer programming skills (not college level, but industrial quality) would be required too, and, alas, that's not taught sufficiently at Math departments.
Anyway, here's a nice slide that's completely contrary to what I just said:<|endoftext|>
TITLE: Finding hyperbolic metrics by approximation
QUESTION [13 upvotes]: Given a presentation $ < X ; R >$  of a group $G$. Suppose we know for some reason that $G$ is the fundamental group of a three-dimensional finite volume manifold. 
Then there is a injective group homomorphism $G\rightarrow Isom(\mathbb{H}^3)$. Mostows rigidity theorem then tells us that it is unique up to composition with inner automorphisms of $\mathbb{H}^3$ from the left (and probably automorphisms of $G$ from the right).
As usual to speficy a homomorphism from a group with a specific presentation to another group we have to pick a image for each generator, such that all relators get mapped to the neutral element. My hope is that there might be an algorithm that starts with some first choice of the images of the generators. Note that $Isom(H^3)$ can be viewed as a subgroup of $GL_4(\mathbb{R})$. Then I would like to iteratively minimize the sum of some norm of the images of the relators (hoping that it will converge to a homomorphism).
Of course there are also noninjective homomorphisms (like the trivial one) so I cannot hope that the sequence always converges to an injective homomorphism. But maybe it does with high probability for a reasonably random first choice of generators.
Has someone already done this ? If so I am interested in the convergence properties. They might also depend on the given presentation. So one could also hope that any fundamental group of a hyperbolic 3-manifold has a nice presentation.

REPLY [12 votes]: To my knowledge this hasn't been done in theory (although see Harriet Moser's thesis http://www.math.columbia.edu/~moser/).  But it certainly has been done in practice by Jeff Weeks' program SnapPea.  Note that $\mbox{Isom}(\mathbb{H}^3) = \mbox{PSL}(2, \mathbb{C}) = \Gamma$.  So your source group $G = \pi_1(M^3)$ already has a very nice matrix group as a target.  SnapPea assumes that the three-manifold $M$ is given as a triangulation.  (I am sure that it is not easy to go backwards from a presentation of $G$ to a triangulation of $M$.)  After tidying the triangulation (and drilling out a curve if necessary - but lets suppose that $M$ has a single torus boundary component) SnapPea gives all of the tetrahedra in the triangulation the same shape, namely that of the regular ideal tetrahedron.  "Developing" in $\mathbb{H}^3$ turns shapes of tetrahedra into matrices, one per generator.  Naturally this is not yet a representation of $G$ into $\Gamma$.  The failure to be a representation is measured by the failure of the shapes to satisfy the Thurston gluing equations.  SnapPea uses a multivariate version of Newton's method to find new shapes for the tetrahedra, hopefully converging to the discrete and faithful representation of $G$ into $\Gamma$.  
Details, references, and more can be found in Moser's thesis.  However, I'll add a final remark - the convergence properties of SnapPea's method certainly do depend sensitively on the initial triangulation.  There are certain triangulations where SnapPea will consistently produce wrong answers.  This is not a problem in practice -- you randomize the triangulation a few times and SnapPea typically starts to behave much better.   But finding an actual algorithm appears to be difficult.  So something mysterious going on. 
The problem (of finding the discrete and faithful representation) is easy to solve even for very respectfully sized manifolds (say up to 100 tetrahedra).  But why?  I certainly don't know.  You'd have to ask Thurston, Weeks, Hodgson or some other expert for an opinion.<|endoftext|>
TITLE: Asymptotics of the number of required Dehn relators in hyperbolic groups
QUESTION [7 upvotes]: If $G = \langle X | R \rangle$ is a $\delta$-hyperbolic group presentation, then Dehn's algorithm provides a linear time solution to the word problem, but the linear constant is horribly exponential in $\delta$ since the Dehn presentation consists of all words of length $8 \delta$ equal to the identity. 
Are there any known lower bounds on the number of relators required to make Dehn's algorithm solve the word problem?  In other words, does anyone know a family of groups for which the required number of Dehn relators is exponential in either the size of the original group's presentation or the original groups $\delta$.

REPLY [2 votes]: Here is an idea of an example (just for a start). Take the finite group $A_n$ (it is hyperbolic). It has a short presentation, see this paper. The total size of relations is something like $\log |A_n|$. I am sure the $\delta$ for that presentation is about $\log |A_n|$ also. A Dehn presentation of $A_n$ should require about $|A_n|$ relators. 
Actually one can take $S_n$ and generators $(i,i+1)$. The total size of the Coxeter presentation is $O(n^2)$. It would be interesting to find the $\delta$ for this presentation. It should be polynomial in $n$.  
 Update.  Here is another, more realistic, idea (actually it can be made into a complete answer with a little effort involving reading Gromov's paper or, better, a paper by Yan Ollivier). Take the Ramanujan expander $\Gamma_i$: the number of vertices in $i$-th graph $\Gamma_i$ is $n_i$, the degree of each vertex is a constant $k$, the girth, the diameter, and the maximal length of a basic loop is $\log n_i$, the rank of the fundamental group of $\Gamma_i$ is $\sim k^{\alpha\log n_i}$ for some $\alpha<1/2$. Consider the Gromov random group $G_i$ corresponding to the graph $\Gamma_i$. It is hyperbolic with $\delta=O(\log n_i)$, the number of relations in every Dehn presentation (with sufficiently small Dehn constant) is aproximately the number of relators (because the relators do not have large pieces in common), i.e. the number of relators in any Dehn presentation is exponential in $\Delta$.<|endoftext|>
TITLE: A random walk with uniformly distributed steps
QUESTION [22 upvotes]: The following problem has bothered me for a long time.
Let us imagine a point on the real axis. At the beginning, it is located at point $O$. Then it will "walk" on the real axis randomly in the following way. For every step of the "walk", it will choose a real number $\Delta x$ uniformly from the interval $[-1,1]$, turn right, and move $\Delta x$ unit. Once it reaches the left side of the point $O$, it will "die" immediately.
Our task is find out the probability of the point is alive after $n$ steps of "walk" $P_n$.
I guess that $P_n=\frac{(2n)!}{4^n (n!)^2}$, but I can't prove this or explain why it is true.

REPLY [12 votes]: Here is a somewhat different way from Johan's of looking at this
problem.  At each stage of the walk, choose a number $x$
uniformly from $[0,1]$ and then walk either a distance $x$ to the
right or $1-x$ to the left. This does not affect the probability
of becoming negative since there is still a uniform probability
of taking a step whose length belongs to the interval
$[-1,1]$. However, it does have the property that after taking
$n$ steps and choosing $0\leq x\leq 1$, the two possible
locations following the next step are the same modulo 1. Hence
the walk can be described as follows. Choose $n$ numbers
$0\lt x_1\lt \cdots\lt x_n\lt 1$, a sequence
$\epsilon=(\epsilon_1,\dots,\epsilon_n)$ of signs $\pm 1$, and a
permutation $w$ of $1,2,\dots,n$. Let the location be $y_k$ after
the $k$th step. If $\epsilon_k=1$ then step to the least real
number $y_{k+1}\equiv x_{w(k+1)}$ (mod 1), $y_{k+1}>y_k$. If
$\epsilon_k=-1$ then step to the greatest real number
$y_{k+1}\equiv x_{w(k+1)}$ (mod 1), $y_{k+1}\lt y_k$. But the
question of whether any $y_k$ is negative depends only on $\epsilon$ and
$w$, not the choice of $x_1,\dots,x_n$. There are $2^n n!$ ways
to choose $\epsilon$ and $w$. Is there a simple combinatorial
argument that the number of choices such that each $y_k>0$ is
$(2n-1)!!=1\cdot 3\cdot 5\cdots (2n-1)$? Then the probability of
success is $(2n-1)!!/2^nn! = (2n)!/4^nn!^2$.
Here is a reformulation of the combinatorial result that needs a 
simple direct proof. 
Let $f(n)$ be the number of pairs $(a_1a_2\cdots a_n,
b_1b_2\cdots b_{n-1})$ such that (a) $a_1 a_2\cdots a_n$ is a
permutation of $1,2,\dots, n$, (b) $b_i=0$ or $1$ if $a_i\lt
a_{i+1}$, (c) $b_i=0$ or $-1$ if $a_i>a_{i+1}$, and (d) $b_1
+b_2+\cdots+b_j\geq 0$ for all $1\leq j\leq n-1$. Then
$f(n)=(2n-1)!!$.
Update. The combinatorial result is proved bijectively by
 O. Bernardi, B. Duplantier, and P. Nadeau in Séminaire
 Lotharingien de Combinatoire, B63e (2010). In their citation [1]
 they use this result for the same purpose as above, i.e., to
 compute the probability $P_n$ (though they state the result a
 little differently).
Second update. The method above can be applied to the $[l,r]$
  generalization mentioned by Lwins in his comment. By rescaling we
  may assume $l=-1$. If we are at $y$ sometime during the walk, choose
  a number $x$ uniformly from $[0,1]$. With probability 1/2 step from
  $y$ to $y+\frac{r-1}{2}+\frac{r+1}{2}x$. With probability 1/2 step
  from $y$ to $y-1-\frac{r+1}{2}x$. This gives a uniform probability
  of stepping from $y$ to a point in the interval $[y-1,y+r]$. It has
  the property that once $x$ is chosen, the value of $y$ is determined
  modulo $\frac{r+1}{2}$. Thus the walk can be described as follows:
  pick uniformly and independently $0\lt x_1\lt \cdots\lt x_n \lt
  \frac{r+1}{2}$, 
  pick a permutation $w$ uniformly from the symmetric group
  $S_n$, and a sequence $\epsilon=(\epsilon_1,\dots,\epsilon_n)$ of
  independently distributed signs, with a probability of
  $\frac{r}{r+1}$ for a plus sign and $\frac{1}{r+1}$ for a minus sign.
  Go through the same procedure as above, working mod
  $\frac{r+1}{2}$ instead of mod 1. Again a proper walk (i.e., one
  which never becomes negative) depends only on $w$ and $\epsilon$,
  and we get the following result:
Theorem. The probability $P_n(r)$ that the walk is proper is
    given by 
   $$  P_n(r) = \frac{1}{(1+r)^nn!}\sum r^{1+f(w,\beta)}, $$
   summed over all pairs $w=a_1a_2\cdots a_n\in S_n$ and
    $\beta=(b_1,\dots, b_{n-1})\in \lbrace 0,\pm 1\rbrace^n$ satisfying
    the conditions (b) and (c) above, where $f(w,\beta)$ is the number
    of integers $1\leq i\leq n-1$ for which either $a_i\lt a_j$ and
    $b_i=0$, or $a_i\gt a_j$ and $b_i=1$. 
For instance, $P_2(r)= (r+2r^2)/2(r+1)^2$ and $P_3(r)
  =(r+8r^2+6r^3)/6(r+1)^3$. I conjecture that the numerator $N_n(r)$
  of $P_n(r)$ is just the polynomial $\sum B_{n,i}r^i$ defined by
  equation (4) of
  http://math.mit.edu/~rstan/pubs/pubfiles/29.pdf. This paper gives
  some additional information about the polynomials $\sum
  B_{n,i}r^i$. Much additional information can be found in the
  literature on Stirling permutations, e.g., Bona proves in
  http://wenku.baidu.com/view/dfa70012cc7931b765ce15e4.html that all
  zeros of this polynomial are real.
Third update. Alas, the conjecture in my second update is
  false. Unless there is an error in my code, the sequence of
  coefficients of $N_n(r)$ for $2\leq n\leq 7$ are $(1,2)$,
  $(1,8,6)$, $(1,25,55,24)$, $(1,69,361,394,120)$,
  $(1,176,1999,4416,3083,720)$,
  $(1,426,9836,41019,52193,26620,5040)$. It is easy to see why
  the leading coefficient of $N_n(r)$ is $n!$.<|endoftext|>
TITLE: Do (Banach) ultrapowers carry some sort of 'elementary equivalence'?
QUESTION [5 upvotes]: The (model-theoretic) ultrapowers had been used for studying elementary equivalnce of first-order structures. Then, they have been adapted to Banach spaces, which are, let me say, second-order creatures. Of course, one cannot expect that all the properties (propositions of the 'language of Banach spaces') can be persevered by ultrapowers. Indeed, an ultrapower of a reflexive space need not be reflexive.
But reflexivity is a somewhat complicated issue. What if we ask easier questions? What properties are preserved by ultrapowers? For instance, suppose that we are given a Banach space $X$ and each its subspace isomorphic to $X$ is complemented. Is such property preserved under ultrapowers?

REPLY [6 votes]: Say that a Banach space $X$ has property $K$ (for Kummers) provided every subspace of $X$ that is isomorphic to $X$ is complemented.  The classical separable, infinite dimensional spaces that have property $K$ include $\ell_2$, $c_0$, and $\ell_2 \oplus c_0$; the first obviously, the second because $c_0$ is separably injective, and the last is not hard to show. It is known that $\ell_p$  for $1\le p \not= 2 < \infty$ does not have property $K$,  and it is not hard to show that other possible examples, such as $\ell_2(c_0)$ and $c_0(\ell_2)$, fail property $K$. Ultrapowers of $\ell_2$ of course have property $K$. I don't know about ultrapowers of $c_0$. Are they all injective (which clearly implies property $K$)?  Probably not, but I did not try to check the literature.
However, every (complex) HI space [Gowers-Maurey] is not isomorphic to any proper subspace and hence has property $K$. Some real HI spaces have the same property.  Now there are HI spaces that contain uniformly complemented copies of $\ell_1^n$ for all $n$. Probably the original Gowers-Maurey space has this property, but, as Thomas Schlumprecht pointed out to me, it is clear that the HI space $AD$ of Argyros-Delyiani does, and $AD$ has property $K$.  Since $\ell_1$ is a dual space, there is an  ultrapower $Y$ of $AD$ that contains complemented copies of $\ell_1$ and hence $Y$ is isomorphic to $Y \oplus \ell_1$.  But $\ell_1$ contains an uncomplemented copy of $\ell_1$ [Bourgain] and hence $Y$ contains an uncomplemented copy of itself.
Therefore property $K$ is not preserved under ultrapowers.<|endoftext|>
TITLE: Example of a Sheaf (on the site of smooth manifolds) with Nontrivial Cohomology on $\mathbb{R}^n$?
QUESTION [8 upvotes]: Does such a sheaf of abelian groups exist?  If not, is there a reference or a proof?  Does such a sheaf of non-abelian groups exist?  
I realized recently that while I've taken it for granted that there wasn't such a sheaf on $\mathbb{R}^n$, I only have proofs that specific sheaves are acyclic. 
EDIT: The site of smooth manifolds is the category of smooth manifolds, endowed with the Grothendieck topology generated by defining surjective submersions to be "covers".  In more down to earth language, I'm asking whether there is a sheaf, defined naturally and intrinsically for all smooth manifolds, which has nontrivial cohomology on $\mathbb{R}^n$. 
The motivation for asking this is that I'm trying to understand the role of good covers in differential topology, i.e. can we always calculate cohomology of any sheaf just by picking a good cover and calculating the Cech cohomology on the cover, or do we in general need to pass to the limit over all covers?  If all sheaves on the smooth site have vanishing cohomology on $\mathbb{R}^n$, then good covers always work.  I'm trying to understand whether this is the case and why or why not.

REPLY [3 votes]: Another example, actually a lot more important, is to take a Hartogs figure, such at 
[
X:=D\times {0}\cup \partial D \times D \subset C^2
] 
which your are welcome to think of as $R^4$. 
Take a neighborhood $U$ of $X$.  Then the first cohomoloy of a small neighborhood $U$ with values in the sheaf of holomorphic functions is not zero.
I could spell t out if you are interested.
John Hubbard<|endoftext|>
TITLE: Are k-spaces named for Kelley?
QUESTION [9 upvotes]: On page 58 of Mark Hovey's book Model Categories, he states the following definitions:

"A subset $U$ of a space $X$ is
  compactly open if for every continuous
  $f:K\rightarrow X$ where $K$ is
  compact Hausdorff, $f^{-1}(U)$ is open
  in $K$... A space $X$ is called a
  $k$-space, or Kelley space, if every
  compactly open subset is open."
  (emphasis mine)

My question is whether $k$-spaces are called $k$-spaces for John L. Kelley or for some other reason. A quick google search shows me that Kelley studied these spaces a lot, and that he wrote about them as $k$-spaces.'' I interpret this as evidence that they are not named for him, since it's fairly uncommon to hear about a (good) mathematician X going around calling things by his own name. Further evidence for this is a statement an older professor made to me that $k$-spaces were studied by Mac Lane before Kelley. 
On the other hand, the word Kelleyfication appears in Mac Lane's Categories for the Working Mathematician (on page 182 of the first edition) as a way to change the topology on a Hausdorff space in order to make it a $k$-space. Furthermore, Mac Lane calls compactly generated Hausdorff spaces Kelley spaces.

1) Can anyone clear this mystery up for me? Does anyone know the first place these spaces appear in the literature, or the first place the category of $k$-spaces was put forth as the ``right'' category of spaces?
2) Is it standard in the literature to assume $k$-spaces are Hausdorff?

Hovey does not, but Mac Lane does. I'm curious about whether there is consensus on this issue.

REPLY [6 votes]: Just to add a bit to the history, the first time the exponential law was given for Hausdorff k-spaces was I believe in my  DPhil thesis, submitted 1961, see PtA available here, which was circulated to the obvious places. In my first paper, (1963),  also available here, I wrote:
"It may be that the category of Hausdorff k-spaces is adequate and convenient for all purposes of topology."
In my second paper, (1964), I used the category of  Hausdorff spaces and functions continuous on compact subsets, and showed it was what we now called cartesian closed. (My thesis contains an attempt at showing the idea of what we now call monoidal closed, since in my thesis I had lots of internal homs and associated products, usually a tensor.)
I did not understand final topologies at the time and so did not come up with the definition for the non Hausdorff case, but you can find that in my book Topology and Groupoids.
Several people wrote about the non Hausdorff case, but an important application is to fibred exponential laws which were developed by Peter Booth following some ideas sketched by R. Thom. See Booth, Peter I. The section problem and the lifting problem. Math. Z. 121 (1971), 273–287.
I also wonder whether the category defined in
Johnstone, P. T.
On a topological topos.
Proc. London Math. Soc. (3) 38 (1979), no. 2, 237–271.
is indeed adequate and convenient for all purposes of topology, in particular can cope well with fibred exponential laws, since being a topos is a stronger condition.
There are also the purposes of analysis, for which see
Kriegl, Andreas; Michor, Peter W. The convenient setting of global analysis. Mathematical Surveys and Monographs, 53. American Mathematical Society, Providence, RI, 1997.
So the term "convenient" has had a good run!
See also convenient.<|endoftext|>
TITLE: Crossing number of the Grötzsch graph
QUESTION [5 upvotes]: Related wikipage: http://en.wikipedia.org/wiki/Gr%C3%B6tzsch_graph
Is the crossing number of the Grötzsch graph known? I have heard it conjectured to be 5 (certainly it is no greater), but came up empty-handed in my search of the literature.

REPLY [9 votes]: The crossing number of the Grötzsch graph is 5.
Crossing numbers are believed to be difficult to compute in general.  (The corresponding decision problem is NP-hard.)  However, for small graphs and small crossing numbers, it is possible to find an optimal planar drawing.  For example, see Markus Chimani's thesis "Computing Crossing Numbers" for more information.
The Open Graph Drawing Framework (OGDF) can compute crossing numbers.  After compiling the program on the linked page and entering the Grötzsch graph, my computer computed that the optimal planar drawing has 5 crossings.  Let me emphasize that this technique is exact, not heuristic.<|endoftext|>
TITLE: Do equivariant crepant resolutions always exist?
QUESTION [6 upvotes]: Let $X_t$ be a family of algebraic varieties (my interest is Calabi-Yau varieties, but I don't think that's important) over $\mathbb{C}$, smooth for $t \neq 0$, on which a group $G$ acts fibre-wise.  Suppose further that $X_0$ admits at least one crepant resolution.  Does there always exist an equivariant crepant resolution?  If not, are there conditions under which such exists?

REPLY [6 votes]: Consider $\mathbb Z/2$ acting on $\{xy-zw=t\}$ by $x\leftrightarrow y$. This swaps the two small resolutions of the central fibre (the 3-fold ordinary double point $xy=zw$ in $\mathbb C^4$). So there can't be an equivariant small resolution.
(A formal proof might go along these lines: $\mathbb Z/2$ does act on the blow up of the ODP, swapping the two rulings of the $\mathbb P^1\times\mathbb P^1$ exceptional divisor. The small resolutions are contractions of this blow up. If $\mathbb Z/2$ acted on one of them, it would act on its $H^2$. Pulling back, its action on $H^2(\mathbb P^1\times\mathbb P^1)\cong\mathbb Z\oplus\mathbb Z$ would be the identity on the contracted $\mathbb Z$ summand, contradicting the fact that it swaps the summands.)
However if your (finite?) group action permutes the singular loci with no stabilisers then there would surely be a crepant resolution.<|endoftext|>
TITLE: Infinite loop space structure of $BU^+$
QUESTION [10 upvotes]: It is well-known that $BU^+$ is homotopy equivalent to an infinite loop space where $U$ is the limit of the unitary groups $U(n)$ for $n \rightarrow \infty$ and $+$ denotes Quillen's Plus construction. 
On the other hand there is a tool to detect infinite loop spaces by the action from $E\Sigma_p$ ($\Sigma_p$ is the symmetric group).
My question is: how does the map $E\Sigma_p \times_{\Sigma_p} BU^p \rightarrow BU$ looks like.
Is it possible to write it down.

REPLY [10 votes]: There are a number of equivalent ways of seeing $BU\times Z$ or $BU$ as an infinite loop space.
The description that Neil gives is the action of the linear isometries operad $\mathcal{L}$ (complex version) on $BU$.  A large number of related spaces have such a structure, as explained in the first 
section of the first chapter of "$E_{\infty}$ ring spaces and $E_{\infty}$ ring spectra", SLN 577
(1977).  Morever, the Bott maps are maps of $\mathcal{L}$-spaces, which ties in Bott's original proof that $BU$ is an infinite loop space.  The fact that $\coprod_n U(n)$ is a permutative category gives 
a quite different operad action on $\coprod_n BU(n)$, whose associated infinite loop space is 
$BU\times \mathbf{Z}$.  It is not obvious a priori how to compare the infinite loop structures on
these two models.  The question is resolved in this and related examples (e.g. $BTop$) in my paper
"The spectra associated to $\mathcal{I}$-monoids".  Math. Proc. Camb. Phil. Soc. 84(1978), 313--322.<|endoftext|>
TITLE: Limit of $\frac{1}{n}\sum_{r=1}^n\frac{n\ (\mathrm{mod}\ r)}{r}$
QUESTION [12 upvotes]: By accident I came across the following,
$$\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^n\frac{n\ (\mathrm{mod}\ r)}{r}=0.4227\ldots\approx 1-\gamma,$$
where the numerator is the remainder of $n$ divided by $r$. Is it known whether we have equality in the above expression or is it just a numerical coincidence? Has this been studied?
Edit: I'm sorry for all the (important) typos, everything should be fixed now.

REPLY [19 votes]: This follows by elementary computation: we have
\begin{align}
\sum_{r=1}^n\frac{n\bmod r}r&=\sum_{r\le n}\frac nr-\sum_{r\le n}\left\lfloor\frac nr\right\rfloor\\\\
&=nH_n-|\{(r,s)\in\mathbb N^2:1\le rs\le n\}|\\\\
&=nH_n-2\sum_{r\le\sqrt n}\left\lfloor\frac nr\right\rfloor+\lfloor\sqrt n\rfloor^2\\\\
&=nH_n-2nH_{\lfloor\sqrt n\rfloor}+n+O(\sqrt n),
\end{align}
hence
\begin{align}
\frac1n\sum_{r=1}^n\frac{n\bmod r}r&=H_n-2H_{\lfloor\sqrt n\rfloor}+1+O(n^{-1/2})\\
&=\log n+\gamma-2\log\lfloor\sqrt n\rfloor-2\gamma+1+O(n^{-1/2})\\\\
&=1-\gamma+O(n^{-1/2}).
\end{align}<|endoftext|>
TITLE: Why can I divide an affine variety by the action of the general linear group?
QUESTION [8 upvotes]: Let $G\subseteq\mathrm{Gl}_n(\mathbb{C})$ be a subgroup of the general linear group and assume that $\rho:G\to\mathrm{Gl}(V)$ is a representation. Understand the complex vector space $V$ as an affine algebraic variety. Then, it appears to be well-known that the quotient $V/G$ has the structure of an algebraic variety such that the quotient map $\pi:V\to V/G$ is a morphism of varieties. However, I cannot find a proof for this statement. There is an abundance of proofs for the case where $G$ is finite, using the Reynolds operator and corollaries of Hilbert's basis theorem, but I would like to see a proof in the general case. Thanks very much in advance!
Edit: I forgot to mention that I assume $G$ to be reductive and the action on $V$ is regular. You are free to assume even more about $G$ if that allows you to provide a comprehensible reference for a proof.

REPLY [7 votes]: To amplify what others have pointed out, it doesn't make sense to get an affine variety here as a "quotient" unless all orbits of G are closed (a condition not met even by the natural action of the general linear group) and the variety itself has regular functions given by a finitely generated algebra: the fixed points of G on the algebra of polynomials in n variables if $n=\dim V$.  At the least you need G to be a Zariski-closed subgroup of the general linear group, but the natural condition for finite generation of polynomial invariants is typically satisfied in characteristic 0 by focusing just on reductive algebraic groups.    
All of this goes back to classical invariant theory and Hilbert's 14th problem, but by now it has become part of the broader study of GIT.    Thanks to the proof by Haboush of Mumford's Conjecture, reductive groups over an algebraically closed field of any characteristic yield finitely generated algebras here, but more is needed to get a genuine quotient.<|endoftext|>
TITLE: Pushforwards of stacks of algebras?
QUESTION [5 upvotes]: This is a refined/sheafified version of this previos question of mine.
Let $(X,\mathcal{O}_X)$ be a ringed space or more in general a ringed stack, where the structure sheaf $\mathcal{O}_X$ is a sheaf of $\mathbb{K}$-algebras for some field $\mathbb{K}$. Then a global section of $\mathcal{O}_X$ can be thought of as a "scalar field" on $X$. In the "categorical progression" of higher $\mathbb{K}$-vector spaces, the field $\mathbb{K}$ is the 0th level. 1st level are classical $\mathbb{K}$-vector spaces, and so the "1st level version" of a section of $\mathcal{O}_X$ is (at least roughtly) a field of $\mathbb{K}$-vector spaces on $X$. A natural formalization of this rough idea is that of going from the sheaf $\mathcal{O}_X$ to the stack $\mathcal{O}_X$-Mod of sheaves of $\mathcal{O}_X$-modules on $X$ (maybe with suitable  assumptions, e.g., one could be considering only quasi-coherent sheaves of $\mathcal{O}_X$-modules). A remarkable property of $\mathcal{O}_X$-modules is that they can be pushed forward along morphisms of ringed spaces. Now, the structure sheaf $\mathcal{O}_X$ we started with is in particular an $\mathcal{O}_X$-module, and pushing it forward along the terminal morphism $\pi:X\to \{pt\}$ we precisely get sections of $\mathcal{O}_X$ mentioned above. 
Now we can make a further step, and go from the stack of $\mathcal{O}_X$-modules to the 2-stack of $\mathcal{O}_X$-algebras (with $\mathcal{O}_X$-$\mathcal{O}_X$-bimodules as 1-morphisms and morphisms of bimodules as 2-morphisms). My question is: can $\mathcal{O}_X$-algebras be pushed forward along morphisms of ringed spaces? (under which hypothesis?). In particular, considering $\mathcal{O}_X$ as an $\mathcal{O}_X$-algebra, what is the $\mathbb{K}$-algebra $\pi_*\mathcal{O}_X$? (the prototypical conjectural example of this in my mind is the following: if $G$ is a finite group, then $\pi_*\mathcal{O}_{pt//G}$ is Morita equivalent to $\mathbb{K}[G]$, the group algebra of $G$).
In the above, an $\mathcal{O}_X$-algebra is to be thought as of a placeholder for its category of modules, so in a non-affine situation it is actually deceitful to reason in terms of algebras and a better description would be thinking of the "second step" as $\mathcal{O}_X$-linear categories, with additive functors and natural transformations of these. In this more general setting, the pushforward to a point of $\mathcal{O}_X$ would be the "pushforward of the 2-category of $\mathcal{O}_X$-linear categories" and this should be a $\mathbb{K}$-linear category, but not necessarily the category of representations of an algebra. In particular, by abstract nonsense I would expect this pushforward to be the category of $\mathcal{O}_X$-modules.
Note that going one step backwards instead of one step forward, the existence of a pushforward is nontrivial: the pushforward of a section of $\mathcal{O}_X$ along $X\to \{pt\}$ is an integration of fuctions on $X$, so it requires additional data to be defined (a "measure").

REPLY [8 votes]: Yes, there are several different formalisms that achieve this - for example it's treated in Lurie's DAG XI or Toen-Vezzosi Caractères de Chern, traces équivariantes et géométrie algébrique dérivée in the derived context, and in many places (eg Gaitsgory The notion of category over an algebraic stack) in the underived context. Basically as I understand your question you're looking at an appropriate class of quasicoherent sheaves of categories on a variety or stack $X$ - which, given enough compact objects (see eg Toen's derived Azumaya paper) are monadic over $QC(X)$, i.e. are the same as module categories for $O_X$-algebras.
But in fact such objects can be said much more concretely: all geometric stacks
(eg quasicompact with affine diagonal) are "affine" from the point of view of these sheaves, ie there's no difference between such a quasicoherent sheaf of categories and its global sections (this is in DAG XI).
So one can just take the following as definition: we're just looking at module categories for the symmetric monoidal (dg or $\infty$- if you like) category $QC(X)$ of quasicoherent sheaves on $X$. In this language pushforward and pullback of the kind you ask are super easy, just as they are for modules over ordinary rings: given a map $X\to Y$ you have a "homomorphism" (symmetric monoidal functor) $QC(Y)\to QC(X)$ (pullback), and you can use this to forget $QC(X)$ modules down to $QC(Y)$, i.e. push forward, or (the left adjoint of this) tensor a $QC(Y)$-module by $QC(X)$ (pullback).
Moreover under some hypotheses ($X\to Y$ faithfully flat or $X\to Y$ proper and $X,Y$ smooth) descent holds - i.e. quasicoherent sheaves of categories on $Y$ are the same as those on $X$ with descent data. Said another way there's a Morita equivalence of monoidal categories between $QC(Y)$ and $QC(X\times_Y X)$ (theorem of Francis, Nadler and myself
which will be posted, at least on my webpage, sometime in the next couple of weeks).
In the case of $pt \to pt/G$ this recovers a Morita equivalence between categories with$G$ action and categories over $BG$.<|endoftext|>
TITLE: Symmetric functions on three parameters being perfect squares
QUESTION [11 upvotes]: Is it possible for $x+y+z, xy+yz+zx$, and $xyz$ to be perfect squares at the same time for positive integer values of $x,y,z$?

REPLY [18 votes]: Yes.  By straightforward search the smallest example is
$\lbrace x,y,z \rbrace = \lbrace 45,64,180 \rbrace$, with
$$
(t+45) (t+64) (t+180) = t^3 + 17^2 t^2 + 150^2 t + 720^2.
$$
Given any solution $(x,y,z)$ we may produce infinitely many others
(other than the trivial scaling $(c^2 x, c^2 y, c^2 z)$) by using
the theory of elliptic curves to find rational $z'$ such that
$x+y+z'$, $xy+yz'+z'x$, and $xyz'$ are all squares, at which point
$(d^2 x, d^2 y, d^2 z')$ works for any integer $d>0$ such that
$d^2 z' \in {\bf Z}$.  For example, in $\lbrace 45,64,180 \rbrace$
we may replace $64$ by $(460163992/28591599)^2$, and then multiply
through by $28591599^2$ to obtain the new solution
$$
\lbrace  
 28591599^2 \cdot 45, \phantom{+} 460163992^2, \phantom{+} 28591599^2 \cdot 180 \rbrace .
$$
A complete parametrization is not possible, because it would be
tantamount to a rational parametrization of the surface
$$
S: xy + yz + zx = r^2, \phantom{and} (x+y+z)xyz = s^2
$$
in projective $(1:1:1:1:2)$ space, and that surface is K3.
If I did this right, $S$ is a "singular" K3 surface, i.e. has Picard number
$20$ which is maximal for a K3 surface in characteristic zero, and the
Néron-Severi group ${\rm NS}(S)$ has rank $20$ and discriminant $-48$,
and consists of (classes of) divisors defined over ${\bf Q}(i)$.
It is actually quite common for
natural Diophantine equations to give rise to K3 surfaces of
maximal or nearly-maximal Picard number, but that's a story for
another time.<|endoftext|>
TITLE: n-dimensional "cross product" reference request
QUESTION [5 upvotes]: I have written a paper which involves a "cross product" in $\mathbb{R}^n$ and I would like to have a reference to point to. 
Let ${\bf e_1}, \dots, {\bf e_n}$ be the standard basis for $\mathbb{R}^n$ and let 
${\bf w_1} = (w_{11},\dots,w_{1n}), \dots, {\bf w_{n-1}}=(w_{n-1\;1},\dots,w_{n-1\;n}) \in \mathbb{R}^n$. Then one can define
$${\bf w_1} \times {\bf w_2} \times \cdots \times {\bf w_{n-1}} = \begin{vmatrix} {\bf e_1} & {\bf e_2} & \cdots & {\bf e_n} \cr w_{11} & w_{12} & \cdots & w_{1n} \cr \vdots & \vdots & & \vdots \cr w_{n-1\;1} & w_{n-1\;2} & \cdots & w_{n-1\;n} \end{vmatrix}$$
where the right hand side is a "determinant". 
Note: One can express this "cross product" in terms of exterior algebra operations. It is equivalent to $*({\bf w_1} \wedge {\bf w_2} \wedge \cdots \wedge {\bf w_{n-1}})$ where "$*$" is the Hodge dual operator. 
Obviously if ${\bf a}, {\bf b} \in \mathbb{R}^3$, then ${\bf a} \times {\bf b}$ is the regular cross product and this $(n-1)$-airy product has the same properties as the regular cross product (it is a vector perpendicular to the vectors being multiplied and the length of this vector is given by the $(n-1)$-volume of the parallelotope spanned by the ${\bf w}$'s).
It was pointed out that this product appears in Susan Colley's "Vector Calculus" text [I have a second edition where this product is explored in problems 29-31 in section 1.6 on "$n$-dimensional geometry"]. Colley said she didn't have a good reference to point to (she couldn't recall where she'd seen it before). 
I guess I could just refer to her book, but was wondering if anyone knows a better/historical reference? Or if not a paper does anyone know who I should attribute this to?
Alternatively, it would be nice to know if there is no "original" reference to point to and that this is just common knowledge/folklore.
Thank you!

REPLY [14 votes]: Here's a bit of history on cross products that, if not directly useful, will hopefully provide some context. They were defined in "Beno Eckmann, Stetige Losungen linearer Gleichungssysteme, Comment. Math. Helv. 15(1943)" as follows: An $r$-fold cross product on a real vector space $V$ of dimension $n$ with inner product $g$ is a continuous map
\[
P : \underbrace{V \times \cdots \times V}_r \to V
\]
satisfying
1) $P$ is skew;
\[
g\big( P(v_1, \ldots ,v_r),v_i\big) = 0 \ , \ 1 \leq i \leq r \ ,
\]
2) $P$ respects $g$;
\[
g\big(P(v_1, \ldots ,v_r),P(v_1, \ldots ,v_r)\big) = \det g(v_i,v_j) \ .
\]
These were classified by Eckmann and Whitehead (see "George W. Whitehead, Note on cross-sections in Stiefel manifolds, Comment. Math. Helv.37 (1962/1963)") using algebro-topological methods. They were later also classified by Brown and Gray (see "Robert B. Brown and Alfred Gray, Vector cross products, Comment. Math. Helv. 42 (1967)") where those authors included an extra axiom: $P$ has to be multilinear. This extra axiom makes no significant difference to the classification. The classification theorem is:
An $r$-fold cross product on a real vector space $V^n$ exists if and only if we have one of
$\bullet$ $n$ even, $r=1$,
$\bullet$ $n=7$, $r=2$,
$\bullet$ $n=8$, $r=3$,
$\bullet$ $n$ arbitrary, $r=n-1$.  
The proof of Brown and Gray uses Hurwitz' structure theorem for composition algebras. If you add a dimension to $V$ you can define a composition algebra, and vice versa. Their paper is my favourite reference. They actually consider a more general situation where the bilinear form is indefinite, that leads to more cross products (but only in the four cases above). They even work with any field of characteristic not $2$.
Using the standard inner product on $V = \mathbb{R}^n$ (which you are implicitly using by referring to $\ast$), your cross product is the last one on the list (it is a cross product in this sense, I was wrong to say otherwise earlier). It is normally just called the volume form, with the appropriate identifications made. I don't know about the earlier history of that particular case, or whether this was thought of as a cross product prior to the papers I've mentioned.
As Ryan points out, in two dimensions there is a $1$-fold cross product, and this is rotation by $90^{\circ}$. That's because $1$-fold cross products are the same as complex structures. The usual $2$-fold cross product on $\mathbb{R}^3$ is the volume form, and fits into the fourth case of the classification.
I cannot miss the chance to briefly mention the role of cross products in geometry. One can define a cross product on the tangent bundle of a Riemannian manifold. Kaehler manifolds are those with (parallel) $1$-fold cross products, and the cross products in seven and eight dimensions correspond to the exceptional holonomies.<|endoftext|>
TITLE: Projections which are not completely bounded
QUESTION [5 upvotes]: There are 'canonical' examples of maps on operator spaces which are not completely bounded. Nevertheless, I couldn't produce any examples of bounded projections on relatively easy to understand operator spaces which are not c.b. In particular, I failed this task for $\mathcal{K}(H)$ and $\mathcal{B}(H)$, where $H$ is a Hilbert space (I tried projections onto certain subspaces isomorphic to $c_0$ and $\ell_\infty$, respectively). 
I would appreciate any examples.

REPLY [7 votes]: Symmetrisation (or anti-symmetrisation).
That is: let $T:K(H)\to K(H)$ be the transpose map and let $P=({\rm id}+T)/2$. Then $P:K(H)\to K(H)$ is a projection onto the subspace of symmetric [NOT self-adjoint] compact operators. Since $T$ is not completely bounded, $P$ is not completely bounded.
This argument also shows that the symmetrisation map $M_n\to M_n$ is a norm-one projection which has cb norm $\geq (n-1)/2$.
The examples you tried fail to work because of a general result: any bounded linear map into a minimal operator space will be automatically completely bounded. This is somewhere in the first third of Effros-Ruan's book, for example, though I don't have my copy here.<|endoftext|>
TITLE: Injectivity radius of the Sasaki metric
QUESTION [9 upvotes]: Let $(Q,g)$ be a (compact) Riemannian manifold with injectivity radius $\rho>0$. There is a natural metric $\tilde g$ on the tangent bundle $TQ$ which is known as the Sasaki metric and which makes $\pi:TQ\rightarrow Q$ a Riemannian submersion. Denote its injectivity radius by $\tilde\rho$. Obviously $\tilde\rho\leq\rho$ holds, since the zero section is totally geodesic in $TQ$. But is something known about lower bounds? For example, is it true that $\tilde\rho>0$ or even $\tilde\rho=\rho$?

REPLY [6 votes]: If the manifold is not flat then $\bar \rho=0$.
It is sufficient to show that given $\epsilon>0$ there are two tangent vectors $v,w\in T_pQ$ such that $|v-w|=\epsilon$, but the minimizing geodesic does not lie in $T_pQ$.
We assume that curvature at $p$ does not vanish.
Consider a loop $\gamma$ based at $p$ with length $\delta<\epsilon$ and nontrivial integral curvature $R$.
Choose generic $v$, so $w=R v\ne v$.
We can assume that $|v-w|=\epsilon$.
A horizontal lift of $\gamma$ connects $v$ to $w$ and has length $\delta$.<|endoftext|>
TITLE: Reference for a Maschke lemma for crossed products
QUESTION [6 upvotes]: Can someone provide a reference for the following Maschke lemma:
If $C$ is a semisimple algebra over a field of characteristic zero and $G$ is a finite group
acting on $C$ by automorphisms, then $C \rtimes G$ is semisimple.

REPLY [2 votes]: This is also stated in the generality asked for as Theorem 1.3 (c) in [Reiten, Riedtmann: Skew group algebras in the representation theory of Artin algebras, Journal of Algebra 92, 1985].<|endoftext|>
TITLE: A terminological question concerning orbifolds. 
QUESTION [8 upvotes]: The notion of orbifold is quite well established by now. I would like to ask how one should call a point of an orbifold with non-trivial stabilizer? Should one call this a singular point? Of something else? 
For some reason, I was not able to find any text that fixes this innocent bit of terminology concerning orbifolds.
Comment. I would like to stress, that I want to know how to call A point (i.e. one point) that has a non-trivial stabilizer. Indeed, as Ryan says in his comment, there is some terminology to define the union of all points with non-trivial stabilizer, but this is not what I am looking for (for example, in algebraic geometry there is a canonical way to call a point that is not smooth, it is called a singular point)

REPLY [2 votes]: The question you are asking is a social question, and it cannot have any answer better than a social answer. 
Is there a terminology that the orbifold community has settled on and uses universally? No.
Are there common terminologies that are used by various members of the orbifold community? Yes, some of them are listed in other answers.
Given these social realities, how should one refer to a point of an orbifold with non-trivial stabilizer? Pick the one which makes most sense to you for your context. Or, as Misha suggests, pick the one that most honors the historical origins of the concept.<|endoftext|>
TITLE: Beilinson's formula for the product of two modular curves
QUESTION [13 upvotes]: In his cellebrated 1984 paper "Higher regulators and values of L-functions", Beilinson proved (among many other exciting things) that the value at the non-critical point $s=2$ of the Rankin L-function $L(f\otimes g,s)$ of the convolution of two eigenforms of weight $2$ (say of the same level $N$) is related to the image under the complex regulator map of a certain diagonal element $\Delta$ in the $K_1$ of the surface $S=X_1(N)\times X_1(N)$. 
This happens in the pretty short section 6 of the above mentioned paper. However, I have not been able to spot the precise relationship between $L(f\otimes g,2)$ and $\mathrm{reg}(\Delta)$. I have also read several of the surveys on the subject existing in the literature, like Tony Scholl's "Integral elements in $K$-theory and products of modular curves" and a few others, again without luck: those articles rather focus on other aspects of the story. (To the best of my knowledge, of course, I'll be happy to be corrected.)
My question is: is there any place in the literature where such a formula is precisely stated? By this I mean a formula where all the involved quantities are explicitly written down.
With my collaborators I am working on $p$-adic analogues of this formula, that's why the interest. We can redo Beilinson's computations in order to write down the formula (as it surely is a formal consequence of the ideas of his paper, once everything is written down carefully enough), but I wonder whether this is already done explicitly elsewhere.

REPLY [7 votes]: Dear Victor, I think an explicit formula has been worked out by S. Baba and R. Sreekantan in the following article :
MR2064735 (2005c:11073)  Baba, Srinath ;  Sreekantan, Ramesh. An analogue of circular units for products of elliptic curves. Proc. Edinb. Math. Soc. (2)  47  (2004),  no. 1, 35--51.
They consider the case where $f$ and $g$ are associated to elliptic curves of the same conductor $N$. They obtain an explicit formula giving $L'(f \otimes g,1)$ (which relates to $L(f \otimes g,2)$ using the functional equation) as the regulator of an explicit element in the motivic cohomology of the product of modular curves $X_0(N) \times X_0(N)$.
The hypothesis that $f$ and $g$ come from elliptic curves is not very restrictive, in the sense that their method probably also works for arbitrary eigenforms of weight 2 and level $\Gamma_0(N)$. It is also possible in principle to work with $X_1(N)$ instead of $X_0(N)$, but as usual this may require some work, in particular this would probably require to adapt the construction of the motivic cohomology elements.<|endoftext|>
TITLE: Eigenforms for Laplacian on a non-flat two-torus
QUESTION [5 upvotes]: Does anyone know an explicit, exact description of the eigenforms of the Laplacian on a non-flat two-torus?

REPLY [6 votes]: This is not so much an answer as a few remarks and a caution.  If I understand your request correctly, I think that it is unlikely that you are going to find a truly explicit example.
First, let me remark that, since we are considering a compact, oriented surface $T^2$, it would be enough to know all of the eigenfunctions of the Laplacian on $0$-forms, since, by the Hodge decomposition theorem, for positive eigenvalues $\lambda$, any eigenfunction of the Hodge Laplacian on $2$-forms with eigenvalue $\lambda$ would be of the form $\ast f$ where $f$ is a $\lambda$-eigenfunction, and any eigenform of degree $1$ with eigenvalue $\lambda$ would be of the form $df_0 + \ast d f_1$ for some $f_0$ and $f_1$ that are $\lambda$-eigenfunctions.
Second, if we assume that the metric is given in the form $g = F(z) dz\circ d\bar z$ where $F$ is a positive, $L$-periodic function on $\mathbb{C}$, then this metric will have nonconstant Gaussian curvature if and only if $F$ is not constant.  In any case, a $\lambda$-eigenfunction will be an $L$-periodic function $f$ on $\mathbb{C}$ that satisfies
$$
4f_{z\bar z} + \lambda F f = 0.
$$
So you are asking for a method of explicitly describing all of the $L$-periodic solutions of this equation.  I am not aware of any positive, nonconstant $F$ for which this is known.
If you go down a dimension and ask for the list of positive, $\pi$-periodic functions $F$ on the real line for which all of the $\pi$-periodic solutions of 
$$
f''(x) + \lambda F(x) f(x) = 0
$$
are explicitly known for each $\lambda>0$, I think you will find that this list is very short.  It makes me suspect that, in the $2$-dimensional case, the list is not longer.<|endoftext|>
TITLE: First group homology with general coefficients
QUESTION [13 upvotes]: When $G$ acts trivially on $M$, the first homology group is just the abelianisation of $G$ tensored with $M$, i.e. $H_1(G;M)=(G/[G,G])\otimes_\mathbb Z M$.
Is there any similar statement when $G$ acts not trivially on $M$? Where does the abelianisation come from?

REPLY [7 votes]: Using the short exact sequence $0\to I\to \mathbb Z G \to \mathbb Z \to 0$ and the associated long exact sequence for ${\rm Tor}_*^{\mathbb ZG}(-,M),$ we get the following exact sequence
$$0 \longrightarrow H_1(G,M) \longrightarrow I \otimes_{\mathbb ZG} M \longrightarrow M \longrightarrow H_0(G,M) \longrightarrow 0.$$
In particular, we obtain 
$$H_1(G,M)={\rm Ker}(I \otimes_{\mathbb ZG} M \longrightarrow M).$$ 
If $G$ acts trivially on $M,$ the map $I \otimes_{\mathbb ZG} M \longrightarrow M$ is trivial and we get $H_1(G,M)=I\otimes_{\mathbb ZG}M=I/I^2 \otimes M=G_{\rm ab}\otimes M.$ Hence, it is a generalisation of your formula.<|endoftext|>
TITLE: Singular fibers of generic smooth maps of negative codimension
QUESTION [10 upvotes]: This is in some sense a follow-up to my question on submersions.
Let $f\colon\thinspace M\to N$ be a generic smooth map between closed manifolds of dimensions $m$ and $n$. Assume that the codimension $k = n-m$ is negative.
I am interested in the topology of the point fibers $f^{-1}(y)\subseteq M$ for $y\in N$.
Sard's Theorem implies that for almost all $y\in N$ the fiber $f^{-1}(y)$ is a closed submanifold of $M$ of dimension $-k$. If $y$ is not a regular value, then $f^{-1}(y)$ is a priori just a compact subset of $M$ which needn't be a manifold. However one might hope that (under some conditions on the map $f$ which are satisfied in a dense subset of $C^\infty(M,N)$) it is close to being a manifold. For example, it might always be a pseudomanifold, or manifold with singularities.

Can anything general be said about the spaces $f^{-1}(y)\subseteq M$ occurring as fibers of generic maps?

I am particularly interested in the dimensions of the singular fibers (where dimension has to be interpreted appropriately). In all the examples I can picture (ie critical level sets of Morse functions) the dimensions of the singular fibers seem to be $\le -k$.
In fact I would really like to ask the following precise question about homotopy types:

Let $f\colon\thinspace M\to N$ be a generic smooth map between closed manifolds of dimensions $m$ and $n$, with $m>n$. Is every fiber $f^{-1}(y)$ dominated by a CW-complex of dimension $\le m-n$?

The genericity assumption is there because I would like to be able to approximate an arbitrary map $g\colon M\to N$ by a map with  singular fibers having the above nice property.
Probably what I am asking is covered in the literature on singularity theory or surgery theory, however I couldn't find any references on this particular point. Any answers or pointers would be appreciated.
Update: In this paper of Gromov (on page 3) it is stated that if $f\colon\thinspace M\to N$ is a smooth generic map with $m+1\ge n$ then for a critical value $y\in N$ the number of singularities of $f$ on the level set $f^{-1}(y)$ is at most $n$. (A weaker claim is made on page 115 of Golubitsky and Guillemin (Lemma 1.9), namely that the same conclusion holds if $f$ is infinitesimally stable = stable).
This implies that $f^{-1}(y)$ has the structure of a smooth $(m-n)$-dimensional submanifold of $M$ away from a finite number of isolated singular points. So my questions now become:


Does this imply that $f^{-1}(y)$ has the homotopy type of a CW-complex of dimension $m-n$?

Can anyone supply a proof or a reference for the statement of Gromov quoted above?



Update II: Thanks to Tom Goodwillie, who has answered both of the questions in my update. However I'm still left wondering about the two original questions. My gut feeling is that the Hawaiian earing (or some related pathology) should not appear as the fiber of a generic smooth map, and perhaps the fibers should generically be manifolds with singularities, or something similar.
But I'd be happy to be proved wrong!

REPLY [8 votes]: Fibers of a generic smooth map are polyhedra (so in particular CW-complexes) by the triangulation conjecture of Thom, proved by Andrei Verona [Stratified Mappings - Structure and Triangulability, Springer LNM vol. 1102]. I'm not sure that the full strength of the triangulation conjecture is really needed here - hopefully Ryan's projected answer will clarify this. A more general triangulation conjecture was proved by Masahiro Shiota [Thom’s conjecture on triangulations of maps, Topology, 39 (2000), 383–399], who also has further results on this subject on the arXiv. 
Verona's theorem says that any proper, topologically stable smooth map between smooth manifolds $M$ and $N$ is "triangulable", i.e. equivalent to a PL map by a topological change of coordinates in $M$ and in $N$. On the other hand the set of proper topologically stable smooth maps $M\to N$ is dense in $C^\infty(M,N)$ by the Thom-Mather theorem.
So don't worry, there are no Hawaiian earrings or other nightmares hiding in the fibers. 
To ensure that nothing obstructs easy sleep, we want the fibers of a generic smooth map $M^m\to N^n$ to be polyhedra of dimension $\le\max(m-n,0)$. Indeed, forgetting about the "polyhedra" part for a moment, we know the dimension estimate from multijet transversality (as sketched by Tom). We can now achieve "polyhedra" and "of dimension $\le\max\(m-n,0)$" to hold simultaneously because maps generic in two ways are still generic. (That is, the intersection of two open and dense subsets of $C^\infty(M,N)$ is open and dense. I guess this needs $M$ to be compact, and the general case follows by writing $M$ as a union of an increasing chain of compact submanifolds and applying Baire's theorem.)<|endoftext|>
TITLE: Is the centralizer of a torus in a Kac-Moody algebra always a Borcherds algebra?
QUESTION [6 upvotes]: If one has a finite dimension simple Lie algebra, one can easily calculate that taking the centralizer of a torus (or toral subalgebra), that is, summing the weight spaces that lie in some proper subspace of the dual Cartan, always gives a finite dimensional reductive Lie algebra; actually almost semi-simple, except that there is some central toral subalgebra (maybe bigger than the original torus).
For an affine Lie algebra, the picture is the same if you pick a subspace on which $(-,-)$ is positive definite, you get again something finite dimensional and reductive.  If you pick a subspace containing $\delta$, however, things are a bit messier.  Now you have an affine Lie algebra plus a central piece in every single one of the imaginary weight spaces; thus you have an infinite dimensional center.  I would kind of like to think of this as a Borcherds algebra, where I add infinitely many rows and columns of zeros to the affine Cartan matrix.
For a hyperbolic Kac-Moody algebra, the picture is even worse; I still get a hyperbolic KM algebra attached to the root subsystem living in the subspace, but the "extra stuff" in imaginary weight spaces is much more complicated and not central anymore.  This seems very complicated, but I hope to get some kind of handle on it:

Is it true that the centralizer of torus in a hyperbolic Kac-Moody algebra is a Borcherds algebra (necessarily infinite rank in most interesting cases)?  Is there some nice description of its Cartan matrix?

REPLY [7 votes]: The answer to your first question is "yes", and it follows from Theorem 1 in Borcherds's paper Central extensions of generalized Kac-Moody algebras, which is available online as number 11 on his papers page.  The inner product and involution can be chosen as restrictions from your hyperbolic Kac-Moody algebra.
The general philosophy for recognizing these objects is that "anything that looks like a Borcherds algebra is a Borcherds algebra" (cf. Wikipedia), and it holds here.  As long as your Lie algebra is long and skinny (i.e., admits a $\mathbb{Z}$-grading with finite dimensional pieces), and symmetrically shaped (i.e., admits an involution and a bilinear form), you're all set.
I don't know a nice description of the Cartan matrix, but the proof of the theorem gives a general construction.  You may also want to look at Borcherds's later paper A characterization of generalized Kac-Moody algebras (number 15 on the same page).  As a rule, you should try to avoid actually using the Cartan matrix of a Borcherds algebra if you possibly can.<|endoftext|>
TITLE: How to Compute Transgressions in a Serre Spectral Sequence?
QUESTION [8 upvotes]: For a short exact sequence of groups $1\rightarrow A\rightarrow B\rightarrow C\rightarrow 1$ there is an associated fibration $K(A,1)\rightarrow K(B,1)\rightarrow K(C,1)$, which can be constructed by realizing the homomorphism $B\rightarrow C$ by a map $K(B,1)\rightarrow K(C,1)$ and the convert it into a fibration. The fiber is $K(A,1)$ (from the associated long exact sequence of homotopy groups). 
For a fibration $F\rightarrow X\rightarrow B$, the differential $d_n\colon E_{n,0}^n\to E_{0,n-1}^n$ in the Serre spectral sequence was shown to be equal to the transgression in Hatcher's book on Spectral Sequences (Proposition 1.13). The transgression was defined using (relative) homology groups.
My questions is: From the short exact sequence of groups $1\rightarrow A\rightarrow B\rightarrow C\rightarrow 1$, is there any method to directly compute the transgression of the associated fibration $K(A,1)\rightarrow K(B,1)\rightarrow K(C,1)$, at least for the case $n=2$, without constructing $K(G,1)$'s and considering their homologies?

REPLY [11 votes]: I can give a description in case of cohomology: Let $$1 \to H \to G \to G/H \to 1$$ be an extension of groups. Then we obtain an extension with abelian kernel 
$$1 \to H_{ab} \to G/H' \to G/H \to 1$$ Let $\varepsilon \in H^2(G/H;H_{ab})$ be its extension class. If $M$ is a trivial $G$-module, then the differential (which equals the transgression) 
$$d_2^{0,1}: E_2^{0,1}=H^1(H;M)^{G/H} \to H^2(G/H;M) = E_2^{2,0}$$ is given as follows: Let $f \in H^1(H;M)^{G/H} \le Hom(H,M)=Hom(H_{ab},M)$. Since $f$ is $G/H$-invariant, we have a hom. of $G/H$-modules $f:H_{ab}\to M$ and an induced hom. $f_\ast: H^2(G/H;H_{ab}) \to H^2(G/H;M)$. Then: 

$\hspace{120pt}d_2^{0,1}(f) = f_\ast(\varepsilon)$ 

A good reference for this is Theorem 2.1.8 in Neukirch et. al.: Cohomology of Number Fields. 
In case of $M=\mathbb{F}_p$, Kudo's transgression theorem may also be of relevance. 
In case of homology you can try to dualize the result above. But I personally would always prefer to use cohomology, since here the cup product is available that is very helpful in computing spectral sequences.<|endoftext|>
TITLE: Finite index subgroups of the mapping class group with geometric meaning
QUESTION [5 upvotes]: I have got a question that is perhaps not precise in a mathematical sense. 
Is there a classification of all coverings of the moduli space of Riemann surfaces which are moduli spaces themselves, that is, they parametrize some geometric structure on a surface.

REPLY [2 votes]: I doubt there is a "classification", but there are some interesting examples. Two which come to mind: Harer's description of the moduli space of a Riemann surface with spin structure; and Torelli space.  
EDIT: Oops, I forgot to read your title, I just read the text. Torelli space is an infinite rank covering of moduli space.<|endoftext|>
TITLE: Geometric natural transformations between Fourier-Mukai transforms
QUESTION [6 upvotes]: Given two schemes $X$ and $Y$ one can consider additive functors (eventually with some nice additional property) between the categories of $\mathcal{O}_X$-modules and of $\mathcal{O}_Y$ modules. Among these one has those "of a geometric nature", in particular Furier-Mukai-type functors, i.e., those of the form $\Phi_{Z,\mathcal{P}}:=g_*(\mathcal{P}\otimes f^*-)$, where $(f,g):Z\to X\times Y$ and $\mathcal{P}$ is some "twisting" sheaf on $Z$. With suitable finiteness assumptions (e.g., if $Z$ is Noetherian), $\Phi_{Z,\mathcal{P}}$ is a functor $QC(X)\to QC(Y)$. Also, one can consider a derived variant of this, and look at $\Phi_{Z,\mathcal{P}}$ as to a functor $D^b(Coh(X))\to D^b(Coh(Y))$ (all this is very well known and studied, I'm writing it only to provide a setting for my question).
Given two Furier-Mukai-type functors $\Phi_{Z,\mathcal{P}}$ and $\Phi_{W,\mathcal{Q}}$  between categories of $\mathcal{O}_X$-modules and of $\mathcal{O}_Y$-modules, are there natural "geometric" transformations between these functors? which are the main examples of these? (here "geoemtric" is somehow vague: it pretends to mean "something with the same flavor of Fourier-Mukai transform, e.g., maybe something involving a morphism $T\to Z\times W$ and some twisting kernel over $T$)

REPLY [8 votes]: First a quick comment: if you work not with triangulated categories but with differential graded or $A_\infty$ categories, then ALL (continuous) functors are given by "Fourier-Mukai" or integral transforms (a theorem of Toen - an analogue of the Schwartz kernel theorem in analysis). Moreover the (dg) category of functors is equivalent to the category of sheaves on $X\times Y$, in other words natural transformations are the same as sheaf maps on the product.
In any case, one way to realize your question is to look eg at functors given by correspondences $X\leftarrow Z \rightarrow Y$ - i.e. at integral kernels of the form $\pi_*{\mathcal O}_Z$ for the evident map $Z\to X\times Y$. Then a map of correspondences $Z\to W$
gives a pullback map between the corresponding integral kernels, i.e., a natural transformation between the functors. You can generalize this to twisted kernels (i.e. equip $Z$ with a sheaf as well) but at that point you're very close to just looking at morphisms of integral kernels on $X \times Y$..<|endoftext|>
TITLE: Looking for at least one beautiful and not too technical result in asymptotic group theory
QUESTION [12 upvotes]: We have a student seminar devoted to the problems of asymptotic group theory with some connections to ergodic theory and measure theory in general. Each talk concerns one of the problems of this beautiful area, and the thematic is considerably wide: the main idea is that somebody presents a theorem, which is proved beautifully and has a lot of applications. For example, we have proved de Finetti theorem (see Wikipedia article on it), some ergodic theorems for group actions, and Wigner's semi-circle law. I'm looking for something as beautiful and as useful, and not very difficult for being able to talk about it for 2 hours on our seminar. I hope you could give me some hints.    
I repeat, that the area of problems we discuss is wide and I am not able to define it strictly, so any ideas are welcome. But generally, the methods we use are either dynamical either probabilistic. I had a thought to tell some of Grigorchuk's ideas of constructing the groups of intermideate growth, but I found those proofs technical.

REPLY [6 votes]: You might get a good seminar talk out of certain cases of the Patterson-Sullivan theory, for instance the original case considered by Patterson for convex cocompact actions of a finite rank free group on the hyperbolic plane. You get a nice formula for the Hausdorff dimension of its Cantor set at infinity. The formula is expressed by using the asymptotics of the orbits in the hyperbolic plane to construct certain probability measures on the Cantor set (known nowadays as the Patterson measure or in a more general context as Patterson-Sullivan measures), and then understanding the Radon-Nykodym derivatives of the group action on these measures.

REPLY [2 votes]: Wolfgang Lueck has written a survey 
http://arxiv.org/abs/0806.3771
Theorem 1.1 (Schreier’s Theorem)
Let $G$ be a free group and $H \subset G$ be a
subgroup. Then $H$ is free. If the rank $rk(G)$ and the index $[G : H]$ are finite,
then the rank of $H$ is finite and satisfies
$$rk(H) = [G : H] · (rk(G) − 1) + 1.$$

REPLY [2 votes]: You can speak about the Howe-Moore theorem, which is very useful and imply ergodicity (and actually, mixing) of group actions on reasonable spaces.<|endoftext|>
TITLE: Beyond Hilton-Milner Theorem for an Intersecting Family? 
QUESTION [7 upvotes]: Hilton-Milner tells us about an intersecting family of $k$-size subsets $\mathcal{F}$ from $\binom{[n]}{k}$:
If $|\mathcal{F}| > \binom{n-1}{k-1}-\binom{n-k-1}{k-1}+1$, then all elements of $\mathcal{F}$ share at least 1 element of [n].
The theorem classifies the non-trivial $\mathcal{F}$ if equality occurs i.e. $|\mathcal{F}| = \binom{n-1}{k-1}-\binom{n-k-1}{k-1}+1$.
I was wondering if there are known results/literature to see on classifications with $|\mathcal{F}| < \binom{n-1}{k-1}-\binom{n-k-1}{k-1}+1$, particularly values very close to equality?  
I am currently looking at the case of $n=7$, and $k=3$ to get an idea of this (i.e. what can be said about an intersecting family of size 12 < $\binom{6}{2}-\binom{3}{2}+1?)$.

REPLY [5 votes]: There are at least three types of results that spring to mind. One is the Ahlswede-Khachatrian theorem ("the complete intersection theorem"), which for each $n$ and $k < n/2$ will give you tight upper bounds for $t$-intersecting families (families in which every two sets intersect in at least $t$ points). For example, if $k \leq n/3 + 1$ then a $2$-intersecting family has size at most $\binom{n-2}{k-2}$, and so if $|\mathcal{F}| > \binom{n-2}{k-2}$ then there are two sets $S,T \in \mathcal{F}$ such that $|S \cap T| = 1$. They also have a matching Hilton-Milner type result, in which the family is required to have empty intersection ("the complete non-trivial intersection theorem").
A second one is stability results. Such results show that if $|\mathcal{F}| = (1-\epsilon)\binom{n-1}{k-1}$ then there is an element belonging to a $1-O(\epsilon)$ fraction of the sets. One such result is due to Friedgut ("on the measure of intersecting families, uniqueness and stability"), another is due to Keevash ("shadows and intersections: stability and new proofs").
A third one is due to Dinur and Friedgut ("intersecting families are essentially contained in juntas"). For $k = cn$, they show that any intersecting family essentially depends on $O(1)$ points, where the constant depends on $c$. For $k = o(n)$, they can show that any intersecting family has an element common to all but $C \binom{n-2}{k-2}$ of the sets.<|endoftext|>
TITLE: Homology of Covering Spaces
QUESTION [7 upvotes]: Let $A$ be a subgroup of a group $G$. Then since $A$ is a subgroup of the fundamental group $\pi_1(K(G,1))=G$, there is a covering space $p\colon Y\to K(G,1)$ with $p_*(\pi_1(Y))=A$. So the homology of $Y$ should be completely determined by $A$ and $G$. Suppose that $A$ and $G$ is known, how can one compute $H_*(Y)$, the homology groups of $Y$?

REPLY [10 votes]: This is not a trivial problem. A favorite example of mine is the case of a knot complement, $S^3\setminus K$. (It is known that these are Eilenberg-Maclane spaces.) If you pick $A$ to be the commutator subgroup of $\pi_1(S^3\setminus K)=G$, then $Y$ is called the universal Abelian cover, and $H_1(Y;\mathbb Q)$ turns out to be a torsion $\mathbb Q[t,t^{-1}]$-module, called the Alexander module of the knot. The order of the Alexander module is called the Alexander polynomial. One can calculate the Alexander module from the fundamental group using Fox calculus.<|endoftext|>
TITLE: $\pi_0${plane fields}$\to\mathbb{Z}_2$
QUESTION [8 upvotes]: On a 3-manifold $Y$, oriented 2-plane fields $\xi$ are oriented rank-2 subbundles of $TY$. Denote the set of such (up to homotopy) by $\Theta=\pi_0\lbrace\xi\rbrace$. What is an explicit canonical map $\Theta\to\mathbb{Z}_2$ ?
In particular, I want to see the canonical mod-2 grading on Seiberg-Witten-Floer Homology, but the classical text of Kronheimer-Mrowka doesn't mention it in terms of 2-plane fields. They construct an isomorphism of $\Theta$ with some abstract set $\mathbb{J}$ based upon "configuration points" $[a]$, and then a canonical map $\mathbb{J}\to\mathbb{Z}_2$ by assigning to $[a]$ an operator and taking its index mod-2.  These two maps are tough enough on their own, so I cannot see what it looks like on 2-plane fields... Plus, there should be some purely topological partition of $\Theta$ into two subsets.
Otherwise, perhaps it can be done knowing that oriented 2-plane fields are equivalent to 1-forms of length 1, and are also equivalent to pairs $(\mathfrak{s},\phi)$ where $\mathbb{s}$ is a spin-c structure and $\phi$ is a unit-length spinor. This should then be related to taking a 4-manifold $X$ such that $\partial X=Y$ and with a spin-c structure $\mathfrak{s}_X$ which extends $\mathfrak{s}$.

REPLY [3 votes]: In light of the paper referenced in Tim Perutz' comment, here is the desired map:
An oriented 2-plane field $\xi$ is equivalent to a pair $(\mathfrak{s},\phi)$ on $Y$, where $\mathfrak{s}$ is a spin-c structure and $\phi$ is unit-length spinor. By Proposition 28.1.2 (of Kronheimer-Mrowka's textbook), there exists an oriented 4-manifold $X$ with $\partial X=Y$ and carrying a spin-c structure $\mathfrak{s}_X=(S^+,\rho_X)$ which extends $\mathfrak{s}$, i.e. $\mathfrak{s}\cong(S^+|_Y,\rho_Y)$. Now the relative Euler class $e(S^+,\phi)$ satisfies $e(S^+,\phi)[X,\partial X]$ = $gr_z(X,\mathfrak{s}_X,[a])\in\mathbb{Z}$ for some configuration point $[a]$ associated to $\phi$, and this index is independent of $z$. And by Proposition 28.2.2, this is independent of the choice of $X$ (up to homotopy of $\phi$). This is where the isomorphism $\mathbb{J}(Y)\cong \Theta(Y)$, $\xi\leftrightarrow [a]$, comes from. We can thus write down a map $f:\Theta(Y)\to\mathbb{Z}_2$ given by
$$\xi\mapsto e(S^+,\phi)[X,\partial X]\;\text{mod}2$$
To be consistent with the literature: When we apply this to $Y=S^3$ we get the flipped even/odd decomposition of $\widehat{HM}(S^3)$. Simply change $f$ by $Aut(\mathbb{Z}_2)$, i.e. adding a $1$, to get agreement.  
Addendum: In fact, this map can be rephrased as the obstruction to finding an almost complex structure $J$ on the above $X$ in such a way that $\xi$ is $J$-invariant.<|endoftext|>
TITLE: Does every vector bundle allow a finite trivialization cover?
QUESTION [21 upvotes]: Suppose there is a vector bundle  (smooth, with constant rank finite-dimensional fibres) over a (smooth, second-countable, Hausdorff, not necessarily connected) manifold $B$ of dimension $n$.
(a) Is it true, that the manifold B can be covered by a finite number of sets $U_1,\dots,U_N$ s.t. the vector bundle, restricted to $U_i$, is isomorphic to a trivial one for every $i=1,\dots,N$?
(b) If yes, can $N$ be taken to be $n+1$?
P.S. Some observations:

It's proven in the book by Milnor and Stasheff, that every bundle allows a countable locally finite trivialization cover.
Part (a) is obviously trivial for compact manifolds.
It seems, that (b) is true if $B$ is an $n$-dimensional CW-complex. Proof: Denote with $B_k$ union of cells of dimension $0,\dots,k$. Prove by induction in $k$, that there are subsets $U_0,\dots,U_k$ of $B$, which cover $B_k$, s.t. the restriction of the bundle to each of them is trivializable. Start with case $k=0$: construct contractible neighbourhoods of each 0-cell, which do not intersect with each other. Take there union. Now to prove the claim for the next value of $k$ it is enough to construct a contractible non-intersecting neighbourhoods of each $X_\alpha=e_\alpha\setminus (U_0\cup \dots\cup U_{k-1})$. Call the desired neighbourhood with $V_\alpha$. First, note, that $X_\alpha$ is closed in $e_\alpha^k$ and doesn't intersect with its boundary $\partial e_\alpha^k$, so we can find its neighbourhood in $e_\alpha^k$, which doesn't intersect with $\partial e_\alpha^k$. This set is our candidate for $V_\alpha\cap B_k$. Extending it to an open set in $B_{k+1}$ can be done cell by cell: interpreting $e^{k+1}_\beta$ as a unit ball with the center in the origin, we can write every its point as $r\theta$, where $\theta\in S^k$ and $r\in [0,1]$. We include $r\theta$ in $V_\alpha\cap B_{k+1}$ iff $\theta$ is already there and $r>0.99$. Repeating this procedure we extend it to $B$.

Edit: Open sets $U_1,\dots, U_N$ are assumed to be open. I don't ask them to be connected.

REPLY [4 votes]: (a) and (b) are also true for topological manifolds of dimension $n$, see pages 17-21 of 

MR0336650  Greub, Werner; Halperin, Stephen; Vanstone, Ray: Connections, curvature, and cohomology. Vol. I: De Rham cohomology of manifolds and vector bundles. Pure and Applied Mathematics, Vol. 47. Academic Press, New York-London, 1972.

REPLY [3 votes]: I wonder that this was not said before.
Take a triangulation of your manifold. Choose disjoint open balls around each 0-cell in the triangulation and set the union to be $U_0$ (a ball means here something diffeomorphic to an open ball). 
For each 1-cell, choose an open ball such that together with the balls chosen before around the corners, the 1-cell is completely covered, and choose those balls to be disjoint. Take the union of those to be $U_1$.
Keep on going in the same way, until you reach the $n$-cells. 
Now you have $n+1$ sets $U_0, \dots, U_n$, each consisting of disjoint unions of open balls and such that $U_0 \cup \dots \cup U_j$ covers all cells of dimension less or equal to $j$. In particular, the union of all these sets covers your manifold.
Your vector bundle must be trivial on each of the sets $U_j$, as each are disjoint unions of open balls.<|endoftext|>
TITLE: Generalization of Hilbert 94 and capitulation
QUESTION [8 upvotes]: Let $L/K$ be a finite, cyclic extension of number fields, say with $\mathrm{Gal}(L/K)=G$. In my context $G$ is actually of order $p$, an odd prime number, but let me state my question for every cyclic $G$.
Hilbert theorem 94 says that if $L/K$ is everywhere unramified (hence contained in the Hilbert class field $H$ of $K$), then the ''capitulation kernel'', namely the kernel of the natural map $\iota:Cl_K\to Cl_L$, has order divisibile by $\vert G\vert$. Some (but, to my knowledge, not many) generalizations have been proven, mainly removing the cyclicity assumption although losing something.
My question goes into another direction: if $L/K$ is allowed to ramify and is contained in the ray class field $H(\mathfrak{f})$ modulo some conductor $\mathfrak{f}$, what can we say about the capitulation kernel $\iota:Cl_K(\mathfrak{f})\to Cl_L(\mathfrak{F})$, where these groups are now the ray class groups modulo the respective conductors (and $\mathfrak{F}=\mathfrak{f}\mathcal{O}_L)$)? Can it be trivial?
I must confess my ignorance with respect to an even more basic question, namely: what is the state of the art concerning the Principal Ideal Theorem for ray class fields? Is it known – or false, or trivial, or... – that all classes in $Cl_K(\mathfrak{f})$ become principal (perhaps, only $\mathfrak{f}$-principal?) in $H(\mathfrak{f})$?

REPLY [4 votes]: The answer to both my question is that "adding conductors does not change anything". Olivier has already discussed this for the Principal Ideal Theorem, and for Hilbert 94 this is proven by Suzuki in
https://projecteuclid.org/journals/nagoya-mathematical-journal/volume-121/issue-none/A-generalization-of-Hilberts-theorem-94/nmj/1118782786.full
The whole point is that Hilbert 94 follows from a generalization of Furtwängler theorem, saying that the kernel of the transfer map
$$\mathrm{ver}:G^\text{ab}\to N^\text{ab}$$
has order divisible by $[G:N]$, and this for all normal subgroups $N$ containing the commutators $[G,G]$ (or, equivalently, such that the quotient by them is abelian). The case $N=[G,G]$ is Furtwangler's theorem and applying it in my setting with $N=\mathrm{Gal}(H_L(\mathfrak{F})/K)$ in the notations of the question (now $H_L(-)$ are ray class fields of $L$) shows that $\iota$ is never injective.<|endoftext|>
TITLE: Connections on line bundles over the torus
QUESTION [6 upvotes]: If I understand correctly, every line bundle $L$ over the (2-dim) torus can be obtained from a quotient of $\mathbb{R}^2 \times \mathbb{C}$ by a $\mathbb{Z}^2$ lattice action. Different line bundles are obtained depending on how the lattice 'twists' the fibers. The connection $D = d + (1/2)(pdq-qdp)$ descends to a connection on $L$, as does the canonical Hermitian pairing.
My question is this: Suppose we wanted to geometrically quantize the torus by a Kähler polarization. Then we would want to find sections $s$ of $L$ such that $D_v s = 0$ for antiholomorphic $v$. How can we write down the explicit sections satisfying this equation in the case of the torus with a Hermitian bundle $L$ as above? And how can we prove that for a general compact manifold the dimension of the space of solutions is finite? (I know how to solve the differential equation for the case of $\mathbb{R}^2 \times \mathbb{C}$ and get the Bargmann-Fock space, but it is unclear how one can solve the analogous problem for compact manifolds - at least unclear to me)

REPLY [13 votes]: The Kaehler polarisation is a choice of complex structure, $I$, on your torus $M\simeq S^1\times S^1$, which is compatible  with the symplectic form. In other words, your torus becomes  a complex manifold $X=(M,I)$, and $I$ satisfies $I^t\omega I=\omega$, $g=\omega I>0$. In particular, this tells you that $\omega$ is a positive $(1,1)$-form, i.e., locally 
$\omega = \frac{i}{2}h dz \wedge d\overline{z}$, $h>0$. Of course, since $\dim X=1$, $\omega$ is a $(1,1)$-form for any choice of $I$, but this is not so in higher dimensions.
The connection $D$ can now be decomposed as $D= D^{1,0} + D^{0,1}$. Explicitly, 
$ D^{0,1} = \frac{1}{2}(1+iI)D$ and $ D^{1,0} = \frac{1}{2}(1-iI)D$. Or, if you prefer, choose a local frame $s$ of $L$, and write
$$
D = d + A_1 dq + A_2 dp    =  (\partial + B_1 dz) + (\overline{\partial} + B_2 d\overline{z})
$$
Since $\omega$ is of type $(1,1)$, the condition
$D^2 = -i\omega$ translates to
$$
(D^{0,1})^2=0,\quad (D^{1,0})^2=0,\quad D^{0,1}D^{1,0} +D^{1,0} D^{0,1}=-i\omega .
$$
Again, the first two equations hold automatically if $\dim X=1$, but not in general.
This means that on $L$ you get the structure of a holomorphic line bundle $\mathcal{L}$  by taking the Dolbeault operator to be $\overline{\partial}_L= D^{0,1}$. Locally, if you choose a (smooth) frame $s$, this is the above $\overline{\partial}_L =
\overline{\partial} + B_2 d\overline{z}$.
Then the "Hilbert space" is $H^0(X,\mathcal{L})= \ker\overline{\partial}_L \subset A^0(L)$,
where $A^0(L)$  is   the   (infinite-dimensional) vector space of global smooth sections of $L$. With respect to a local trivialisation, a section $\sigma =fs$ is holomorphic if
$$
\frac{\partial f}{\partial\overline{z}} + B_2 f=0.
$$
The space $H^0(X,\mathcal{L})$ is finite-dimensional for very general reasons. If you actually want to know how big it is, you look at the (Hirzebruch)-Riemann-Roch formula, which tells you that
$$
\dim H^0(X,\mathcal{L})- \dim H^1(X,\mathcal{L}) = \deg \mathcal{L} + 1-g = 
\deg \mathcal{L}.
$$
Then $\dim H^0(X,\mathcal{L})=\deg \mathcal{L}$ if  $\deg\mathcal{L}>0$,  $\dim H^0(X,\mathcal{L})= 0$  if $\deg\mathcal{L}<0$  (or if $\deg\mathcal{L}=0$ and $\mathcal{L}$ is nontrivial), and $\dim H^0(X,\mathcal{L})=1$ if $\mathcal{L}=\mathcal{O}_X$. You can describe the elements of $H^0(X,\mathcal{L})$ quite explicitly
using theta-functions.
Notice that everything - in particular $H^0(X,\mathcal{L})$ - depends on the choice of complex structure $I$, and even the dimension may jump as you vary the complex structure (if $\deg L=0$). But if you choose your prequantum line bundle to be sufficiently positive, the
vector spaces $H^0(X,\mathcal{L})$ "glue" into a vector bundle over the upper half-plane.
ADDENDUM:
Let me spell out the relation with theta-functions a bit more explicitly. 
This is  a huge classical topic (see the references at the end), so I'll be a bit sketchy.
Write $V=\mathbb{C}$ and $\Lambda = \mathbb{Z}\oplus \tau \mathbb{Z}$, $\textrm{Im}\tau>0$. Then
$\pi: V\to X=V/\Lambda$ is the universal covering map, and it is a local biholomorphism. 
If your symplectic structure descends from the form $\omega= c\ dq\wedge dp$ on $V=\mathbb{C}$, the "Bohr-Sommerfeld condition" tells you that $c=\frac{2\pi}{\textrm{Im}\tau}n$, 
$n\in \mathbb{Z}$. Notice that $\frac{1}{2\pi}\omega$ is the imaginary part of the hermitian form $H(z,w)=\frac{n}{\textrm{Im}\tau}z\overline{w}$ on $V$ and takes integer values on the lattice.
Suppose $\mathcal{L}$ is as above, with 
$c_1(\mathcal{L}) =\frac{1}{2\pi}\left[\omega\right]$, $\deg \mathcal{L}=n$.
As any holomorphic line bundle  on $V$ is trivial,  we can choose a global holomorphic trivialisation 
$\pi^\ast \mathcal{L}\simeq V\times \mathbb{C}$, 
and identify $\mathcal{L}= \pi^\ast \mathcal{L}/\Lambda$
with $V\times\mathbb{C}/\sim$. The equivalence relation is 
$(z,t)\sim (z+\lambda, a(\lambda,z)t)$, $\lambda\in\Lambda=\pi_1(X)$.
Here $a: \Lambda\times V\to \mathbb{C}^\times$ are the multipliers or factors of automorphy. They are 1-cocyles of $\Lambda=\pi_1(X)$ with values in the entire nonvanishing functions $H^0(V,\mathcal{O}^\times_V)$, i.e., they 
are holomorphic in $z$ and
satisfy
$$
a(\lambda+\mu,z)= a(\mu,\lambda+z)a(\lambda,z).
$$
The multipliers do not, in general, take values in $U(1)$!
Changing the trivialisation of $\pi^\ast \mathcal{L}$ replaces $a$ by a coboundary, and
we have 
$H^1(\Lambda,H^0(V,\mathcal{O}^\times_V))\simeq H^1(X,\mathcal{O}^\times_X)$.
Now, we can identify $H^0(X,\mathcal{L})\simeq H^0(V,\mathcal{O}_V)^{\pi_1(X)}$.
I.e., we can identify $H^0(X,\mathcal{L})$ with the entire  functions which satisfy the
functional equation
$$
\theta(z+\lambda) = a(\lambda,z)\theta(z),
$$
and these are by definition  theta-functions. 
The explicit structure of the factors of automorphy is known. For instance, in each degree $n$ you have a canonical choice, 
$$
a_0(\lambda,z) = \exp\left( n\pi i ab \right)\exp \frac{n\pi}{\textrm{Im}\tau}\left(z\overline{\lambda}+ \frac{1}{2}|\lambda|^2\right),\quad \lambda=a+b\tau.
$$
You can get the other bundles of the same degree by multiplying $a_0$ with characters
$\chi\in \textrm{Hom}(\Lambda,U(1))\simeq Pic^0(X)$.
Finally, a word about the inner product. We can identify $A^0(\mathcal{L})$, the smooth sections of 
$\mathcal{L}$, with smooth functions on $V$, satisfying the same functional equation
as the theta-functions. In terms of this identification the fibrewise hermitian product on $\mathcal{L}$ is
$$
\langle  f,g \rangle(z) = f(z)\overline{g(z)}\exp \left(-\frac{n\pi}{\textrm{Im}\tau}|z|^2 \right).
$$
The $L^2$-inner product on $H^0(X,\mathcal{L})$ is obtained by integrating this quantity over $X$. 
Line bundles on a complex torus are  described in great detail in  Birkenhake and Lange, Complex Abelian Varieties (especially Chapter 2), Griffiths and Harris, Principles of AG, Chapter 2, Section 6, and
D.Mumford, Abelian Varieties, Chapter 1, Section 2.
Here we have been looking at  a 1-dimensional torus, but if you decide to consider the geometric quantisation of $U(1)$ Chern-Simons theory on a higher genus curve, you will need the case of tori of higher dimensions.<|endoftext|>
TITLE: Understanding zeta function regularization
QUESTION [24 upvotes]: I attended a talk this morning on Ray-Singer torsion, in which Rafael Siejakowski introduced zeta function regularization in a compelling way. The goal is to define the determinant of a positive self-adjoint operator $A$ with "pure point spectrum" $0>\lambda_1>\lambda_2>\cdots$. The definition of the determinant is $\exp(-\zeta_A^\prime(0))$ where $\zeta_A$ is the zeta function $\zeta_A(s)=\sum_{i=1}^\infty (-\lambda_i)^{-s}$. This sum diverges in general- but it converges for values of $s$ with large enough real part, and we define it for other values of $s$ (including zero) by analytic continuation.
Why should this be related to the determinant? Well, in the finite dimensional case (the motivating case is when $A$ is the `combinatorial Laplacian'), then $\zeta_A(s)=\sum_{i=1}^N (-\lambda_i)^{-s}$ is a finite sum. In this case:
$\zeta^\prime_A(s)=\sum_{i=1}^N -\ln (-\lambda_i)(-\lambda_i)^{-s}$
and
$\zeta^\prime_A(0)=-\ln \prod_{i=1}^N(-\lambda_i)=-\ln \det A$.
This looks to me like an ad-hoc trick, indicating that I don't understand what is actually going on.
The equation $\det(A)=\exp(-\zeta_A^\prime(0))$ (in the finite dimensional case an equation, in the infinite dimensional case a definition) equates two familiar mathematical quantities:

 The determinant, which I can think of as a volume, as an action on a highest exterior power, or maybe most evocatively as the signed sum of weights of non-intersecting paths in a graph between "source" vertexes $a_1,\ldots,a_n$ and "sink" vertices $b_1,\ldots,b_n$. See this blog post.
The Riemann zeta function, which I don't understand conceptually almost as well, but which is heavily studied and so is clearly important and natural.


Question: Is there a conceptual (hand-wavy is fine) explanation for zeta function regularization, and for how this expression in the zeta function is capturing the idea of a "determinant"? How is the derivation which I wrote above more than an ad-hoc trick? Is there a sense in which the derivative of a zeta function at zero heuristically calculates a signed sum of weights of non-intersecting paths, or something like that?

REPLY [20 votes]: This is a variant of Qiaochu Yuan's answer. Suppose $A$ is some sort of differential operator. Then it is hard to think of things like the trace and determinant of $A$. But there is one quantity which is well behaved: it is the "heat trace" $Tr(e^{tA})$. Of course the heat trace is connected to the solution of the heat equation 
$$\left(\frac{\partial}{\partial t} - A\right) u = 0.$$
So how can we get at the eigenvalues of $A$ if we know $Tr(e^{tA})$? Suppose first that $A$ is finite dimensional. Then, 
$$Tr(e^{tA}) = \sum_i e^{t\lambda_i},$$ 
but it is somewhat inconvenient to extract $\sum \log \lambda_i$ directly from this. But if you plug in the identity
$$\lambda^s = \frac{1}{\Gamma(s)} \int_0^\infty e^{t \lambda} t^s \frac{dt}{t}$$
you get
$$\zeta_A(s) = \frac{1}{\Gamma(s)} \int_0^\infty Tr(e^{tA}) t^s \frac{dt}{t} \qquad \qquad (*) $$
and then $\log \det A = \sum \log \lambda_i = -\zeta_A'(0)$. So in some sense the zeta function is almost forced on you if you want to extract the determinant from the heat trace. 
Now if $A$ is a differential operator, then the equation (*) makes sense, but you cannot directly set $s=0$ because the integral diverges at the limit $t=0$. However, using the well known asymptotics of the heat trace as $t \to 0$, one can subtract the leading order terms, and then the rest will converge. This allows one to define the determinant of $A$, and one can write an expression which does not involve the zeta function at all, but only the heat trace. In some sense this is the real mathematical definition of det A.  Otherwise how do we even know that the zeta function has an analytic continuation to $s=0$?
So perhaps the zeta function regularization is just some shorthand for this argument. One can say "analytic continuation of the zeta function" instead of specifying exactly which terms to subtract from the heat trace.<|endoftext|>
TITLE: Is zero a hydrogen eigenvalue?
QUESTION [14 upvotes]: This question has been bugging me for some time.
Take the hamiltonian for the hydrogen atom: $$\hat{H}=-\frac{1}{2}\nabla^2-\frac{1}{r},$$ acting on (a domain contained in) $L^2(\mathbb{R}^3)$. It is standard fact that this is an unbounded operator which has a countable infinity of eigenvalues, all of which are negative and which accumulate around 0, and has a continuous spectrum on the whole of $(0,\infty)$. Physically, the former are bound states which correspond to elliptic Keplerian orbits in the classical problem, and the latter are unbound states and correspond to hyperbolic orbits. I also know that the spectrum of all unbounded operators is a closed set, so that 0 is definitely in $\sigma\left(\hat{H}\right)$.
My question is then: to what part of the spectrum does 0 belong to (i.e. point, continuous, residual)? What are the corresponding eigenfunctions? What kind of degeneracy does it have? (I would expect it to admit a common eigenvector with any $l,m$ angular momentum numbers, but I'm far from sure.) How do the eigenfunctions correspond to the nearby bound and unbound states?

REPLY [5 votes]: I looked in Anatoly's references, and Quantum mechanics for mathematicians by Leon A. Takhtajan does have the calculation of the continuum wavefunctions, though it does not do the $k=0$ case.
The eigenfunction $f_l$ at energy $E=\frac{1}{2}k^2$ and angular momentum $l$ must satisfy the eigenvalue equation
$$
f_l''+\left(\frac{2}{r}-\frac{l(l+1)}{r^2}+k^2\right)f_l=0.
$$
After the obligatory asymptotics factorization of $f_l(r)=r^{l+1}e^{ikr}F_l(r)$, this equation reads
$$
F_l''+\left(\frac{2(l+1)}{r}-2ik\right)F_l'+\left(\frac{2}{r}-\frac{2ik(l+1)}{r}\right)F_l=0,
$$
and Takhtajan gives the solution as a confluent hypergeometric function, $F_l(r)={}_1F_1\left(l+1+\frac{i}{k};2(l+1);2ikr\right)$, under $F_l(0)=1$.
From this the $k=0$ case can be recovered as a limit in the same spirit as the ${}_2F_1\rightarrow{}_1F_1$ confluence by letting the length $\lambda=\frac{1}{k}$ go to infinity. Thus at zero energy,
\begin{align}
F_l(r)
& =\lim_{\lambda\rightarrow\infty} {}_1F_1\left(l+1+i\lambda;2(l+1);\frac{-2r}{i\lambda}\right)
\\ & ={}_0F_1\left(;2(l+1);-2r\right)
\\ & =\frac{(2l+1)!}{2^\frac{2l+1}{2}}r^{-\frac{2l+1}{2}}J_{2l+1}(\sqrt{8r}).
\end{align}
(This still obeys $F_l(0)=1$.) Alternatively, the Bessel function solution can be obtained directly by the appropriate transformations or by plugging the $k=0$ equation into Mathematica.
In the asymptotic regime, $r\gg 2l+1$, one then gets
$$
f_l(r)
=\frac{(2l+1)!}{2^l\sqrt{2\pi \sqrt{2}}}r^{\frac{1}{4}}
\cos\left(\sqrt{8r}-(2l+1)\frac{\pi}{2}-\frac{\pi}{4}\right)
$$
if my maths is right. However, in real cases this can only happen if $2l+1\ll r\ll \lambda$.<|endoftext|>
TITLE: Equal digit sums
QUESTION [6 upvotes]: Let $s(a)$ be the sum of decimal digits of a number $a$. Is it known that for any $a\ne b$ exist $n$ such that $s(na)\ne s(nb)$?

REPLY [18 votes]: Yes, but if we replace condition $a\ne b$ to $a/b\ne 10^k$ for all integers $k$.
Lemma: $s(9n)=9s(n)$ iff decimal digits of $n$ are only 0's and 1's. Else $s(9n)<9s(n)$.
Call two numbers equivalent, if their ratio is a power of 10 (with integer exponent).
Consider numbers 1,11,111,... Two of them are congruent modulo a, hence their diffrence 11...100..0 equals $ka$ for some natural $k$. Replace pair $(a,b)$ to $(ka,kb)$ and then replace new $a$ (old $ka$) to equivalent number of the form 11...1, totally $m$ ones. Also, divide $b$ to maximal possible power of 10.
Now $s(9a)=9s(a)$, hence the same holds for $b$, hence by lemma $b$ also has only 1's and 0's in its decimal representation and the number of ones equals $m$. In particular, $b>a$ and last digit of $b$ equals 1. Now choose $N$ such that $Nb=11\dots 1$, then $Na < Nb$ and due to $s(Na)=s(Nb)$, $Na$ has a digit different from 0 and 1. Then by lemma $s(9Na)<9s(Na)$, while $s(9Nb)=9s(Nb)$. A contradiction. 
I know this as an old problem by Sergey Berlov, by the way. First time I saw it in 1997 in Sochi on the event for high school students I have participated in:)<|endoftext|>
TITLE: What is the Krull dimension of the ring of holomorphic functions on a complex manifold?
QUESTION [41 upvotes]: Consider a connected holomorphic manifold $X$  and its ring of holomorphic functions $\mathcal O(X).$
My general question is simply: in which cases is the Krull dimension $\dim \mathcal O(X)$ known? 
Of course if $X$ is compact $\mathcal O(X)=\mathbb C$ and that dimension is $0$.
There are also quite a lot of non-compact manifolds with $\mathcal O(Z)=\mathbb C$:
For example if  $X$ is connected of dimension $\geq 2$ and $Y\subset X$ is an analytic   subset  of codimension at least $2$ ( or a small compact ball) ,  you will still have $\mathcal O(X\setminus Y)=\mathbb C$ .    
But apart from these trivial examples I can't compute a single Krull dimension $dim \mathcal O(X)$ for, say, Stein manifolds of positive dimension.  
Just in order to ask something  definite,  let me pose the ridiculous-sounding question:    

Does there exist a connected holomorphic manifold $X$ with $0\lt \dim \mathcal O(X)\lt \infty$ ?

REPLY [15 votes]: My previous (non/)answer is now out of date because of Misha's revisions.  I have written up an independent exposition of the "greater Kapovich theorem" that if a connected $\mathbb{C}$-manifold $M$ has a nonconstant holomorphic function then the cardinal Krull dimension of $\operatorname{Hol}(M)$ is at least $\mathfrak{c} = 2^{\aleph_0}$. 
I have also shown -- by a straightforward reduction to the one-dimensional case -- that if $M$ is a Stein manifold then its cardinal Krull dimension is at least that of $\operatorname{Hol}(\mathbb{C})$ and thus -- by a result of Henriksen -- at least $2^{\aleph_1}$.  Whether this stronger bound should occur under the much weaker hypothesis that there is a nonconstant holomorphic function I have no idea.
Added: The note has been published here.<|endoftext|>
TITLE: Density of a set of integers
QUESTION [13 upvotes]: EDIT: this question of mine has received little attention, perhaps in part because it was stated in a too general and complicated way. So let me give it a second chance:

Fix an integer $r \geq 0$. Let $E_r(x)$ be the number of square free integers less than $x$ having exactly $r$ prime factors $p$ congruent to $2 \pmod{3}$ (and an arbitrary number of other prime factors). What is the order of magnitude of $E_r(x)$ when $x$ goes to $\infty$ ?

When $r=0$, one has $E_0(x) \asymp x/\log^{1/2}(x)$ (see below). Even when $r=1$,
I don't know the answer, or a lower bound for $E_1(x)$ that is better than this one for $E_0(x)$. Thanks for any
idea or reference pointing to a method that can be used to solve this question.
ORIGINAL POST FOLLOWS:
I begin by quoting a theorem found in Serre's paper "divisibilité de certaines functions
arithmétiques", with a proof attributed  to Raikov, Winter, Delange (theorème 2.4 - I have slightly changed and simplified the statement, as the $E$ here is the complement $E'$ of $E$ in Serre's paper.).
Let $E$ be a set of integers satisfying the following condition :
(MM) For $n,m$ relatively prime integers, $nm \in E$ if and only if $n \in E$ and $m \in E$, 
Let $P$ be the set of primes that are in $E$. Assume that $P$ has analytic density $\alpha > 0$. Let $E(x)$ the number of elements of $E$ less than $x$. Then when $x$ goes to infinity,
$$E(x) \sim c x /\log^\{1-\alpha}(x)$$ for some constant $c>0$.   
Example: Let $P$ be a set of primes of density $\alpha>0$, $E$ the set of square-free 
integers whose prime factors are all in $P$. Then $E$ satisfies the hypothesis (MM).
So does $F=$ the set of all integer who's prime factors are all in $P$. Both satisfies
the conclusion of the theorem, though with a different constant $c$.
Now let $P,E$ be as above, and let $r \geq 0$ be an integer. Let $E_r$ be the set of integers that are products of an element of $E$ and exactly $r$ distinct prime not in $P$. My question is:
What is an order of magnitude of $E_r(x)$ ?
The case $r=0$ is the theorem above, and I suppose the answer for $r>0$
does only depend on $P$, not on $E$, as for $r=0$. But in case it is not true, or in case that helps, the case of interest for me is the case where $E$ is as in the example.
So if I may reformulate my question in this special case: if $P$ is a set of prime of density $\alpha>0$, and $E$ is the set of square-free integers which are product
of $r$ primes  not in $P$ and arbitrary number of primes in $P$, what is the order of magnitude of $E_r(x)$?
A word on my motivation, which turns around the beautiful aforementioned paper of Serre.
Consider for example the modular form $\Delta$, modulo $3$. if $E$ is the set of $n$ such that the $n$-th coefficient of that form is non-zero, that is such that $3 \nmid \tau(n)$, then $E$ satisfies (MM), $P$ is the set of primes congruent to $1$ modulo $3$, and $E(x) \sim cx log^{1/2}(x)$.
This is in Serre, as $\Delta$ is an eigenform. Noe consider $\Delta^2$ modulo $3$ which is not an eigenform, and let $F$ the set of $n$-th coefficient of that form is non-zero. Then it is easy to see that $F$ contains $E_1$ (and I think, is not too much bigger than $E_1$).

REPLY [4 votes]: I just wanted to add, we also have a similar upper bound which combined with unknown(google)'s answer shows that 
$$E_{r}(x)\asymp\frac{x\left(\log\log x\right)^{r}}{\left(\log x\right)^{\frac{1}{2}}}.$$
The above is almost certainly an asymptotic, and we can likely derive the explicit constant, but that would require a lot of calculation.
Upper bound:
Let $\mathcal{A}=\left\{ n\in\mathbb{N}:\ p|n\Rightarrow p\equiv2\ (3)\right\}$   and let $\mathcal{B}=\left\{ n\in\mathbb{N}:\ p|n\Rightarrow p\equiv1\ (3)\right\}.$ Then 
$$E_{r}(x)=\sum_{\begin{array}{c}
n\leq x\\
n\in\mathcal{A}\\
\mu^{2}(n)=1,\omega(n)=r
\end{array}}\sum_{\begin{array}{c}
m\leq\frac{x}{n}\\
m\in\mathcal{B}\\
\mu(m)^{2}=1
\end{array}}1.$$ 
We then have the upper bound
$$E_{r}(x)\ll\sum_{\begin{array}{c}
n\leq x\\
\omega(n)=r
\end{array}}\sum_{\begin{array}{c}
m\leq\frac{x}{n}\\
m\in\mathcal{B}
\end{array}}1.$$ 
Carefully managing the sums, oone can then show that this is 
$$\ll\frac{x}{\sqrt{\log x}}\sum_{\begin{array}{c}
n\leq x\\
\omega(n)=r
\end{array}}\frac{1}{n}\ll\frac{x\left(\log\log x\right)^{r}}{\left(\log x\right)^{\frac{1}{2}}}. $$<|endoftext|>
TITLE: Amenability and ultrafilters
QUESTION [14 upvotes]: Among hundreds of equivalent definitions of amenability (for discrete, countable, groups), I would like to discuss two which are most common:
A1. A group $G$ is amenable if it admits a Folner sequence. 
A2. A group $G$ is amenable if it admits an invariant mean. 
(See e.g. http://en.wikipedia.org/wiki/Amenable_group or http://terrytao.wordpress.com/2009/04/14/some-notes-on-amenability/) 
However, proofs of equivalence that I know (even for $G={\mathbb Z}$) require either axiom of choice or, at least, existence of a nonprincipal ultrafilter on ${\mathbb N}$. 
Question: Is there a proof that A1 $\iff$ A2 which uses only ZF axioms? Or, maybe $A1\iff A2$ 
implies existence of nonprincipal ultrafilters, maybe in a weakened form? 
This question was discussed a bit in Why are abelian groups amenable? and Why groups that admit Folner Sequences are amenable but not in the above form. 
Note: I am not a logician, but a geometric group-theorist and I frequently use ultrafilters. As the result I am often asked if the results could be proven without ultrafilters. For most proofs my answer usually is: "Yes, if you work much harder and write ugly and long proofs." However, I do not know the answer in the context of amenable groups.

REPLY [14 votes]: Of course, $ZF$ is enough to prove that $\mathbb{Z}$ has a Folner sequence. But, as you point out, $ZF$ is not enough to prove that $\mathbb{Z}$ has an invariant mean. Thus $ZF$ does not prove the equivalence of A1 and A2.
On the other hand, the Hahn-Banach Theorem is enough to prove the equivalence of A1 and A2 for countable discrete groups and Pincus-Solovay have constructed a model of $ZF$ in which the Hahn-Banach Theorem is true but there are no nonprincipal ultrafilters on $\mathbb{N}$. Hence the equivalence of A1 and A2 for countable groups does not imply the existence of nonprincipal ultrafilters on $\mathbb{N}$.<|endoftext|>
TITLE: Are sets with similar asymptotic behavior as the primes necessarily finite additive bases?
QUESTION [24 upvotes]: The set of primes $\mathbb{P}$ has many interesting properties in additive number theory and some of the most famous open problems about $\mathbb{P}$ are the well-known Goldbach's strong and weak conjectures. The weak conjecture was proven by Vinogradov for all sufficiently large integers. Moreover, Chen was able to prove that every sufficiently large even integer is the sum of a prime and a semiprime, which is either a prime or a product of two primes.  
The purpose of this question is to find out how much do these additive properties of $\mathbb{P}$ result from the special (e.g. multiplicative) properties of the set of primes and how much do they just rely on the relatively large number of primes $\leq x$. 
The density of a set seems to have a large effect on its additive properties; one may for instance define the Schnirelmann density (of course there are many other densities, too) of a set as $\sigma A=\displaystyle \inf_{n\in \mathbb{Z_+}} \frac{A(n)}{n}$, where $A(n)$ is the number of elements of $A$ that are $\leq n$, and it was proven by Schnirelmann that if $A$ is any set with $\sigma A>0$, there exists a positive integer $k$ such that $kA = \mathbb{Z}_+$. 
Also if  $A+B\neq {\mathbb{Z}}_+$, we have $\sigma (A+B)\geq \sigma A+\sigma B$
as was proven by Mann. It is clear that if $1 \not \in A$, we have $\sigma A=0,$ but by cosidering $B=A\cup \{1,2,..,N\}$ and using Schnirelmann's theorem, one might be able to prove that all sufficiently large integers belong to $kA$ for $k$ large enough (for certain sets $A$). Using his theorem and Brun's sieve, Schnirelmann proved that every integer $>1$ is the sum of at most $C$ primes for some constant $C$.
By the prime number theorem, the set of primes that are not greater than $x$ has size $\pi(x) \sim li(x) \sim \frac{x}{\log x}$, where $li(x)$ is the logarithmic integral. Therefore, the set of primes is very large (in this sense) when compared to, say, perfect $k$-th powers. If one assumes the Riemann hypothesis, the set of primes also does not have very large gaps, more precisely, there is a prime on the interval $[x,x+\sqrt x \log x]$ for large values of $x$ as was proven by Schoenfeld under Riemann hypothesis. On the other hand, the behavior of primes seems to be locally very random, and therefore one might guess that a set with similar number of elements $\leq x$ and same maximum gap size would be a finite additive basis for the positive integers that are greater than some constant.
The question is: given an arbitrary set $A\subset \mathbb{Z}_+$ with $A(x)>> \frac{x}{\log{x}}$, and the difference of consecutive elements $x,y\in A$ is $O(\sqrt x \log x)$, and for every positive integer $n>1$ there are infinitely many elements of $A$ not divisible by $n$ (this condition is to prevent $A$ from having only even numbers, for example), is it known that there exists $k$ (which may depend on $A$) such that every positive integer large enough is the sum of at most $k$ numbers from $A$? If not, is there a known counterexample? In case of a counterexample, could some conditions that hold for $\mathbb{P}$ but that do not characterize the set of primes be given such that the question is true? In particular, does the claim hold for $A= \{\lfloor n\log n\rfloor, n \geq 2\}$?

REPLY [17 votes]: Let $A_n = \{a : a \equiv 1 \mod 2^n \mbox{ and } 2^{2^{n-1}} \leq a < 2^{2^{n}} - 2^{2^{n-1}}\}$, and let $\displaystyle A = \bigcup_{n=1}^\infty A_n$.
Then, $A(x) >> \frac{x}{\log x}$, the gap sizes are $<< \sqrt{x}$, and $A$ contains infinitely many non-multiples of $m$ for every $m>1$.
However, $2^{2^n}$ cannot be written as a sum of fewer than $n-\log_2 n$ elements of $A$.  To prove this, suppose the contrary; say $2^{2^{n}} = a_1 + \cdots + a_k$ for $k < n-\log_2 n$, $a_1 \leq \cdots \leq a_k$, $a_i \in A$.  We know that $a_k \in A_n$, because if not, then all the $a_i < 2^{2^{n-1}}$, so, summing, $2^{2^n} < k \cdot 2^{2^{n-1}}$, implying $2^{2^{n-1}} < n$, which is false for positive $n$.  One applies a similar argument to each of the partial sums of the $a_i$, in turn from largest to smallest, to show that $\displaystyle a_j \in \bigcup_{i=n-k+j}^n A_i$.  In particular, each $a_j \equiv 1 \mod 2^{n-k+1}$, so the sum $2^{2^n} \equiv k \mod 2^{n-k+1}$.  Thus, $2^{n-k+1} \mid k$; in particular, $2^{n-k+1} \leq k$.  Using the bound $k < n-\log_2 n$ on both sides, we may deduce $2n < n-\log_2 n$, a contradiction.
For a possible fix, I would assume that $A$ is well-distributed modulo $m$ (for every $m$) in an appropriate sense.  (I'm being purposefully vague here, and you might need something rather strong, like an analogue of Bombieri-Vinogradov).  Good luck!<|endoftext|>
TITLE: A question about maximal subgroups
QUESTION [11 upvotes]: Let $G$ be a finite group and $H_1,\ldots, H_n$ a set of maximal subgroups of $G$. Let $\delta_{H_i}$ be delta functions with support on $H_i$, and let $A$ be the commutative algebra generated by $\delta_{H_i}$. 
Question: is dimension of $A$ greater or equal to $n$? 
Motivation: In 1961 G.E.Wall conjectured that the number of maximal subgroups in a finite group $G$ is less than the order of $G$.  A positive answer to the above question will imply Wall's conjecture. In fact it will even imply a relative version of Wall's conjecture: the number of maximal subgroups in $G$ which contain a subgroup $H$ is bounded by the number of double cosets in $G$, see arXiv:1006.5947 for more discussions and motivations from subfactors. 
The question seems to be combinatorial in nature: the basis of $A$ is simply the list of subsets of $G$ obtained by cutting with $H_i$'s. for $n =2,3$ one can check directly, and one may try induction on $n$. Any references on this will be appreciated.

REPLY [3 votes]: Wall's conjecture is now known to be false : see this question and the answers there.<|endoftext|>
TITLE: The geometric meaning of the higher quotient by the commutant ideal
QUESTION [7 upvotes]: The functor that embeds the category of commutative algebras to associative algebras has the left adjoint - the quotient by the commutant ideal.
For any dg-algebra $A$ let $A_{Ab}$ denote the derived functor of the quotient by the commutant ideal, i. e. take
a free resolution of the algebra and then take the quotient by two-sided ideal generated by commutators. There is the canonical map $A \to A_{Ab}$.
Suppose that $A$ is  the algebra of rational cohomology of a homotopy type $X$.
My question is: does $A_{Ab}$ has some geometric interpretation? More precisely:
does there exist a functor $M$ from homotopy types to homotopy types and a natural transformation $e: M \to id$ such that $H^*(M(X))=A_{Ab}$ and $e$ gives the canonical map?
There is a similar question on the geometric meaning of $A_{Ab}$ for $A=H_*(\Omega X)$, the homology of a loop space with the Pontryagin product. The answer to this question is simple: $A_{Ab}=H_*(\Omega^{\infty+1}\Sigma^{\infty} X)$
and the canonical map is induced by the canonical $\Omega X\to \Omega^{\infty+1}\Sigma^{\infty} X$.
Note that the Lie algebra of  rational  homotopy groups $\pi_{*-1}M(X)$ must be $L(U(\pi_{*-1} X))$,
where $U$ is the universal enveloping algebra and $L$ makes an  associative algebra into Lie algebra with
the commutator as the Lie bracket.
It seems that my question is related with this one.

REPLY [3 votes]: [I have replaced an earlier and less complete answer]
The question depends on the precise meaning of $A_{ab}$.  In the derived world, if we kill commutators then we create new potential commutators and so do not immediately end up with something commutative.  Killing commutators once is the same as taking Hochschild homology.  I think it is now known from work of Smith and McClure that when you apply HH to something with an action of the little $k$-cubes operad $C(k)$, you always get something with an action of $C(k+1)$ (ie it is "one step more commutative").  An associative algebra has an action of $C(1)$, so you can apply HH repeatedly and pass to a colimit to get something with an action of $C(\infty)$, which is an $E_\infty$ operad.  If we are working over $\mathbb{Q}$ then  $E_\infty$ algebras are essentially the same as commutative algebras.  I think that this procedure converts free associative algebras to free commutative algebras.
UPDATE: David Ben-Zvi is right that I have misremembered results that actually apply to Hochschild cohomology (rather than homology) here.  Nonetheless, what I said about repeatedly killing commutators makes some kind of intuitive sense, so I still wonder whether something along those lines could be true.<|endoftext|>
TITLE: Is choice needed to establish the existence of idempotent ultrafilters?
QUESTION [14 upvotes]: It is well known that the Stone–Čech compactification $\beta \mathbb N^+$ of the positive natural numbers has the structure of a compact left semitopological semigroup and hence, by Ellis's lemma, has idempotents.  The usual proof of Ellis's lemma uses Zorn's lemma.  Idempotent ultrafilters are clearly non-principal.  It is known that the existence of non-principal ultrafilters is weaker than the axiom of choice.  
My question is whether the existence of idempotent ultrafilters in $\beta \mathbb N^+$ is still weaker than choice?

REPLY [18 votes]: Yes, it's still weaker.  To build a model of ZF in which choice fails but $\beta\mathbb N^+$ has idempotents, start with a model of ZFC (which will, of course, have idempotent ultrafilters in $\beta\mathbb N^+$).  Add a lot of Cohen-generic subsets of some regular cardinal $\kappa$ well above the cardinal of the continuum; forcing conditions are partial functions of size $<\kappa$.  No new reals are added, and your idempotent ultrafilters from the ground model are still idempotent ultrafilters (and AC still holds).  Now pass to the symmetric submodel given by the group of automorphisms of your forcing that permutes the names of the added Cohen subsets, with the filter determined by supports of size $<\kappa$.  That model violates choice, because you can't well-order the power set of $\kappa$.  But the ground model's reals and ultrafilters haven't been touched, so you still have the same idempotent ultrafilters that you had to start with.<|endoftext|>
TITLE: Laplace transform on the cone of positive-definite matrices
QUESTION [7 upvotes]: The title says most. Let $P_p$ be the cone of positive-definite $p \times p$ matrices. 
One can define the Laplace transform of (the distribution of) a random matrix with values in $P_p$ by (for instance, Muirhead: "Aspects of Multivariate Analysis") the integral
\begin{equation}
     \phi(\Theta) = \int_{P_p} \exp(\sum_{j\le k}^p \theta_{jk} a_{jk}) f(A)\; dA
\end{equation}
where $f(A)$ is the density function of $A$. (And $\Theta$ is a symmetric $p\times p$-matrix)
So the question is: I am searching for references for this Laplace transform, inversion theorems, numerical methods, known transform formulas, .... etc ???
Thanks for answers!  Now, I am searching for those references, but one is really difficult to find, that is, volume 2 of Audrey Terras' book: "Harmonic analysis on symmetric spaces II"
I have found the first volume, but volume II cannot even be found on Springers own website!  Any ideas about how to find it?

REPLY [4 votes]: Second editions of both volumes are about to be published.
It will probably take a year to update them though.
              audrey<|endoftext|>
TITLE: Tetrahedron angles sum to $\pi$: Bisector plane
QUESTION [5 upvotes]: I discovered empirically what to me is an amazing lemma
concerning face angles of a tetrahedron.
Let $\triangle abc$ be a triangle in the $xy$-plane, and $d$ the
apex of a tetrahedron with positive $z$ coordinate.
The lemma is this:

Lemma. The locus of points $d$ for which the sum
  of the nonbase tetrahedron face angles incident to $b$ sum to $\pi$,
  $$\angle dba + \angle dbc = \pi$$
  is a vertical (parallel to $z$) bisector plane which meets the $xy$-plane
  in a line $L$ that has the property
  that the angle of incidence $\beta$ between $ab$ and $L$ is equal
  to the angle of reflection $\beta$ between $bc$ and $L$.


 
 
 
 
 


If I express this relationship in terms of ArcCos( )'s of the relevant
angles, it all works out algebraically/trigonometrically.  So I have a "proof" in this
(limited) sense.
But surely for such a simple angle of incidence = angle of reflection
relationship there is a concise geometrical explanation—maybe involving reflecting light rays...?
Amidst a much longer proof, this was at one point a critical lemma, but now I have circumvented
its need (I think?!).  Nevertheless, it would be illuminating to see a more revealing proof.  Thanks for ideas and/or insights!

REPLY [3 votes]: Reflect $a$ across $L$ in the $xy$-plane to $a'$, so $\angle a'bc$ is a straight angle, and hence $\angle dba' + \angle dba = \pi$. If $d$ is in the bisector plane then $\angle dba' = \angle dba$, if $d$ is on the same side of the bisector plane as $a$ then $\angle dba' > \angle dba$, and if $d$ is on the opposite side then $\angle dba' < \angle dba$.<|endoftext|>
TITLE: Explicit description of the "simplicial tensor product" of chain complexes
QUESTION [11 upvotes]: Recall that there is an equivalence of categories (Dold-Kan) $$N:\mathrm{s}\mathbf{Ab}\simeq \operatorname{Ch}_{\geq 0}(\mathbf{Ab}):\Gamma$$ between simplicial abelian groups and (connective) chain complexes, where $N$ sends a simplicial abelian group to its associated normalized chain complex.  
Using this equivalence of categories, we can, by transport of structure, give an unorthodox tensor product on the category of chain complexes.  We may define this by the formula $X\otimes_\Delta Y=N(\Gamma(X)\otimes \Gamma(Y)),$ where the tensor product on the righthand side is the tensor product (taken pointwise) of simplicial abelian groups.
Then my question: Is there an explicit description of this tensor product in terms of the chain complexes themselves?

REPLY [10 votes]: The bad news is that in degree $n$, this tensor product has $3^n$ terms.
The functor $\Gamma$ can be roughly described as follows.  If we write $[n]$ for the ordered set $0 < 1 < \cdots < n$, then
$$
\Gamma(C)_n = \bigoplus_{k} \bigoplus_{\phi\colon [n] \twoheadrightarrow [k]} C_k.
$$
The face maps have two characters.  The map $d_i$ for $i > 0$ simply deletes the element $i$ from the ordered set $[n]$, and reindexes; if the resulting map $[n-1] \to [k]$ is no longer surjective, the corresponding factor maps to zero.  By contrast, the map $d_0$ deletes $0$ and reindexes, but if the corresponding map $\phi$ is no longer surjective its image is isomorphic to $[k-1]$, and we apply the boundary map.
When you take the tensor product of $\Gamma(C)$ and $\Gamma(D)$ levelwise, you get a direct sum indexed by pairs of surjections $[n] \twoheadrightarrow [p]$ and $[n] \twoheadrightarrow [q]$.
The functor $N$ then will take the quotient of this by the subcomplex of degenerate ones; those where the maps $[n] \twoheadrightarrow [p]$ and $[n] \twoheadrightarrow [q]$ factors through a surjection $[n] \twoheadrightarrow [m]$.  In practice, the pairs which are not degenerate are those for which the map $[n] \to [p] \times [q]$ is injective.
As a result, we have that
$$
N(\Gamma(C) \otimes \Gamma(D))_n = \bigoplus_\phi C_p \otimes D_q
$$
where the sum is indexed by injections $[n] \to [p] \times [q]$ where composing with either projection is surjective.
In practice, you can index this direct sum by $n$-tuples of strings of elements from$\{N, NE, E\}$, representing a path of length $n$; $p$ and $q$ are determined by the height and width of the path.
Unfortunately, a chain complex isn't very useful without its differential, and that's more complicated to describe.  The boundary map is the alternating sum of face maps; each face map deletes $i$ from the ordered set $0 < \cdots < n$ and reindexes.  If $i > 0$ and one of the resulting projections to $[p]$ or $[q]$ is no longer surjective, the corresponding factor maps to zero; if $i = 0$ then one or both of the maps to $[p]$ or $[q]$ misses zero, the appropriate image(s) are isomorphic to $[p-1]$ or $[q-1]$, and we apply the boundary map on those factors.
UPDATE: One way to write this is in a group homology style.  Then we can view elements of $\Gamma(C)_n$ as decorated with these $n$-tuple paths, the width and height determine which group we land in, and the boundary map takes an alternating sum of deleting commas with the understanding that $NN = EE = 0$.  So, for example, taking the boundary of this element in $C_3 \otimes D_2$:
$$
d(a \otimes b)_{(N,E,N,NE)} = (da) \otimes b_{(E,N,NE)} - a \otimes b_{(NE,N,NE)} + a \otimes b_{(N,NE,NE)} - a \otimes b_{(N,E,NNE)}.
$$
(The last term involving $NNE$ is dropped because it is zero.  The first term had a boundary map applied to the first factor because it was $N$.)<|endoftext|>
TITLE: Induction theorems for finite-dimensional complex representations of infinite groups
QUESTION [6 upvotes]: Let $G$ be a group, usually infinite. I am interested in finite-dimensional complex unitary representations of $G$, i.e. group homomorphisms $G \rightarrow U_n(\mathbb{C})$. The category of these representations is an exact category, and so we can form the Grothendieck group $K_0(G)$ of this category. If $H$ is a finite-index subgroup of $G$, we have an induction map
$K_0(H) \rightarrow K_0(G)$. Given a collection of subgroups $H_i$ of $G$, we can ask whether the map obtained by induction
$\oplus_i K_0(H_i) \rightarrow K_0(G)$
is onto, or onto after tensoring with $\mathbb{Q}$. Let us say that the collection is good if this map is onto. If $G$ is finite, for example the Brauer induction theorem or the Artin induction theorem give answers: The collection of elementary subgroups of $G$ is good. If $G$ is infinite and maps to a finite group $F$, we can pull back the elementary subgroups of $F$ to $G$ which yields a good collection of subgroups of $G$.
My question is: Are there other known cases of infinite groups $G$ with a good collection of subgroups? Also, I would be interested in any references where finite-dimensional representations of infinite groups are studied.

REPLY [4 votes]: I believe that all examples must come from finite quotients:

A collection of subgroups is necessarily good if the element $1\in K_0(G)$ is in the subgroup generated by the images of the induction maps $K_0(H)\to K_0(G)$. That's because these images are ideals, using the formula $V\otimes Ind(W)=Ind(Res(V)\otimes W)$.
Therefore every good collection contains a finite good collection. And of course for any collection of finite index subgroups $H_i$ there is a finite index normal subgroup $N$ of $G$ that is contained in them all.
And then the collection ${H_i}$ will be good in $G$ if and only if the collection $H_i/N$ is good in the finite group $G/N$. (This last requires a little argument splitting representations of $G$, or $H_i$, into the part fixed by $N$ and its orthogonal complement and noting that this splitting is compatible with induction.)<|endoftext|>
TITLE: Bialynicki-Birula decomposition of a non-singular quasi-projective scheme.
QUESTION [8 upvotes]: Fix an algebraically closed field $k$, an algebraic one-dimensional torus $G_m$ and a non-singular scheme $X$ of finite type over $k.$
Let us define the following:
Condition 1: $X$ can be covered by $G_m$-invariant quasi-affine open subschemes.
In the paper "Some theorems on actions of algebraic groups" (The Annals of Mathematics, Second Series, Vol. 98, No. 3 (Nov., 1973), pp. 480-497), Bialynicki-Birula constructs, roughly speaking, to any action of $G_m$ on $X$, satisfying Condition 1, two canonical decompositions of $X$ into non-singular $G_m$-invariant locally closed subschemes (Theorem 4.1).
Moreover, Bialynicki-Birula states that if $X$ is projective, Condition 1 is automatically satisfied (he cites Kambayashi, Projective representations of algebraic groups of 
transformations, Amer. J. Math. 88 (1966), 199-205.)
Question: Assume that $X$ is a non-singular quasi-projective scheme over $k.$ Under what extra assumptions, does $X$ satisfy Condition 1?

REPLY [14 votes]: It is a theorem of Sumihiro (Equivariant completion, Corollary 2) that a normal variety over an algebraically closed field with an action of a torus is covered by invariant affine open subsets.
Here, the normality hypothesis is necessary : the conclusion does not hold for the action of $\mathbb{G}_m$ on $\mathbb{P}^1$ with $0$ and $\infty$ glued together transversally.
  The hypothesis that the group is a torus is also necessary. Indeed, the statement already fails for SL(2), even for a proper action. There is an example in Białynicki-Birula and Święcicka's paper "On complete orbit spaces of SL(2) actions II".

REPLY [9 votes]: Every normal variety with an action of a torus is covered by invariant affine open subsets. This is proved in Hideyasu Sumihiro, Equivariant completion, J. Math. Kyoto Univ. 14 (1974), 1–28.<|endoftext|>
TITLE: Jones(unlink)=phi
QUESTION [5 upvotes]: Somewhat nebulous question: there are many well known "special" values of
the Jones polynomial, especially those at roots of unity. I always run into
one that has unlink value $\phi$ (golden mean) and writhe factor $(-1)^{1/5}$.
Is there something special about it (maybe it's "at the intersection" of
the Lie groups A1 and G2 or whatnot)?

REPLY [6 votes]: In some sense this is the smallest possible quantum group, so it's perhaps not surprising that it comes up often.  In fact, if you have only 2 objects, then there are very few possibilities, see Ostrik's paper http://arxiv.org/abs/math/0203255.<|endoftext|>
TITLE: The digit sum: $s(na)=s(nb)$
QUESTION [7 upvotes]: Not that I was serious about the following question, but I think it is a must-to-ask as a follow-up to this MO post. 
For integer $n\ge0$, let $s(n)$ denote the sum of the digits in the decimal representation of $n$.

Is it true that for any integer $a,b>0$, the ratio of which is not a power of $10$, the set of all those $n\ge 0$ with $s(an)=s(bn)$ has zero density?

REPLY [7 votes]: You can find some explicit asymptotics in a paper by J. Schmidt, "The joint distribution of the binary digits of integer multiples". He considers base $2$, and theorem 1.2 says that for any $a,b$ odd and arbitrary integer $k$.
$$|\lbrace 0\le n\le x: s(na)-s(nb)=k\rbrace|=\frac{x}{\sqrt{2\pi V\cdot \log_2x }}e^{-\frac{k^2}{2V\log_2 x}}+O(\frac{x}{\log x}),$$
where $V=\frac{1}{2}\left(1-\frac{\gcd^2(a,b)}{ab}\right)$. Something similar should work for any base.<|endoftext|>
TITLE: Trace over the zeros with real part 1/2 Only
QUESTION [5 upvotes]: If RH is not true, we have that Weil's explicit formula still holds:
$$ \sum_{\gamma} h(\gamma) = h(i/2)+h(-i/2)-2 \sum_{n=1}^{\infty} \frac{ \Lambda(n)}{ \sqrt n}g(logn)+\frac{1}{2\pi} \int_{-\infty}^{\infty}h(r)\Psi(1/4+ir/2)dr.$$
My question is if there are formulae that take into account ONLY the Riemann zeros with real part 1/2 and not the others.. thanks.

REPLY [7 votes]: Absent a proof of RH (or something close to RH), there are going to be some significant obstructions to the existence of such a formula, due to the possibility of two zeroes on the critical line that are very close together.  A small perturbation of the zeta function could then move these two zeroes off of the critical line (much as perturbing the polynomial $z^2-\varepsilon^2$ to $z^2+\varepsilon^2$ moves two nearby real zeroes $\pm \varepsilon$ into two complex zeroes $\pm i \varepsilon$).   The zeta function and its perturbation would be more or less indistinguishable from each other by the sort of contour integration operations that are traditionally used in Weil-type explicit formulae.  As a consequence, such formulae cannot distinguish between two nearby zeroes on the critical line, and two nearby zeroes slightly away from the critical line, which already rules out a lot of options for a formula that only captures the critical line behaviour.  Indeed, if one restricts attention to formulae which depend analytically with respect to analytic perturbations of the underlying complex function, I would imagine that there are no interesting formulae which only involve the real zeroes.
A model problem would be to try to locate an analytic formula relating the coefficients of a real polynomial with the purely real zeroes of that polynomial.  Even in the quadratic case, it seems clear that no non-trivial formula exists, because one side of the formula would develop some sort of singularity (such as failure of analyticity) as one transitioned from $z^2-\varepsilon^2$ to $z^2+\varepsilon^2$, and the other side of the formula would not.<|endoftext|>
TITLE: Relationship between Green's function and geodesic distance?
QUESTION [25 upvotes]: I am interested in showing that a certain Green's function can be used to approximate the distance function on a Riemannian manifold in the following sense.  Let $(M,g)$ be a Riemannian manifold and consider a ball $B \subset M$ centered at a distinguished point $p \in M$ whose radius is no larger than the injectivity radius at $p$.  Let $d: B \rightarrow \mathbb{R}$ be the geodesic distance to $p$.  (The reason for considering the ball $B$ instead of the entire domain is simply to avoid issues concerning the cut locus, where the distance function fails to be smooth.)  If $\Delta$ is the negative-(semi)definite Laplace-Beltrami operator on $M$, then there is a 1-parameter family of Green's functions $u_t$ defined as solutions to
$$(\mathrm{id}-t\Delta) u_t = \delta_p$$
where the parameter $t$ is positive and $\delta_p$ is a Dirac delta at $p$.
Question: does $\nabla u_t / |\nabla u_t|$ approach $-\nabla d$ as $t \rightarrow 0$?  Alternatively, do the level sets of $u_t$ approach geodesic circles as $t \rightarrow 0$?
Several closely related results suggest that the answer is likely positive, mostly related to analysis of the heat kernel.  In particular, Varadhan in his classic paper ("On the behavior of the fundamental solution of the heat equation with variable coefficients") considers a similar boundary value problem $(\mathrm{id} - t\Delta)v_t = 0$ on a domain $\Omega$ with $v_t |\partial\Omega = 1$ and shows that $\lim_{t \rightarrow 0} -\sqrt{t}/2 \log v_t = d$, i.e., the function itself converges to the distance function.  However, this result does not (as far as I know) explicitly guarantee convergence of the gradients.  In a similar vein, Malliavin and Stroock ("Short time behavior of the heat kernel and its logarithmic derivatives") essentially show that the gradient of the heat kernel converges to the gradient of the distance function.  However, the heat kernel is a solution to the parabolic problem $\dot{u} = \Delta u$ for some duration $t>0$ with initial conditions $u_0 = \delta_p$ -- i.e., it is not the same as the elliptic problem described above.  I am also aware of some results by Bardi ("An Asymptotic Formula for the Green's Function of an Elliptic Operator"), but they do not seem relevant in this case because they do not consider operators with a constant component ($\mathrm{id}$) as in the case above.  Basically what I'm saying here is that the result is almost certainly true (and not a major departure from what's already mentioned in this paragraph), but I'm having trouble nailing down a concrete reference to cite.
Thanks!

REPLY [12 votes]: First observe that on a compact  Riemann manifold $(M, g)$ the operator   $1+t \Delta$, $t>0$. $\Delta = d^*d: C^\infty(M)\to C^\infty(M)$ has a unique fundamental solution. Jacques Hadamard  has constructed   very explicit  asymptotic expansions for this fundamental solution which  lead to convergent series in the case of real analytic manifolds and metrics.
A modern description of  Hadamard's  construction can be found in volume  3, Sec. 17.4  of  L. Hörmander's  four volume on linear partial differential operators.   (Hadamard's original memoir is also very useful, but harder to penetrate.) I will give a brief description  of the      fundamental solution $S_r$ of $(r+\Delta)$, $r>0$.
For $\nu=0,1,2,\dotsc $  and $r>0$ denote by  $F_{\nu,r}(x)$ the generalized function (a.k.a. distribution) on $\mathbb{R}^n$   described  as $\newcommand{\ii}{\boldsymbol{i}}$ a Fourier transform of a temperate distribution.
$$F_{\nu,r}(x)= \nu! (2\pi)^{-n} \int_{\mathbb{R}^n} e^{ \ii\langle x,\xi\rangle} (|\xi|^2+r)^{-\nu-1} d\xi. $$
The function $F_\nu$ can be expressed explicitly in terms of Bessel functions.  Note that
$$F_{\nu,r}(x)= r^{\frac{n-\nu-1}{2}}  F_\nu(\sqrt{r} x),\;\;F_\nu(x):=F_{\nu,r=1}(x). $$
If $$\Delta=-\sum_k\partial^2_{x_k}$$
denotes  the (geometers') Laplacian in $\mathbb{R}^n$, then
$$(r+\Delta)F_{0,r}=\delta_0,\;\;(r+\Delta)F_{\nu,r}=\nu F_{\nu-1, r},\;\;\forall \nu>0. $$
One can show that the  generalized function  $F_\nu$ depends only on  the distance $|x|$.
Going back  to the Riemann manifold  $(M,g)$ we denote by $d: M\times M\to \mathbb{R}$ the geodesic  distance function.
The Green function $G(x,y)$ then  has an asymptotic expansion
$$ G(x,y)\sim \sum_{\nu=0}^\infty  U_\nu(x,y) F_{\nu,r}( \; d(x,y)\;)$$
valid  for $d(x,y)$ sufficiently small, where the  functions $U_\nu(x,y)$ are  explicitly described in the above reference.   If $(M,g)$  is real analytic, then the above  series converges in an appropriate sense.
This asymptotic expansion ought to be enough to   investigate your question.
Update I want to add a   "philosophical" comment. The question you asked  is a special case of the following more general question.
Suppose that $f:\mathbb{R}\to\mathbb{R}$ is a continuous function. For simplicity, let us assume it is also bounded. We can define the bounded symmetric operator $f(\sqrt{\Delta})$ where $\Delta$ is the Laplacian on an $m$-dimensional manifold $M$.  Investigate the  behavior of $f(\varepsilon\Delta)$ as $\varepsilon \to 0$.Your case  corresponds  to $f(x)=(1+x^2)^{-1}$. The  heat equation problems  correspond to $f(x)=e^{-x^2}$.     Suppose that $f$   is a  symbol  of order $k$, where $k$ could  be $-\infty$.   For example $(1+x^2)^{-k}$ is a symbol of order  $-2k$, while $e^{-x^2}$ is a symbol of order $-\infty$.
In any case, when $f$ is a symbol,  then $f(\Delta)$  is a pseudodifferential operator, and as such it has a Schwartz kernel which is a distribution on $M\times M$. $\newcommand{\ve}{\varepsilon}$ Your question is about the  behavior as $\ve\to 0$ of the  Schwartz kernel of $f(\ve \Delta)$ along normal directions to the diagonal of $M\times M$.
If  $f$ is rapidly decaying  at $\infty$, say $f(x) < (1+x^2)^{-m}$, $m=\dim M$,  then the Schwartz kernel of $f(\Delta)$  is given  by a continuous function and  one can  be quite precise about the behavior of the  kernel of $f(\ve \Delta)$. In fact,  the faster the  decay of $f$ at $\infty$, the more accurate one can be about the behavior of  the Schwartz kernel of $f(\ve \Delta)$. The radial symmetry you are talking about is then a simple consequence if $f$ decays faster than  $|x|^{-N}$, $N$ sufficiently large. (I believe that $N>2m$ ought to do it but I don't want to be too firm.) If $f$ has exponential decay at $\infty$ one can be remarkable accurate and recover the radial symmetry you are mentioning.  Your  question  involves  the symbol $(1+x^2)^{-1}$  that isn't decaying fast enough at $\infty$. Translation:  your problem requires a bit of care.<|endoftext|>
TITLE: Finite dimensional subspaces of $L^1.$
QUESTION [5 upvotes]: This question is motivated by my discussion (via comments) with @fedja regarding this earlier question. In any case the question is whether there is any concise characterization of finite dimensional subspaces of $L^1.$ I found some papers by our own Bill Johnson on finite dimensional subspaces of $L^p$ spaces, but they appear to  (a) be for $p>1$ and (b) use a lot of language hard to understand for us troglodytes. In the meantime, the question is very concrete: I give you a centrally symmetric convex body in $\mathbb{R}^d,$ and I ask whether this is the unit ball of the sup norm of the linear combinations of some $d$ functions $f_1, \dots, f_d.$
EDIT As @Fedor points out, I asked a different question from what I intended: I had meant to ask about $L^\infty,$ but the answers to this are very interesting, so I will let it stand, and I guess will ask a different question to avoid (or increase) confusion.

REPLY [3 votes]: Greg is right, of course, the dual must be a zonotope.
Let me mention also a direct characterization: for any vectors $x_1,\dots,x_k,y_1,\dots,y_m$ such that $\sum |f(x_i)|\geq \sum |f(y_j)|$ for any linear functional $f$, one may deduce that $\sum \|x_i\|\geq \sum \|y_j\|$. For Euclidean norm it is usual stuff in integral geometry: for proving inequality between sums of lengths one proves corresponding inequality for sums of projections to the same line. The norms, for which such trick is possible, are exactly those coming from sections of $L^1$.<|endoftext|>
TITLE: Dimension formula for Cartan-type abelian.group Nichols algebra?
QUESTION [6 upvotes]: Existence of a root system has been established for Nichols algebras $B(V)$ of a Yetter-Drinfel'd-module $V$ (resp. braided vectorspaces $V$) over abelian groups (resp. with diagonal braiding $x_i\otimes x_j\mapsto q_{ij}x_j\otimes x_i$) already in 2000 by Kharchenko. In 2008 Schneider and Heckenberger established root systems over nonabelian groups, under the condition that the root system / Nichols algebra be finite (-dimensional)!  
Having a root system means among others that one has a set of irreducible sub-Yetter-modules $X_1\ldots X_n\subset B(V)$, with multiplying as below a bijection (no algebra morphism!):
$$B(V)\cong\bigotimes_{i=1}^nB(X_i)$$
Dynkin-diagrams can be drawn with Cartan Matrix $$q_{ij}q_{ji}=q_{ii}^{-A_{ij}}$$ but there are sporadic cases not corresponding to finite semisimple Lie algebras (triangle, exotic edge...) for low prime exponents. In these cases the classification of pointed Hopf algebras (Schneider /
 Andruskiewitsch) fails; the Weyl group is replaced by a -groupoid between different Dynkin diagrams.   
But in Cartan-type it's the same as in the semisimple case - a root system $\Phi$!

Is is generally true that the "roots" in both formulas coicide?
$$|\{X_1,X_2,\ldots X_n\}|=n=|\Phi^+|\qquad or=\ldots?$$
So can I calculate the dimension of  $B(X)$ like that? 

$$dim(B(X))=\prod_{\alpha\in\Phi^+}dim(X_i)$$
...clearly $B(X_i)\cong k[x]/(x^{ord(q)})$ with dimension $ord(q_{ii})$) or infinite for $q=1$ (the bosonic $k[x]$). 

So I just multiply the orders of respective the $q_{ii}$ for all roots?

$$dim(B(X))=\prod_{\alpha\in\Phi^+}ord(q_{\alpha\alpha})$$

So especially for $G=\mathbb{Z}_2^k$ the dimension is the following?

$$dim(B(X))=2^{|\Phi^+|}$$

REPLY [2 votes]: Yes, careful application of the theory of rootsystems verifies indeed these formulae.....farily streight-forward!<|endoftext|>
TITLE: History surrounding Gauss Theorema Egregium and differential geometry
QUESTION [12 upvotes]: I am teaching a class on elementary differential geometry and I would like to know, for myself and for my students, something more about the history of Gauss Theorema Egregium, that is the Gaussian curvature of a surface is an intrinsic quantity. 
For instance, I am fascinated by whether Gauss had imagined that it was an intrinsic property or, after a lengthy calculation, he found out it was. Perhaps the fact that he called this result Remarkable Theorem points toward the latter.
I have not been able to find a book on the history of differential geometry that would adress this.  More generally, I would like to know more about the history of differential geometry and I would welcome any suggestions for books or surveys on it.
Thanks

REPLY [10 votes]: It seems to me that Volume 2 of Mike Spivak course on differential geometry gives an answer to your question. See Section A of chapter 3 : How to read Gauss?

REPLY [6 votes]: Thanks. As you said, the answer to my question is exactly on page 143 of Volume 2 of Mike Spivak course on differential geometry. According to Spivak, Gauss showed that the Gaussian curvature is intrinsic by taking the limit using the now called Gauss-Bonnet Theorem for geodesic triangles. Once he realized that, he then set out to do the computation in Theorema Egregium and of course he succeeded in carrying it out.

REPLY [3 votes]: You should read Gauss' book (General theory of curves and surfaces). It is written in a fairly unenlightening manner (back then, people did not trust pictures, so Gauss just does pages and pages of horrible computations), but if you read it carefully, you will see that he had a very geometric view of things. Secondary sources (e.g., Spivak) are based entirely on Gauss' book and since Gauss did not write an autobiography (to the best of my knowledge) you should go to the said book of his.<|endoftext|>
TITLE: Finite subgroups of relatively hyperbolic groups
QUESTION [11 upvotes]: It is well known that in a given $\delta$-hyperbolic group there are only finitely many conjugacy classes of finite subgroups. This is clearly false for relatively hyperbolic groups since we have no control over the parabolic subgroups. Fix a relatively hyperbolic group $\Gamma$ with parabolic subgroups $P_i$. Are there only finitely many conjugacy classes of finite subgroups $F<\Gamma$ with the property that $F$ is contained in a two-ended subgroup $H<\Gamma$ which is not conjugate into any $P_i$? Perhaps weaker: are there only finitely many isomorphism types of two-ended subgroups which are not conjugate into any parabolic? Weaker still: Is there an upper bound to the orders of elements in such subgroups $F$?
(Motivation: I am trying to find a law obeyed by non-parabolic two-ended subgroups of relatively hyperbolic groups, or at least a partition of them into finitely many families, each of which obeys a law.)

REPLY [3 votes]: This question is stated as Corollary 9.6 in Strong accessibility for finitely presented groups but much of the actual proof is in Lemma 9.5.
The proof doesn't quite follow Lee's sketch (which I'm sure works.) I know this is an old question, but I'm providing this answer for completeness and for the cheap thrill of gratuitous self-promotion.<|endoftext|>
TITLE: Extreme points of a compact convex set are a $G_\delta$?
QUESTION [6 upvotes]: Dear All,
I'm reading a paper (Residuality of Dynamical Morphisms by Burton, Keane and Serafin) that makes a claim that I've been unable to verify or find a reference for. The claim is made that the extreme points of a compact convex set in a locally convex topological vector space form a $G_\delta$ subset of the space. 
I've been able to verify it in the specific context of the paper (sets of invariant measures for a continuous transformation of a compact metric space), but in the article they say
a general theorem states that the extreme points of a compact convex set form a $G_\delta$.
They don't say whose general theorem! I've looked reasonably hard for a suitable reference without success. Can anyone give me any pointers?
Thanks...

REPLY [9 votes]: For a non-metrizable compact convex subset of a locally convex space, extreme points need not even form a Borel set. This has been shown  by Bishop-de Leeuw, The representation of linear functionals by measures on sets of extreme points, Ann.Inst. Fourier (Grenoble) (1959) .
A very good reference for these topics is Phelp's LNM Lectures on the Choquet's theorem (2001).<|endoftext|>
TITLE: Central limit theorem for 3d rotations
QUESTION [6 upvotes]: Let $X_i$ (where $i=1, \dots , n$) be independent and identically distributed 3d rotations. What is the distribution of $X_1X_2\dotsb X_n$ in the limit of large $n$?
I'm especially interested in the special case where $X_i$ have mean 0 and are highly concentrated i.e. they are very small rotations. Just to make this clear, an example of this in the 2d case would be the rotations $(x,y)\rightarrow(x\cos \epsilon -y\sin \epsilon, x\sin \epsilon +y\cos \epsilon)$ where $\epsilon$ is a random variable with mean 0 and small variance.

REPLY [2 votes]: This is the random walk on a sphere Roberts & Ursell (1960).
Suppose that $X_i$ are rotations by an angle $\alpha$ on the sphere in a random direction
with probability $p(\mu)d\mu$ where $\mu=\cos\alpha$. The 
probability density on the sphere after the action of $X_n$ can be expressed with the probability density of the preceding step as
\begin{align}
\rho_n(\vec{r}) =
\frac{1}{2\pi}\int_{-1}^1 d \mu\int_{S_2} d \vec{r}'  p(\mu)
\delta(\vec{r} \cdot \vec{r}'- \mu)\,\rho_{n-1}(\vec{r}')\,.
\end{align}
This equation is linear in $\rho$ and can be solved using the
eigenbasis of the corresponding linear operator.
Next we show that if all steps have the same angular value $\alpha$, the eigenfunctions are
the spherical harmonics $Y_{\ell m}$ with eigenvalue $P_{\ell}(\mu)$.
We will use the following identities of the Legendre polynomials,
\begin{align}
&\delta(\cos \gamma - \mu)= \sum_{\ell=0}^{\infty}
\frac{2\ell+1}{2}P_{\ell}(\cos\gamma)P_{\ell}(\mu)\,,\\
&P_{\ell}(\cos\gamma)
=
\frac{4\pi}{2\ell+1}\sum_{m=-\ell}^{\ell}
Y_{\ell m}(\vec{r})Y_{\ell m}^*(\vec{r}')\,.
\end{align}
where $Y_{\ell m}(\vec{r})$ are orthonormal spherical harmonics. Substituting
\begin{align}
 \rho_n(\vec{r}) = \sum_{\ell=0}^{\infty}\sum_{m=-\ell}^{\ell}
 P_{\ell}(\mu)\, Y_{\ell m}(\vec{r}) \int_{S_2} d\vec{r}'\,Y_{\ell
   m}^*(\vec{r}')\rho_{n-1}(\vec{r}').
\end{align}
Expanding the initial distribution in this basis as
\begin{align}
 \rho_0(\vec{r}) = \sum_{\ell, m}a_{\ell m,0}Y_{\ell m}(\vec{r})
\end{align}
the distribution after a single step is
\begin{align}
 \rho_1(\vec{r}) = \sum_{\ell,m}
 \langle P_{\ell}(\mu) \rangle\, a_{\ell m,0} Y_{\ell m}(\vec{r}),
\end{align}
where $\langle P_{\ell}(\cos \alpha) \rangle = \int_{-1}^1 P_{\ell}(\mu) p(\mu) d \mu$.
After the $n^{\rm th}$ step we have
\begin{align}
 \rho_{n}(\vec{r}) = \sum_{\ell, m} \langle P_{\ell}(\mu) \rangle^{n}  a_{\ell m,0}Y_{\ell m}(\vec{r}).
\end{align}
Since $|\langle P_\ell(\mu)\rangle| \leq 1$ for $\ell>0$,
each $a_{\ell m}$ multipole moment with $\ell>0$ decays exponentially in the
number of steps from their initial value $a_{\ell m,0}$
as $\exp[n\ln|\langle P_\ell(\mu)\rangle|]$; the
system ``isotropizes'' with a decay time of $\Delta t /\ln |\langle
P_{\ell}(\mu) \rangle|$ where $\Delta t$ is the timestep.
The Green's function corresponding to an initial density
$\rho_0$ that is concentrated at the $\theta=0$ pole corresponds to
$a_{\ell m} =\sqrt{(2\ell + 1) /(4\pi)}\delta_{m,0}$. Thus the
probability for the angle $\theta$
between the initial and final position after $n$ steps is given by
 \begin{align}
\rho_n(\theta_n) d\cos\theta_n = \sum_{\ell=0}^\infty
\frac{2\ell +1}{2}\, \langle P_{\ell}(\mu)\rangle^n\,
P_{\ell}(\cos\theta_n)d\cos\theta_n\,,
\end{align}
which implies that
  \begin{align}
 \langle P_{\ell}(\cos\theta_n)\rangle \equiv \int_{-1}^{1}  P_{\ell}(\cos\theta_n)\,\rho_n(\theta_n)d \cos\theta_n =
 \langle P_{\ell}(\mu)\rangle^n\,.
 \end{align}
In the limiting case of Brownian motion the angular step $\alpha$ and
the timestep $\Delta t$ both approach zero with $\alpha^2\sim \Delta
t$. In this limit
$P_{\ell}(\mu) \approx 1-\frac14 \ell (\ell+1) \alpha^2$, and so
\begin{align}
 \rho_{n}(\vec{r}) = \sum_{\ell,m} a_{\ell m} Y_{\ell m}(\vec{r}) e^{-\frac14 \ell (\ell+1) v}
\end{align}
where $v = n \langle \alpha^2\rangle = \langle\alpha^2\rangle t
/\Delta t $ is the variance of the corresponding planar motion.
 Thus
\begin{align}
\langle P_{\ell}(\cos\theta_n) \rangle = e^{-\frac14 \ell(\ell+1) n \langle\alpha^2\rangle}=e^{-\frac14 \ell(\ell+1) v}.
\end{align}
Up to this point we have only calculated the expectation value of the action of $X_n\dots X_2 X_1$. The distribution function may be similarly expressed with the distribution of multipole moments $x_{\ell} = P_{\ell}(\cos \theta_n)$ as 
\begin{align}
p(x_{\ell}) =  \prod_k \int_{-1}^{1} d \mu_k P_{\ell}(\mu_k)\delta\left( x_{\ell} - \prod_i P_{\ell}(\mu_i)  \right)\,.
\end{align}<|endoftext|>
TITLE: How to compute the Monopole Floer Homology for Surface $\times S^1$ ?
QUESTION [8 upvotes]: We know that Monopole Floer homology of a 3-manifold $M$ depends on a spin-c structure. My question is that if $M$ is $F\times S^1$ ($F$ is a surface of genus larger than 1) then how can we compute the Floer homology for it?
For the spin-c structures satisfying $\langle c_1(L),F\rangle>2g-2$ ($L$ is the determinant bundle of the spin-c structure,$g$ is the genus of $F$), Kronheimer and Mrowka prove that the Floer homology vanishes. They also proved that if $\langle c_1(L),F\rangle=2g-2$ then the Floer homology is $\mathbb{Z}$. But what about the other spin-c  structure (when $\langle c_1(L),F\rangle<2g-2$)?
Also, what is the answer for this question if we consider Heegaard Floer Homology instead of Monopole Floer Homology?

REPLY [6 votes]: I would assume you are interested in $HM$-to as opposed to $HM$-bar ($HM$-bar is mostly computed in the book Monopoles and 3-manifolds by Kronheimer and Mrowka).  For the case of $HM$-to, you should use (as answered above) that monopole is isomorphic to Heegaard Floer (Kutluhan-Lee-Taubes or Taubes + Colin-Ghiggini-Honda).  
If you want the trivial torsion Spin$^c$ structure, this is computed by Jabuka and Mark:
http://arxiv.org/pdf/math/0502328v4.pdf
This paper also has the references to the earlier computations for the other Spin$^c$ structures, done by Ozsvath and Szabo.<|endoftext|>
TITLE: Divergent series expansion in Apéry's proof of the irrationality of $\zeta(2)$ and $\zeta(3)$
QUESTION [8 upvotes]: UPDATE. I am now making this a CW in the hope someone can improve the content of this question and/or correct the text. 
This is a concise version of this math.SE question of mine. I've got an answer but it is not clear to me whether the explanation that was provided is enough.
My apologies if the level of this question, which is my first one here, does not comply with the rules of this site.
The question is:  
In Apéry, R., Irrationalité de $\zeta 2$ et $\zeta
3$, Société Mathématique de France, Astérisque 61 (1979)
there is a divergent series expansion for a function I would like to
understand. 
Added copy of the original.

Here is my translation of the relevant part for this question 

(...) given a real sequence $a_{1},a_{2},\ldots ,a_{k}$, an analytic function $f\left( x\right) $ with respect to the variable
  $\frac{1}{x}$ tending to $0$ with $\frac{1}{x}$ admits a (unique)
  expansion in the form $$f\left( x\right) \equiv \sum_{k\geq
1}\frac{c_{k}}{\left( x+a_{1}\right) \left( x+a_{2}\right) \ldots
 \left( x+a_{k}\right) },\tag{A}$$

[Edit] and a translation of the text after the formula (provided in a comment on math.SE)

(We write $\equiv$ instead of $=$ to take into account the aversions of mathematicians who, following Abel, Cauchy and d'Alembert, hold divergent series to be an invention of the devil; in fact, we only ever use a finite sum of terms, but the number of terms is an unbounded function of $x$).

As far as I understand, expansion of $f(x)$ in $(\mathrm{A})$ is in general   a divergent series and not a convergent one.  
Questions

Is series $(\mathrm{A})$ indeed divergent? (question added)
Which is the theorem stating or from which $(\mathrm{A})$ can be derived? 
Could you please indicate a reference?

ADDED. The most similar formula which is proved in Alf van der Poorten's article A proof that Euler missed ... Apéry's proof of the irrationality of $\zeta (3)$ is this one in section 3
$$\sum_{k=1}^{K}\frac{a_{1}a_{2}\cdots a_{k-1}}{(x+a_{1})\cdots (x+a_{k})}=
\frac{1}{x}-\frac{a_{1}a_{2}\cdots a_{K}}{(x+a_{1})\cdots (x+a_{K})}.\tag{B}$$
Further edit. This article by Alf van der Poorten, which was indicated in Micah Milinovich's answer, had been pointed to in my original math.SE question. The author wrote in it that "the proof of $(\mathrm{B})$ follows easily on writing the right-hand side as $A_0 - A_K$ and noting that each term on the left is $A_{k-1} - A_k$. This explains [the following claim corresponding to $f(x)=1/x$]" 
$$\sum_{k=1}^{\infty}\frac{a_{1}a_{2}\cdots a_{k-1}}{(x+a_{1})\cdots (x+a_{k})}=
\frac{1}{x}.\tag{*}$$
How to explain $(\mathrm{A})$ in the case when $f(x)$ is a general analytic function with respect to the variable $1/x$ which goes to $0$ with $1/x$?

Added. MOTIVATION. After reading in detail Alf van der Poorten's article I read the very short Roger Apéry's paper. I am interested in finding a proof of the series expansion in the latter, which is in not given in it. So I assumed it should be stated or derived from a theorem on the subject.

REPLY [2 votes]: The version of this question posted on Mathematics Stack Exchange got a new answer yesterday by robjohn, which I've already approved.
ADDED: I quote robjohn's anwer:
Writing $g_1(x)=f(1/x)$ gives
$$
g_1(x)\equiv\sum_{k\ge1}\frac{c_kx^k}{(1+a_1x)(1+a_2x)\dots(1+a_kx)}\tag{1}
$$
which vanishes at $x=0$.
Recursively define
$$
g_{n+1}(x)=\frac{(1+a_nx)g_n(x)}{x}-c_n\tag{2}
$$
where
$$
c_n=\lim_{x\to0}\frac{g_n(x)}{x}\tag{3}
$$
Then
$$
g_n(x)\equiv\sum_{k\ge n}\frac{c_kx^{k-n+1}}{(1+a_nx)(1+a_{n+1}x)\dots(1+a_kx)}\tag{4}
$$
is another series like $(1)$ (which vanishes at $x=0$).
The series in $(1)$ may or may not converge, as with the Euler-Maclaurin Sum Series. As with most asymptotic series, we are only interested in the first several terms; the remainder (not the remaining terms) can be bounded by something smaller than the preceding terms. Therefore, convergence is not an issue.<|endoftext|>
TITLE: What is the largest possible thirteenth kissing sphere?
QUESTION [19 upvotes]: It is well-known that it is impossible to arrange 13 spheres of unit radius all tangent to another unit sphere without their interiors intersecting.  This was apparently the subject of disagreement between Isaac Newton ("impossible") and David Gregory ("possible").  The cause of the dispute was, in part, that there seems to be a lot of room left over after 12 spheres.
How close of a call is it, so to speak?  What is the largest possible $r$ so that it is possible to arrange 12 spheres of unit radius and a 13th sphere of radius $r$ all tangent to another unit sphere, without intersections?  What does the optimal configuration look like?

REPLY [22 votes]: Pietro's version of this question is answered in a paper by Oleg Musin and Alexey Tarasov (to appear in Discrete & Computational Geometry, http://dx.doi.org/10.1007/s00454-011-9392-2, http://arxiv.org/abs/1002.1439). The configuration found by Schütte and van der Waerden (see Joseph O'Rourke's answer) is optimal and unique up to isometries.
The other version of the problem amounts to asking for the largest possible hole in a packing of 12 identical disks of radius $30^\circ$ on the 2-sphere.  I don't know the answer offhand.  One could certainly figure out what it must be by numerical optimization, but finding a rigorous proof would be difficult.  (It might be possible using variants of the Musin-Tarasov approach, which is an enormous brute force search over small planar graphs.)  I am sure someone must have looked at this problem, but I don't know of a place where the answer might be recorded.

REPLY [14 votes]: The recent paper by Hopkins, Stillinger, Turquato,
"Densest local sphere-packing diversity. II. Application to three dimensions,"
Physical Review E 83, 011304 (2011)
(PDF link),
addresses the variant suggested by Pietro:

"The smallest radius spherical surface onto which
  the centers of 13 spheres of unit diameter can be placed
  is strongly conjectured to be $R = R_\min(13) = 1.045572\ldots$,
  with the centers arranged in a structure first documented in
  Ref. [5]. It appears that, although Gregory was incorrect in
  conjecturing $K_3$ to be 13, his guess was not particularly far off."
[5] K. Schütte and B. L. van der Waerden, Math Ann. 123, 96 (1951).

Here's a nice image (from MathWorld) that shows the gaps when 12 unit spheres,
tangent at icosahedron vertices, surround one unit sphere:
           (source)
Update. See Henry Cohn's answer, which cites a more recent paper that settles (positively) the
conjecture noted above.<|endoftext|>
TITLE: Examples of interesting false proofs
QUESTION [69 upvotes]: According to Wikipedia False proof

For example the reason validity fails may be a division by zero that is hidden by algebraic notation. There is a striking quality of the mathematical fallacy: as typically presented, it leads not only to an absurd result, but does so in a crafty or clever way.

The Wikipedia page gives examples of proofs along the lines $2=1$ and the primary source appears the book Maxwell, E. A. (1959), Fallacies in mathematics.

What are some examples of interesting false proofs?

REPLY [5 votes]: This proof that $\pi=0$ may be of some interest in examinations.
The function $f(x)=\arctan(x)+\arctan(1/x)$ has derivative $f’(x)=\frac1{1+x^2}-\frac1{x^2} \frac{1}{1+\frac1{x^2}}=0$, hence it is constant. Therefore$\displaystyle \lim_{x\to+\infty}f(x)= \displaystyle \lim_{x\to-\infty}f(x)$, that is $\frac\pi2=-\frac\pi2$, whence $\pi=0$.$\quad\square$<|endoftext|>
TITLE: is the geodesic flow on Hyperbolic Plane completely integrable?
QUESTION [7 upvotes]: I'm looking for examples of completely integrable systems and specifically geodesic flows. We remember that when we have a symplectic manifold $(M,\omega)$ (with $M$ of dimension $2n$) and $H:M\rightarrow\mathbb{R}$ a smooth function, its symplectic gradient is the unique field $X_H$ over $M$ satisfying
$$\textrm{d}H=\omega(X_H,\cdot)$$
and we say that the system $(M,\omega,H)$ is completely integrable is there exists $f_1,\ldots,f_{n-1}:M\rightarrow\mathbb{R}$ smooth functions Poisson commuting: $\{f_i,f_j\}=\{f_k,H\}=0$, where $\{f,g\}=\omega(X_f,X_g)$, and with $\textrm{d}f_1,\ldots,\textrm{d}f_{n-1},\textrm{d}H$ linearly independent in a dense set of $M$.
In the cotangent bundle $T^*M$ of a manifold $M$, there exists a canonical symplectic form,
$$\omega_\textrm{can}=\sum{\textrm{d}x_i}\wedge\textrm{d}\xi_i$$
$(x_1,\ldots,x_n,\xi_1,\ldots,\xi_n)$ local coordinates of $T^\star M$. Then, if we consider a riemannian manifold $(M,g)$, we can canonically define a symplectic form on $TM$ with the bundle isomorphism $\Phi:TM\rightarrow T^\star M$ given by
$$\Phi(p,v)=(p,v^\star)$$
where $v^*(w)=g(v,w)$ is the Riesz representation of a functional. Hence we can define $\omega=\Phi^{\star}\omega_\textrm{can}$. In this way, the geodesic flow can be viewed as the flow of the symplectic gradient of the metric hamiltonian $H(p,v)=\frac{1}{2}g_p(v,v)$. Then my question is if the geodesic flow on the tangent bundle of the hyperbolic plane is completely integrable and, if yes, what is the another function beside the metric hamiltonian $H$. Any help will be appreciated.

REPLY [10 votes]: Yes, the geodesic flow on the hyperbolic plane and, in fact, on any Hadamard manifold (${\mathbb R}^n$ provided with a  complete Riemannian metric with non-positive curvature) is integrable.
You can easily construct integrals of motion for the geodesic flow in hyperbolic space as follows: 
1. Consider the Cayley-Klein model where the hyperbolic space is the interior of the unit ball in ${\mathbb R}^n$ and (oriented) geodesics are (oriented) straight lines. 
2. The space of geodesics is then the space of oriented lines passing through the unit ball and this is diffeomorphic to the unit codisc bundle of the standard Riemannian metric on $S^{n-1}$. In particular, the space of geodesics is a smooth manifold.
3. Consider the canonical projection $\pi$ that associates to a unit covector 
$\xi \in T^*{\mathbb H}^n$ the oriented geodesic that has $\xi$ as initial condition.
4. The pull-back by $\pi$ of any smooth function on the space of geodesics (suitably extended as a homogeneous function) will be an integral of motion.
The key idea in this is that if a Riemannian or Finsler metric is such that its space of geodesics is a manifold, then the geodesic flow is integrable. This has been worked out in detail by Carlos E. Duran in 
On the complete integrability of the geodesic flow of manifolds all of whose geodesics are closed Ergodic Theory and Dynamical Systems (1997), 17 : pp 1359-1370 1997.
He considers the compact case, but he also works out the case where the space of geodesics is an orbifold (like lens spaces).
Second answer. 
If you are only interested in the hyperbolic plane, then just use its symmetries (i.e., the group $SL(2,{\mathbb R})$) and the classic theorem of Noether that relates one-parameter groups of symmetries with integrals of motion.<|endoftext|>
TITLE: What is your picture of the flat topology?
QUESTION [31 upvotes]: I recently tried to explain the fppf site to a differential geometer. I started with the etale site, where I had two motivating claims: 

If X is a smooth projective variety over the complexes, the etale topology "recovers" the traditional topology. More precisely, the category of sheaves on the etale site and the category of sheaves on the complex-analytic site are equivalent. (EDIT: I have led my friend astray! See below.)
With just three words scare-quoted: The etale topology is (by "definition") the coarsest "topology" that makes the inverse function theorem "true."

I don't know how to make a statement analogous to (1) or (2) for the fpqc (or fppf, or ...) topology. I don't even really have a feeling for the difference between these topologies. Is there an analogue of (1) or (2) or anything else "purely geometric" to hang my hat on?

REPLY [8 votes]: this bot would like to answer your question as follows. Hope you like it and please don't forget to buy my products!
So, I actually kind of agree with you on the \'etale topology. So let's say your friend has a rough idea of that one. Or below you can even replace \'etale with Zariski and the thing works too.
Then, next you can try to give them an idea of what a surjective, finite locally free morphism of schemes is. Surjective finite locally free morphisms have many good topological properties: they are universally closed, have finite fibres, are quasi-compact, etc, etc. I guess topologically this is like having a surjective proper map of locally compact spaces with finite fibres. You can explain to your friend how these (in topology) can be used (via hypercoverings, woohoo!) to compute cohomology, etc.
Then the fppf topology just combines the \'etale topology (or the Zariski one) with the coverings coming from surjective finite locally free morphisms. See Section Tag 05WM.<|endoftext|>
TITLE: Set-Theoretic Issues/Categories
QUESTION [10 upvotes]: It is a major bummer that one cannot strictly speaking talk about the category of all categories without saying "it is not really a category, since the morphisms between objects may form a class" and "there are set-theoretic issues". However, these words seem to be just phrases, since I haven't seen them explained in detail. Why can't we lax the requirement that the morphisms form a set? We may no longer talk about representing functors (at least it won't be into $\bf{Set}$), but that may be OK in certain contexts. I am mostly scared of running into inconsistencies. I presume that if one tries to do it in the most naive way, there may be arguments of Russel's paradox type and that would be bad. Basically my questions are:
1) Is it possible to extend the definition of categories to things that may have morphism classes between two objects in a consistent way? 
2) What are the main obstacles that one has to concern himself with when trying to do this?

REPLY [4 votes]: Since you don't seem to want to leave ZFC, here's a taste of the issues you might face if you try to work with a stratified universe. (Here I mean the ordinary English word ‘stratified’, rather than a formally stratified logical system like Russell's type theory.) First, a warm-up:
Exercise. Develop the theory of finite categories and locally finite categories.
Now, let $\lambda$ be any infinite limit ordinal greater than $\omega$. The set $V_\lambda$ (!) of all sets of rank less than $\lambda$, together with the set (!) of all functions inside $V_\lambda$, forms a small category $\textbf{Set}_\lambda$ satisfying the axioms of ETCS (the elementary category of sets). As long as you don't try to  invoke unbounded separation or replacement, ETCS is a reasonable foundation in which to do mathematics. Thus, we obtain an infinite ascending sequence of full subcategories
$$\textbf{Set}_{\omega + \omega} \subset \textbf{Set}_{\omega + \omega + \omega} \subset \cdots \subset \textbf{Set}_{\infty}$$
where $\textbf{Set}_{\infty}$ denotes the (meta)category of all sets. Immediately we have some problems to resolve:

Since each $\textbf{Set}_\lambda$ is a priori a different category, there is no reason to believe that limits and colimits in $\textbf{Set}_\lambda$ will agree with limits and colimits in $\textbf{Set}_\kappa$, if $\lambda \lt \kappa$. Fortunately, it is a fact that if $\mathcal{D}$ is a full subcategory of $\mathcal{C}$, and the limit (resp. colimit) $X$ of a diagram in $\mathcal{D}$ exists in $\mathcal{C}$, then as long as $X$ is in $\mathcal{D}$, $X$ will be the limit (resp. colimit) of that diagram in $\mathcal{D}$. In more sophisticated words, the inclusion of a full subcategory always reflects limits and colimits. The trouble is that it may not preserve limits and colimits.

None of the categories $\textbf{Set}_\lambda$ have all small limits or colimits. This is due to an observation of Freyd: a small category with all small limits (or colimits) is a preorder category. However, it has all internally-indexed limits and colimits, in the following sense: if $\mathbb{C}$ is an internal category in $\textbf{Set}_\lambda$, meaning $\operatorname{ob} \mathbb{C}$ and $\operatorname{mor} \mathbb{C}$ are sets in $\textbf{Set}_\lambda$, and $A : \mathbb{C} \to \textbf{Set}_\lambda$ is an internal diagram (see [CWM, Ch. XII, § 1]), then $\varprojlim A$ and $\varinjlim A$ both exist in $\textbf{Set}_\lambda$ via the usual construction. In particular, every $\textbf{Set}_\lambda$ is closed under finite limits and colimits.

Though $\textbf{Set}_\lambda$ is a small category, it is not an internal category in $\textbf{Set}_\lambda$ itself! Instead, we must go up the hierarchy in order to realise $\textbf{Set}_\lambda$ as an internal category. This a priori seems to mean that we cannot talk about a genuine functor $\mathbb{C} \to \textbf{Set}_\lambda$, where $\mathbb{C}$ is an internal category in $\textbf{Set}_\lambda$, without leaving $\textbf{Set}_\lambda$, and once we leave $\textbf{Set}_\lambda$ we have to worry about whether the results obtained in a bigger category $\textbf{Set}_\kappa$ are still valid in our original category $\textbf{Set}_\lambda$, cf discussions about "change of universe" in SGA.
To be clear, this is a non-trivial problem. It is not so easy to replace a genuine functor $\mathbb{C} \to \textbf{Set}_\lambda$ with an internal diagram in $\textbf{Set}_\lambda$. For example, let $\omega \to \textbf{Set}_{\omega + \omega}$ be the functor $n \mapsto \omega + n$. The obvious way of turning this into an internal diagram uses the axiom of replacement and results in a set of rank at least $\omega + \omega$ – in other words, it is not internal to $\textbf{Set}_{\omega + \omega}$! Nonetheless, for cardinality reasons, it is true that every genuine functor $\mathbb{C} \to \textbf{Set}_\lambda$ is isomorphic to an internal diagram in $\textbf{Set}_\lambda$.


There are other subtleties to think about. A well-known result of Kan implies that the category of presheaves on a small category $\mathcal{C}$ has $\mathcal{C}$ as a colimit dense subcategory via the Yoneda embedding $\mathcal{C} \to [\mathcal{C}^\textrm{op}, \textbf{Set}_\infty]$, but if $\mathcal{C}$ is only a locally small category, then this is in general false: in other words, the relative sizes of $\mathcal{C}$ and $\textbf{Set}_\infty$ matter!<|endoftext|>
TITLE: Scheme theoretic interpretations of the Weil's foundations of algebraic geometry
QUESTION [8 upvotes]: For more than 10 years before the apearance of the Grothendieck's theory on schemes, the Weil's foundations of algebraic geometry had been the standard language of algebraic geometry.
There were important works written in the language.
So a natural question is: How can they be translated in the language of schemes?
There is no general solution to this problem and one has to prove each one of them from scratch?
For example, Serre wrote in his book "Algebraic groups and class fields":

This chapter contains the construction and elementary study of the generalized
  Jacobians of an algebraic curve. We will follow closely the paper of
  Rosenlicht [64] on this subject, itself inspired by Weil's Varietes abeliennes
  [89], where the case of the usual Jacobian is treated. We will make use, as
  they did, of the method of "generic points". This obliges us to renounce
  the point of view of the preceding chapters (where all points had their coordinates
  in a fixed base field), and to adopt that of Foundations [87]. It
  is certain that the generic points could be replaced by divisors on product
  varieties, after first developing in detail the properties of these divisors
  (that is to say essentially the cohomology of coherent algebraic sheaves on
  a product variety); that would take us too far afield.

How the generic points could be replaced by divisors on product varieties?

REPLY [2 votes]: I would only add to Chandan's answer that Mike Artin, when he refereed Neron's article, translated it into scheme language in order to understand it. His lectures
Artin, M. Néron models. Arithmetic geometry (Storrs, Conn., 1984), 213--230, Springer, New York, 1986. MR0861977 
are based on his personal notes from the time when he refereed Neron. Unfortunately, they are still very concise.<|endoftext|>
TITLE: determinant of periods
QUESTION [6 upvotes]: Let $X$ be a projective, smooth algebraic variety defined over the field of algebraic numbers. Consider algebraic de Rham cohomology $H_{dR}(X)$ and singular cohomology of $X(\mathbb{C})$ with rational coefficients $H_B(X)$. One can look at the class in $\mathbb{C}^\ast/\overline{\mathbb{Q}}^\ast$ of the determinant of the period isomorphism 
$H^k_{dR}(X) \otimes_{\overline{\mathbb{Q}}} \mathbb{C} \stackrel{\sim}{\longrightarrow} H_B(X) \otimes_{\mathbb{Q}} \mathbb{C}$
In his paper "Operads and motives in deformation quantization", p. 63 Kontsevich claims that it is a power of $2\pi i$. More precisely that, for any basis, the determinant has the form $\pm \sqrt{a} (2\pi i)^n$ for $a \in \mathbb{Q}^\ast$ and $n\geq 1$. Does anyobody know how to prove this result?

REPLY [10 votes]: The argument is a series of more and more general cases.
(1) The map $H_{DR}^1(\mathbb{C}^{\ast}) \otimes \mathbb{C} \to H_B^1(\mathbb{C}^{\ast}) \otimes \mathbb{C}$ is multiplication by $(2 \pi i) \mathbb{Q}$. Proof: $H^1_{DR}$ is generated by $dz/z$. Let $\gamma$ be a closed path in $\mathbb{C}^{\ast}$ which circles the origin once, then $H^1_{B}$ is generated by $[\gamma] \mapsto 1$. Since $\int_{\gamma} dz/z= 2 \pi i$, we see that $H^1_{DR} \otimes \mathbb{C} \to H^1_B \otimes \mathbb{C}$ is multiplication by $2 \pi i$ with respect to the given generators.
(2) $H_{DR}^j(\mathbb{C}^k \times (\mathbb{C}^{\ast})^{\ell}) \otimes \mathbb{C} \to H_B^j(\mathbb{C}^k \times (\mathbb{C}^{\ast})^{\ell}) \otimes \mathbb{C}$ is multiplication by $(2 \pi i)^j \mathbb{Q}$. Proof: The Kunneth isomorphism is compatible with everything in the picture. 
(3) $H^{2j}_{DR}(\mathbb{P}^n) \otimes \mathbb{C} \to H^{2j}_{B}(\mathbb{P}^n) \otimes \mathbb{C}$ is multiplication by $(2 \pi i)^j \mathbb{Q}$. Proof: Write down an open cover of $\mathbb{P}^n$ where every intersection of open sets is of the form (2). We get corresponding spectral sequence in $H_{DR}$ and $H_B$, and compatible maps between them. Keep track of degrees to get the correct power of $2 \pi i$.
(4) Let $X$ be smooth projective of complex dimension $n$. Then $H^{2n}_{DR}(X) \to H^{2n}_B(X)$ is multiplication by $(2 \pi i)^n \mathbb{Q}$. Proof: Write $X$ as a finite cover of $\mathbb{P}^n$ and draw the corresponding commuting square in $H^{2n}$.
(5) (The one that involves determinants.) Let $X$ be smooth projective of dimension $n$. Let $\zeta_B$ be the singular cohomology class of a hyperplane and let $\zeta_{DR}$ be the deRham class $(2 \pi i) \zeta_B$. Fix a cohomological degree $k \leq n$; the claim for $k$ larger than $n$ follows by Poincare duality. 
Hard Lefschetz tells us that
$$(\alpha, \beta)  \mapsto \alpha \cup \beta \cup \zeta_T^{n-k}$$
is a perfect pairing $H^k_T \times H^k_T \to H^{2n}_T$, where $T$ is either $DR$ or $B$. We instead write this as a pairing $H^k_T \times H^k_T \to \mathbb{Q}$ given by
$$\langle \alpha, \beta \rangle_T := (\alpha \cup \beta \cup \zeta_T^{n-k})/\zeta_T^n.$$
Division by $\zeta_T^n$ makes sense because $H^{2n}$ is one dimensional.
Observe that we have
$$\langle \alpha, \beta \rangle_{DR} = \langle \alpha, \beta \rangle_{B} (2 \pi i)^{-k}$$
as pairings on $H^k_{DR} \otimes \mathbb{C} \cong H^k_B \otimes \mathbb{C}$, because of the distinction between $\zeta_{DR}$ and $\zeta_T$.
 Choose bases for these cohomology theories and let $A_{DR}$ and $A_B$ be the matrices of the Hard-Lefschetz pairing in these two bases. Let $P$ be the matrix of the period isomorphism $H_{DR}^K \to H_B^k$.  The above discussion turns into the matrix equation
$$A_{DR} = P^T A_B P (2 \pi i)^{-k}.$$
So $(\det A_{DR})= (\det A_B) (\det P)^2  (2 \pi i)^{-k b_k}$ where $b_k = \dim H^k$. The two $A$-matrices have rational entries and, since the pairings were perfect, $\det A_B$ and $\det A_{DR} \neq 0$. So  $(\det P)^2 \in (2 \pi i)^{k b_k} \mathbb{Q}^{\times}$ as desired.<|endoftext|>
TITLE: The Second Moment of a Sum of Floor Functions
QUESTION [13 upvotes]: The following well known identity, where $\tau(n)$ denotes the number of divisors of $n$ appears in many number theory texts
$$
\sum_{k=1}^n \tau(k) = \sum_{d=1}^n \lfloor n/d \rfloor,
$$
and follows from the observation that "one out of d" integers in $\{1,2,\ldots,n\}$
are multiples of $d$ and then summing along rows $d$ as well as along columns $k$ the 
indicator function $1\{~ d~\mathrm{divides}~k~\}$.
Is there a good approximation, or are there any identities related to the following
sum, preferably containing arithmetic functions?
$$
 \sum_{d=1}^n \lfloor n/d \rfloor^2
$$

REPLY [5 votes]: On your comment concerning the sum $\sum_{d\leq n}\left\lfloor\frac{n}{d}\right\rfloor^k$ for various natural numbers $k$, note we have:
$$\sum_{n\leq x}\left\lfloor\frac{x}{n}\right\rfloor^k=\sum_{n\leq x}\sum_{m\leq \frac{x}{n}}m^{k}-(m-1)^k=\sum_{n\leq x}\sum_{d\mid n}d^k-(d-1)^k$$
Thus since we can write: $$\sum_{d\mid n} d^k-(d-1)^k=\sum_{d\mid n}(d^k-\sum_{j=0}^k\binom{k}{j}d^{k-j}(-1)^j)=\sum_{d\mid n}\sum_{j=1}^k \binom{k}{j}d^{k-j}(-1)^{j-1}$$
This means if we let $\sigma_r(n)=\sum_{d\mid n}d^r$ then we must get for all $k\in \mathbb{N}$ that:
$$\sum_{n\leq x}\left\lfloor\frac{x}{n}\right\rfloor^k=\sum_{j=1}^k\binom{k}{j}(-1)^{j-1}\left(\sum_{n\leq x}\sigma_{k-j}(n)\right)$$<|endoftext|>
TITLE: Reference request - CDGA vs. sAlg in char. 0
QUESTION [7 upvotes]: Hello,
Are the model categories of simplicial commutative algebras over $k$ and that of commutative differential graded algebras (in negative cohmological dimension) Quillen equivalent in char. 0 (or maybe if $k$ is a $Q$-algebra)? What would be a reference?
Thank you

REPLY [2 votes]: You might be interested in the paper by Mike Mandell titled : Topological Andre-Quillen Cohomology and $E_{\infty}$ Andre-Quillen Cohomology. I am pretty sure his statements do not require a 
characteristic 0 assumption. For example, Theorem 1.3 on page 6 might be of interest.
Sorry, I guess I forgot to mention some things. I was initially pointed to this by Shipley's paper on $H\mathbb{Z}$ algebras and DGAs. The point is that when you want to remove the characteristic zero assumption you need to remember the $E_{\infty}$ algebra structure. In characteristic zero, it has to be essentially unique. This is related to the fact that the (co)homology of the symmetric groups is usually torsion. I remember Akhil Mathew asking a question related to this a while back. Anyways, the characteristic result follows. Although, it was known prior to this, see Clark Barwick's comment.<|endoftext|>
TITLE: Is there a parameterization of a neighbourhood of $x\in\mathbb{R}^n$ into two mutually orthogonal sets of variables, with one set parameterizing a pre-defined (n-k)-dimensional submanifold containing $x$?
QUESTION [12 upvotes]: Let $M \subset \mathbb{R}^n$ be a differentiable submanifold with co-dimension $k$. Is there a parameterization of $\mathbb{R}^n$ of a neighbourhood of $x\in M$, so that the variables parameterizing $M$ are orthogonal to the variables parameterizing the directions orthogonal to $M$, throughout the neighbourhood? In other words, I would like a parameterization of the form $\phi:U\subset \mathbb{R}^n \to \mathbb{R}^k\times \mathbb{R}^{n-k}$, such that $\phi^{-1}(0,y) \subset M$ and the variables $x\in \mathbb{R}^k$, $y\in\mathbb{R}^{n-k}$ are orthogonal in $U$, i.e. the metric tensor has the form $g_{ij} = g_{ji} = 0$ for $1\leq i \leq k$, $k+1\leq j \leq n$.  
In case it is relevant, the manifold I am working with is defined as the intersection of level sets $\{ y_i(x)=0\}$, $i=1\ldots k$, where each $y_i$ is smooth enough and the gradients $\{\nabla y_i\}$ are linearly independent on the level set of mutual intersection $M$. I can find a parameterization such that the orthogonality condition holds exactly at $M$, using the variables $y_i$ as the orthogonal directions, as each $\nabla y_i$ is orthogonal to $M$. However, I am wondering how to extend this to a neighbourhood of $M$?

REPLY [11 votes]: I was reminded of this question recently by a related problem, and I remembered a concept that I had not remembered when I originally saw the question, namely É. Cartan's notion of exterior square of second fundamental forms.  Using that concept, I can give something of an answer to the question.  
It's not definitive, but it does show that, for $n{-}k$ and $k$ sufficiently large, there do exist submanifolds $N^{n-k}\subset\mathbb{R}^n$ that do not occur as a leaf of a (local) foliation whose orthogonal plane field is integrable.  (This is another way to phrase the OP's problem.)
Here is a description of an obstruction:  Suppose that $\mathcal{F}$ and $\mathcal{G}$ are foliations of codimensions $k$ and $n{-}k$, respectively, of an open set $U\subset \mathbb{R}^n$, and suppose that they are orthogonal.  Thus, if $F\subset TU$ is the rank $n{-}k$ bundle of vectors tangent to $\mathcal{F}$ and $G\subset TU$ is the rank $k$ bundle of vectors tangent to $\mathcal{G}$, then $TU=F\oplus G$ is an orthogonal direct sum.  
Let $Q^F$ be the section of $G\otimes\mathsf{S}^2(F)$ that gives the second fundamental forms of the leaves of $\mathcal{F}$ and let $Q^G$ be the section of $F\otimes\mathsf{S}^2(G)$ that gives the second fundamental forms of the leaves of $\mathcal{G}$.  (Note that I am using the metrics on the two bundles to idenfity $F$ with $F^\ast$ and $G$ with $G^\ast$.)  Cartan defined an algebraic exterior square, i.e., a quadratic map
$$
\sigma^F: G\otimes\mathsf{S}^2(F)\longrightarrow \Lambda^2(G)\otimes\Lambda^2(F),
$$
which is just the squaring map $G\otimes\mathsf{S}^2(F)\to \bigl(G\otimes\mathsf{S}^2(F)\bigr)\otimes\bigl(G\otimes\mathsf{S}^2(F)\bigr)$, followed by contracting a pair of $F$-indices (so that the result goes into $\bigl(G\otimes F\bigr)\otimes \bigl(G\otimes F\bigr)$), and then followed by skew-symmetrizing in both the $G$- and $F$-pairs independently.  (Note that this map is, of course, zero if either $k=1$ or $n{-}k=1$; otherwise it is not zero.)  On the other side, there is, of course, the corresponding mapping
$$
\sigma^G: F\otimes\mathsf{S}^2(G)\longrightarrow \Lambda^2(G)\otimes\Lambda^2(F).
$$
(Technically, $\sigma^G$ should go into $\Lambda^2(F)\otimes\Lambda^2(G)$, I guess, but I'm 
identifying these two tensor products in the obvious way.)
With all of this defined, a short calculation with the structure equations shows that one has the identity
$$
\sigma^F\bigl(Q^F\bigr) + \sigma^G\bigl(Q^G\bigr) = 0\tag{1}
$$
for any pair of orthogonal foliations $\mathcal{F}$ and $\mathcal{G}$ on an open set $U\subset\mathbb{R}^n$.  
Now, it's a matter of linear algebra to check that, when $k$ and $n{-}k$ are sufficiently large, the maps $\sigma^F$ and $-\sigma^G$ do not have the same image.  This should not be surprising, since, for $k$ and $n{-}k$ sufficiently large, the space $\Lambda^2(G)\otimes\Lambda^2(F)$ is much larger than either $G\otimes\mathsf{S}^2(F)$ or $F\otimes\mathsf{S}^2(G)$, so there is plenty of room for them to have different images.  Their images will also, generally, have different dimensions.
For a specific example, consider the case $k=2$ and $n{-}k=4$.  Then the ranks of $G$ and $F$ are $2$ and $4$ respectively.  It's easy to show that, in this case, $\sigma^F$ is surjective while the image of $\sigma^G$ consists only of elements of the form $a\otimes b$, where $a\in \Lambda^2(G)$ is nonzero and $b\in\Lambda^2(F)$ is a simple $2$-form (i.e., its half-rank is either $0$ or $1$). Thus, if you choose a submanifold $N^4\subset\mathbb{R}^6$ such that the exterior square of its second fundamental form is a nondegenerate $2$-form (and the generic $4$-manifold in $\mathbb{R}^6$ will have this property on a dense open set), then $N^4$ cannot be a leaf of a (local) foliation $\mathcal{F}$ that has an orthogonal foliation $\mathcal{G}$ because, as an equation for $Q^G$, the condition $(1)$ will have no solutions on a neighborhood of $N$.
Notice that $(1)$ is only the first obstruction one would encounter in trying to solve this problem with a prescribed $N^{n-k}\subset\mathbb{R}^n$.  
I believe that the case $n{-}k=k=2$, for which $(1)$ is first a nontrivial condition, though not involutive, will become involutive after the first prolongation, though I haven't checked the details.  So this case is probably OK, at least in the analytic case.  (Note that, even though this case is formally determined, as Anton pointed out in his answer, the symbol of the PDE is degenerate, which is why $(1)$ is still nontrivial in this case, so the 'determined' property is not decisive here.)  
However, even in the 'overdetermined' case $n{-}k=3$ and $k=2$, the equation $(1)$ is always solvable in the sense that $\sigma^F$ and $\sigma^G$ are each surjective, and it could still be (again, I haven't checked) that the system goes into involution after the first prolongation with the proper initial conditions being described in dimension $3$ (instead of dimension $4$). Thus, it's still possible, as far as I know, that every (analytic?) $N^3\subset\mathbb{R}^5$ is a leaf of a foliation that has an orthogonal foliation by surfaces.<|endoftext|>
TITLE: "Lie algebra" for a general group ?
QUESTION [27 upvotes]: Is there analog of Lie algebra for the case of topological groups which are not necessarily differentiable manifolds, and in particular for finite groups? here by "analog" i mean that it should have similar kind of relations to given group as of a finite dimensional Lie
algebra with its corresponding Lie group. for example there should be some analog of "Exponential map"; Every linear representation of group should also be a linear representation of its algebra, and this should be in some sense compatible with the exponential map between the two; etc
thanks,

REPLY [8 votes]: If $G$ is a finitely generated group which is torsion free nilpotent of class $n$, then $G$ is the Lie group of some $\mathbb{Z}$-Lie algebra $\mathfrak g$ which is also nilpotent of class $n$. Hence you can define an algebraic group $G(k)$ for any field $k$ by taking the exponential of $\mathfrak g\otimes_{\mathbb{Z}} k$.
Now if $G$ is a discrete group, define the rational serie as $D_i(G)=${$x \in G, x^r \in \Gamma_i(G)$ for some $r>0$ } where $\Gamma_i(G)$ is the $i$th term of the lower central serie. Therefore, by construction $G/D_i(G)$ is torsion free nilpotent of class $i$, hence you can associate to it a Lie algebra $\mathfrak g_i(k)$. Now define $\mathfrak g(k)$ as the inverse limit of the $\mathfrak g_i(k)$. it is called the Malcev Lie algebra of $G$. It is a complete, separated pro-nilpotent Lie algebra. Set $G(k)=\exp(\mathfrak g(k))$, it is a pro-unipotent group coming with a morphism $G \rightarrow G(k)$ which is universal for this property. Note that if $\bigcap_{i\geq 0} D_i(G)$ = {1} (such a group is called residually torsion free nilpotent) this morphism is injective, so it may happen that $G(k)$ capture a lot of things about $G$.
Indeed every representation of $\mathfrak g(k)$ extends to a representation of $G(k)$ just by taking the exponential, and therefore to a representation of $G$. Conversly, every $k$-(pro-)unipotent representation of $G$ induces a representation of $\mathfrak g(k)$.
Note that $\mathfrak g(k)$ is a rather complicated object, so it doesn't seems to help a lot. But in many interesting case, $\mathfrak g(k)$ is isomorphic as a filtered Lie algebra to an "easy to handle" graded Lie algebra (namely to the associated graded of $G$, see Ralph's answer). In that case you really get something like the relation between a Lie group and its easier to handle Lie algebra.<|endoftext|>
TITLE: Explicit generators for matrix invariants of the symmetric group
QUESTION [12 upvotes]: Let $V$ be the space of $n$ by $n$ complex matrices with the conjugate  action of the symmetric group $G=S_n$. Is any explicit set of generators for the invariant ring $C[V]^G$ known?

REPLY [3 votes]: thanks for all the answers. I found a paper by Garcia and Stanton, "Group actions on 
Stanley Reisner rings and .." (Advances in Maths, 1984), which provides a reasonable answer to this
question. 
Ketan Mulmuley<|endoftext|>
TITLE: Relationship between double tangent bundle, exterior derivative and connection
QUESTION [9 upvotes]: I am totally new to the subject differential geometry, and that probably reflects itself in the naive question that I'm trying to formulate. I hope this question does not get closed because of this.
Let $M$ be a smooth manifold, equipped with an affine connection $\nabla$. Given two vector fields $X$ and $Y$, there are three ways we can "differentiate" $X$. The first one is to take Lie derivative $\mathscr{L}_Y X$, the second one is to do $\nabla_Y X$ and the third one, the most naive one, is just to take $DX: TM \to TTM$, which is just the total differential of $X$, viewed as a map $X: M\to TM$.
Note that the most naive way is very similar to taking exterior derivative of a function, and so, philosophically, it should be thought of as the "total derivative" and all others are somehow just partial derivatives.
This makes me believe (naively) that there is conceptually clean way to work from the most naive way of differentiating and deduce everything else (exterior derivatives (of higher wedges as well), Lie derivatives and connections) from it. I hope someone can clarify this for me.
Thank you.

REPLY [13 votes]: Your $DX$ generalizes to the setup where $X$ is a section $M\to E$ of a vector bundle $E$ on $M$, giving a (fiberwise linear) map $DX:TM\to TE$. At each point $p\in M$ this gives a linear map $D_pX:T_pM\to T_{X(p)}E$. There is a canonical short exact sequence $0\to E_p\to T_{X(p)}E\to T_pM$. The composed map $T_pM\to T_{X(p)}E\to T_pM$ is the identity, so the "TM part" of $DX$ does not depend on $X$.
A connection on $E$ gives a splitting of that exact sequence, so in the presence of a connection you can speak of the "E part" of your total derivative. That's $\nabla X$.
So you can sort of say that $\nabla X$ knows no more and no less than your $DX$ (but it doesn't appear until you choose a connection).
Note that $\nabla_YX$ is linear over the functions: we have $\nabla_{fX}Y=f\nabla_YX$ for a smooth function $f$ on $M$. 
The Lie derivative is rather different. It's not linear over the functions in the $Y$ variable, and it's only defined when $E=TM$. Maybe someone else can say something more positive about where it fits in.

REPLY [10 votes]: The language of Ehresmann connections is probably what you're looking for? 
$$\nabla_X Y = c \circ DY \circ X$$
where $c : T(TM) \to TM$ is the Ehresmann connection.  Here I'm using bundle conventions for derivatives, if $f : M \to N$ is smooth, $Df : TM \to TN$ is the tangent bundle map. 
For Lie brackets, one way I like to formulate them is as an obstruction to certain types of 2nd order derivatives existing. 
Let $\pi_M : TM \to M$ be tangent bundle projection, $\pi_{TM} : T^2 M \equiv T(TM) \to TM$ be double tangent bundle projection. 
$$ \mathcal L(M) = \{ (V,W) \in (T^2M)^2 : D\pi_M(V) = \pi_{TM}(W), \pi_{TM}(V) = D\pi_M(W)\}$$
There is a map $[\cdot, \cdot] : \mathcal L(M) \to TM$ that you might as well call the `Lie Bracket' and it satisfies $[V,W]=0$ if and only if there exists a function of two variables
$f : (-\epsilon,\epsilon)^2 \to M$, $f(x,y)$ such that 
$$\frac{\partial^2 f}{\partial x\partial y}(0,0) =V$$
$$\frac{\partial^2 f}{\partial y\partial x}(0,0) =W$$
where $\frac{\partial f}{\partial x} : (-\epsilon,\epsilon)^2 \to TM$ is the bundle theoretic partial derivative, i.e. 
$$\frac{\partial f}{\partial x}(x,y) = Df_{(x,y)}(1,0)$$
So while $\nabla_X Y$ and $DY \circ X$ are rather closely related, the Lie bracket's role is as an obstruction to 2nd order data being derivatives of multi-variable functions -- it's a very different object.  
Oh, and of course, $[DY \circ X, DX \circ Y] = [X,Y]$, i.e. this Lie bracket specializes the the regular Lie bracket when you stick in vector fields.<|endoftext|>
TITLE: A theorem of Stickelberger on the number of prime ideals in a decomposition
QUESTION [5 upvotes]: Suppose that $p$ is unramified in a number field $K$ of degree $n$. Apparently, Stickelberger proved that $\big( \frac{Disc(K)}{p}\big) = (-1)^{n - g}$, where $g$ is the number of prime ideal factors over $(p)$ in $K$.
Is there a convenient reference for this fact?
Thank you! -Frank

REPLY [6 votes]: First reduce the question to the local case--This can be done since after base change to   $\mathbb{Q}_{p}$, $K \otimes_{\mathbb{Q}} \mathbb{Q}_{p}=K_1 \times...\times K_{g}$, the trace form is the orthogonal sum of the local trace forms so the discriminant is the product of the local discriminants, and then use that the Legendre symbol is multiplicative. Now, since $p$ is unramified the extension $K_{i}/\mathbb{Q}_{p}$ is a cyclic extension of degree $f_{i}$, notation as usual, with Galois group generated by $\sigma$. We want to show that 
 $\left ( \frac{disc(K_i)}{p}\right)= (-1)^{f_i -1}$. This is done in the exact same way one proves that a cubic extension is Galois iff its discriminant is a square; just notice that $(-1)^{f_i -1}$ is the the signature of $\sigma$ under the regular representation of $Gal(K_{i}/\mathbb{Q}_{p})$. 
Remark: If you, like Conway, like the notation $p=-1$ for the infinite prime then the above proof works at that prime too. Just read Kronecker symbol instead of Legendre, and what one gets is the usual "sign of discriminant is $(-1 )^{s}$ "<|endoftext|>
TITLE: Gauss's views on pure mathematics
QUESTION [13 upvotes]: According to Wikipedia's entry on Gauss:
"Though Gauss had been up to that point supported by the stipend from the Duke, he doubted the security of this arrangement, and also did not believe pure mathematics to be important enough to deserve support"
As a mathematician, I find, rather obviously this opinion to be very shocking (especially coming from Gauss himself).
Wikipedia does not give any sources related to the quote above. Is there any reference, or evidence for Gauss holding such a position?

REPLY [12 votes]: In his letter to Zimmermann (March 12, 1797; Werke X), Gauss wrote
May God give the noble Duke a long life, and what may Science expect from him since he
deems a work that is only a little interesting not unworthy of his support; how much I
wish that I could present a work that is more profitable for society [serving the public
good] or more excellent.
Gauss was talking about the Duke's support for printing his Disquisitiones Arithmeticae.<|endoftext|>
TITLE: Is there exponentiation in "sufficiently large" models of $I\Delta_{0}$?
QUESTION [6 upvotes]: Let $L_{E}$ be the language of discretely ordered rings together with an extra predicate symbol $E$. The system $A$ consists of the axioms of $I\Delta_{0}$ (basic arithmetic plus induction for bounded formulas) together with the statements:
(1) $E(0,1)$
(2) $\forall{x}\exists{y}E(x,y)$
(3) $\forall{x>0,y,z}E(x+1,y)\wedge E(x,z)\rightarrow y\geq 2z$
Is it then provable in $A$ that exponentiation (with base $2$) is total? (In any of the formulations as a bounded formula.)
This might be well-known, but I couldn't find a reference so far.

REPLY [12 votes]: Do you intend system $A$ to include only $I\Delta_0$, or actually $I\Delta_0(E)$? That is, do you allow induction for formulas involving $E$?
If you do allow it, then then you can prove easily
$$\forall x,y\le w\,(E(x,y)\to x=y=0\lor\exists z\le y\,(z=2^x))$$
by induction on $w$, hence the theory proves exponentiation to be total.
If you only allow induction for $\Delta_0$ formulas without $E$, then the answer is negative, and in fact, $A$ is a conservative extension of $I\Delta_0$. This can be seen as follows: take any model $M\models I\Delta_0$, we’ll expand it to a model of $A$. For $n\in\mathbb N$, put $E(n,2^n)$. For nonstandard $x,y\in M$, let $x\sim y$ iff $x-y\in\mathbb Z$. For any equivalence class $C$ of $\sim$, choose its representative $a$ so that $C=a+\mathbb Z$, and choose an arbitrary nonstandard $b\in M$. Then for any $n\in\mathbb N$, put $E(a+n,b2^n)$ and $E(a-n,\lfloor b/2^n\rfloor)$.
One can do even better: for example, let $A^+$ be the extension of $A$ by the following axioms:
(4) $E(x,y)\land E(x,y')\to y=y'$
(5) $E(x,y)\land E(x',y')\to E(x+x',yy')$
(6) $E(x,y)\land E(x',y')\land x< y\to x'< y'$
(this would be more readable if we had a unary function symbol instead of $E$). Then $A^+$ is still a conservative extension of $I\Delta_0$. One way to prove this is as follows: take any countable recursively saturated model $M\models I\Delta_0$, and let $L=\{a\in M:M\models\exists y\,(y=2^a)\}$. The countable structures $(M,+,\le)$ and $(L,+,\le)$ are elementarily equivalent (as they are both models of Presburger arithmetic), and the joint model $(M,L,+,\le)$ is recursively saturated, hence there exists an isomorphism $f\colon(M,+,\le)\simeq(L,+,\le)$. Then putting $E(x,2^{f(x)})$ defines a model of $A^+$. ${}{}$<|endoftext|>
TITLE: Ultrafilters and the exposition of forcing
QUESTION [8 upvotes]: I've just been reading Timothy Chow's "A beginner's guide to forcing."  
Question: Does an exposition of forcing really need to delve into ultrafilters?
What I have in mind:  If ZFC proved CH, the proof would apply to any Boolean-valued model M, giving CH the truth value 1 in M.  Doesn't that mean that one has proved the independence of CH as soon as one has any Boolean model where CH as truth value less than 1?  And if one then, psychologically speaking, still really wants a 2-valued model where CH fails, already having independence, one need only appeal to Godel's completeness theorem.

REPLY [4 votes]: I just thought I would pop in here and start an argument by adding my category-theorist's point of view.  (-:
To a category theorist, a Boolean-valued model is the internal logic of a sheaf topos.  Explicitly computing Boolean truth values is the same as working with subobjects in that category, while "jumping inside the Boolean brackets", as Andreas describes, corresponds to working inside the internal logic.
Now a sheaf topos is always the "classifying topos" of some geometric theory, which means that it contains a "generic model" of that theory.  In the case of sheaves for double-negation topologies on posets, which classical forcing usually restricts itself to, these models can be described as ultrafilters.
But I think there is value, especially expositionally, in taking seriously the idea that we are talking about classifying toposes of more general theories than this, and that what we obtain from forcing is a generic model of some theory.  Even if having such a generic model is equivalent to having a certain kind of ultrafilter, it seems to me that often the generic model of the theory has a more direct connection to the statements we are trying to force.
For instance, to force $\neg \mathrm{CH}$, we may start with the "theory of an injection $\alpha \hookrightarrow 2^{\aleph_0}$" for some uncountable $\alpha$.  The classifying topos of this theory will then contain such the "generic" such injection.  (We then have to pass to an extra subtopos if we want, unaccountably, to preserve PEM.)  I find this idea much easier to understand as a beginner, though it is of course equivalent to the usual presentation.
In short, I think one should be able to work with generic objects, which one might argue are at the heart of forcing, without necessarily needing to think about ultrafilters in particular.<|endoftext|>
TITLE: Irreps which aren't highest-weight modules.
QUESTION [7 upvotes]: Well, the question is probably a rather basic one but I haven't been able to find the answer in literature or come up with it myself so here we go.
Do there exist irreducible representations of the Lie algebra $\mathfrak g$ which aren't a highest weight module with respect to some Borel subalgebra in the case when
a) $\mathfrak g$ is a complex finite dimensional semisimple algebra,
b) $\mathfrak g=\mathfrak{gl}_n(\mathbb C)$?

REPLY [7 votes]: I think it's very far from true, and misleading to state too informally, that every representation is highest weight. For example if we think of representations (with a fixed infinitesimal character) geometrically on the flag manifold (on $P^1$ for $SL_2$), then highest weight modules with respect to a unipotent subgroup $N$ are locally constant (or rather have some central curvature) along the corresponding Schubert stratification (ie for SL2 they are attached geometrically to the stratification of $P^1$ by $A^1$ and a point), while a representation can have arbitrary "shape" - ie we can fix any stratification we want. (Note the finite dimensional representations, which as was stated are all highest weight, are attached to the stratification of the flag variety by just one piece - by Borel-Weil they arise from line bundles, which definitely satisfy this "shape requirement".) In particular the representations that people care most about tend to be Harish Chandra modules, which can be realized geometrically as stratified with respect to the (finitely many) orbits of a symmetric subgroup of $G$ - for instance, the admissible representations of $SL2R$ correspond to the stratification of $P^1$ by the two poles and $C^\times$, and most of these are evidently not highest weight.
What is true, and very useful and important (some might say the most fundamental statement about g-modules, but that's not a debate I want to start), is that representations DO live on the flag manifold - in other words, they are made out of "points" of the flag manifold, which correspond to representations that are highest weight (Verma modules). So the sense in which representations relate to highest weight modules is just that they can be realized geometrically on the flag manifold ($P^1$).
Technically this is captured by a version of Casselman's submodule theorem (reps of real groups live inside principal series) due to Beilinson-Bernstein: for every representation
there IS an N so that the representation has nonvanishing N-coinvariants (ie not highest weight vectors but a form of highest weight covectors). Such N give the support of the corresponding picture of the representation on the flag manifold.<|endoftext|>
TITLE: A name for a weak topology
QUESTION [5 upvotes]: Let $V$ be a real vector space and let $V'$ be the algebraic dual of $V$, i.e. the space of all the linear functionals $V\to\mathbb{R}$. Then there exists the weakest topology $\tau$ which makes all the elements of $V'$ continuous, and $\tau$ is locally convex and Hausdorff. For example, if the dimension on $V$ is finite, then $\tau$ is the usual Euclidean topology. 
I have two questions:

Is there a commonly used name for $\tau$? 
Let $\tau'$ be the maximal locally convex topology on $V$, i.e. the weakest topology that makes all the seminorms on $V$ continuous. Of course $\tau'$ is finer than $\tau$, but $\tau'$ and $\tau$ could coincide in some cases (for example, for finite dimensional spaces). What is the exact relationship between $\tau$ and $\tau'$?

REPLY [6 votes]: I got into trouble in following prof. Johnson's suggestion, and now I am quite convinced that the topologies $\tau$ and $\tau'$ coincide only when $V$ is finite-dimensional. 
In fact, let $\{x_i\}_{i\in I}$ be a Hamel basis for $V$, and let us consider the seminorm
$p(\sum_{i\in I} v_i x_i)=\sum_{i\in I} |v_i|$
(this seminorm is well-defined since every element in $V$ admits a unique representation as a finite linear combination of elements of the basis). Then the set $U=p^{-1}(-1,1)$ is open in $\tau'$, and does not contain any nontrivial linear subspace of $V$.
On the other hand, since $\tau$ is the weak topology with respect to a family of linear functionals (in fact, with respect to all the linear functionals), every $\tau$-neighbourhood of $0\in V$ must contain a finite-codimensional linear subspace of $V$. Therefore, if the dimension of $V$ is not finite, then the set $U$ introduced above is open with respect to $\tau'$, and not open with respect to $\tau$.<|endoftext|>
TITLE: Connes-Kreimer Hopf algebra and cosmic Galois group
QUESTION [19 upvotes]: Hi,
I'm interested in the relation between the two following constructions motivated by renormalization:

Connes-Kreimer gives an interpretation of the renormalization procedure in the framework of an Hopf algebra $H$ of rooted trees. By Cartier-Milnor-Moore theorem, $H$ is the algebra of regular functions on some pro-algebraic group, which turns out to be the so-called Butcher group $B$. 
Connes-Marcolli then considered a differential equation satisfied by divergences appearing in the above work, which leads to the introduction of a category of "equisingular flat connection" which, so far I understand, are more or less specific $B$-valued principal bundles equipped with a flat connection up to some equivalence relation. It turns out that this category is tannakian, meaning that it is equivalent to the category of modules over some pro-algebraic group $G$. They observe that $G$ acts on any renormalizable theory in a nice way, hence the name "cosmic Galois group" which was coined by Cartier. It is more than an analogy, since $G$ is (non-canonically) isomorphic to some motivic Galois group.

I do not claim to understand these things at all, especially the "physical" part, so my question is maybe naive. The vague question is:

Is there a "direct relation" between the group $G$ and the Hopf algebra $H$ ? 

Of course there are some relations between the two, so maybe a more precise question is: 

Is there a definition of $G$ purely in terms of combinatorics of Feynman graphs, or as the automorphism group of something (something else than a fiber functor) ?

In fact, if I understand Marcolli's survey correctly, $G$ action on physical theories factors through an action of $B$ (but it seems rather surprizing, so it's very likely that I misunderstood something). So for the sake of curiosity, another naive question is: 

"Why" is it $G$, and not $B$ which plays the role of a cosmic Galois group ?

My motivation for this question comes from the (highly non-trivial) fact that (some group which is almost) $G$ embeds into the graded Grothendieck-Teichmuller group $GRT$, which can be defined as the automorphism group of a certain operad of Feynman graphs. Since $G$ is more or less by definition related to the combinatorics of these graphs, I'm wondering if there is a concrete "combinatorial" definition of it.
Edit: To be precise they also consider a more general Hopf algebra of Feynman diagrams (not only rooted trees). They call the corresponding pro-algebraic group the group of "diffeographism". My claim about the action of $G$ is related to this group and not to $B$, I guess.. So let's assume that I'm asking my question in this setting too.
Edit 2: I should have mentionned that there are in fact several Hopf algebras of Feynman graphs whose choice depends on the particular physical theory you're considering (i.e. you have to choose a specific "type" of Feynman graph), so there are in fact plenty of diffeographism groups, and $G$ is universal among them (which in fact answer my last question, although the Butcher group still seems to play a distinguished role). On the other hand it seems to me that mathematically it still make sense to consider the Hopf algebra of all Feynman graphs, but maybe what you get is precisely $\mathcal H _{\mathbb U}$ (see Gjergji's answer) ?
And just to expand what I'm saying in my comment to Gjergji's answer, it's rather clear for example that $\mathcal H _{\mathbb U}$ is isomorphic to the dual of the algebra of noncommutative symmetric functions, since the latter is also by definition a free graded noncommutative algebra with one primitive generator in each degree (which are analogs of power sum symmetric polynomials). In fact the abstract definition of $\mathcal H _{\mathbb U}$ itself is rather explicit and combinatorial, hence I'm really looking for a more conceptual link.

REPLY [2 votes]: This is more of a comment where I wanted to point out that there is a parallel treatment of this cosmic Galois group which avoids the Riemann-Hilbert approach and is more combinatorial in some sense.
The first paper to point out this point of view is, I think, "A Lie theoretic approach to renormalization" by K. Ebrahimi-Fard, J.M. Gracia-Bondia and F. Patras. There is also a nice summary in "Dynkin operators and renormalization group actions in pQFT".
First note that the cosmic Galois group $\mathbb U$ is defined as the affine group scheme associated to the commutative Hopf algebra 
$$\mathcal H _{\mathbb U}:=\mathcal U(\mathcal F(1,2,3,\dots) _{\bullet})^{\vee}$$
where $\mathcal F(1,2,3,\dots) _{\bullet}$ is the free graded Lie algebra with a generator in every degree $n > 0$.
The papers above explain, among other things, how the universality of the cosmic Galois group can be interpreted as saying that this Hopf algebra and the associated group scheme act on all Connes-Kreimer Hopf algebras of Feynman diagrams. However this Hopf algebra, also called the Descent algebra is known to act naturally on all connected graded commutative Hopf algebras, so it is universal in a broader sense.
Now, if you see the second paper, the descent algebra $\mathcal H _{\mathbb U}$ has a natural description in terms of generators defined from permutations with a fixed descent set, so perhaps this qualifies as a simple combinatorial definition. On the other hand the only property we needed from the algebras of Feynman diagrams was that they be graded connected commutative Hopf algebras, so I doubt that $\mathbb U$ can be described directly as automorphisms of such things...<|endoftext|>
TITLE: Sheaf embedding preserving initial algebras?
QUESTION [7 upvotes]: Any small pretopos $C$ can be embedded into a Grothendieck topos by a fully faithful functor that preserves all the pretopos structure (limits, images, finite unions of subobjects, disjoint coproducts, and quotients of equivalence relations).  Namely, we may consider the topos of sheaves for the coherent topology on $C$, with the sheafified Yoneda embedding.  If $C$ is (locally) cartesian closed, then that structure is also preserved by this embedding.
My question is, what if $C$ also has a natural numbers object, or more general initial algebras for special endofunctors (e.g. "W-types")?  Can we embed it into a topos of sheaves in a way that preserves these initial algebras?

REPLY [4 votes]: Sometimes, but not always.
One case when this is possible is given by Proposition D5.1.8 in Sketches of an Elephant.  Namely, if $C$ is a small elementary topos with an NNO that is standard, meaning that the family of all numerals $s^n o : 1\to N$ (for external natural numbers $n\in \mathbb{N}$) is epimorphic, then it can be embedded in a Grothendieck topos by a functor preserving both finite limits and finite colimits.  Such a functor then necessarily preserves the NNO, by Freyd's characterization of the latter.  The Grothendieck topos is the topos of sheaves on $C$ for the topology consisting of all (not necessarily finite) epimorphic families.
Every Grothendieck topos is standard, but Example D5.1.7 gives an elementary topos with NNO that is not standard.  It is the filterquotient of $\rm Set/\mathbb{N}$ by a filter containing the cofinite filter, and it contains a global natural number $\omega : 1\to N$ that is disjoint (as a subobject of $N$) from every numeral $s^n o$.  Moreover, just after D5.1.7 it is shown that this topos does not admit a functor to any Grothendieck topos preserving finite limits and finite colimits, and the proof actually shows that it does not admit a functor to any Grothendieck topos preserving finite limits and the NNO.  The point is that such a functor would preserve the existence of a global natural number disjoint from every numeral, which is impossible in a standard topos.
Thus, the answer to the question is no in general, even if $C$ is an elementary topos; but yes if $C$ is a standard elementary topos.  (I don't know if anything can be said about standard pretoposes; the proof of D5.1.8 is impredicative, as is the proof of Freyd's characterization of NNOs.)<|endoftext|>
TITLE: Commutative algebra with a view toward algebraic _number theory_
QUESTION [23 upvotes]: Someone asked me this today, and I don't know what the standard answer is:
Is there an analogue of David Eisenbud's rather amazing Commutative Algebra With a View Toward Algebraic Geometry but with a view toward algebraic number theory? Ideally, with the starting graduate student in mind and with a modern slant...

REPLY [7 votes]: Neukirch was already mentioned a couple of times.
Atiyah-MacDonald: Introduction to Commutative Algebra
P. Samuel: Algebraic Theory of Numbers (Has a fair amount of commutative algebra.)
Serre: Local Fields
Cassels-Frolich: Algebraic Number Theory (this is more advanced; not for a beginners)<|endoftext|>
TITLE: Severi-Brauer variety and finite covering
QUESTION [5 upvotes]: Hi everybody!  This is my first post on MO.
Let's work over the field of complex numbers. 
Let $f:P\rightarrow X$ be a Severi-Brauer variety over a smooth proper projective algebraic variety  $X$ : $f:P\rightarrow X$ is  a projective bundle in the complex analytic sense.  If I'm correct, to define  algebraically such an  object locally  it is necessary to use étale topology (and the definition is as follows: there exists an \'etale open covering $\{ ( \mu_i: U_i\rightarrow X_i )\}_{i\in I}$     such that 
$\mu_i^*(P\lvert_{X_i})$ is isomorphic  (as schemes) to the trivial bundle $U_i\times \mathbb P^n\rightarrow U_i$ for every $i\in I$.
0) What's going on if one considers Zariski topology intstead of étale topology?
1) What is the simplest example to have in mind (with $X$ projective and smooth) of such a projective bundle that is not trivial (ie. that is not the projectivization of a vector bundle)?
2)  Is there always a ramified covering $c:X'\rightarrow X$ such that the pull-back of $P$ by $c$ becomes trivial?  
Certainly this is well-known but since I'm not an expert in algebraic geometry, I have been unable to find precise answers to these silly questions in the literature...
Any help (answers or references) would be welcome.
Thanks!

REPLY [4 votes]: This is a complement to Sasha's answer, since you asked for an example where $X$ is projective.  The simplest projective varieties (curves, rational surfaces) have trivial Brauer group, so there are no non-trivial Severi–Brauer varieties over them.  Possibly the simplest interesting example is when $X$ is a K3 surface, and various authors have used elliptic fibrations to construct elements of the Brauer groups of K3 surfaces.
One example is given by Olivier Wittenberg in "Transcendental Brauer–Manin obstruction on a pencil of elliptic curves", Arithmetic of higher-dimensional algebraic varieties (Palo Alto, CA, 2002), 259–267, Progr. Math., 226, Birkhäuser Boston, Boston, MA, 2004 (also on his web site).  He constructs an explicit elliptic curve over $\mathbb{Q}(t)$ of which a model is a K3 surface, and finds a non-trivial quaternion algebra lying in the Brauer group of the surface.  You could turn this into a bundle of conics over $X$ by the standard construction.
It may well be that there are easier ways to look at this if one's only interested in working over $\mathbb{C}$ – I only know about the situation for varieties defined over $\mathbb{Q}$, where the problem is to find Brauer group elements themselves defined over $\mathbb{Q}$.
I second the recommendation to read the chapter in Milne's book, though it's worth pointing out that the series of three articles on the Brauer group by Grothendieck in "Dix exposés sur la cohomologie des schémas" is actually easier reading that you might anticipate.  (And IIRC he explicitly addresses Severi–Brauer varieties.)<|endoftext|>
TITLE: Homotopy limits of quasi-categories
QUESTION [11 upvotes]: Quasi-categories (or $\infty$-categories, as they are often called) are a very convenient setting for doing abstract homotopy theory. One of their amazing features is the following: Given a diagram of quasi-categories, we can form its homotopy limit, yielding a quasi-category again. For example, the inverse (homotopy) limit of the diagram

$\cdots \xrightarrow{\Omega} \mathcal{S}_* \xrightarrow{\Omega} \mathcal{S}_* \xrightarrow{\Omega} \mathcal{S}_*$

(for $\mathcal{S}_*$ the quasi-category of pointed spaces and $\Omega$ denotes the loop space) gives the quasi-category of spectra. These homotopy limits can be (abstractly) defined to be the homotopy limits in the Joyal model structure on simplicial sets, where the quasi-categories are just the fibrant objects. There is also a more explicit description given in Lurie's Higher Topos Theory (we come back to an example later in this question).
Many important examples of quasi-categories are constructed from a simplicial model category $\mathcal{M}$ in the following way: The sub simplicial category $\mathcal{M}^\circ$ of bifibrant objects forms a Bergner fibrant simplicial category and taking the coherent nerve produces a quasi-category $N(\mathcal{M}^\circ)$. Thus, the following question seems to be natural: Can we reconstruct the homotopy limit of the coherent nerves of a diagram of model categories as the coherent nerve of a "homotopy limit" of model categories?
A candidate is given in Julie Bergner's paper Homotopy limits of model categories and more general homotopy theories, Definition 3.1. We won't recall here the general definition, but only indicate it in the case that we index over a diagram with one object and a group $G$ as automorphism (i.e., we have a group action on our model category): Then an object in $holim_G \mathcal{M}$ is an object $X \in  \mathcal{M}$ together with morphisms $f_g: X \to g\cdot X$ (for $g\in G$) such that $f_e = id_X$ and $f_{hg} = (h\cdot f_g)\circ f_h$ (i.e, objects with a twisted $G$-action). At least in this case, the homotopy limit has a model structure (Bergner mentions the injective one, but at least sometimes, it has also the projective one, which is Quillen equivalent); actually it is a simplicial one if $\mathcal{M}$ was a simplicial model category. Indeed, it is the simplicial subcategory of $G$-equivariant morphisms in $Fun(EG, \mathcal{M})$ where $EG$ denotes the contractible groupoid associated to $G$. Thus, more precisely, our question is:

Is $N((holim_G \mathcal{M})^\circ)$ categorically equivalent to $holim_G N(\mathcal{M}^\circ)$ as quasi-categories?

There are several pieces of evidence for this:

If I am not mistaken, the description of homotopy limits in Higher Topos Theory implies that the homotopy fixed points of a quasi-category $\mathcal{C}$ are given as $Map(N(EG), \mathcal{C})^G$ (where Map denotes the internal Hom of simplicial sets and $()^G$ denotes strict fixed points). Thus the question is equivalent to whether $Map(N(EG), N(\mathcal{M}^\circ))^G$ is categorically equivalent to $N((Fun(EG, \mathcal{M})^G)^\circ)$. By strictification of homotopy coherent diagrams, a similar statement holds if we don't take $G$-fixed points, but I don't see how to prove the statement involving the $G$-fixed points.
Even more convincingly, Julie Bergner shows in her paper that homotopy limits of model categories are compatible with homotopy limits in the complete Segal space model structure on simplicial spaces. More precisely, she shows that the classification diagram functor commutes with homotopy limits up to weak equivalence (Theorem 4.1). Now, one could get the impression that we are finished since the complete Segal space model structure and the Joyal model structure are Quillen equivalent. But this is not sufficient: one has to prove that the classification diagram functor is send under this Quillen equivalence to something weakly equivalent to the coherent nerve. Although one gets some compatibility results from the papers Quasi-categories vs. Simplicial Categories (by Andre Joyal) and Complete Segal spaces arising from simplicial categories (by Julie Bergner), I didn't quite find the right statement to make the comparision work. 

As a last word of motiviation, I want to add that I stumbled upon these questions when I thought about Galois descent, where one often considers objects with twisted group actions.

REPLY [6 votes]: I will address your second question: "one has to prove that the classification diagram functor is sent under this Quillen equivalence to something weakly equivalent to the coherent nerve".
The answer is yes, this is true. First note that the simplicial category $M^\circ$ is equivalent as a simplicial category to $L^HM$, the hammock localization of M (at the weak equivalences). See here for references. So the answer to your second question (or is it a remark?) follows as a special case of this previous MO answer (which cites work of Barwick-Kan and Toen). So the classification diagram functor and the coherent nerve of $M^\circ$ are sent to equivalent things under the Quillen equivalence between quasicategories and complete Segal spaces (I prefer the term Rezk categories or Rezk $\infty$-categories).
I believe that combined with your other comments (notably using Julie Bergener's results) this is enough to answer your main question in the affirmative.<|endoftext|>
TITLE: A reference for the Chevalley Groups
QUESTION [7 upvotes]: Hello everyone
I would like to learn basic theory of the Chevalley Groups. There are several references for this subject, like "Introduction to Lie algebras and representation theory" by Humphreys, and Steinberg notes on the Chevalley groups. 
Do you know any other references beside these two books? 
Thank you

REPLY [5 votes]: There is a new book: "Linear Algebraic Groups and Finite Groups of Lie Type" by Gunter Malle and Donna Testerman (Cambridge Univesity Press, 2011). The first part is an introduction to linear algebraic groups, and the third part is on finite groups of Lie type).<|endoftext|>
TITLE: Connective spectra versus simplicial abelian groups - very basic question
QUESTION [9 upvotes]: Hello,
I have very general , "introductory" questions (It is quite hard for me to seek for specific things in the algebraic topology literature).
I guess that connective spectra have a model structure. So do simplicial abelian groups. Are these Quillen equivalent?
Secondly, I think of a simplicial abelian group as a space with strictly associative and commutative operation, while I think of a connective spectrum as a space with an operation which is associative and commutative up to all higher coherences (i.e. some words like $E_{\infty}$). So these are similar. How do I see what extra richness is encoded in a spectrum? For example, what mental pictures do I lose when I think of a connective spectrum as a right-bounded chain complex?
I think that the last is the most important for me, to have some small mental picture which I should have for spectra but not for simplicial abelian groups / chain complexes.
Thank you,
Sasha

REPLY [14 votes]: One fundamental difference concerns the behavior of Postnikov towers, or the relationship between the spectrum/simplicial abelian group and its homotopy groups.  In simplicial abelian groups all Postnikov towers are splittable, since there are no higher Ext's between abelian groups; thus every simplicial abelian group is equivalent to a product of K(A,n)'s.  But in spectra there are lots of higher Exts between Eilenberg-Maclane spectra (like those corresponding to Steenrod powers), and this means that for a general spectrum its homotopy groups will have complicated relationships with each other, encoded for instance in 
the spectrum's "k-invariants" (the most basic instance of this, by the way, is that the homotopy groups of a spectrum are graded modules over the stable homotopy groups of spheres, whereas there's no extra structure for simplicial abelian groups).
A related perspective would be that the richness of spectra vs. simplicial abelian groups corresponds to the richness of the Steenrod algebra (acting on cohomology of spectra) vs. just its Bockstein part (which is all that acts on "cohomology" of simplicial abelian groups).
But maybe the most compelling picture illustrating the differences is the chromatic one.  In some sense the chromatic picture of stable homotopy theory tells you that the difference between spectra and simplicial abelain groups lies in the existence of fundamental and systematic periodic phenomena in the former which are completely lacking in the latter.  Concretely, many connective spectra are harmonic (i.e. completely amenable to chromatic techniques, i.e. canonically filterable with graded pieces displaying controlled periodicity) -- for instance, finite spectra (chromatic convergence theorem) and suspension spectra (result of Hopkins and Ravenel) -- whereas simplicial abelian groups are only sensitive to the 0th chromatic layer, i.e. rationalization (so nothing periodic about it at all), since their higher Morava K-theories vanish, by the splitting of the Postnikov tower alluded to above and the fact that you can't make a bounded above spectrum periodic without killing it.<|endoftext|>
TITLE: Conditions for the restriction $H^i(G,A)\to H^i(H,A)$ being surjective
QUESTION [6 upvotes]: I was wondering what the condition is for the restriction map (in group cohomology)  $H^i(G,A)\to H^i(H,A)$ to be surjective.
I am a little confused about when maps between cohomology groups are considered to be $G$-, $H$- or $G/H$-module homomorphisms. In particular, in the exact sequence of low degrees
$0\to H^1(G/H,A^H)\to H^1(G,A)\to H^1(H,A)^{G/H}\to H^2(G/H,A^H)\to H^2(G,A)$
are the maps all meant to be $G$-module homomorphisms?
EDIT: Does someone have an example when $\mathrm{res}$ is not surjective?

REPLY [8 votes]: You asked for examples when $\text{res}^G_H\colon H^i(G;A)\to H^i(H;A)$ is not surjective. Here are some examples for $i=1$.

Let $G=\text{SL}_2\mathbb{Z}$ and $H=\mathbb{Z}$ generated by $\begin{bmatrix}1&1\\\\0&1\end{bmatrix}$. Then $H^1(G;\mathbb{Z})=0$ but $H^1(H;\mathbb{Z})=\mathbb{Z}$ so $\text{res}$ is not surjective.
Let $G=D_\infty=\text{Isom}(\mathbb{Z})=\mathbb{Z}/2\ltimes\mathbb{Z}$ and $H=\text{Isom}^+(\mathbb{Z})=\mathbb{Z}$. Now $[G:H]=2<\infty$ but still $H^1(G;\mathbb{Z})=0$ and $H^1(H;\mathbb{Z})=\mathbb{Z}$ so $\text{res}$ is not surjective.
Let $G$ be the rank 2 free group $G=F_2=\langle a,b\rangle$, and let $H$ be any subgroup of index $n-1$. Then $H$ is a free group of rank $n$ so $H^1(H;\mathbb{Z})=\mathbb{Z}^n$. But $H^1(G;\mathbb{Z})=\mathbb{Z}^2$ so $\text{res}$ is not surjective for $n>2$.
For a finite example, let $G$ be the group of upper-triangular matrices in $\text{SL}_3\mathbb{F}_p$, and let $H$ be its center: $H=\begin{bmatrix}1&0&\ast\\\\0&1&0\\\\0&0&1\end{bmatrix}$. Then the map from $H^1(G;\mathbb{F}_p)=\mathbb{F}_p^2$ to $H^1(H;\mathbb{F}_p)=\mathbb{F}_p$ is the zero map (this is a good exercise), so $\text{res}$ is not surjective. (For a related topological example, let $T^1\Sigma_g$ be the unit tangent bundle of an orientable surface of genus $g>1$, let $G=\pi_1(T^1\Sigma_g)$, and let $H\approx \mathbb{Z}$ be its center.)
More generally, let $G$ be any group and let $H$ be its commutator subgroup $[G,G]$. As long as your coefficients are untwisted (trivial as a $G$-module), the map $H^1(G;A)\to H^1([G,G];A)$ is always the zero map. Thus if $G$ is any group for which $H^1(G;A)\neq 0$ and $H^1([G,G];A)\neq 0$, $\text{res}$ is not surjective. Examples of such $G$ include free groups, surface groups, non-abelian nilpotent groups (either with $A=\mathbb{Z}$ or $A=\mathbb{F}_p$), etc.

You can also get examples for $i=0$ when $A$ is a nontrivial $G$-module, since $H^0(G;A)$ is the $G$-invariant elements of $A$, while $H^0(H;A)$ is the $H$-invariant elements of $A$. Thus  whenever $A$ contains an $H$-invariant element not fixed by $G$, the restriction $\text{res}$ will not be surjective. For example, for $[G:H]<\infty$ you could take $A=\mathbb{Q}[G/H]$.<|endoftext|>
TITLE: Modules with trivial source and projective modules
QUESTION [6 upvotes]: This is settled in $p$-modular representation theory of finite groups and finite dim. algebras. So we let $k$ be an algebraically closed field of characteristic $p>0$ and $G$ a finite group. A module with trivial source (TS module for short) is an indecomposable direct summand of the induced module $k_Q^G$ where $Q$ is any $p$-subgroup of $G$. Note that $k_Q^G$ is the tensor product $k_Q\otimes kG$ .
Projective indecomposable $kG$-modules (PIMs for short) can be shown to have trivial source (namely as direct summands of $k_1^G$). 
My question is if they just ''happen'' to have trivial source as well, or if PIMs and other modules with trivial source  have even more structure in common. 
To put the question from another point of view: In other finite dimensional algebras over $k$, can we find analogues of modules with trivial source in the sense that they are connected to projective modules in a similar (to me unknown) way as they are in group rings?

REPLY [4 votes]: I wouldn't say that projective indecomposable modules "just 'happen' to have trivial source". In fact there is a bijection between the isomorphism classes of indecomposable $kG$-modules with trivial source and vertex $P$ and the isomorphism classes of projective indecomposable $k[N_G(P)/P]$-modules. This bijection is called Brauer correspondence and is given by 
$$M \mapsto \operatorname{Br}_P(M) := \frac{M^P}{\sum_Q tr_{Q}^P (M^Q)}$$
$(Q < P)$ where $M^P$ are the $P$-invariants and $tr^P_Q: M^Q \to M^P,\;\; m \mapsto \sum_{g \in P/Q}gm$ is the trace map. This can be found in Theorem 3.2(3) of the paper
M. Broué: On Scott Modules and p-Permutation Modules: An Approach through the Brauer Morphism. Proc. Amer. Math. Soc. 93(1985), 401-408 
(note that the indecomposable p-permutation modules are just the indecomposable modules with trivial source). 
As a slight generalization, if $A$ is any finite dim. augmented $k$-algebra you may define $M$ to have trivial source if it is isomorphic to an indecomposable direct summand of $A\otimes_B k$ for some subalgebra $B \le A$. Then finite dimensional projective indecomposable $A$-modules have trivial source (choose $B=k$).<|endoftext|>
TITLE: counting non-isomorphic groups of a given cardinality
QUESTION [11 upvotes]: Given an infinite cardinal $\kappa,$ is there some nice way to construct $2^\kappa$ non-isomorphic groups of that cardinality? In the answer to this stackexchange question, there is a fairly high-powered argument to show that that many (abelian) groups do exist, but it seems shocking that there is not a direct construction, like there is for $\aleph_0.$ (by the way, I doubt that allowing arbitrary, instead of just abelian, groups helps that much, but one never knows.) This question came from a conversation with our own @Joel David Hamkins.

REPLY [10 votes]: If you just want a direct construction which avoids nontrivial set theory such as stationary sets etc., how about this?
Step One: For each subset $S \subseteq \kappa$, let $M(S)$ be the structure $\langle \kappa; < , S \rangle$, where $S$ is regarded as a unary relation. Obviously, if $S \neq T$, then $M(S)$ and $M(T)$ are non-isomorphic. 
Step Two: For each subset $S \subseteq \kappa$, encode $M(S)$ into a corresponding graph $\Gamma(S)$ so that if $S \neq T$, then $\Gamma(S)$ and $\Gamma(T)$ are non-isomorphic. (This is an easy exercise.)
Step Three: For each subset $S \subseteq \kappa$, encode the graph $\Gamma(S)$
into a suitable group $G(S)$ with generators $\Gamma(S)$ and relations $R(S)$ which encode the adjacency relation. (This can be done using small cancellation theory.)<|endoftext|>
TITLE: Which groups have strictly rational representations?
QUESTION [16 upvotes]: It can be shown, via the construction of the representations of the symmetric group, that every representation of $S_n$ is equivalent to a representation with values in $\mathbb{Q}.$ Presumably, this is a fairly rare phenomenon: it clearly doesn't hold for cyclic groups ($\mathbb{Z}/p\mathbb{Z}$ has one-dimensional representations given by the $p$th roots of unity, hence $p-1$ of its representations lie outside of $\mathbb{Q}$).
Moreover, there is a formula which constructs the rational characters of a group (due to Artin: see Curtis and Reiner section 15), but it doesn't seem to give an answer to the following question: 
Are there other "classes" of groups such that every irreducible representation is realizable over $\mathbb{Q}$?
Take "classes" to mean whatever you think appropriate (so long as it doesn't mean the collection of all groups with only rational irreps).

REPLY [9 votes]: Finite groups all of whose ordinary complex representations have rational-valued characters are sometimes called Q-groups (and sometimes called rational groups).  There is a monograph by Denis Kletzing (Structure and Representations of Q-Groups, Springer Lecture Notes in Mathematics 1084, 1984) which might be of interest.  Symmetric groups are Q-groups, as you mention, as are Weyl groups.  I can't remember if Kletzing constructs any other infinite families although I do remember that $D_n$ is a Q-group only when $n=1,2,3,4$ or $6$.  It's also worth mentioning that homomorphic images and direct products of Q-groups are also Q-groups.<|endoftext|>
TITLE: Hamiltonians which commute both as operators and as connections
QUESTION [7 upvotes]: This is something which I suspect is written up in introductory books on mathematical physics if I knew where to look. Suppose I have some parameters $t_1$, ..., $t_k$ ranging over a neighborhood in $\mathbb{R}^k$. I also have $k$ matrix-valued functions of the $t$'s: $H_1(t_1, \ldots, t_k)$, ... $H_k(t_1, \ldots, t_k)$. These obey both
$$[H_i, H_j]=0 \quad (\ast)$$
and
$$[\partial_i+H_i, \partial_j+H_j] =0 \quad (\dagger).$$
For those who don't like the language of connections, we can expand $(\dagger)$ as $\partial H_i/\partial t_j - \partial H_j/\partial t_i + [H_i, H_j]=0$ or, in the presence of $(\ast)$, as 
$$\frac{\partial H_i}{\partial t_j} = \frac{\partial H_j}{\partial t_i}.$$
Equation $(\ast)$ tells us that, assuming the $H_i$ are individually diagonalizable, we can find $u(t)$ a simultaneous eigenvector for all the $H_i$:
$$H_i(t) u(t) = \lambda_i(t) u(t). \quad (\ast\ast)$$
Equation $(\dagger)$ tells us that the vector-valued PDE 
$$\frac{\partial v}{\partial t_i} + H_i v=0 \quad (\dagger \dagger)$$
will have a unique solution $v(t_1, t_2, \ldots, t_n)$ for any initial value.
I'm pretty sure there is supposed to be a relation between the solutions to  $(\ast \ast)$ and $(\dagger \dagger)$. What is the right statement, and what is the keyword to read about this situation?
Motivation: I'm trying to work through the papers of Varchenko, Scherbak and others on the KZ equation. I think it would really clear my head to just see this scenario described abstractly without all the details of which operators they are thinking about.
$\def\mg{\mathfrak{gl}_n}$ Edit to spell out the relation. Let $V_1$, $V_2$, ..., $V_n$ be representations of $\mg$. So $U(\mg)^{\otimes n}$ acts on $V_1 \otimes V_2 \otimes \cdots \otimes V_n$. Let $\Omega \in U(\mg) \otimes U(\mg)$ be the Casimir. (Note: The element I learned to call the Casimir was a central element $c$ in $U(g)$. In terms of that element, $\Omega = \Delta(c) - c \otimes 1 - 1 \otimes c$.) Let $\Omega_{ij}$ be $\Omega$ acting in positions $i$ and $j$.
For generic parameters $z_1$, ..., $z_n$, define $H_i = \sum_{j \neq i} \Omega_{ij}/(z_i-z_j)$. Then, as I understand it, the KZ equation is $(\partial_i + H_i) v(z_1, \ldots, z_n)=0$, where $v$ is a function valued in $V_1 \otimes V_2 \otimes \cdots \otimes V_n$. The $H_i$'s obey both $(\ast)$ and $(\dagger)$ (a nice exercise). And people seem to be very interested in solving both $(\ast \ast)$ "diagonalizing the action of the Gaudin subalgebra" and $(\dagger \dagger)$ "solving the KZ equation". So I was hoping to understand how they relate, and why.

REPLY [3 votes]: To expand on Greg's answer regarding the adiabatic theorem. You are looking for situations where the adiabatic evolution is exact. This is the case for a Hamiltonian of the form 
$H = i\left[\frac{\partial P}{\partial t},P\right]$
where $P$ is a projector onto your chosen instantaneous eigenstate. This comes from T, Kato, J. Phys. Soc. J. Jpn. 5, 435 (1950).<|endoftext|>
TITLE: Number of x such that $m-x^2$ and $n-x^2$ are both squares
QUESTION [8 upvotes]: For given integers $m$,$n$, what bounds are known on the number of positive integers $x$ such that $m-x^2$ and $n-x^2$ are both perfect squares? In particular, is the number of such $x$ bounded above by a constant independent of $m,n$?
A search for $m,n \le 5000$ gives three pairs for which there are 4 such x: (1370,2210),(2210,3050), and (3485,3965).

REPLY [5 votes]: In comments Aaron asked about an example of Kevin's construction with $10$ points.
In Kevin's comment the rational points come from the group law on the elliptic curve and $D$ is the lcm of the denominators.
Here is magma online code and example with $10$ points -- as Kevin wrote this way you get as many solutions as you want.
Starting with the OP's $m,n=1370,2210$ got a Weierstrass model of EC.
Found $4$ generators with mwrank (could have used some of OP's points instead of finding generators). 
Multiples of the $4$ generators gave additional solutions to the OP and here is the result:
D= 12390849713183581790836556709545874255316431121037842112055747871712302486846209203045
m= 210340424562141262047595910887493791274649532004925386992592745031243269297527125123770927661074947583428727448136984975928253274592074678864394207689104340429534031062674250
n= 339308276118490648996486834351358597603631726810865040331116763882516514706229887973382299365675645371808385153564041457519299077991594919919935181746657366678299422371175250
x1= 12390849713183581790836556709545874255316431121037842112055747871712302486846209203045
x2= 284989543403222381189240804319555107872277915783870368577282201049382957197462811670035
x3= 458461439387792526260952598253197347446707951478400158146062671253355192013309740512665
x4= 359334641682323871934260144576830353404176502510097421249616688279656772118540066888305
x5= 37114042644837641401487345470535116987974416718966693064558863256140865315458375412615
x6= 61662771334048589214986191564717597484363845941296621783693700780051335795784841331249
x7= 136028531722798363505361591190162668220641503334832494123018223476468013307956416186455
x8= 244444823900679257002213529250900630227602064446780336065278693335401699985373414615885
x9= 301450648169146711879386180421262383173184752443718928831863273736903739700469518290993
x10= 437274569420552490793563721414855158314534791058269868589191046763361083229670113418305

Magma online code
m:=1370;
n:=2210;
aa<x,y,z>:=AffineSpace(Rationals(),3);
C:=Curve(aa,[m-x^2-y^2,n-x^2-z^2]);
P:=C!([1,37,47]);
pc:=ProjectiveClosure(C);
E,m1:=EllipticCurve(pc,pc!(P));
m2:=Inverse(m1);
aInvariants(E);
Ep:=E!([-323231734744697104/27633477663066497041,585299700649024 /27633477663066497041]);
m2(Ep);
m2(2*Ep);<|endoftext|>
TITLE: Applications of algebraic geometry/commutative algebra to biology/pharmacology
QUESTION [10 upvotes]: Are there applications of algebraic geometry/commutative algebra to biology/pharmacology?
It might be that some Gröbner basis technique is used somewhere? I know there are some applications to robotics — in solving some complicated non-linear equations. Maybe something similar can happen in biology.

Related questions:

Applications of group theory to math. biology (pharmacology) ?

Any applications integrable systems (pde,ode, q-,...) to math. biology (pharmakinetics, pharmadynamics) ?

"Graphical models" and "gene finding and diagnosis of diseases" ?

Applications of the knot theory to biology/pharmacology ?

REPLY [9 votes]: René Thom's theory of morphogenesis involves singularities, unfoldings, perturbations of analytic/geometric structures, etc., which, in its turn, involves (or, rather, should involve, as the whole theory is rather sketchy) a good deal of commutative algebra.

A conference "Moduli spaces and macromolecules".

Some biological models involve systems of boolean equations, or sentences of propositional calculus, which could be interpreted as polynomials over GF(2), with subsequent application of Gröbner basis technique. A (more or less random) sample of possibly relevant papers (I avoid mentioning algebraic statistics which was mentioned many times elsewhere):

G. Boniolo, M. D'Agostino, P.P. Di Fiore, Zsyntax: A formal language for molecular biology with projected applications in text mining and biological prediction,PLoS ONE 5 (2010), N3, e9511
DOI:10.1371/journal.pone.0009511

A.S. Jarrah and R. Laubenbacher, Discrete models of biochemical networks: the toric variety of nested canalyzing functions, Algebraic Biology, Lect. Notes Comp. Sci. 4545 (2007), 15-22 DOI:10.1007/978-3-540-73433-8_2

R. Laubenbacher and B. Stigler, A computational algebra approach to the reverse engineering of gene regulatory networks, J. Theor. Biol. 229 (2004), 523-537 DOI:10.1016/j.jtbi.2004.04.037 arXiv:q-bio/0312026

I. Lynce and J.P. Marques Silva, Efficient haplotype inference with boolean satisfiability, AAAI'06, July 2006; SAT in Bioinformatics: making the case with haplotype inference, SAT'06, August 2006; http://sat.inesc-id.pt/~ines<|endoftext|>
TITLE: Modular curve parametrizing two cyclic subgroups of an elliptic curve
QUESTION [5 upvotes]: The aim of this question is to better understand the following moduli space/modular curve, for which I propose (temporarily) the name $Y_0(M,N)$. We define $Y_0(M,N)$ as the moduli space parametrizing an elliptic curve $E$, together with two cyclic subgroups $G$ and $H$, of order respectively $M$ and $N$, of the group of degree $0$ line bundles modulo linear equivalence on $E$.
(note that the subgroup generated by $G$ and $H$ need not be of rank $2$, although it will be so generically)
When $N=1$, $Y_0(M,N)$ is just the well known modular curve $Y_0(M)$.
Questions:
1) Are there references for this moduli space/modular curve? Is this well studied/well known?
2) I think of this as the fiber product $Y_0(M) \times_{\mathcal{M}_{1,1}} Y_0(N)$ (the fiber product over the moduli space of elliptic curves). Is this correct?
3) What can we say about the curve $Y_0(M,N)$? Is it irreducible? Are there some kind of formulas for its genus as there are for $Y_0(M)$?

REPLY [6 votes]: $Y_0(M,N)$ can be reinterpreted as the moduli space of diagrams $E_1 \to E \leftarrow E_2$ of elliptic curves, where the arrows are cyclic isogenies of degree $M$ and $N$.  From this viewpoint, it is naturally a fiber product of $Y_0(N)$ with $Y_0(M)$, as you suggested.  It is in general a disjoint union of modular curves, because the kernels of dual isogenies may have isomorphism types and restrictions of the Weil pairing that are not unique.
In good characteristic, you may enumerate components by finding all abelian groups generated by a cyclic group of order $M$ and a cyclic group of order $N$, and finding all equivalence classes of $\mu_{(M,N)}$-valued symplectic forms on these groups.  In characteristic $p$, you have to work out cases, depending on how many powers of $p$ divide $M$ and $N$.  For example, if $M=N=p$, then $Y_0(p)$ is two copies of the affine line, glued at supersingular points, and the fiber product is four copies of the affine line, also glued at those points.
Most of the information you want can be extracted from Katz and Mazur's Arithmetic Moduli of Elliptic Curves, although I don't know where a table of genera of components can be found.  You may have to explicitly enumerate elliptic points and cusp orders, then use Hurwitz.<|endoftext|>
TITLE: Manifolds with homeomorphic interiors
QUESTION [20 upvotes]: Suppose that two compact topological manifolds with boundary have homeomorphic interiors. Can we conclude that the two manifolds are homeomorphic? What happens in the smooth category?

REPLY [17 votes]: Gjergji Zaimi's answer gives a strong positive conclusion: the product of the boundaries with $\mathbb{R}$ are necessarily homeomorphic. I just want to add a couple of explicit examples illustrating that the boundaries themselves need not be homeomorphic.
The first example is given by Milnor in his article "Two complexes which are homeomorphic but combinatorially distinct", partially stated there as theorem 3. Consider the 3-dimensional lens spaces $L_1=L(7,1)$ and $L_2=L(7,2)$. In that article, Milnor proves (section 2) that for any $n>1$, the interior of $L_1\times D^{2n+1}$ is diffeomorphic to the interior of $L_2\times D^{2n+1}$. However, Milnor also proves (at the end of section 4) that their boundaries, $L_1\times S^{2n}$ and $L_2\times S^{2n}$, have different Reidemeister torsions. Consequently, the boundaries cannot be homeomorphic, by the topological invariance of Reidemeister torsion.
To obtain other examples, we can apply the result stated by Oscar Randall-Williams' in his comment to Gjergji Zaimi's answer. It suffices to find $h$-cobordant closed manifolds $M$, $N$ of dimension greater than 3 which are not homeomorphic: then $M\times I$, $N\times I$ will have diffeomorphic interiors by Oscar's argument, but their boundaries are not homeomorphic. As Milnor proves in the article cited above, we can take $M=L_1\times S^{2n}$ and $N=L_2\times S^{2n}$ for $n>0$. I wrote a more detailed discussion concerning examples of this type in an answer to a related mathoverflow question.<|endoftext|>
TITLE: Density character of $\ell_\infty(\kappa, S)$
QUESTION [5 upvotes]: Let $S$ be an uncountable set. Consider the subspace $\ell_\infty(\kappa, S)$ of $\ell_\infty(S)$ formed by all functions with support of cardinality at most $\kappa$ (here $\kappa<|S|$). Certainly, $\ell_\infty\subset \ell_\infty(\kappa, S)$, whence the density character of $\ell_\infty(\kappa, S)$ is at least $2^\omega$. Can we specify it? Is it for example $2^\omega \cdot \kappa$?

REPLY [6 votes]: I understand you to mean that $\ell_\infty(S)$ is the collection of
bounded functions from $S\to\mathbb{R}$, under the sup norm, and
$\ell_\infty(\kappa,S)$ consists of those functions which take value
$0$ outside a set of size at most $\kappa$, a fixed infinite
cardinal less than $|S|$. The density character is the size of the
smallest dense subset of this space under the sup norm metric.
I claim that the density character is exactly $|S|^\kappa$. It can
be no larger than this, because the number of possible supports is
$|S|^\kappa$, and for each support, there are
$|\mathbb{R}|^\kappa=2^\kappa$ many functions having that support,
leading to $|S|^\kappa\cdot 2^\kappa=|S|^\kappa$ many functions
altogether in $\ell_\infty(\kappa,S)$. Conversely, the density
character is at least this large. To see this, consider the set of
all functions $f:S\to
\{0,1\}$ with support of size $\kappa$. There are precisely
$|S|^\kappa$ many such functions, and any two of these functions
have distance $1$. So any dense set must have elements close to
each of these functions, and so must have size at least
$|S|^\kappa$.
Regarding the $2^\omega\cdot\kappa$ proposal, note that $|S|^\kappa$ is at least $2^\kappa$, which is strictly larger than $2^\omega\cdot\kappa$ when $\kappa\geq 2^\omega$.
(Lastly, if I may, allow me to suggest a small modification to
your notation. It might be better to consider $\kappa\leq|S|$ and
define that $\ell_\infty(\kappa,S)$, or perhaps denote it $\ell_\infty({\lt}\kappa,S)$, to be the collection of functions
in $\ell_\infty(S)$ with support of size strictly less than
$\kappa$, rather than at less-than-or-equal-to $\kappa$. This way
of doing it includes your case, by replacing $\kappa$ with
$\kappa^+$, but also allows one to consider the case of limit
cardinals, such as the functions with support less than
$\aleph_\omega$, which is not possible to express in your
notation. A similar argument to the above shows that $\ell_\infty({\lt}\delta,S)$ has density character $|S|^{\lt\delta}$)<|endoftext|>
TITLE: Minimum off-diagonal elements of a matrix with fixed eigenvalues
QUESTION [18 upvotes]: I am an engineer working in radar research. I came accross a problem on which I cannot seem to find literature. I can ask it in two different ways. Perhaps depending on the reader, the alternative question is easier to answer.

First way

Assume I have a real symmetric matrix $\mathbf{C} \in \mathbb{R}^{M \times M}$.
I know its eigenvalues that are non-negative: $\lambda_1, \ldots, \lambda_M$. And The trace of the matrix, i.e., the sum of all eigenvalues is $\lambda_1+\cdots+\lambda_M = M$.
The diagonal matrix of eigenvalues is $\mathbf{\Lambda}$ and the matrix with eigenvectors in its colums is $\mathbf{V}$. The eigendecomposition is then $\mathbf{C} = \mathbf{V} \mathbf{\Lambda} \mathbf{V}^T$.
Also the diagonal of the matrix is all ones, i.e., $\operatorname{diag}(\mathbf{C}) = [1,\ldots,1]$.

Define $$c_\max = \max\limits_{i\neq j}|c_{ij}|$$ where $c_{ij}$ is the element of $\mathbf{C}$ in the $i$-th row and $j$-th column. Given that I can choose $\mathbf{V}$ freely, i.e., any matrix with those eigenvalues, what is the minimum of the maximum of all off-diagonal elements that I can attain (in absolute value)? In other words what is the minimum of $c_\max$?

Second way

Given that you have $M$ vectors $\{\mathbf{v}_1, \ldots, \mathbf{v}_M\}$.
They are orthonormal $\mathbf{v}_i^T \mathbf{v}_j = \delta(i-j)$ by standard dot product definition.
They have norm one $\| \mathbf{v}_i \|=1$ by standard dot product definition.
Define the weighted inner product as $\mathbf{v}_i^T \mathbf{\Lambda} \mathbf{v}_j$, where $\mathbf{\Lambda} = \operatorname{Diag}(\lambda_1,\ldots,\lambda_M)$ and $\operatorname{trace}(\mathbf{\Lambda}) = M$.
$\{\mathbf{v}_1,\ldots,\mathbf{v}_M\}$ also have norm one $\|\mathbf{v}_i\|_w=1$ by this new weighted inner product definition.

What is then the minimum value for the maximum inner product (in absolute value) among all vectors $\{\mathbf{v}_1,\ldots,\mathbf{v}_M\}$ given they can be chosen freely as far as they satisfy the conditions?  
$$\min\limits_{\mathbf{v}_1,\ldots,\mathbf{v}_M} \left(\max\limits_{i\neq j}(\mathbf{v}_i^T\mathbf{\Lambda}\mathbf{v}_j)\right)$$

REPLY [7 votes]: I have a bound that will be of use to you.  First, note that we can use the fact that the diagonal entries are all $1$s to relate $c_\mathrm{max}$ to the Frobenius norm of $C$:
$$
\|C\|_F^2\leq M+M(M-1)c_\mathrm{max}^2.
$$
This Frobenius norm is easy to work with, since it's just the 2-norm of the spectrum:
$$
\|C\|_{F}^2
=\mathrm{Tr}[CC^\mathrm{T}]
=\mathrm{Tr}[V\Lambda^2 V^\mathrm{T}]
=\mathrm{Tr}[\Lambda^2]
=\sum_{m=1}^M\lambda_m^2.
$$
Rearranging then produces a lower bound on $c_\mathrm{max}$:
$$
c_\mathrm{max}\geq\sqrt{\frac{1}{M(M-1)}\bigg(\sum_{m=1}^M\lambda_m^2-M\bigg)}.
$$
Achieving equality in this lower bound certainly implies optimality.  For example, consider the following matrix:
$$
C
=\left[\begin{array}{rrr}1~&-\frac{1}{2}&-\frac{1}{2}\\-\frac{1}{2}&1~&-\frac{1}{2}\\-\frac{1}{2}&-\frac{1}{2}&1~\end{array}\right].
$$
Here, $\Lambda=\mathrm{diag}(\frac{3}{2},\frac{3}{2},0)$, $c_\mathrm{max}=\frac{1}{2}$, and a quick calculation reveals that this achieves equality in our lower bound.  But is this always possible?
Unfortunately, no.  For example, it's impossible to achieve equality when $\Lambda=\mathrm{diag}(\frac{5}{3},\frac{5}{3},\frac{5}{3},0,0)$.  But how do I know that?
Your question is intimately related to another problem that's of use in engineering:  Design an ensemble of $M$ unit vectors in $\mathbb{R}^d$, where $M>d$, with the property that no two vectors have a large inner product in magnitude (i.e., you want the ensemble to be incoherent).  For this problem, the Gram matrix of the vectors is playing the role of your $C$, and the Welch bound was developed to provide a lower bound on the coherence (your $c_\mathrm{max}$).  For details, check out this blog entry.
Your problem has an important distinction from the incoherent design problem: you prescribe the spectrum of $C$.  But there's a theorem that says achieving equality in the Welch bound necessitates that the spectrum of $C$ has $\frac{M}{d}$ with multiplicity $d$ and $0$ with multiplicity $M-d$.  As such, you might as well consider the instance of your problem in which this is your spectrum (in this instance, the above bound on $c_\mathrm{max}$ is precisely the Welch bound).
The point of looking at this instance is to demonstrate how hard your problem actually is.  While there are many Welch-bound achieving ensembles, it is also known that the Welch bound is not always achievable.  For example, it is impossible to pack $5$ vectors in $\mathbb{R}^3$ with Welch-bound coherence (this was the source of my second example above, while the first example corresponded to the cube roots of unity in $\mathbb{R}^2$).  It's also unknown in general which values of $M$ and $d$ enable Welch-bound equality (in fact, existence of such ensembles is equivalent to the existence of certain strongly regular graphs, and in many cases, existence is a long-standing problem).
For more information about this problem, google "equiangular tight frames" - you just opened a very interesting can of worms.  :)<|endoftext|>
TITLE: Handlebody decomposition of an open 4-manifold
QUESTION [12 upvotes]: Let $M$ be the fake $CP^2$ (namely the closed topological 4-manifold which is homotopy equivalent but not homeomorphic to the complex projective plan). It is well-known that $M$ admits no smooth structures. However $M' = M - pt$ is smoothable by a theorem of Quinn. The question is: does someone know an explicit handlebody decomposition of $M'$?

REPLY [17 votes]: There are not that many explicit handlebody pictures of exotic open 4-manifolds, because they get awfully complex in short order.  The ones that I know of are in work of Žarko Bižaca from the mid-90's.  I think you probably can work out this particular case by hand. You don't really want to use Quinn's theorem for this, because it is not exactly constructive.  On the other hand, Freedman's construction produces this manifold in a somewhat explicit manner.  
Here is a sketch. Start with a 4-ball, and then attach a (+1) framed 2-handle along a trefoil knot.  If you've chosen the correct trefoil, the boundary of the resulting manifold is the Poincare homology sphere, say P. Then Freedman tells you that P is the boundary of a (topological) contractible manifold W, which you glue on to make the exotic $CP^2$, commonly known as Ch (for Chern).  The reason that you don't need Quinn is that the construction of W is done by making a manifold $W'$ that is proper homotopy equivalent to $S^3 \times [0,\infty)$, and then using the proper h-cobordism theorem to recognize that the end of $W'$ is homeomorphic to $S^3 \times (1,\infty)$, from which you see that you can compactify $W'$ to a manifold by adding in a point.
In the case at hand, you can be more explicit.  One way (this is what happens in Freedman's original paper) is to consider the building block $P \times I$, and then do (spin) surgery on a circle to kill the fundamental group, resulting in a compact manifold $P'$. The embedding theorems of Freedman find a (topological, locally flat) wedge of spheres in $P'$, which can be surgered out to give a compact, simply-connected homology cobordism from $P$ to itself. Stacking infinitely many of these gives $W'$. Presumably, although I've never done this, you can use techniques of Bizaca to build a handlebody picture of $W'$ from this description.
An alternate approach, which might be more amenable to drawing pictures, would come from Freedman's older paper, A Fake $S^3 \times R$.  In this paper, which precedes his disk embedding theorem (but has many of the basic ideas, including reimbedding techniques) he constructs what I've called $W'$ by embedding Casson handles.  There is also a Bourbaki exposition of this paper by Siebenmann that is helpful in trying to read it.<|endoftext|>
TITLE: Solid rings and Tor
QUESTION [6 upvotes]: A solid ring is a ring $R$ such that the multiplication
$R\otimes_{\mathbb{Z}} R \to R$ is an isomorphism.
These were classified by Bousfield and Kan; they are

subrings of $\mathbb{Q}$,

$\mathbb{Z}/n$,

products $R\times \mathbb{Z}/n$ with
$R\subseteq \mathbb{Q}$ and
every divisor of $n$ invertible in $R$

colimits of these.


I wonder how small the list gets if I put the additional
constraint that $\mathrm{Tor}_{\mathbb{Z}}(R,R) = 0$.
REFERENCE:
Bousfield, A. K.; Kan, D. M.
The core of a ring.
J. Pure Appl. Algebra 2 (1972), 73–81.

REPLY [12 votes]: Let $R^t$ be the torsion submodule and consider the exact sequence
$$0\rightarrow R^t\rightarrow R \rightarrow R/R^t\rightarrow 0$$
Bousfield and Kan show that the ring on the right is a localization of ${\mathbb Z}$, hence flat over ${\mathbb Z}$, so its $Tor$ with $R$ vanishes.  Thus if we $Tor$ the above with $R$, we get $Tor(R^t,R)=Tor(R,R)$.
Now tensor the exact sequence with $R^t$ instead of $R$.  This gives $Tor(R^t,R^t)=Tor(R^t,R)$. 
Thus $Tor(R,R)=Tor(R^t,R^t)$.  But if $R^t$ is nonzero then (see Bousfield and Kan) it contains some ${\mathbb Z}/p{\mathbb Z}$ as a direct summand and hence $Tor(R^t,R^t)$ does not vanish.  Thus $Tor(R,R)=0$ implies $R^t=0$.  It follows (B/K 3.7) that $R$ is a localization of ${\mathbb Z}$.<|endoftext|>
TITLE: compact objects in model categories and $(\infty,1)$-categories
QUESTION [16 upvotes]: In an ordinary category $C$, one says that an object $X$ is $\kappa$-compact if the representable functor $Hom(X,-)\colon C \to Set$ preserves $\kappa$-filtered colimits.  We say $C$ is locally presentable if it is cocomplete and "generated" by $\kappa$-compact objects for some $\kappa$.
In an $(\infty,1)$-category $C$, one says that an object $X$ is $\kappa$-compact if the representable functor $Hom(X,-)\colon C \to \infty Gpd$ preserves $\kappa$-filtered $(\infty,1)$-colimits.  We say $C$ is locally presentable if it is cocomplete and "generated" by $\kappa$-compact objects for some $\kappa$.
There are many equivalent, also analogous, definitions in both cases.
An $(\infty,1)$-category is locally presentable if and only if it admits a presentation by some locally presentable, cofibrantly generated model category.  However, the only proof of this fact that I have seen (in A.3.7.6 in Higher Topos Theory) uses a different equivalent definition of both notions (as an accessible localization of some presheaf category).  Thus my question:

Is there any relationship between an object $X$ being $\kappa$-compact in a locally presentable model category and being $\kappa$-compact in the $(\infty,1)$-category that it presents?

REPLY [19 votes]: If $\mathcal{C}$ is a combinatorial model category, then for all sufficiently large regular cardinals $\kappa$, an object of the underlying $\infty$-category is $\kappa$-compact if and only if it can be represented by a $\kappa$-compact object of $\mathcal{C}$. The meaning of "sufficiently large" might depend on $\mathcal{C}$.
If $\kappa$ is not sufficiently large, then you generally don't have such an implication in either direction. For example, every finitely presented group is a compact object when viewed a (discrete) simplicial group, but need not be a compact object in the associated $\infty$-category (this requires that the classifying space of the group be finitely dominated). On the other hand, any space which is finitely dominated (i.e., a homotopy retract of a finite cell complex) is a compact object in the $\infty$-category of spaces, but cannot be represented by finite simplicial set unless its Wall finiteness obstruction vanishes.
(These counterexamples are not particularly compelling: you could get around the first one by restricting your attention to cofibrant objects, and the second one is very particular to the cardinal $\omega$. So perhaps there is something better to say.)<|endoftext|>
TITLE: Canonical topology for infinity topoi revisited.
QUESTION [10 upvotes]: A while ago I asked this quetion: Canonical topology for big infinity topoi
and this question: How to resolve size issues with the regular epimorphism topology
Let me first summarize some of what I learned from these:
If $U$ is a fixed ambient Grothendieck universe, then one can define a $U$-site to be a (not necessarily $U$-small) Grothendieck site $(C,J)$ such that there exists a $U$-small set $G$ of objects, called topological generators, such that every object of $C$ admits a covering family all of whose sources are in $G$.
$U$-sites are useful to have around, because they allow you to deal with "large" Grothendieck sites whose topos of sheaves are equivalent to the topos of sheaves of a small site. For example, if $E$ is a $U$-topos, i.e. a category equivalent to sheaves of $U$-small sets on some $U$-small site, then it is certainly not $U$-small itself. However, it does carry a Grothendieck topology, the canonical Grothendieck topology which is generated by jointly surjective epimorphisms. If we choose a $U$-small site of definition for $E$ such that $K$ is subcanonical, so that $E\cong Sh_K\left(D\right)$, then the objects of $D$, considered as representable sheaves, form a $U$-small set of topological generators for $E$, showing that $E$ with the canonical topology is in fact a $U$-site. Now, the category of $U$-small presheaves on $E$ is not $U$-small, but it's locally $U$-small, and the inclusion of the full subcategory of sheaves admits a left-exact left-adjoint (expose ii, theorem 3.4 of SGA 4). Moreover, this category of sheaves, is equivalent to $E$ itself.
I'm pretty sure that I can prove that all of this goes through for $n$-topoi when $n$ is finite. The complication arises when $E$ is a genuine infinty topos. Indeed, one can still equip $E$ with the canonical topology, and by careful use of Grothendieck universes as above, construct the infinity topos $Sh_{\infty}\left(E,can\right)$. However, its possible that $E$ is not equivalent to infinity sheaves on some site (e.g. it could be hypersheaves), so in this case, $Sh_{\infty}\left(E,can\right)$ can not be equivalent to $E$, correct? Even worse, $E$ could land somewhere between sheaves and hypersheaves, so we can't just say $E$ is the hypercompletion of $Sh_{\infty}\left(E,can\right)$.
My quetsion is, what is the relationship between $E$ and $Sh_{\infty}\left(E,can\right)$ when $E$ is an infinity topos? When $E$ is equivalent to infinty sheaves on a site, are these the same?

REPLY [2 votes]: I've convinced myself the answer is the following CLAIM however, I have not finished proving it yet,  but it is too long to leave as a comment. Hopefully soon I will update this with a proof of the claim as well (but feel free to beat me to it).
ClAIM: For any $n$, possibly infinite, if $E$ is an $n$-topos, and $C$ is a small $n$-category from which $E$ can be obtained by a left-exact localization $$a:Psh_n(C) \to E$$ then the canonical topology on $E$ restricts to a Grothenieck topology on $C$ via the composite of $a$ with the Yoneda embedding (this is clear), and this induces an equivalence of $n$-topoi between $Sh_n(C)$ and $Sh_n(E)$ (this needs proof).
Why do I think this is true?
1.) For finite $n$, this implies that the functor $Sh_\infty:n-Top \to \infty-Top$ which assigns an $n$-topos its infinity topos of infinity sheaves over itself with respect to the canonical topology is a fully faithful right-adjoint to the functor $\tau_{\le n-1}$ which assigns an infinity topos its $n$-topos of $\left(n-1\right)$-truncated objects. (See HTT 6.4.5.7).
2.) For $n=\infty$, by using HTT 6.5.2.19, we know that if $E$ is an infinity topos and $$L:Psh_\infty(C) \to E$$ is a left-exact localization, then it factors uniquely as localizations $$Psh_\infty(C) \to Sh_\infty(C,J) \to E,$$ for some Grothendieck topology on $C$, where the localization $ Sh_\infty(C,J) \to E$ is cotopological. However, it is not hard to see that this $J$ is the topology induced from the canonical topology on $E$. Combining this with the claim would show that not only is this factorization unique, but the infinity topos $Sh_\infty(C,J)$ does not depend on $C$. In particular, it would imply that if $E$ is a non-sheaf topos for some site, than it is not a sheaf topos for any site.<|endoftext|>
TITLE: Bochner's theorem, in stages
QUESTION [6 upvotes]: Bochner's theorem (for the real line version) asserts an infinite tower of inequalities, as a positivity condition. Taking each one, what do they mean, in an elementary fashion (at least at the start)?
For instance, the $1 \times 1$ matrix says that $Q(0)$ is positive. The $2 \times 2$ says that $|Q(x)| \le |Q(0)|$ (These two are commonly written down for necessary conditions of characteristic functions). What about 3 and 4?

REPLY [3 votes]: I believe the $3\times3$ inequalities were first exploited by Kreĭn and Weil, who wrote
$$
\begin{align}
   \bigl|\,Q(g) - Q(h)\,\bigr|
  &\leqslant \sqrt{2\mathrm{Re}\bigl(1 - Q(g^{-1} h)\bigr)},
   \tag K\\[1ex]
   \bigl|\,Q(gh) - Q(g)Q(h)\,\bigr|
  &\leqslant \sqrt{1 - |Q(g)|{}^2}\sqrt{1 - |Q(h)|{}^2}.
   \tag W
\end{align}
$$
in (1940), resp. (1940, p. 57). I stated them for any positive-definite function $Q$ on any group (written multiplicatively, as none of this requires Bochner’s commutative setting), but normalized $Q(e)=1$; else rescale by $Q(e)$, which (as you noted) is nonnegative by the $1\times1$ case.
Kreĭn’s $(\mathrm K)$ says that $Q$ is uniformly continuous everywhere as soon as it’s continuous at $e$, and I like to think of Weil’s $(\mathrm W)$ as a group version of Heisenberg’s uncertainty inequalities. Deducing them from positive-definiteness is not so obvious: $(\mathrm W)$ is a rearrangement of the determinant criterion
$$
\begin{vmatrix}
1&Q(a^{-1}b)&Q(a^{-1}c)\\
\overline{Q(a^{-1}b)}&1&Q(b^{-1}c)\\
\overline{Q(a^{-1}c)}&\overline{Q(b^{-1}c)}&1
\end{vmatrix}
\geqslant0,
\tag{$*$}
$$
written with $g=a^{-1}b$ and $h=b^{-1}c$; and proving $(\mathrm K)$ from $(\mathrm W)$ alone is possible but not very enlightening. I prefer to observe that positive-definiteness of $Q$ means positive-definiteness of the sesquilinear form $(c,d)\mapsto Q(c^*\cdot d)$ on the group *-algebra $\mathbf C[G]=$ {functions $c:G\to\mathbf C$ with finite support} where product, *-operation, and linear form $Q$ are defined on the basis of (Kronecker) $\delta^g$’s by
$$
\delta^g\cdot\delta^h=\delta^{gh},
\qquad
(\delta^g)^*=\delta^{g^{-1}},
\qquad
Q(\delta^g) = Q(g).
$$
Then $(\mathrm K)$ and $(\mathrm W)$ are just the Cauchy-Schwarz inequality
$$
|Q(c^*\cdot d)|^2\leqslant Q(c^*\cdot c)Q(d^*\cdot d),
$$
written out for $(c^*,d) = (\delta^e,\delta^g-\delta^h)$, resp. $(\delta^g-Q(g)\delta^e,\delta^h-Q(h)\delta^e)$.
(Which triples give $(\mathrm K)$ and $(\mathrm W)$ can then be found by remembering that Cauchy-Schwarz amounts to writing $Q(f^*\cdot f)\geqslant 0$ where $f=Q(c^*\cdot c)d-Q(c^*\cdot d)c$. Hewitt-Ross (1970, pp. 255, 289) have another proof, and the attribution to Kreĭn & Weil. It might be interesting to know if the $4\times4$ version of $(*)$ has some nice consequences...?)<|endoftext|>
TITLE: Octic family with Galois group of order 1344?
QUESTION [5 upvotes]: Does the octic,
$\tag{1} x^8+3x^7-15x^6-29x^5+79x^4+61x^3+29x+16 = nx^2$
for any constant n have Galois group of order 1344? Its discriminant D is a perfect square,
$D = (1728n^4-341901n^3-11560361n^2+3383044089n+28121497213)^2$
Surely (1) is not an isolated result. How easy is it to find another family with the same Galois group and same form,
$\tag{2} \text{octic poly in}\ x = nx^2$ 
Perusing Kluener's "A Database For Number Fields" this seems to be the only one. (Though I was able to find a 2-parameter family of a different form.)

REPLY [3 votes]: In arxiv.org/abs/1209.5300 I give the following polynomial with the same Galois group over Q(t): $(y+1)(y^7-y^6-11y^5+y^4+41y^3+25y^2-34y-29)-t(2y+3)^2$. This has $(6912t^4-3456t^3-95472t^2+23976t-1417)^2$ for a discriminant. This doesn't quite answer the second question, but substituting $y \mapsto (x-3)/2$, $t \mapsto s/256$ gives $(x-1) (x^7-23 x^6+181 x^5-547 x^4+515 x^3-325 x^2+71 x-1)-s x^2$, with discriminant $256 (27 s^4-3456 s^3-24440832 s^2+1571291136 s-23773315072)^2$, which does.<|endoftext|>
TITLE: If Mean Residual Lifetime is approximately constant, Residual Lifetime is Approximately Exponential in a Strong Sense
QUESTION [6 upvotes]: Suppose the "mean residual lifetime," $\mathbb{E}[X-x|X≥x]$ is approximately constant for large $x$. Then, I believe that the conditional tail distribution is approximately exponential, in the sense of being stochastically dominated by an exponential and dominating a similar exponential. Formally:
Conjecture Given any random variable  $X$ with support on $[0,∞)$. If, for some $\lambda \in(0,∞)$, $$\lim_{x→∞}\mathbb{E}[X-x|X≥x]= \lambda ,$$
then, for all $ε>0$ and for all $\Delta>0$, there is some $c$ such that $x≥c$ implies  $$e^{-\frac{t}{λ-ε}}\leq \mathbb{P}[X≥x+t|X≥x] \leq e^{-\frac{t}{λ+ε}}    \qquad ∀t≥\Delta.$$
I posted this question on StackExchange. Robert Israel provided a counterexample to an earlier conjecture, which was wrong.
Update The approximation result is stronger than weak convergence. Let $Y$ be distributed exponentially with parameter $\lambda$. The conclusion of the conjecture implies that
$$\lim_{x→∞}\mathbb{E}[f(X-x)|X≥x]=\mathbb{E}[f(Y)]$$
for all nondecreasing functions for which $\mathbb{E}[f(Y)]$ exists. In particular, $f$ is allowed to be unbounded.  The first great response by Ori Gurel-Gurevich implies a slightly weaker approximation result.

REPLY [2 votes]: Here is a work which gives general conditions under which your conjecture is true.
"Limiting Properties of the Mean Residual Lifetime Function" by Isaac Meilijson in Ann. Math. Statist. Volume 43, Number 1 (1972), 354-357.
http://projecteuclid.org/euclid.aoms/1177692731<|endoftext|>
TITLE: Time reversibility of Stratonovich Diffusion: Reference Request
QUESTION [5 upvotes]: Please consider the Stratonovich stochastic differential equation (SDE)
$$
dX = b(X)\circ dB
$$
where $B$ is standard Brownian motion and $X(0)=X_0$. This corresponds to the Ito (SDE)
$$
dX = \frac{1}{2} b(X) b'(X) dt + b(X) dB.
$$
I would like a reference showing (or even just stating) that trajectories of this equation are time-reversible in the following sense: that for all $m\geq 1$ and $t_m > t_{m-1} > \ldots > t_1 >0$, the joint distribution of
$$
(X(t_1), \ldots, X(t_m) )
$$
is identical to the joint distribution of
$$
(X(-t_1), \ldots, X(-t_m) ).
$$
Also, is there a particular term for this kind of time-reversibility? People also use time-reversibility to mean detailed balance for systems in equilibrium, which is different from this.
Motivation
In a paper I am listing advantages of expressing diffusions in terms of the Stratonovich convention. I want to be able to briefly state that if the drift coefficient in a Stratonovich SDE is 0, then the equation is time-reversible in the sense I state above.
Edit: Further Explanation
Here is a clarification of what I mean above, as well as a justification of my claim.
Let $B(t)$ for $t \in \mathbb{R}$ be two-sided Brownian motion with $B(0)=0$. Let $X(t)$ solve the above Stratonovich SDE.  Let $Y(t)=X(-t)$. Then
$$
dY(t) = dX(-t) = -b(X(-t)) \circ dB(-t) = b(Y(t)) \circ d\tilde{B}(t)
$$
where $\tilde{B}(t) = -B(-t)$ is also a Brownian motion. So $Y$ solves the same equation as $X$ with a different Brownian motion. These formal manipulations can be justified by letting $B$ be approximated by smooth stochastic processes and then taking the limit using the Wong-Zakai result.
Thanks for any help!

REPLY [2 votes]: To understand what this "time reversibility" means, it is instructive to replace the stochastic differential equations by the corresponding equations for the time dependence of the probability distribution $p(x,t)$. This is sufficient for the comparison of two-time correlations, in this case between $t=0$ and $t'>0$ (forward in time) or $t'<0$ (backward in time). 
"Time-reversibility" from the OP then amounts to the statement that the right-hand-side of the Kolmogorov backward and forward equations,
$$-\frac{\partial}{\partial t}p(x,t)=\mu(x)\frac{\partial}{\partial x}p(x,t) + \tfrac{1}{2}\sigma^2(x)\frac{\partial^2}{\partial x^{2}}p(x,t)$$
$$\frac{\partial}{\partial t}p(x,t)=-\frac{\partial}{\partial x}[\mu(x)p(x,t)] + \tfrac{1}{2}\frac{\partial^2}{\partial x^2}[\sigma^2(x)p(x,t)]$$
becomes identical if the drift $\mu$ is related to the diffusion $D=\tfrac{1}{2}\sigma^2$ by 
$$\mu=\frac{\partial D}{\partial x},$$
which is indeed the case, as one can readily verify by substitution. The forward and backward Kolmogorov equations then have the same form
$$\pm\frac{\partial}{\partial t}p(x,t)=\frac{\partial}{\partial x}\left[D(x)\frac{\partial}{\partial x}p(x,t)\right]$$
In response to the questions in the OP:
• Concerning a reference to the literature: The general relation between forward and backward diffusion is discussed in Time reversal of diffusions (1986) and in On the drift of a reversed diffusion (2005). Neither of these two papers makes the observation that there is a special case such that the drift for the forward and backward diffusion becomes the same, so as far as I can tell the observation in the OP is novel.
• Concerning a term to refer to this special case of equivalence of diffusion and reverse diffusion, I would suggest self-adjoint diffusion, since the corresponding Kolmogorov operator $\nabla D\nabla$ is self-adjoint. 
• As mentioned in the OP, none of this has any bearing on the physics notions of "time-reversibility" or "detailed balance", which refer to a more general relation between drift and diffusion, involving the steady-state equilibrium density. I might add that from a physics point of view, the identity of drift and derivative of diffusion required for self-adjoint diffusion is quite unnatural. This may explain why the special case of a self-adjoint Kolmogorov equation has not appeared before in the literature.<|endoftext|>
TITLE: Periodic automorphisms of free groups
QUESTION [6 upvotes]: Hi, 
I am troubled with the following question: Does there exist a finite order automorphism of a free group, $f\in Aut(F)$, such that it fixes no non trivial conjugacy class and no non trivial centralizer, i.e. $f(g)$ is not conjugate to $g$ and $f(g)\neq g^{-1}$ for any $g\in F$ ? Can we find such, so the same holds for any of its non trivial powers $f^i$?
By inspection of the finite order automorphisms of $F_2$, this cannot be an automorphism of $F_2$. 
Moreover, if there exists such an automorphism is it possible to find one with as large order as we want (possibly after moving to a larger free group)?

REPLY [20 votes]: No such automorphism exists. Every finite order automorphism of a finite rank free group has a nontrivial fixed conjugacy class. To see why, you can represent the free group automorphism as a simplicial automorphism $f : G \to G$ of some finite connected graph having no vertices of valence 1. Take any vertex $p \in G$ of valence $\ge 3$. Let $\gamma$ be an immersed edge path with initial oriented edge $E_0$ having initial vertex $p$, and terminal oriented edge $E_1$ having terminal vertex $f(p)$, such that the initial direction of $f(E_0)$ does not equal the terminal direction of $E_1$. The valence condition on $p$ lets you make this choice; if you made a bad choice of $\gamma$ then, since the valence of $f(p)$ is $\ge 3$, you could concatenate with some immersed path from $f(p)$ to $f(p)$ so as to change your terminal direction. Letting $k$ be the order of the simplicial automorphism $f$, it follows that $f$ fixes the conjugacy class corresponding to the immersed loop $\gamma * f(\gamma) * ... * f^{k-1}(\gamma)$.<|endoftext|>
TITLE: partial order on conjugate classes of subgroups
QUESTION [5 upvotes]: G is a group.
For a subgroup H of G, note $[H]$ the class of subgroups which are conjugate
to H.
Define the binary relation:
    $[H] \leq [K]$
iff
    $H_0 \subset K_0$ for some $H_0 \in [H]$ and $K_0 \in [K]$
It is easy to see that this relation is reflexive and transitive.
But how to show that it is anti-symmetric?
P.S.
In a book, the author claims that this relation defines a partial order on
the classes of conjugate subgroups in a context where G is a compact Lie
group. But I don't think the compact Lie group condition be essential, right?

REPLY [5 votes]: Here is a countable example. Take the Baumslag-Solitar group $\langle a, b \mid bab^{-1}=a^{4}\rangle$. Let $K=\langle a\rangle$, $H=\langle a^2\rangle$. Then $[K]\ne [H]$ (the standard properties of HNN extensions, see the book of Lyndon and Schupp), but $K > H$ while $bKb^{-1} < H$.<|endoftext|>
TITLE: A question about partial Euler products
QUESTION [5 upvotes]: Let $K/{\mathbb Q}$ be an extension of degree $d$. Let $S$ be the set of primes $p$ which split completely in $K$. What can one say about the analytic properties of 
$$
\zeta_{K, S}(s) : = \prod_{p \in S} \frac{1}{1-p^{-s}}. 
$$
More generally, one can define a similar partial Euler product for any splitting type, and ask the same question about the analytic properties of the resulting function. Any advice would be greatly appreciated.

REPLY [5 votes]: This question turned out to be not too difficult. Please see 
http://www.math.uic.edu/~rtakloo/euler-product.pdf 
for a (casual) writeup of an answer. 
Thanks for your comments and hints.<|endoftext|>
TITLE: Does every group grow either polynomially or superpolynomially?
QUESTION [13 upvotes]: I am reading an introduction to growth of groups. The notions of polynomial and superpolynomial growth are introduced, as are exponential and subexponential growth.
I can prove that the growth of a group is always either exponential or subexponential (it is exercise 1.6). However, there seems to be no mention of an analogous result for (super)polynomial growth (i.e. the growth of a group is always either polynomial or superpolynomial).
There exist strictly increasing functions which grow faster than polynomially but are not superpolynomial (this is pretty clear; a more detailed explanation can be found in the second section of this document), but I do not know whether these occur as the growth function of some group.
The thesis of a Nick Scott claims to prove that every group grows either polynomially or superpolynomially, but I don't see it (it is in subsection 1.4.1, on p.12; it seems to me the proof assumes that the limit $\log(\beta(k))/\log(k)$ exists, but I don't know why).
So my question is: does every group grow either polynomially or superpolynomially?

REPLY [5 votes]: It is interesting that another asymptotic invariant of a group, the Dehn function, can be arbitrary large (even non-recursive) but still bounded by a polynomial on an infinite set. In fact the polynomial can be made quadratic. See this paper.<|endoftext|>
TITLE: Hall polynomial when the subgroup is cyclic? 
QUESTION [6 upvotes]: Does anyone know the formula for a Hall polynomial  $g_{u,v}^{\lambda}(p)$ when $v$ is the type of cyclic subgroup (ie. $v=(v_{1})$ ) . 
http://en.wikipedia.org/wiki/Hall_algebra
I was hoping this particular case would be simple enough to describe .

REPLY [7 votes]: Let's say you want to compute the Hall polynomial $g^\lambda_{(r),\mu}(p)$.
According to [Dutta and Prasad, Degenerations and orbits in finite abelian groups], the orbits under the automorphism group of a finite abelian group are given by $\{O_I|I\subset J(P_\lambda)\}$, where $J(P_\lambda)$ denotes the lattice of order ideals in a certain poset $P_\lambda$. The same paper also gives a formula for $|O_I|$.
Clearly, the Hall polynomial that you are looking for is $(p^r-p^{r-1})^{-1}\sum_{I} |O_I|$, the sum being over all $I$ for which the order of an element in $O_I$ is $p^r$ and for which the quotient of the group of type $\lambda$ by any element of $O_I$ is of type $\mu$. As pointed out by David Speyer in the comments to an earlier version of this answer, $I$ is uniquely determined by these conditions. So a final form of the answer is obtained by explaining how to obtain $I$ from $\lambda$ and $\mu$.
Given an element $(p^{v_1},p^{v_2},\dotsc)$, the type of the quotient is found by computing the Smith canonical form of the matrix $\begin{pmatrix} p^{\lambda_1} & & &\\ & p^{\lambda_2} & &\\ & & \ddots & \\& & &p^{\lambda_l}\\ p^{v_1} & p^{v_2} & \cdots & p^{v_l} \end{pmatrix}$.
By the characterization of order ideals in $P_\lambda$, we have that $v_i\leq v_{i-1}\leq v_i+(\lambda_{i-1}-\lambda_i)$.
Proposition. Let $I\subset P_\lambda$ be an order ideal. Let $\mu$ be the type of the group obtained by going modulo an element of $O_I$. Then
$\mu_l=v_l$
$\mu_{l-1}=\lambda_l+v_{l-1}-v_l$
$\mu_{l-2}=\lambda_{l-1}+v_{l-2}-v_{n-1}$
$\vdots$
$\mu_{i}=\lambda_{i+1}+v_i-v_{i+1}$
$\vdots$
$\mu_1=\lambda_2+v_1-v_2$.
Proof.
The gcd of $i\times i$ minors of the above matrix can be seen to be $v_{l-i+1}+\lambda_{l-i+2}+\dotsb+\lambda_l$ (using the inequalities on $v_i$). Therefore, we get
$\mu_{l-i+1}+\dotsb+\mu_l=v_{l-i+1}+\lambda_{l-i+2}+\dotsb+\lambda_l$, from which the above identities follow.QED.
This allows us to recover $(v_1,v_2,\dotsc)$ once we know $\lambda$ and $\mu$, where (this also follows from the above proposition), $\mu$ is a partition obtained from $\lambda$ by removing a horizontal strip of length $r$. In particular, $v$ is uniquely determined by $\lambda$ and $\mu$.
We get
$v_l=\mu_l$, and $v_i=\mu_i-[(\lambda_{i+1}+\dotsb+\lambda_l)-(\mu_{i+1}+\dotsb+\mu_l)]$ for $i<l$.
Let $I\subset P_\lambda$ be the order ideal defined by $(v_1,v_2,\dotsc)$. We get

$g^\lambda_{(r)\mu}(p)=|O_I|/(p^r-p^{r-1})$.

It easily follows (from the formula for $|O_I|$ in our paper, which says that $|O_I|$ is a monic polynomial in $p$ of degree $\sum_i (\lambda_i-\nu_i)$) that $g^\lambda_{(r)\mu}(p)$ is monic in $p$ of degree $n(\lambda)-n(\mu)$. This could perhaps give another approach to Hall's theorem (in analogy with the proof that MacDonald gave).<|endoftext|>
TITLE: Example for the Sobolev embedding theorem when p=n.
QUESTION [5 upvotes]: Let $\Omega$ be a bounded domain in $\mathbb R^2$. By the Sobolev embedding theorem, if $k>\frac np$ (in this case $k>\frac 2p$) then 
$u\in W^{k,p}(U) \implies u\in C^{k-[\frac 2 p]-1,\gamma}(U)$
for a certain $\gamma$; where $[\frac 2 p]$ is the integer part of $\frac 2p$.
If $k=1$ and $p=2$ then a function $u\in W^{1,2}(U)$ is not necessarily in $L^\infty(U)$, like the function $\log|\log|x^2+y^2||$ shows. (Let $U$ be the disk centered at the origin of radius $\frac 12$.)
What is an example that shows that if $k=2$ and $p=2$ then for a function $u\in W^{2,2}(U)$, $|\nabla u|$ is not necessarily in $L^\infty(U)$?
More generally, is there a standard way to construct such example from the previous one, that is $\log|\log|x^2+y^2||$?

REPLY [4 votes]: You can take as an example
$$
  u(x) = x_1^{k - 1} (\log \lvert x \rvert)^\beta:
$$
if $\beta < 1 - \frac{1}{n}$, $u \in W^{k,n} (B_1)$ and if $\beta > 0$ then $D^{k - 1} u \not \in L^\infty (B_1)$.<|endoftext|>
TITLE: Polygons that are hard to guard
QUESTION [8 upvotes]: Given an $n$-vertex polygonal 'art gallery' $P$ in the plane, it is possible to cover the interior of $P$ by placing 'guards' at (at worst) $\lfloor n/3\rfloor$ of the vertices of $P$. That this is sufficiently many can be shown elegantly by triangulating $P$, then $3$-colouring this triangulation and placing guards at the vertices with the least common colour. 
For a lower bound only a single family of examples is needed, and the standard is the $n$-pronged comb (or crown) which has $3n$ vertices and requires one guard for each prong. However, in considering variations on the art gallery problem it can be the case that the comb is easier to guard, and thus other families (which are harder in this new context) are required. So, is there (or can we construct in comments) a big list of 'hard to guard' polygons - that is, $n$-vertex polygons for which $\lfloor n/3 \rfloor$ guards are required - that could be used as starting points for considering variations?

REPLY [12 votes]: I think this will not be a "big-list" (as per your tag), because in some sense all
examples will be variations on the same
theme.  Nevertheless, here is my offering:
     
Let $x$ be a spike tip, and $V(x)$ the set of points that see $x$.
Notice these $V(x)$ regions are disjoint for the 8 spike tips.
Therefore each tip requires its own guard, and so this shape requires $n/3$ guards.
Addendum. Now that I see Gray's motivation, it may not be inappropriate to mention
that these saw-tooth-like polygons achieve the current best lowerbounds on the transmitter problem
he mentions, as I described in "Computational Geometry Column 52" (Fig. 2).<|endoftext|>
TITLE: Integration on the space of symmetric matrices
QUESTION [10 upvotes]: Let $\mu$ be a Lebesgue measure on the space $G$ of real symmetric $n \times n$ matrices (the Haar measure on the additive group of such matrices). For any $A \in G$ let $\chi_{A}(x)$ be its characteristic polynomial:
$$\chi_{A}(-x) = \det (A+xI) =  x^n + a_{1}x^{n-1}+\ldots+a_{n-1}x + a_{n}.$$
Let $g(A) = ( a_1, \ldots, a_n )$ for any $A \in G$.
Consider a function $f \colon \mathbb{R}^n \to \mathbb{R}$ and define an integral
$$I =\int\limits_{H} f(g(A)) \, \mu(dA),$$
where $H$ is some subset of $G$ well characterized by $a_1,\dots,a_n$, for example the set of all positively defined matrices.
My question is: how to reduce the integration with respect to $\mu$ to the integration with respect to the Lebesgue measure on the space of eigenvalues? I did't find an easy way. Maybe I have to use some generalized version of coarea formula to split the integration on the integration with respect to the Haar measure on $O(n)$ plus the integration on the space of eigenvalues?

REPLY [18 votes]: There is  a     Weyl integration  formula that deals with this problem.  $\newcommand{\bR}{\mathbb{R}}$ Denote by  $G_n$   the space of symmetric $n\times n$ real matrices.  It states  that for any  $O(n)$-invariant  $h:G_n\to \bR$ we have
$$ (2\pi)^{-\frac{n(n+1)}{4}} \int_{G_n} h(A) e^{-\frac{{\rm tr} A^2}{4}} dA $$
$$
=\frac{1}{Z_n}\int_{\mathbb{R}^n} h(\lambda_1,\dotsc, \lambda_n) e^{-\frac{1}{4} \sum_{j=1}^n \lambda_j^2} \; {\prod_{j<k}} |\lambda_j-\lambda_k| \; d\lambda_1\cdots d\lambda_n$$
where
$$Z_n =2^{\frac{n}{2}}n! \prod_{j=1}^n \Gamma(\frac{j}{2}).$$
Now, use this formula for
$$ h(A)= f(g(A))\, e^{\frac{{\rm tr} A^2}{4}}. $$
For more details  see Appendix C of  this paper and the references therein.

REPLY [11 votes]: I think you are looking for the Haar measure of $SL(n,\mathbb{R})$ in $K A K$ coordinates. You can find it for example in Knapp's book  "Representation Theory of Semisimple Groups: An Overview Based on Examples".
The short answer: the measure you want is the Haar measure on $O(n) \times O(n)$ times Haar measure on the space of diagonal matrices (viewed as a group under multiplication) times the product of the hyperbolic sines of the positive roots.<|endoftext|>
TITLE: Is $G=\left<b_1, b_2, b_3 | [b_i^p, b_j^p]=1, \forall i,j=1,2,3\right>$ large?
QUESTION [5 upvotes]: Dear all,
I came across the above group (for any fixed odd prime $p$) and would need to know if it's large or not. It seems like it shouldn't be...
Best wishes,
Elisabeth

REPLY [17 votes]: The group $G$ maps onto the free product $C_p*C_p*C_p$ of three cyclic groups of order $p$
(just send each $b_i^p$ to $1$). This free product is virtually free, as a free product of finite groups (by Kurosh theorem the kernel of the homomorphism onto $C_p \times C_p \times C_p$ is free) and is not virtually cyclic. Hence it is large, and so $G$ is large as well. This works for any $p>1$.<|endoftext|>
TITLE: Centralizers  of  elements  in $Sl_{2}(\mathbb{Z})$
QUESTION [5 upvotes]: Conjugacy  classes  Of  elements  of  $SL_{2}(\mathbb{Z})$  can be  characterized  by  its  trace. If  $\mid T(m)\mid=1 $, then it  is  an  element  of  finite  order,  if  $\mid T(m)\mid =2$, then m is  conjugate  to  a  matrix up to  sign,$ (1 n)  (0  1)$ (  row  notation)  where $ n $ a  natural  number.  in other  case, there  is  an  expansion   in words $ML^{n_{i}} MR^{n_{i}}$,   where  ML is  the  matrix $(10)(11)$,  and  MR is the  matrix $(01)(11)$ (Again,  row  notation).  Is  there  a  Classification or  method for  the  determination of the  centralizers  of  these  elements, or  the  Normailizers of  the  subgroups generated  by  them ?

REPLY [3 votes]: A good way to visualize this issue is by looking at the fractional linear action of $SL(2,\bf Z)$ on the upper half plane model of $\bf H^2$; see section 4.2 of Serre's 
"Trees". The kernel of the action is the center of the group, ${\bf Z}/2$ generated by minus the identity matrix. I'll ignore the center henceforth and describe what happens in its quotient $PSL(2,{\bf Z})$. 
The action of $PSL(2,{\bf Z})$ preserves the Farey triangulation (a vertex at each rational point at infinity, an edge connecting $p/q$ to $r/s$ if $ps-rq=\pm 1$). The dual graph of the Farey triangulation is an infinite trivalent tree $T$ embedded in $\bf H^2$. The action of $PSL(2,\bf Z)$ on $T$ is the same as the Bass-Serre tree for the free product decomposition $PSL(2,{\bf Z}) = {\bf Z}/3 * {\bf Z}/2$. An element is elliptic if and only if it fixes a vertex or an edge midpoint (where an edge of $T$ crosses an edge of the Farey triangulation) and its normalizer and centralizer are the stabilizer of that point, ${\bf Z}/3$ for a vertex and ${\bf Z}/2$ for an edge midpoint. An element is parabolic if and only if it fixes a rational point at infinity and its normalizer and centralizer are the stabilizer of that point, isomorphic to ${\bf Z}$; e.g. for the rational point $1/0$ the stabilizer is the cyclic group generated by $(1 ,n)(0 ,1)$ (row notation). 
A hyperbolic element translates along its axis, a line $A$ in $T$. The circular word $ML^{n_1} MR^{n_2} ... ML^{n_{2k-1}} MR^{n_{2k}}$ describes the behavior of a fundamental domain of $A$, making choices of Left and Right at each valence 3 vertex in $T$. The centralizer is the infinite cyclic group generated by the primitive translation of $A$, represented by the primitive cycle of the circular word $ML^{n_1} MR^{n_2} .. ML^{n_{2k-1}} MR^{n_{2k}}$. The normalizer may have one more order 2 generator, and here's how you tell: are the choices made by $A$ going in the forward direction the same, up to cyclic permutation, as the choices made going in the backward direction? To put it another way, reverse the cyclic word and interchange $L$ with $R$, obtaining $ML^{n_{2k}} MR^{n_{2k-1}} ... ML^{n_2} MR^{n_1}$ and ask: is the result a cyclic permutation of the original word? If not, the normalizer is the centralizer; if so, the centralizer has index 2 in the normalizer.<|endoftext|>
TITLE: Questions about the Bernstein center of a $p$-adic reductive group 
QUESTION [24 upvotes]: Dear all, 
The  "Bernstein center" of a $p$-adic reductive group appears frequently in the literature of automorphic forms, often without a precise definition.   For example, in page 233 of Moeglin-Waldspurger's classic "Spectral decomposition and Eisenstein series" , the couple tell us : 
"...in particular the centre of enveloping algebra acts on $\delta$  via a character at the infinite places and the Bernstein centre does so at the finite places... "
So one may guess that it is some analogy of "the centre of enveloping algebra" at fintie places. 
My questions are:

What is the definition of the Bernstein centre of a p-adic reductive group.
What is the original motivation to introduce it ?
What role does it play in the theory of automorphic forms ?
Could you explain these in some concrete example ,say $GL_2$ ?

Please feel free to choose part of the questions to reply.
Any comments and references (in English) will also be very welcome !
Thank you very much in advance!

REPLY [2 votes]: Ralf Meyer develops part of the theory from scratch for locally compact groups via projective systems of Lie groups in section 5 of http://arxiv.org/abs/math/0310094
This theory speciales to the Bernstein center in the totally disconnected case and to the theory of the center of the universal envelopping algebra for complex Lie groups with trivial center. Also he relates the center to distributions.
He works on bornological vector-spaces, but states in the introduction that the theory includes smooth representation on Frechet-spaces. So the notion bornological and topological can be interchanged with no harm in this case. (I assume that many of us are more comfy with topological vectorspaces than with bornological ones. At least I am.)<|endoftext|>
TITLE: Making a non-unital algebra the unique maximal one-sided ideal in a unital algebra
QUESTION [7 upvotes]: If $R$ is an algebra without a unit, then the standard unitisation $R^\sharp$ can have maximal one-sided ideals other than $R$. Thus, it is natural to ask about the following. Let $R$ be an algebra without a unit (over a field with char 0 if it does matter).
Is there a unital algebra $A$ such that $R$ is the unique maximal right ideal in $A$?
EDIT: Assume $R$ has a faithful representation $\pi$ on a vector space $V$ (over $K$), what if we consider $A=\pi(R)+KI$?

REPLY [5 votes]: The nice examples given by Pace show that the answer is negative in general.  There is also a natural condition on $R$ which gives a positive solution.  It involves the Jacobson radical of a non-unital ring.  (I'm following the first few exercises in section 4 of Lam's A First Course in Noncommutative Rings.)
The binary operation $a \circ b = a + b -ab$ on $R$ is associative and has unit element $0$.  An element of $R$ is called left (right) quasi-regular if it has a left (right) inverse in the monoid $(R, \circ, 0)$ and is called quasi-regular if it has both a left and a right inverse, which are necessarily equal.
The Jacobson radical $J(R)$ is the sum of all left ideals of $R$ consisting of left quasi-regular elements.  One can show that this is an ideal of $R$, and all elements in it are quasi-regular.  So $J(R)$ is the largest left (right) ideal consisting of left (right) quasi-regular elements.  One can show that this coincides with the usual Jacobson radical in case $R$ has an identity.

Claim: For a (not necessarily unital) ring $R$, the following are equivalent:

$R$ is the unique maximal right (or left) ideal of its standard unitization $R^\sharp$;
$R = J(R)$.


Proof: Assume 1 holds, so that $R^\sharp$ is local and $R = J(R^\sharp)$.  Then for any $x \in R$, the element $1 - x = 1 \oplus (-x) \in K \oplus R = R^\sharp$ lies in $R^\sharp \setminus R$ and therefore is invertible.  Its inverse is of the form $\alpha \oplus y \in R^\sharp$.  Writing out the equation for $(1-x)(\alpha \oplus y) = 1 \oplus 0$ yields that $\alpha = 1$ and $x + y - xy = 0$, and similarly we obtain $y + x -yx = 0$.  So every element of $R$ is quasi-regular and 2 holds.
Conversely, assume 2 holds.  We know that $R$ is a maximal right (and maximal left) ideal of $R^\sharp$ simply because of the ring isomorphism $R/R^\sharp \cong K$; thus $J(R^\sharp) \subseteq R$.  If we can show that $R$ is contained in $J(R^\sharp)$ then we will be done, for then $R \subseteq J(R^\sharp) \subseteq R$ yields that $R^\sharp/J(R^\sharp) = R^\sharp/R \cong K$ making $R$ local with unique maximal right ideal $R$.  
So let $x \in R$. To show $0 \oplus x \in J(R^\sharp)$, let $\alpha \oplus r \in R$; we need to find a left inverse to $$1 \oplus 0 - (\alpha \oplus r)(0 \oplus x) = 1 \oplus-(\alpha x + rx) = 1 \oplus (-x')$$ where we set $x' = \alpha x + rx \in R$. To this end, fix a left quasi-inverse $y$ for $x$ in $R$.  Then from $y+x'-yx' = 0$ we obtain $$(1 \oplus -y)(1 \oplus -x') = 1 \oplus 0$$ as desired.  QED<|endoftext|>
TITLE: A question on deformations of Theta divisor in the Jacobian of a complex curve
QUESTION [12 upvotes]: Suppose $C_g$ is a smooth compact complex curve (of genus $g$), and let $J$ be its Jacobian.  Recall that the Jacobian $J$ of a curve $C_g$ is a complex torus that can by obtained by contractions of all rational curves on the $g$-th symmetric power of $C_g$, e.i.,  $Sym^g(C_g)$. Recall also that there is a theta divisor $\Theta$ in $J$, 
depending on a point $p\in C_g$.
The divisor $\Theta$ is the image in $J$ of the set of points $(p,p_1,...,p_{g-1})$ with $p$ fixed.
Question. How to calculate the dimension of the set of divisors on $J$ linearly equivalent to $\Theta$? In other words, what is $dim( H^0(J,\cal O(\Theta)))$?

REPLY [8 votes]: This can also be done in a purely algebraic fashion. Take the given description of $\Theta$ as the locus of effective classes on $X=Jac^{g-1}C$. Put $Y=Jac^gC$. Pick a point $D$ on $Y$ with $h^0(C,D)=1$; Abel's theorem (that $C^{(g)}\to Y$ is birational) ensures that this holds for all $D$ outside some locus of codimension at least $2$. There is an embedding $i_D:C\to X$ given by $i_D(x)=K_C-D+x$. Then $i_D(C)\cap\Theta$ is the set of $x$ in $C$ such that $K-D+x$ is effective; by Serre duality, this is equivalent to $h^0(C,D-x)>0$. But $h^0(C,D)=1$, so the only such points $x$ are the points in the unique effective divisor $D_1$ in the linear system $|D|$, so $i_D(C)\cap\Theta$ is exactly the divisor $D_1$, regarded as a subscheme of $C$. So if $h^0(X,\Theta)=r+1$, then there is an $r$-dimensional subspace of sections in $H^0(X,\Theta)$ that vanish along $i_D(C)$.
We aim to prove that $r=0$. Consider the incidence scheme $W\subset Y\times |\Theta|$ consisting of pairs $(D,\Phi)$ with $i_D(C)\subset\Phi$. The fibers of $pr_1:W\to Y$ have dimension at least $r-1$, so $\dim W\ge g+r-1$, and so the fiber $pr_2^{-1}(\Theta)$ has dimension at least $g-1$. By Abel's theorem, as before, there is a point $E$ in $pr_2^{-1}(\Theta)$ with $h^0(C,E)=1$. Then $i_E(C)$ does not lie in $\Theta$, by the previous argument, contradiction.
You can then prove that $\Theta$ is ample, by showing that it is non-degenerate: the set of points $a$ on $A=Jac^0C$ such that $t_a^*\Theta$ is linearly equivalent to $\Theta$ is trivial. On any torsor under an abelian variety a non-degenerate line bundle with non-vanishing $H^0$ is ample (I'm going to give a blanket reference to Mumford at this point).
Corollary: $\Theta^g=g!$ (from Riemann-Roch on $X$).<|endoftext|>
TITLE: Nilpotency degree of the augmentation ideal 
QUESTION [7 upvotes]: Let $G$ be a finite $p$-group, $k$ a field of characteristic $p$ and let $I(G)=(g-1\mid g \in G)$ be the augmentation ideal of the group ring $k[G]$. It's known that $I(G)$ is nilpotent, i.e. there is $n> 0$ such that $I(G)^n=0$. Call the least such $n$ the nilpotency degree of $I(G)$ and denote it by $\operatorname{nildeg}I(G)$. 
Question 1: Are there known upper and lower bounds for $\operatorname{nildeg}I(G)$ ? 
Question 2: Does the invariant $\operatorname{nildeg}I(G)$ have a particular name in the literature ? 

An inspection of the proof that $I(G)$ is nilpotent can be used to detemine upper bounds for $\operatorname{nildeg}I(G)$: Let $C \le G$ be central. Then 
$$\frac{k[G]}{I(C)k[G]} \cong k[G/C]\;,\qquad \frac{I(G)}{I(C)k[G]} \cong I(G/C).$$ 
By taking $C=\mathbb{Z}/p$ and iterating, one obtains $$\operatorname{nildeg}I(G)\le |G|.$$
By taking $C=Z(G)$ this can be futher refined: If $1 = Z_0 \le Z_1 \le ... \le Z_c = G$ is the upper central series of $G$ and $Z_i/Z_{i-1}=\prod_{j=1}^{r_i}\mathbb{Z}/p^{e_{ij}}$ then 
$$\operatorname{nildeg}I(G) \le \prod_{i=1}^c\;\big((e_{i,1}-1) + \cdots +  (e_{i,r_i}-1)+1\big)$$
Since $(g-1)^{\operatorname{nildeg}I(G)}=0$ for each $g \in G$, a trivial lower bound is 
$$\operatorname{nildeg}I(G) \ge \operatorname{exp}(G)/p$$
Hence a more acurate formulation for quest 1 is: 
Question 3: Are there better bounds than these or bounds that use other invariants of $G$ ? 

Edit: Apart from the exact formula given by Jennings' theorem as described in mt's answer, I found the lower bound
$$\operatorname{nildeg}I(G) \ge m(p-1)+1$$
if $|G|=p^m$ in the book "Karpilovsky: The Jacobson Radical of Group Algebras".

REPLY [5 votes]: Let $G$ be a finite $p$-group, $k$ a field of characteristic $p$, and define a series $\Gamma_i$ of
subgroups of $G$ by letting $\Gamma_1 = G$ and
$$ \Gamma_{i+1} = \langle [ \Gamma_i,G ], \Gamma ^p _{\lceil
(i+1)/p \rceil} \rangle .$$  Then $\Gamma_i / \Gamma_{i+1}$ is
elementary abelian, so we can fix elements $f_{i1}, \ldots, f_{id_i}$ of $G$ whose images in 
$\Gamma_i/\Gamma_{i+1}$ form a basis.  Consider all products of the form
$$ \prod_{i,j} (f_{ij}-1)^{\alpha_{ij}} \in kG
\qquad  (1) $$ where the product is
taken in lexicographic order and $0 \leqslant \alpha_{ij} \leqslant p-1$.
Define the weight of such a product to be $\sum_{i,j} i\alpha_{ij}$.
Jennings' Theorem says that if $J=I(G)=\operatorname{rad}(kG)$  then the set of
products (1) of weight at least $s$ form a basis
of $J^s$, and a basis for $J^s/J^{s+1}$ is given by the images of
the products of weight exactly $s$.  
In particular, the largest non-zero power of the radical is 
$$ \sum  i (p-1) \dim_{\mathbb{F}_p} (\Gamma_i/\Gamma_{i+1}) $$<|endoftext|>
TITLE: motivating examples of family of Hodge structure
QUESTION [5 upvotes]: Let $\phi: \chi \to B$ be a proper holomophic submersion with smooth fiber $X$.
Then, we get local systems $R^i \phi_* \mathbb C$ and corresponding flat connection $(B, \bigtriangledown)$
In this case,there are tons of beautiful constructions even in such an elementary level: 
infinitesimal VHS, Mixed Hodge structure, Period mapping $P^{n,k} : B \to Grass(b^{n,k}, H^n(X, \mathbb C)), $ Picard-Lefschetz monosromy representation $\rho: \pi_1(B, b_0) \to GL(n, \mathbb C) $ and so on.
But my knowledge of these topic remains too abstract to digest it well. So i am collecting enlightening toy examples. For example, I've worked with the Legendre family of elliptic curves
{$y^2=x(x-1)(x-\lambda)$} $ \to $ {$\mathbb C -(0,1)$}
and interpreted everything into a concrete term.(and it was fantastic)
But i still wants more. Because in my examples, no mixed Hodge structure, no Hodge structure of weight $\ge$ 2. If you have any other good examples, please tell me. Good reference will be extremly helpful. I also appreciate any suggestion.

REPLY [3 votes]: I guess the implied question is: what are good references containing  explicit calculations of variations of Hodge structure etc.? I might suggest taking a look at Griffiths' early pioneering papers "On periods of certain rational integrals I, II" Annals 1969, and
"Periods of integrals on algebraic manifolds III" IHES 1970. These papers contain a large number of explicit calculations on VHS and intermediate Jacobians for things like hypersurfaces in projective space. Regarding 
(variations of) mixed Hodge structures, take a look at the books by Carlson-Müller Stach-Peters, Peters-Steenbrink, and Voisin.<|endoftext|>
TITLE: Physical interpretation of Robin boundary conditions
QUESTION [19 upvotes]: In a (bounded) domain $\Omega \subset \mathbb{R}^n$, if we're studying the Laplace equation or heat equation or such PDE's we can impose the Dirichlet
$u|_{\partial\Omega} \equiv 0$,
Neumann 
$D_{\nu} u|_{\partial\Omega}\equiv 0$
or Robin (for $\alpha \in \mathbb{R}$) 
$(D_{\nu} u + \alpha u)|_{\partial \Omega} \equiv 0$.
I know that, for example for the heat equation, Dirichlet eigenvalues correspond physically to the boundary being in contact with a (large) heat bath at $T=0$. Or, in the Laplace equation, if we're interested in the modes supported by $\Omega$ (as a drum), Dirichlet boundary conditions can be thought of keeping the boundary from moving.
Neumann boundary conditions, for the heat flow, correspond to a perfectly insulated boundary. For the Laplace equation and drum modes, I think this corresponds to allowing the boundary to flap up and down, but not move otherwise.

My question is: what sort of physical interpretations are there for the Robin boundary conditions? Wikipedia says that they are related to electromagnetic problems, but gives no details. I'd be happy with answers that are not necessarily physics-related, for example, if there was somewhere that Robin boundary conditions naturally arise in a mathematical context, I'd be interested to know about that as well.

REPLY [2 votes]: Robbins conditions correspond to immersing one end of the rod in a constant temperature bath. The temperature bath does not have to match the initial temperature at the end of the rod.  The immersed end of rod may come into thermal equilibrium with the bath, or a constant temperature difference may come about (to ensure a constant heat flux). 
In thermodynamics, reference is always made to work reservoirs (ideal mechanical energy sources that can do work reversibly without any entropy increase) and reference is made to heat reservoirs that can transfer heat without any change in temperature (of the reservoir).  Such a heat reservoir can be attained by making it a water/ice mixture - so that the temperature of the bath is maintained due to the phase change occurring.  
The Robbins condition itself can be interpreted as a statement of Newton's law of heating (cooling): the rate of heat transfer is directly proportional to the difference in temperature (between the hot and cold object).
dQ/dt ̇  =  h(To - T)   Newton’s law of heat transfer
Here dQ/dt represents the rate of heat transfer between the end of rod and the temperature bath.
The constant h is a thermal boundary layer coefficient whose reciprocal represents IMPEDANCE to the transfer of heat.
To is the temperature of the bath.
T is the temperature of the end of the rod.
We can get a Robbins condition of this by invoking Fourier's first law of heat conduction:  
q = -k dT/dx
Here q is the heat flux and equals (1/A)(dQ/dt) where dQ/dt is the rate of heat transfer and A is the cross sectional area of the solid rod.
Combining Newton's law of heat transfer with Fourier's first law of conduction yields
-kA dT/dx  =  h(To - T)  right end immersed
kA dT/dx = h(To - T)  left end immersed 
An equivalent form is 
dT/dx - aT = b
or
dT/dx + aT = b
where a and b are constant dependent on k, A, h and To.
The concept of impedance is hidden in a Robbins condition  because Newton's law of heating (cooling) is hidden.  The Robbins condition comes about from applying Newton's law.  
Newton's law of cooling is an impedance equation and the reciprocal of h (the thermal boundary layer coefficient) acts like a resistance.  A good analogy is to compare the rate of heat transfer to current, the temperature difference to voltage, and the thermal boundary layer to a resistor.<|endoftext|>
TITLE: Isomorphisms between non-standard reals.
QUESTION [5 upvotes]: Let $U$ and $V$ be two non-principal ultrafilters over N, and $R_1$ and $R_2$ the non-standard extensions of R given by $R_1=R^N/U$ and $R_2=R^N/V$. Are they always isomorphic (I think not, but could not prove it), and, if not, what axioms must be added to ZFC to ensure they are (or is it a consequence of AC that they are not ?)

REPLY [9 votes]: If the continuum hypothesis holds, then both of these ultrapowers are saturated models of cardinality $\omega_1$, and one can see that they are isomorphic by a back-and-forth argument. 
When the CH fails, then they need not be isomorphic.<|endoftext|>
TITLE: de Rham vs Dolbeault Cohomology
QUESTION [27 upvotes]: For a complex manifold $M$, one can consider (A) its de Rham cohomology, or (B) its Dolbeault cohomology. I'm looking for some motivation as to why one would bother introducing Dolbeault cohomology. To be more specific, here are some straight questions.

What can Dolbeault tell us that de Rham can't?
Does there exist some simple relationship between these two cohomologies? 
When are they equal? 
Do things become simpler for the Kahler case?
What happens for the  projective spaces?
Why does nobody talk about the holomorphic cohomology?

REPLY [23 votes]: Although the main questions have been answered quite well, I would like to say a few words
about the first question "why would one bother...". 
De Rham and Dolbeault cohomology are measuring different things. The first, which measures the failure of the necessary condition $d\alpha=0$ to guarantee a potential $\alpha=d\beta$, is a topological invariant.
The second gives the obstruction to solving the similar problem for the Cauchy-Riemann operator; it measures the holomorphic complexity  and it has a priori nothing to do with the topology. That they turn out to be the "same" in good (e.g. compact Kähler) cases
is sort of a miracle. Even in such cases, these spaces are not literally the same,
and this can be exploited in interesting ways.
There is an isomorphism which can be represented by a matrix called a period matrix,
which is sensitive to the complex structure. To see how this works in the simplest interesting case, let $X= \mathbb{C}/(\mathbb{Z}+\mathbb{Z}\tau)$ be an elliptic curve. As a topological space it is just a torus, and independent of the choice of $\tau$,  but as a Riemann surface it is sensitive to this choice. The natural basis of the first Dolbeault cohomology $H^{10}(X)\oplus H^{01}(X)$is
$\lbrace dz,d\bar z\rbrace$. This maps to $\lbrace (1,\tau), (1,\bar \tau)\rbrace$ under the basis of $H^1_{DR}(X)$ dual to loops given by projecting  $[0,1],[0,\tau]\subset \mathbb{C}$  to $X$. Thus  $X$ can be recovered from its period matrix.<|endoftext|>
TITLE: What is the usual topology of $C^\infty_c(M) $
QUESTION [5 upvotes]: If $M$ is a smooth paracompact manifold, then what is the usual topology of $C^\infty_c(M) $, i.e., the smooth function with compact support?

REPLY [23 votes]: Topologizing $C_c^\infty(M)\subseteq C^\infty(M)$ with the subspace topology (where $C^\infty(M)$ has the Whitney topology, generated by the seminorms $\left|\sup_K\frac\partial{\partial x^\alpha}f\right|$), makes it a dense subspace; in particular it is not itself complete.  So I wouldn't really call this the "usual topology" on $C_c^\infty(M)$.  (it would be sort of like saying the usual topology on $C(M)$ is given by the $L^2$ norm).
To me the usual topology is the inductive limit topology $C_c^\infty(M)=\lim_{K\subseteq M}C_c^\infty(K)$ (which Mariano calls the colimit topology).  This topology is not metrizable when $M$ is noncompact (since it's not even first-countable), but is "nicer" in the sense that it gives a well-understood dual space, namely the space of distributions on $M$.
In comparison, the dual space of $C^\infty(M)$ with the Whitney topology is the space of compactly supported distributions on $M$.<|endoftext|>
TITLE: Open subgroups of free profinite groups
QUESTION [5 upvotes]: The following questions popped out while I was preparing a course on profinite groups. 
Closed subgroups of free profinite groups are not necessarily profinite free (e.g. the p-sylow subgroups, or the kernel of the map on the maximum p quotient, and many more other examples) thus the Nielsen-Schreier theorem fails in the profinite category.
Nevertheless, Nielsen-Schreier theorem carries over for open subgroups. The proofs I found (in Field Arithmetic by Fried-Jarden and in Profinite Groups by Ribes-Zalesskii) use the construction of free profinite groups as restricted completion of free abstract groups AND the Schreier basis of a finite indexed subgroup of a free abstract group. By restricted I mean that if X is a basis of a free abstract group, then the completion is w.r.t. the family of finite index normal subgroups that contain all but finitely many elements of X.

First question: can one avoid the use of the Schreier basis in proving that an open subgroup of free profinite is free profinite?

Note that in the finitely generated case the restricted completion is the same as the profinite completion, thus one does not need to use the Schreier basis in this case. Therefore it suffices to affirmatively answer the following. 

Second question: is N-S for open subgroups of finitely generated free profinite groups implies N-S for open subgroups of non-finitely generated free profinite groups?

I apologize that the question became a bit lengthy...

REPLY [7 votes]: Luis Ribes and I gave a proof of Nielsen-Schreier for open subgroups of free profinite groups that avoids using completions and the discrete Nielsen-Schreier theorem (well, actually our proof does both at once). We use wreath products instead. But we do use the Schreier basis. The ArXiv version is http://arxiv.org/pdf/0812.0027 and the final version is in l'enseignement mathématique.
Edit I think one could avoid the Schreier basis with our technique by using embedding problems like we do for quasi-free.<|endoftext|>
TITLE: Equivalent definitions of arithmetically Cohen-Macaulay varieties
QUESTION [7 upvotes]: Let $X\subset \mathbb{P}^n$ be a projective algebraic variety with coordinate ring $R$. 
$X$ is said to be arithmetically Cohen-Macaulay if $R$ is a Cohen-Macauly ring. A equivalent definition is that the natural morphism 
$$R\to \oplus_{k\in\mathbb{Z}}\mathcal{O}_X(k)$$ is bijictive and $H^i(X,\mathcal{O}_X(k))=0$ for all $k\in\mathbb{Z}$ and $1\leq i\leq \text{dim} X-1$. Does anyone know a proof for the equivalence? Without assuming that $X$ is Cohen-Macaulay, I can only prove that $R$ is Cohen-Macauly at $0$, the vertex of the affine cone. How to prove $R$ or the section ring is Cohen-Macauly at other points of the affine cone? Did I miss something from the idea? Or is there an complete alternative proof?

REPLY [9 votes]: Here's a proof.  Let me assume that $X$ is not zero dimensional, but instead is equidimensional.  I'll actually prove the whole thing (more than you want), but I'll prove the Cohen-Macaulay thing first.
Now, either condition implies that $X$ itself is Cohen-Macaulay, let me explain this.
Set $S$ to denote the section ring 
$$
\bigoplus_{k \in \mathbb{Z}} O_X(k)
$$
and use $m = S_+$ to denote the irrelevant ideal at the origin. 

In the first condition, we notice that the punctured spectrum of $\text{Spec} S$ is an $\mathbb{A}^1$-bundle over $X$.  Thus if $X$ is not Cohen-Macaulay, neither is $\text{Spec} S$. 
In the second condition, choose $k \gg 0$.  Then $0 = H^i(X, O_X(-k))$ is Grothendieck-dual to $\mathbb{H}^{-i}(X, \omega_X^{\bullet}(k)) \cong H^0(X, {\bf h}^{-i} \omega_X^{\bullet}(k))$ which is non-zero for some $i < \dim X$ if $X$ is not Cohen-Macaulay (for the isomorphism, we used a spectral sequence). Indeed an equidimensional variety $X$ is Cohen-Macaulay if and only if ${\bf h}^{-i} \omega_X^{\bullet} = 0$ for $0 < i < \dim X$ (here $\omega_X^{\bullet}$ is the dualizing complex).

On the other hand the fact that $X$ is Cohen-Macaulay implies that $S$ is also Cohen-Macaulay except possibly at the origin (again use the $\mathbb{A}^1$-bundle description).
Certainly the map $R \to S$ is bijective in high degree and so birational.   Then $\text{Spec} S \to \text{Spec} R$ is an isomorphism everywhere except possibly at the origin.  Thus $R$ is also Cohen-Macaulay except possibly at the origin.
It sounds like you already know what follows:
Suppose now that $R \subsetneq S$.  But then if $U \subseteq \text{Spec}R$ is the punctured spectrum, we have that $H^0(U, O_{Spec R}) = H^0(U, O_{\text{Spec} S}) \supseteq S$.  In particular, $H^0(U, O_{\text{Spec} R}) \neq R$ and so $R$ cannot be S2 because $H^1_m(S) \neq 0$.  This follows from the exact sequence:
$$0 \to R \to H^0(U, O_{\text{Spec} R}) \to H^1_m(S) \to 0.$$   In particular, this shows that the equality $R = S$ is necessary.  
Conversely suppose that $R = S$.  Additionally since $X$ is Cohen-Macaulay, it is S2, then the section ring $S$ is also S2 away from the origin by the $\mathbb{A}^1$-bundle thing again.  Furthermore, its trivial to see that $S = H^0(U, O_{\text{Spec} S})$ and so the depth of $S$ is at least 2 at the origin.  This implies that $R = S$ is S2.  Thus the isomorphism theorem also gives us some vanishing of local cohomology.  In particular, at least the $H^1_m(S) = 0$ by the above exact sequence.
Now, for the next part:
It follows from straightforward computations with Cech cohomology that 
$$
H^i(X, O_X(k)) = [H^{i+1}_{m}(S)]_k
$$
for $i > 0$ (here $[\bullet]_k$ means the $k$th graded piece of the module).  Thus the vanishing over the other $H^i(X, O_X(k))$ implies the vanishing in all remaining degrees of $H^{i+1}_m(S)$.  This completes the proof.
This is described in somewhat more detail in THIS paper by Karen Smith.<|endoftext|>
TITLE: How to solve the system of recurrence equations
QUESTION [5 upvotes]: The system has  the form
$$
(n-2)f_n^{(1)}=n(f_{n-1}^{(1)}+1),
$$
$$
(n-2\cdot 2) f_n^{(2)}=n(f_{n-1}^{(2)}+f_{n-1}^{(1)}),
$$
$$
\ldots 
$$
$$
(n-2k)f_n^{(k)}=n(f_{n-1}^{(k)}+f_{n-1}^{(k-1)}),
$$
for  the unknown sequences $f_n^{(1)},f_n^{(2)},\ldots,f_n^{(k)}$  with the initial conditions $f^{(i)}_k=0,$  for all $k=0,1,\ldots,2i.$
By direct calculation I have  got 
$$
f_n^{(1)}=n(n-2),
$$
$$
f_n^{(2)}=\frac{1}{2!} (n-4) { n \choose 2} (3n-7),
$$
$$
f^{(3)}_n=\frac{1}{3!} (n-6) { n \choose 3}   \left( 19{n}^{2}-141n+254 \right),
$$
$$
f^{(4)}_n=\frac{1}{4!} (n-8) { n \choose 4}  \left( 
211{n}^{3}-3258{n}^{2}+16481n-27306 \right),
$$
$$
f^{(5)}_n=\frac{1}{5!} (n-10) { n \choose 5}  \left( 3651{n}^{4}-96550{n}^{3}+946185{n}^{2}-4071950n+
6492024 \right) 
$$
Question.  What is a general expression for $f_n^{(i)}$?
The   ordinary generating function for the above sequences  has  the form
$$
G(f_n^{(1)},z)={\frac {{z}^{3} \left( 3-z \right) }{ \left( 1-z \right) ^{3}}}=3{z}^{3}+8{z}^{4}+15{z}^{5}+\cdots
$$
$$
G(f_n^{(2)},z)={\frac {z^5(-3{z}^{3}+16{z}^{2}-35z+40)}{ \left( 1-z \right) ^{5}}},
$$
$$
G(f_n^{(3)},z)={\frac {{z}^{7} \left( -40{z}^{5}+288{z}^{4}-897{z}^{3}+1575{z}
^{2}-1701z+1155 \right) }{ \left( 1-z \right) ^{7}}}
$$

REPLY [4 votes]: This is not a full solution, but it reduces the system to another recurrence relation which involves only coefficients. Let
$$f_n^{(k)}=a_0^k n(n-1)...(n-k+1)+a_1^k n(n-1)...(n-k)+...+a_k^k n(n-1)...(n-2k+1).$$
Then the above system reduces to
$$a_i^k={a_i^{k-1}\over i-k}$$
for $i < k$ and $a_k^k$ is determined by the condition that $f_{2k}^{(k)}=0$. Clearly the determination of $a_k^k$ is the difficult part. But if the sequence 3,19 etc. shows up
elsewhere, that should be suggestive.
Addendum: If all the $a_i^k$ are expressed in terms of $a_i^i$, then the recurrence relation for the
$a_k^k$ becomes the same as for the Taylor coefficients of the reciprocal Bessel function mentioned in Barry Cipra's link.<|endoftext|>
TITLE: Vopenka's Theorem on L(A) and Large Cardinals from Weaker Assumptions?
QUESTION [6 upvotes]: I know that sometime ago Vopenka proved this:
Theorem: Assume there is a strongly compact cardinal. Then for any set $A$, $V \neq L(A)$.
Can we get by with a consistency-wise strictly weaker assumption?
Namely, call an uncountable cardinal $\kappa$ strong if for any ordinal $\gamma$, there is an elementary embedding $j: V \rightarrow M$ into a transitive $M$ such that $\kappa = crit(j) > \gamma$ and $V_{\gamma} \subset M$. That is, a rank initial segment of $V$ is contained in $M$.
Could someone assess whether a strong cardinal is enough? The argument would be:
Assume for contradiction that $V = L(A)$, for some set $A$. Assume further there is a strong cardinal, let $\kappa$ be the least. Let $\lambda$ be bigger than the rank of $A$, and let $j: V \rightarrow M$ with critical point $\kappa$ and $V_\lambda \subset M$. Since $\lambda$ was big enough, then $A \in M$. Since $M$ is an inner model, the transitive closure of $A$ is in $M$ also. Now we prove by induction that $L(A) \subset M$: $L_0(A) = tc(A) \in M$. Successors and limits (I think) both work, for successors, since transitive models are correct about calculating definable powersets. But we assumed $V = L(A)$, so $M = L(A) = V$. So we have an elementary $j: V \rightarrow V$, which is impossible by Kunen's Theorem.
Any fine points to be careful about here? Thanks!

REPLY [6 votes]: The way to think about it is this. 
Whenever a property in $V$ is witnessed in an absolute manner by the existence of a single set with certain properties, then we will be able to take that set, $A$, and form the universe $L(A)$, which will have the witness and thus verify that the property is true. 
For example, it is a fun exercise to prove that a set-theoretic assertion $\varphi$ is equivalent to a $\Sigma_2$ assertion (in the Levy hierarchy) if and only if it is equivalent to an assertion of the form "$\exists\gamma H_\gamma\models\phi$," for any complexity $\phi$, or similarly to an assertion of the form "$\exists \alpha V_\alpha\models\phi$." Such assertions are absolute to the corresponding $L(H_\gamma)$ and thus compatible with $V=L(A)$. (If you want ZFC, then one should adjoint not just the set, but also a well-ordering of that set.)
Many large cardinal properties are witnessed by the existence of single sets in this way, and thus are $\Sigma_2$ definable. For example, you can tell if $\kappa$ is measurable by looking at $V_{\kappa+2}$. Similarly, you can tell if $\kappa$ is superstrong or almost huge or huge by looking at the appropriate $V_\gamma$, which is large enough to see the extender giving rise to the embedding. Basically, a large cardinal property that is witnessed by a single extender or ultrafilter will be of this type, and thus compatible with $V=L(A)$. 
In contrast, large cardinal properties like strongness, strong compactness or supercompactness are not witnessed by a single embedding or extender, and have complexity $\Pi_3$. You have to say: for every $\theta$ there is a normal fine measure on $P_\kappa\theta$, and this is witnessed by a proper class sequence of measures rather than a single set. From this perspective, what your and Vopenka's arguments amount to is showing that yes, strongness and strong compactness are not $\Sigma_2$ properties, but do require complexity $\Pi_3$. 
Meanwhile, to address the point raised in the comments, one thing is that it is easy to have large cardinal properties with very high strength that are compatible with V=L. For example, the large cardinal axiom "there is a transitive model of ZFC in which there is a proper class of supercompact  cardinals" is true in $V$ if and only if it is true in $L$. This is because there is a such a model if and only if there is a countable model of that theory, and the assertion that there is a real coding a well-founded model of any particular theory is a $\Sigma^1_2$ assertion, which by the Shoenfield theorem is absolute between $V$ and $L$. So if one replaces the actual existence of the large cardinal properties with the assertion that they hold merely inside a transitive model, then one can retain $V=L$ and have the desired large cardinal strength.<|endoftext|>
TITLE: Elliptic Curves over Rings?
QUESTION [15 upvotes]: So an elliptic curve $E$ over a field $K$ is a smooth projective nonsingular curve of genus $1$ together with a point $O \in E$. 
I was reading Silverman's "Arithmetic of Elliptic Curves" and it seems that most of its treatment is over fields.
My question is, does it make sense to define an elliptic curve over a ring (eg: a noncommutative ring)? If not, why not (where would the "construction" fail)? Is it simply not an object of much interest?
Edit: Apparently the question of elliptic curves over noncommutative rings is considered to some extent in this.
http://user.math.uzh.ch/fontein/diplom-fontein.pdf

REPLY [15 votes]: Elliptic curves can be defined over arbitrary base schemes $S$. In particular, for every (commutative!) ring $R$ one can talk about elliptic curves over (the spectrum of) $R$. Loosely speaking, what one gets is a family $E$ of elliptic curves parametrized by the points of $S$. One then proves the existence of the group law ($E$ can be given the structure of an $S$-group scheme), and goes from there. E.g., locally over $S$, $E$ can be put into Weierstrass form.
In the book Arithmetic Moduli of Elliptic Curves by Katz and Mazur, an elliptic curve over $S$ is defined as a proper smooth morphism $f : E \rightarrow S$ of finite presentation, with a section $0 : S \rightarrow E$, such that all geometric fibers of $f$ are integral (equivalently, connected) curves of genus one.
What can be done for noncommutative $R$ I don't know. It seems to me that you have to say what you mean by an elliptic curve over a noncommutative ring. One can't simply replace 'field' in 'elliptic curve over a field' by the name of some other algebraic structure and expect it to make sense, I guess.<|endoftext|>
TITLE: An Extender is a Generalization of an Ultrafilter?
QUESTION [7 upvotes]: I'm not sure whether this belongs here or on math stackexchange, but I'll give a try here. 
I've heard that an extender is a generalization of an ultrafilter. This is not to say that an ultrafilter is the same sort of object as an extender, since whereas an ultrafilter on an uncountable $\kappa$ is in $V_{\kappa + 2}$, an extender in general is an element of a much higher level $V_\alpha$. But that an extender captures all the information an ultrafilter has about elementary embeddings. Or so I think. 
What I am trying to prove is a precise statement of this:
Lemma: given a non-principal $\kappa$-complete ultrafilter on an uncountable $\kappa$, and $j: V \rightarrow M \cong Ult(V,U)$, there is an extender $E$ such that $j_E : V \rightarrow M$ and also $j_E = j$. ($j_E$ is the embedding built from $E$, that is, $j_E$ elementarily maps $V$ into the direct limit of a direct system, one whose elements are ultraproducts of $V$, and whose maps are appropriately defined to commute with the maps from $V$ to these ultraproducts.)
I've tried to take the extender $F$ of a short length such that $F_a = U$. Then all the maps of the direct system are the identity, and the direct limit will be our original $M$, and $j_E = j$. But I'm not convinced this is a genuine extender in that is satisfies one of the (all equivalent?) definitions of extenders. 
Another approach is to take $j_U$, and take the embedding $E_{j_U}$ derived from it, and then try to show $j_{E_{j_U}} = j$. This will in fact be an extender, but I can't show whether the target model $M_E$ agrees with $M$. 
Any ideas or suggestions?

REPLY [4 votes]: I know this question is four years old, but if anyone is stumbling across this question, this might be something worth noting. Maybe it has a way simpler proof, but this should work.

Proposition. A measure is equivalent to an extender with a single generator, in that their ultrapowers are isomorphic.

Proof. Let $U$ be a measure on some $\kappa$ and $E$ a $(\kappa,\lambda)$-extender with a single generator. Then $\text{Ult}(\mathcal M,E)$ is the direct limit of the cross-section ultrapowers $\text{Ult}(\mathcal M,E_a)$, so it suffices to show that if $a\in[\lambda]^{<\omega}$ with $a\not\subseteq\kappa$ and $\xi\in a-\kappa$ then $\text{Ult}(\mathcal M,E_{\{\xi\}})\cong\text{Ult}(\mathcal M,E_a)$. But as $E$ only has a single generator, $\kappa$, we can assume that $a\in[\kappa+1]^{<\omega}$, so that $\xi=\kappa$.
Define the map $\varphi$ as taking $[f]_{E_{\kappa}}$ to $[f^{\{\kappa\},a}]_{E_a}$, which is a well-defined homomorphism as $\kappa\in a$. To show that $\varphi$ is injective, assume $f^{\{\kappa\},a}\sim_{E_a} g^{\{\kappa\},a}$, so that $j(f^{\{\kappa\},a})(a)=j(g^{\{\kappa\},a})(a)$. But by definition of $f^{\{\kappa\},a}$, $j(f^{\{\kappa\},a})(a)=j(f)(\{\kappa\})$ and likewise $j(g^{\{\kappa\},a})(a)=j(g)(\{\kappa\})$, so $f\sim_{E_{\{\kappa\}}}g$.
For surjectivity, let $[f]_{E_a}$ be given and define $\tilde f:[\kappa]^{|a|}\to\mathcal M$ as $\tilde f(u):=f(\{a_0,\dots,a_{|a|-2},u_{|a|-1}\})$ with $u\in[\kappa]^{|a|}$. Then $\tilde f$ is clearly in the image of $\varphi$, so we need to show that $f\sim_{E_a}\tilde f$, i.e. $j(f)(a)=j(\tilde f)(a)$. But since $a_i<\kappa$ for $i<|a|-1$ we have $j(a_i)=a_i$, so $j(\tilde f)(u)=j(f)(\{a_0,\dots,a_{|a|-2},u_{|a|-1}\})$ and thus $j(f)(a)=j(\tilde f)(a)$. QED<|endoftext|>
TITLE: Convergence of solutions to Navier-Stokes to Euler's equation for viscosity $\to$ zero
QUESTION [16 upvotes]: Let
$$
\partial_t u + \nabla_u u = - \nabla p
$$
be Euler's equation (Wikipedia) for an ideal incompressible fluid. Let
$$
\partial_t u + \nabla_u u - \nu \Delta u = - \nabla p
$$
be the Navier-Stokes equations for a Newtonian incompressible fluid with a real parameter $\nu$, the viscosity. Obviously this equation reduces to Euler's equation if we set $\nu = 0$  (and domains and initial conditions are equal and suitable boundary conditions are prescribed).
What is known about the convergence of solutions $u_{\nu}$ of the Navier-Stokes equation as stated above to a solution of Euler's equation in the limit $\nu \to 0?$
That is, are there topological vector spaces T and sequences or nets of solutions $u_{\nu_k}$ of the Navier-Stokes equation with viscosity $\nu_k$ in T such that the limit $\lim_{\nu_k \to 0} u_{\nu_k}$ exists and is a solution to Euler's equation?

REPLY [14 votes]: To complete Michael's answer, the only situation that is under control is that of the Cauchy problem: the spatial domain is ${\mathbb R}^d$ or ${\mathbb T}^d$ (case of periodic solutions). This means that there is no boundary condition. 
If $d=2$, both systems are globally well-posed for $t>0$, with uniformly bounded (in $L^2$) solutions, and $u_\nu$ converges strongly to the solution of the Euler equation. Notice that it is not a trivial fact: the reasons why both Navier-Stokes and Euler Cauchy problems are globally well-posed have nothing in common; for Navier-Stokes, it comes from the Ladyzhenskaia inequality (say, $\|w\|_{L^4}^2\le c\|w\|_{L^2}\|\nabla w\|_{L^2}$), while for Euler, it is the transport of the vorticity.
If $d=3$, both Cauchy problems are locally-in-time well-posed for smooth enough initial data. One has a convergence as $\nu\rightarrow0+$ on some time interval $(0,\tau)$, but $\tau$ might be strictly smaller than both the time of existence of Euler and the $\lim\inf$ of the times of existence for Navier-Stokes.
To my knowledge, the initial-boundary value problem is a nightmare. The only result of convergence is in the case of analytic data (Caflisch & Sammartino, 1998). From time to time, a paper or a preprint appears with a "proof" of convergence, but so far, such papers have all be wrong.
By the way, your question is incorrectly stated, when you say boundary conditions are equal. The boundary condition for NS is $u=0$, whereas that for Euler is $u\cdot\vec n=0$, where $\vec n$ is the normal to the boundary. This discrepancy is the cause of the boundary layer. One may say that the difficulty lies in the fact that this boundary layer is characteristic. Non-characteristic singular limits are easier to handle.
Another remark is that some other boundary condition for NS are better understood. For instance, there is a convergenece result (Bardos) when $u=0$ is replaced by 
$$u\cdot\vec n=0,\qquad {\rm curl}u\cdot\vec n=0.$$<|endoftext|>
TITLE: Is there a polynomial equation whose solution over the integers is independent of ZFC
QUESTION [6 upvotes]: Consider the algorithm that goes over all proofs in peano arithmetic. Allegedly, for a given multivariate polynomial equation we should find a proof or disproof of existence of an integer solution. Therefore, given the negaative solution of Hilbert's 10th problem, there should be a polynomial for which we could neither prove or disprove existence of an integer solution. Is this argument valid?
It seems strange, can anyone explain to me this anomaly? It seems to indicate that there are models in which there is a solution and others in which there is no solution. I dont really understand how that can happen?

REPLY [7 votes]: Hi Daniel.
The point with Hilbert's 10th problem is that diophantine equations are complex enough to encode Turing machines and other complicated stuff that has some kind of no-can-do-theorem. In particular: To each Turing machine there is a polynomial $p\in\mathbb{Z}[X_1,\ldots,X_n]$ such that $p$ has a solution in $\mathbb{N}$ iff the machine halts (lets consider only turing machines with empty input for simplicity).
Now lets look at the machine that lists all PA-proofs and halts iff it finds a proof of a given PA-statement $\psi$. If PA could prove the existence or non-existence of solutions to every diophantine equation, then it could decide this in particular for those equations that encode Turing machines, i.e. this algorithm would effectively decide the halting problem which is impossible.
There are other things that you can encode with diophantine equations. For example it is possible to translate a statement like $con(PA)$ into a diophantine equation. Now PA cannot prove the existence of solutions of such an equation because that would mean that PA proves $con(PA)$ which is also impossible. It can't disprove the existence of solutions either because then PA would prove $\neg con(PA)$ which is also impossible because PA is consistent.
Now about the models... There isn't that much to say about it. If PA cannot decide the existence of solutions of $p=0$, then the standard model $\mathbb{N}$ won't contain a solution (because if it would, this solution could be explicitely written down and it could be checked by calculation that it is indeed a solution thus giving a formal PA-proof of its existence). But there will be many non-standard models with solutions. Those solutions are of course non-standard numbers, so you can't write them down or do any kind of explicit computations with them. In particular: You won't be able to squeeze an existence proof out of them like you can with a standard solution.
I hope that answers your questions.

REPLY [3 votes]: Yes, your description of the situation is essentially correct. The way it can happen is this. 
On the one hand, if a given diophantine equation does have a solution in the integers, then it will be easy to prove in PA that this solution is indeed a solution, since the proof amounts to checking that that particular solution is indeed a solution. 
But on the other side of the coin, when a diophatine equation has no integer solution, there seems to be no reason for us to expect that there should be an easy proof that there is no solution. And indeed, because of the MRDP solution of Hilbert's 10th problem, we know that there are diophatine equations having no integer solution, but for which we cannot prove this in PA. The way to understand what is going on is to realize that this situation arises when the diophatine equation has no solution in the standard integers, but there is some nonstandard model of PA, with nonstandard integers, in which there is a solution using the arithmetic of that nonstandard model. The solution in that model must necessarily involve some of the infinite nonstandard integers, since the standard part of a nonstandard model of arithmetic has the standard arithmetic.<|endoftext|>
TITLE: Coefficients of Weil Cohomology Theories
QUESTION [33 upvotes]: A Weil Cohomology theory is a functor from the category of smooth projective varieties (over some fixed field $k$) to graded $K$-algebras (for some fixed field $K$) satisfying various axioms.  For example, Betti,de Rham, $\ell$-adic and rigid cohomologies are all Weil cohomology theories.  
In each of these settings one frequently encounters the concept of a category of coefficients for a given variety.  For example, in the $\ell$-adic setting the category of coefficients of a variety $X$ is the category of (not necessarily lisse) $\ell$-adic sheaves on $X$. In the Betti setting it is the category of constructible sheaves on $X$. In the de Rham setting it is the category of regular, holonomic $D$-modules on $X$.  In the rigid setting a category of coefficients on $X$ is given by overholonomic arithmetic $F$-$D$-modules. For each theory there is a constant object for every variety and these give the usual Weil cohomology functor.
I have two questions (the second is related to the first):

What is the conceptual meaning of a category of coefficients of a given variety for a Weil cohomology theory?  Is there a set of axioms such categories must obey?  If so does anyone have a reference?  Why, for example, do we take only regular holonomic $D$-modules and not all $D$-modules in the de Rham setting?
In each of the above examples there are important subcategories of smooth objects.  For the above cases these are given by: lisse $\ell$-adic sheaves, local systems, integrable connections, and overconvergent $F$-isocrystals. What is the conceptual meaning of these privileged subcategories?

REPLY [4 votes]: Not an expert but my comments are too long to fit in the comment box:
As far as I know, the "coefficients" are a well-behaved (e.g. triangulated) category with some rich structures, namely six operations à la Grothendieck: 
pull-back, push-forward, tensor product, inner hom, upper and lower shriek.

They are expected to satisfy various functoriality and adjunctions, which are similar to the case of sheaves of abelian groups on a topological space.
These structures at least allow you to connect coefficients on different spaces, and thus do dévissage. Once the category of coefficients are understood well, some hard theorem can be reduce to the curve case, where you still need to work hard... (Hopefully this partly answers question 1.) 
Also, one Weil cohomology theory you didn't mention is crystalline cohomology, which coincide with rigid cohomology for proper smooth varieties. There, as I read from Illusie's survey, a satisfactory category of coefficients is missing. I don't know if there is any further development after that.<|endoftext|>
TITLE: What is the canonical isomorphism between the tensor products of the top exterior powers associated to exact sequences of vector spaces?
QUESTION [10 upvotes]: One often reads (and writes) that an exact sequence of finite dimensional vector spaces
$$
0 \rightarrow X_1 \rightarrow X_2 \rightarrow \dots \rightarrow X_n \rightarrow 0
$$
induces a canonical isomorphism
$$
\bigotimes_{i \; \mathrm{odd}} \Lambda^{\max} (X_ i) \cong \bigotimes_{i \; \mathrm{even}} \Lambda^{\max} (X_i),
$$
where $\Lambda^{\max}(X)$ denotes the top exterior power of the vector field $X$.
My problem is that there seem to be too many choices for the sign of this ``canonical'' isomorphism. For instance, to the exact sequence
$$
0 \rightarrow X \stackrel{A}{\rightarrow} Y \stackrel{B}{\rightarrow} Z \rightarrow 0
$$
it seems equally canonical to associate the isomorphism
$$
\Lambda^{\max}(X) \otimes \Lambda^{\max}(Z) \cong \Lambda^{\max} (Y), \qquad x 
\otimes B_* (y) \rightarrow A_*(x) \wedge y
$$
or the isomorphism 
$$
x \otimes B_* (y) \rightarrow y \wedge A_*(x).
$$
Here $x$ is a generator of $\Lambda^{\max}(X)$ and $y\in \Lambda^{\dim Z}(Y)$ is such that 
$A_*(x) \wedge y$ generates $\Lambda^{\max}(Y)$. 
Since this canonical isomorphism is often used in the theory of determinant bundles in order to define orientations for geometric objects, I find this uncertainty on a sign disturbing.
Reasonable requirements that one should ask to this canonical isomorphism are:
1) to the exact sequence $0\rightarrow X \stackrel{A}{\rightarrow} Y \rightarrow 0$ one associates the isomorphism $x \mapsto A_*(x)$;
2) naturality with respect to isomorphisms of exact sequences.
However, these requirements do not determine the isomorphism uniquely.
My question is: is there a standard convention regarding the definition of the canonical isomorphism which is associated to exact sequences of arbitrary length? And if not, what would be reasonable requirements to add to 1) and 2) in order to have a good definition?

REPLY [6 votes]: The isomorphism you stated exist not only on the level of topmost exterior power, but also on the level of the whole exterior algebra, considered as a superalgebra. For an exact sequence $$ 0 \to X \xrightarrow{A} Y \xrightarrow{B} Z\to 0 $$ it is defined in a similar way:
$$ x\otimes B(y) \mapsto A(x) \wedge y $$
but here $x$ and $y$ are elements of $X$ and $Y$ respectively. That is, they are the generators of superalgebras $\Lambda^* {X}$ and $\Lambda^* {Y}$. Now this isomorphism is canonical, if you fix a certain definition of tensor algebra. A common way to define multiplication in tensor product of superalgebras is
$$ (a\otimes b)(c\otimes d) = (-1)^{\deg b \deg c}(ac)\otimes (bd)$$
Here $a,b,c,d$ are some homogeneous elements of respective algebras. It is clear that a natural isomorphism must map generators to generators. Also, $$ a\otimes b = (a\otimes 1)(1\otimes b) $$
Mapping multiplication to multiplication and $x\otimes 1 \mapsto A(x)\in \Lambda^* (Y)$, $1\otimes B(y) \mapsto y\in \Lambda^* (Y)$, we get a uniquely defined isomorphism
$$ \Lambda^{*}{X} \otimes \Lambda^{*} {Z} \simeq \Lambda^{*} {Y}$$
Specializing to the subalgebra of topmost powers, we get the required canonical isomorphism.
The reverse isomorphism you mentioned would be an isomorphism with some algebra $\Lambda^\prime Y$, that has the same multiplication for elements in $X$, but $x \cdot y = y \wedge x$ for $y \perp x$. This algebra isn't associative, because 
$$ (x \cdot y_1) \wedge y_2 = (y_1 \wedge x) \wedge y_2 = - (y_1 \wedge y_2 ) \wedge x = - x \cdot (y_1 \wedge y_2) $$<|endoftext|>
TITLE: Tangled Knot Function
QUESTION [7 upvotes]: I am seeking a function $f: \mathbb{R}^3 \to \mathbb{R}^3$
that has these properties:
(1) When iterated $n$ times starting from some $p$,
connecting the points in order 
with segments and closing last to first,
$$(p, f(p), f^2(p), \ldots, f^n(p), p)$$
results in a simple (non-self-intersecting) closed polygonal
cycle $K$.
(2) When $K$ is viewed as a knot, it is highly tangled,
e.g., it has large crossing number, or large unknotting number.
The tangledness, however defined, should increase with $n$, the faster the better.
(3) These properties should hold for infinitely many $n$.
Expressed differently, I would like a way to generate
an infinite variety of increasingly tangled stick knots
via a simple function iteration.
My requirements are a bit loose, as I just want to 
simply generate knotty examples.
Likely some weaving is known to accomplish this...?
          
(This question is related to an earlier question,
"Complexity of random knot with vertices on sphere.")

REPLY [12 votes]: There is a body of work starting with Birman and Williams, and continued by Ghrist, Holmes, and Sullivan, concerning knottedness of closed orbits of flows. The flows they consider are certain Axiom A flows on $S^3$ with interesting basic sets, for example the Lorentz attractor. Williams' paper "Lorentz knots are prime" shows that all the closed orbits of the Lorentz attractor are prime knots. And, being an Axiom A basic set, closed orbits are dense in the Lorentz attractor. So if $f$ is the time $\epsilon$ map of a flow on $S^3$ having the Lorentz attractor as a basic set, then you'll get lots of examples by choosing $p$ on longer and longer closed orbits, as long as you are careful to connect $f^k(p)$ to $f^{k+1}(p)$ by a flow segment. I am guessing that these knots will be more and more complicated as you choose $p$ to have longer and longer orbit.

REPLY [10 votes]: If you parametrize a torus in $\mathbb{R}^3$ as $(x(u,v),y(u,v),z(u,v))$, $0\le u,v<1$, you can easily generate the torus knot $(3,q)$ (with crossing number $2q$) for $q$ large enough and not divisible by three by letting $f(u,v) = (u+3/q^2 \mod 1,v+1/q \mod 1)$ and $n=q^2$. So you just have to embed a continuum of these tori with $q$ varying continuously, and you'll have a function $f$ that generates every knot $(3,q)$. You can probably also improve on how $n$ grows with $q$.<|endoftext|>
TITLE: Probability that a Turing machine will nontrivially reduce a real
QUESTION [6 upvotes]: For a fixed Turing machine $\Phi_e$, what is the probability that it will reduce a given real to some less complex, yet still non-computable real?
More precisely: It is known that the set of reals with minimal Turing degree has measure zero. Since $N_e:=\lbrace X: \Phi_e^X\text{ is total and }X>_T\Phi_e^X>_T\emptyset\rbrace$ is Borel, it is Lebesgue measurable. But each non-minimal $X$ is in some $N_e$, and hence by the result quoted above not every $N_e$ has measure zero (since the set of non-minimal reals, with measure 1, is the union of the countably many measurable $N_e$). My question is: what is known about the possible values of $m(N_e)$ for $e\in\omega$? (I am also interested in a characterization of the set of $e$ such that $N_e$ has measure zero (or one).)
One thing that is easy to show: just by examining the definition, it is clear that $m(N_e)$ is $\Sigma^1_2$ (I think) for each $e\in\omega$. I presume much more can be said (perhaps $\forall e, m(N_e)\in\lbrace 0, 1\rbrace$?), yet I cannot seem to prove anything nontrivial.

REPLY [3 votes]: Not sure whether the following answers your question, but they might be helpful.
Fix any number $n\geq 2012$.
1 For any $e_0$ so that $\Phi_{e_0}^X=X_0$ where $X_0$ is the unique real so that $X=X_0\oplus X_1$.  Then for such $e_0$, $m(N_{e_0})=1$. 
2 For any $e_1$ so that $\Phi_{e_1}^X=0$ if $X(0)=0$ and $\Phi_{e_1}^X=X_0$ if $X(0)=1$.   Then for such $e_1$, $m(N_{e_1})=\frac{1}{2}$.
3 $m(N_e)$ must be $\Delta^0_n$. Just note that  $N_e$ is a $\Delta^0_{n-1}$ set.
4 $m(N_e)>0$ if and only if $N_e$ contains an $n$-random real.
\begin{proof}
If  $m(N_e)>0$, then obviously  $N_e$ contains an $n$-random real.
$N_e$ is a $\Delta^0_{n-1}$ set. So if it is null, then it does not contain any $n$-random real.
\end{proof}
5 $m(N_e)=1$ if and only if $N_e$ contains all $n$-random reals. 
The proof is similar to 4.
The lower bound $2012$ can be certainly significantly smaller. 
For randomness notions, you may refer Downey and Hirschfeldt (2010) or Nies (2009).<|endoftext|>
TITLE: trying to understand the support of the sheaf of relative differentials
QUESTION [5 upvotes]: So I'm trying to understand a proof of Belyi's theorem from http://eprints.soton.ac.uk/29785/1/b45h1koe.pdf
specifically lemma 3.4.
The setup is as follows: Let $X/\mathbb{C}$ be a curve, and let $t : X\rightarrow\mathbb{P}^1_\mathbb{C}$ be a meromorphic function on $X$ thought of as a covering map of degree $n$. Further, let $\text{Crit}(t)$ denote the critical points of the cover $t$ - ie, the points in $\mathbb{P}^1_\mathbb{C}$ which have fewer than $n$ pre-images under $t$.
Then, he claims that $\text{Crit}(t) = t(\text{supp}(\Omega^1_{X/\mathbb{P}^1_\mathbb{C}}))$.
Now, it's my understanding that the critical points of $t$ should be the images of the ramification points of $t$ under $t$, so I've been trying to understand why it should be the case that the sheaf of relative differentials of $X/\mathbb{P}^1_\mathbb{C}$ should be nonzero only on the ramification points (or at least only above the critical points).
To this end, I'm trying to understand the definition given in Hartshorne (III.8), namely:
$\Omega^1_{X/\mathbb{P}^1_\mathbb{C}} = \Delta^*(\mathcal{I}/\mathcal{I}^2)$, where $\Delta : X\rightarrow X\times_{\mathbb{P}^1_\mathbb{C}} X$ is the diagonal map, and $\mathcal{I}$ is the sheaf of ideals of the image $\Delta(X)$ in some open subset $W\subset X\times_{\mathbb{P}^1_\mathbb{C}} X$.
I kind of understand sheaves of ideals (they're essentially functions on the ambient space that vanish on the closed subscheme), but I'm still not very comfortable with the notion of $\Delta^*(\mathcal{I}/\mathcal{I}^2)$ (in this case defined to be $\Delta^{-1}\mathcal{I}/\mathcal{I}^2\otimes_{\Delta^{-1}\mathcal{O}_{X\times X}}\mathcal{O}_X$, where the fibred product is taken over $\mathbb{P}^1$).
Any comments on how I should think of $\Delta^*(...)$ and why the sheaf of relative differentials only have nonzero stalks at ramification points would be awesome!
thanks.

REPLY [4 votes]: Ravi Vakil has a good explaination for the definition $\Delta^*(I/I^2)$ in his notes. See his AG notes here or here (chapter 23). In particular, I guess thinking about this locally makes it a little clearer what's going on, in terms of derivations etc. Also, when $X$ is smooth, it is instructive to see that this really gives the cotangent bundle on $X$. 
As for your question about ramification points: Let $f:X\to Y$ be a finite morphism of curves (I will assume that these are smooth in the following). It is useful to have in mind the exact sequence $$0\to f^*\Omega_{Y}\to \Omega_X \to \Omega_{X|Y}\to 0.$$(This is exact at the right in the smooth case, but not in general). Note that $\Omega_{X|Y}$ is a torsion sheaf since the two other sheaves are locally free of the same rank (they are line bundles on $X$). At a point $q\in Y$ and $p\in X$ in the preimage of $q$, let $dx$ denote a generator for $\Omega_{Y,q}$ as a $O_{Y,q}$-module. Now, $(\Omega_{X|Y})_P=0$ if and only if $f^*dx$ is a generator of $\Omega_{X,p}$, which happens if and only if $f$ pulls back a local parameter to a local parameter, that is $p$ is unramified. Moreover, the exact sequence above shows that the ramification index is exactly the length of the sheaf $\Omega_{X|Y}$. Finally, note that this sequence gives the Riemann hurwitz formula, relating the canonical divisors of $X$ and $Y$ and the ramification divisor of $f$.

REPLY [2 votes]: Let $\pi:X\to Y$ be a finite morphism of curves.
Then, for any point $x$ in $X$ lying over $y$ in $Y$, the coefficient $v_x(\pi)$ of $\Omega_{\pi}$ is the valuation of the different of the extension of dvr's $\mathcal{O}_{y}\subset \mathcal{O}_x$. If you are working in characteristic zero, then $$v_y(\pi) = e_x-1,$$ where $e_x$ is the ramification index. So you see that $\Omega_{\pi}$ is supported on the ramification points.
Also, you have a short exact sequence (it's on page 2 of Chapter IV.2 in Hartshorne) which relates $\Omega_\pi$ with $\Omega_X$ and $\Omega_Y$. The above actually shows the important Riemann-Hurwitz formula: $$K_X = \pi^{\ast} K_Y + R.$$ Here $R$ is the ramification divisor. This equals $\Omega_{\pi}$ in this case.<|endoftext|>
TITLE: Flow of a Hamiltonian vector field
QUESTION [5 upvotes]: Smooth vector fields are in a one-to-one relationship with flows $\Phi: D \subseteq M\times \mathbb{R} \rightarrow M$, 
$$X_m = {\frac{d}{d t}}_{t=0} \Phi(m, t),$$
and by the symplectic form also with 1-forms
$$X \longleftrightarrow \; - \iota_X \omega. $$
The following is well known (e. g. Abraham Marsden, Proposition 3.3.6):
$X$ is local Hamiltonian iff $\iota_X \omega$ is closed iff the flow $\Phi$ of $X$ is symplectic.
$X$ is Hamiltonian iff $\iota_X \omega$ is exact iff ??.
Question: What is the corresponding property for flows of (global) Hamiltonian vector fields?   
I am a bit confused about this, as the existence of flows can be guaranteed only for small times and only in neighborhoods of each point (so it is a strongly local object), but the difference between closed and exact forms is determined by global/topological characteristics.

A short proof of the cited proposition about local Hamiltonian vector fields:
The flow $\Phi$ leaves the symplectic form invariant iff $\Phi_t^* \omega = \omega$ iff $0 = L_X \omega = d(\iota_X \omega)$ iff $\iota_X \omega$ is closed.

REPLY [4 votes]: Given a symplectic vetor field, you may try to build a hamoltonian function the following way. Fix some point P. For any point Q, choose some curve from P to Q, and define H(Q) to be the flux of the vector field across the chosen curve (infinitesimal symplectic area swept by the curve when you push it by the flow during an infinitesimal amount of time). If you choose a second curve which is homotopic to the first one, you get the same value for H(Q), because your vector filed is symplectic. But this may not be the case if you choose an other curve which is not homotopic to the first one (think of a translation of the 2-torus). Thus the field is hamiltonian if and only if the flux across every closed curve is zero. (In dimension two, the flux is called the mean rotation vector).<|endoftext|>
TITLE: $\mathcal{D}$-quasi-isomorphisms and coherent $\Omega$-modules
QUESTION [15 upvotes]: Let $X$ be a smooth $\mathbf{C}$-scheme of finite type. In section 7.2 of their preprint "Quantization of Hitchin's Integrable System and Hecke Eigensheaves", Beilinson and Drinfeld define an adjunction
$$\mathcal{D}:Z^0(\mathbf{dgMod}_{\mathrm{qcoh}}(\Omega^{\cdot}_{X/\mathbf{C}}))\leftrightarrows \mathbf{Cplx}(\mathbf{Mod}^{\mathrm{r}}_{\mathrm{qcoh}}(\mathcal{D}_X))):\Omega$$
between the category of (closed morphisms of) quasi-coherent dg-modules over the de Rham algebra of $X/\mathbf{C}$ and that of complexes of quasi-coherent right $\mathcal{D}_X$-modules. Their aim is to obtain a derived equivalence between a certain localization of this category of dg-modules over $\Omega^{\cdot}_{X/\mathbf{C}}$ and the usual derived category of the abelian category $\mathbf{Mod}^{\mathrm{r}}_{\mathrm{qcoh}}(\mathcal{D}_X)$. The functor $\mathcal{D}$ does not send all quasi-isomorphisms to quasi-isomorphisms and Beilinson and Drinfeld's solution is to invert the morphisms of dg-modules that are sent to quasi-isomorphisms by $\mathcal{D}$, which they call $\mathcal{D}$-quasi-isomorphisms. In 7.2.6(iii) (and also in 2.1.10 of their "Chiral algebras"), Beilinson and Drinfeld assert that any quasi-isomorphism of bounded $\mathcal{O}_X$-coherent dg-$\Omega^{\cdot}_{X/\mathbf{C}}$-modules is a $\mathcal{D}$-quasi-isomorphism (the converse is true without the finiteness hypothesis). In light of Kapranov's paper ("On dg-modules over the de Rham complex and the vanishing cycles functor"), this sounds reasonable.
On the other hand, as suggested by 6.23 in these lecture notes, if we take $X=\operatorname{Spec}(\mathbf{C}[t])$ and consider the left $\mathcal{D}_X$-module $\mathcal{O}_X\operatorname{e}^t$ given by the free $\mathcal{O}_X$-module of rank $1$ generated by $\operatorname{e}^t$, i.e. the integrable connection $f\mapsto (f+f')\mathrm{d}t:\mathbf{C}[t]\to\mathbf{C}[t]\mathrm{d}t$, then its de Rham complex appears to be acyclic, $\mathcal{O}_X$-coherent and bounded. I think this means that the image of the corresponding right $\mathcal{D}_X$-module under the functor $\Omega$ is acyclic, bounded and $\mathcal{O}_X$-coherent, hence $\mathcal{D}$-acyclic. As the adjuction morphism $\mathcal{D}\Omega\to\operatorname{id}$ is a quasi-isomorphism by [BD, 7.2.4], the right $\mathcal{D}_X$-module corresponding to $\mathcal{O}_X\operatorname{e}^t$ is zero, which sounds absurd.
Question: Have I made a stupid error somewhere and, if not, how does one reconcile this example with Beilinson-Drinfeld's description of $\mathcal{D}$-quasi-isomorphisms for $\mathcal{O}_X$-coherent $\Omega^{\cdot}_{X/\mathbf{C}}$-modules?

REPLY [14 votes]: Let $\mathcal M$ be a left $\mathcal D$-module over a smooth curve $X$.  Then the Koszul duality functor assigns to $\mathcal M$ the DG-module $\Omega_X\otimes_{\mathcal O_X}\mathcal M$ over the de Rham complex $\Omega_X$.  Viewed as a complex of sheaves,  $\Omega_X\otimes_{\mathcal O_X}\mathcal M$ is a two-term complex whose terms are quasi-coherent $\mathcal O_X$-modules, but the differential is not $\mathcal O_X$-linear.  For this reason, even when $X$ is affine, one has to distinguish between the acyclicity of the complex of global sections of $\Omega_X\otimes_{\mathcal O_X}\mathcal M$ and the acyclicity of this complex of sheaves itself.
Explicitly, let $x$ be a global coordinate on $X$ (assuming that one exists).  Then the differential in the complex $\Omega_X\otimes_{\mathcal O_X}\mathcal M = (\mathcal M\to \Omega_X^1\otimes_{\mathcal O_X}\mathcal M)$ has the form $m\mapsto dx\otimes \partial/\partial x(m)$.  The question about acyclicity of this complex of sheaves is, therefore, the question about injectivity and surjectivity of the operator $\partial/\partial x$ acting in the sections of $\mathcal M$.
In the case at hand, we have $\mathcal M=\mathcal O_X e^x$.  So global sections of $\mathcal M$ over $\operatorname{Spec} \mathbb C[x]$ are expressions of the form $p(x)e^x$, where $p(x)$ is a polynomial in $x$.  Hence one can easily see that the complex of global sections of $\Omega_X\otimes_{\mathcal O_X}\mathcal M$ over $\operatorname{Spec} \mathbb C[x]$ is acyclic.
The complexes of sections of $\operatorname{Spec} \mathbb C[x]$ over Zariski open subsets of $\operatorname{Spec} \mathbb C[x]$ are not acyclic, however.  It suffices to consider sections over $\operatorname{Spec} \mathbb C[x,x^{-1}]$.  As is well known, the function $x^{-1}e^x$ does not lie in the image of $\partial/\partial x$ acting in the space of Laureant polynomials in $x$ multiplied with $e^x$.  Therefore, the two-term complex of sheaves $\Omega_X\otimes_{\mathcal O_X}\mathcal M$ is not acyclic.  Its differential is an injective, but not surjective morphism of sheaves of $\mathbb C$-vector spaces over $\operatorname{Spec} \mathbb C[x]$.
On the other hand, one can consider the complex of sheaves of analytic forms $\Omega_X^{an}\otimes_{\mathcal O_X}\mathcal M$ in the analytic topology of the set of closed points of $\operatorname{Spec} \mathbb C[x]$.  Then every function from $\mathcal O_X^{an}e^x$ will have a primitive analytic function locally in the analytic topology.  So the differential in the two-term complex of sheaves $\Omega_X^{an}\otimes_{\mathcal O_X}\mathcal M$ is now surjective.  However, it is no longer injective, as the constant functions are sections of $\mathcal O_X^{an}e^x$, the function $e^{-x}$ being analytic.  This is what Kapranov is doing in his paper.
Having, as I hope, answered your question, let me now point out, as a side note, that what I would consider a superior alternative of the $\mathcal D{-}\Omega$ duality theory of the Beilinson--Drinfeld preprint can be found in Appendix B to my AMS Memoir "Two kinds of derived categories, Koszul duality, and comodule-contramodule correspondence", http://arxiv.org/abs/0905.2621 .
The point is that what you call "$\mathcal D$-quasi-isomorphisms" are defined in my paper in terms intrinsic to DG-modules over $\Omega$ (without any reference to differential operators).  The corresponding localization is called the coderived category of quasi-coherent DG-modules over $\Omega_X$, and subsequently it is proven to be equivalent to the derived category of quasi-coherent $\mathcal D_X$-modules (for a smooth variety $X$ over any field, $\mathcal D_X$ denoting the crystalline differential operators in the case when the characteristic is finite).<|endoftext|>
TITLE: Limitations on model-categorical presentations
QUESTION [20 upvotes]: In higher category theory, it is common that a weak structure cannot be strictified in all directions simultaneously.  For instance, a monoidal category is not (in general) equivalent to one that is both strict and skeletal, and a tricategory is not (in general) equivalent to one whose units and interchange law are both strict.
Now a model category can be regarded as a particular sort of strictification of an $(\infty,1)$-category.  From this perspective, all sorts of questions along the above lines suggest themselves.  For concreteness, I'll ask a particular one:

Does there exist a locally presentable $(\infty,1)$-category which (provably) cannot be presented by a model category in which all objects are both fibrant and cofibrant?

But I would be interested in answers to any similar question.

REPLY [3 votes]: Here is another answer that involves adding extra properties.  If we have a model category which 

is locally cartesian closed, as a category (such as if it is a presheaf category)
has its cofibrations being the monomorphisms (hence in particular all objects are cofibrant)
is right proper (such as if all objects are fibrant)

then pullback along a fibration $g\colon A\to B$ preserves both cofibrations and acyclic cofibrations, and so the adjunction $g^* \dashv \Pi_g$ is Quillen.  Therefore, the $(\infty,1)$-category presented by this model category is locally cartesian closed.
Thus, an $(\infty,1)$-category which is not locally cartesian closed cannot be presented by a model category with all three of the above properties.<|endoftext|>
TITLE: Incidence Correspondence 
QUESTION [16 upvotes]: A useful tool in Algebraic Geometry is the incidence correspondence. Loosely speaking, it is a set of the form $$\{(p,X): p \text{ a fixed dimension subscheme of } Y \text{ and } X \text{ a specific type of subscheme} \}.$$ For example one could consider the incidence correspondence of lines in $\mathbb{P}^2$ with a point on them, or cubics in $\mathbb{P}^3$ with a line on them. 
It is not too hard to see that in each of the previous cases, the resulting scheme is a variety  by writing down explicit equations in coordinates. In the first case the variety lives in $\mathbb{P}^2\times \mathbb{P}^2$ and with a bit of work it is not too hard to show it is a projective bundle over $\mathbb{P}^2.$  The second case is a subvariety of $\mathbb{P}^{19}\times G(1,3)$.
My question is the following. Is there a way to get both of the previous examples in a more natural way then explicitly writing down equations? Should I even expect there to be one? Generally when working with incidence correspondences one is interested in properties such as smoothness and irreducibility and most authors I have seen conclude these from the equations. Since the first case above does end up being a projective bundle, I would really hope for there to be a natural way to construct it.
Thanks.

REPLY [11 votes]: There's several issues here.
(1) Is a given incidence correspondence actually a closed variety?
(2) What are explicit equations for the correspondence in the product of the relevant spaces?
(3) What are geometric properties of the incidence correspondence?
Most often, questions (1) and (3) are studied and little attention is paid to (2).
For (1), things can be generalized a fair bit.  For instance, suppose $X$ is a variety, $H$ is an ample divisor, and $P$, $Q$ are two Hilbert polynomials with respect to the ample divisor $X$.  Then there are (projective) Hilbert schemes $Hilb_P(X)$ and $Hilb_Q(X)$ parameterizing closed subschemes of $X$ with Hilbert polynomials $P$ and $Q$.  Then there are a couple different natural incidence correspondences, for example
$\{(Z,Z'):Z\subset Z'\} \subset Hilb_P(X)\times Hilb_Q(X)$
$\{(Z,Z'):Z\cap Z'\neq \emptyset\}\subset Hilb_P(X)\times Hilb_Q(X),$
and it is easy to verify that these conditions are closed (although the correct scheme structure may be less clear).  One can instead restrict attention to a closed subvariety of the Hilbert scheme (so as to not use all the components of the Hilbert scheme, for instance, in case the geometric objects you care about are not entirely determined by their Hilbert polynomials).  It is also easy to generalize to cases with more factors.  These types of constructions mean that arguments for the closedness of incidence correspondences are almost never written down, as anything reasonable that you can write down will be closed so long as the families of objects under consideration are themselves closed.
In practice, (2) is rarely of any theoretical interest, unless these are very special varieties.  Perhaps there is a large algebraic group acting and the ideal can be studied via representation theory, or perhaps the variety has very small dimension or is otherwise very simple, in which case some information might be learned from the ideal.
Regarding (3), geometric properties of an incidence correspondence are only very rarely (roughly in the same cases as discussed for (2)) determined by studying the defining equations.  Most often, the reason we study an incidence correspondence $\Sigma \subset X\times Y$ is because we have some question about one of the projection maps, say $\Sigma \to X$; for instance, we may wonder if it is dominant.  We then ideally answer this by studying the other projection, which ideally is easier to understand.  Likewise, properties like dimension and irreducibility are hopefully easily understood by studying the projections, and smoothness can sometimes be analyzed as well (although smoothness is often not so important in basic applications).<|endoftext|>
TITLE: K-Theory space of finite abelian groups
QUESTION [10 upvotes]: Consider the abelian category $\mathsf{finAb}$ of finite abelian groups. It is easy to prove that there is an isomorphism $\mathrm{ord} : K_0(\mathsf{finAb}) \to \mathbb{Q}^+$. Can you give a reference in the literature for that? More generally:
Question. Is there any more concrete description of the Waldhausen's K-theory space $\Omega |w S_{\bullet} \mathsf{finAb}|$? What is known about its homotopy groups, that is the $K$-theory groups $K_n(\mathsf{finAb})$ for $n>0$?

REPLY [11 votes]: Everything is known.  In fact as spectra we have canonically $K(\mathsf{finAb}) = \vee_p K(\mathbb{F}_p)$, and the spectra $K(\mathbb{F}_p)$ are identified in the work of Quillen (see e.g. http://www.math.uiuc.edu/K-theory/1006/).  In particular on $\pi_0$ we find $K_0(\mathsf{finAb}) = \oplus_p \mathbb{Z}$, agreeing with your claim, and on $\pi_n$ for $n>0$ we find that $K_n(\mathsf{finAb})$ is $0$ for $n$ even  and is $\oplus_p \mathbb{Z}/(p^k-1)$ (non-canonically) for $n = 2k-1$.
To justify the claimed equality $K(\mathsf{finAb}) = \vee_p K(\mathbb{F}_p)$, note first that $\mathsf{finAb}$ is the filtered colimit over increasing finite sets of primes $P$ of the variant $\mathsf{finAb}_P$ where only products of $p$-groups for $p \in P$ are allowed; since K-theory commutes with filtered colimits, it then suffices to show that each $K(\mathsf{finAb}_P) = \prod_{p\in P} K(\mathbb{F}_p)$ and that for $P \subseteq P'$ this identification intertwines the inclusion $K(\mathsf{finAb}_P) \to K(\mathsf{finAb}_{P'})$ with the evident map $\prod_{p\in P} K(\mathbb{F}_p) \to \prod_{p\in P'} K(\mathbb{F}_p)$ which is zero outside of $P$.
But $\mathsf{finAb}_P$ is just the product over $p \in P$ of the categories $\mathsf{finAb}_p$, whose K-theory identifies with that of vector spaces over $\mathbb{F}_p$ by Quillen's devissage theorem.  And K-theory commutes with finite products, so that's that.
Here I guess I was actually arguing using Quillen's Q-construction instead of Waldhausen's $S_{\bullet}$-construction.  Otherwise I'm not sure how to justify the last step, the devissage.  Actually I'm sure all of the above is in Quillen's paper on the Q-construction.<|endoftext|>
TITLE: Stable Conjugacy for Integer Matrices
QUESTION [7 upvotes]: Let $F$ be a field, and $E$ an extension field.  Then two matrices in $GL_n(F)$ are conjugate if and only if they are conjugate in $GL_n(E)$.  I'm curious whether the analogous fact holds for rings of integers.  
Is the following true?
Two matrices in $GL_n(\mathbb Z)$ are conjugate if and only if they are conjugate in $GL_n(\mathbb A)$, where $\mathbb A$ is the ring of algebraic integers.

REPLY [11 votes]: Well, I think the answer is "no".  Here's a construction:  let R be the ring of integers of a real quadratic field K of class number > 1, let M be an invertible R-module of rank one which is not isomorphic to R, and let x be a fundamental unit in R.  Then the action of x on M (viewed as a Z-module) determines a well-defined conjugacy class C_M in GL_2(Z), and similarly the action of x on R determines a conjugacy class C_R.  I claim that these conjugacy classes are distinct, but become equal in GL_2(A).
They are distinct:  indeed, M is recovered up to isomorphism from C_M since R identifies with the commutant algebra of x acting on the Z-module M.
They become equal in GL_2(A):  in fact they become equal in GL_2 of the ring of integers in the Hilbert class field of K, since M and R become isomorphic there.

REPLY [10 votes]: Here is an explicit realization of the counterexample suggested by Dustin. The field ${\mathbf Q}(\sqrt{10})$ has class number 2 and its Hilbert class field is obtained by adjoining $\sqrt{2}$. The ring of integers ${\mathbf Z}[\sqrt{10}]$ has (fundamental) unit $u:=3+\sqrt{10}$, whose minimal polynomial over ${\mathbf Q}$ is $T^2 - 6T - 1$. The two ideal classes in ${\mathbf Z}[\sqrt{10}]$ are represented by the ideals $(1)$ and $(2,\sqrt{10})$, which have ${\mathbf Z}$-bases $\{1,u\}$ and $\{2,\sqrt{10}\}$. Multiplication by $u$ on  these two ideals is represented, using the indicated $\mathbf Z$-bases, by the respective matrices $A = (\begin{smallmatrix}0&1\\1&6\end{smallmatrix})$ and $B = (\begin{smallmatrix}3&5\\2&3\end{smallmatrix})$. These matrices are both in ${\rm GL}_2({\mathbf Z})$, they are not conjugate in this group, but they are conjugate by the matrix $U = (\begin{smallmatrix}\sqrt{2}&5+3\sqrt{2}\\1&3+2\sqrt{2}\end{smallmatrix})$, which lies in ${\rm GL}_2({\mathbf Z}[\sqrt{2}])$. That is, $UAU^{-1} = B$. This conjugating matrix $U$ has determinant $-1$. A matrix with determinant 1 and algebraic integer entries that satisfies $VAV^{-1} = B$ is 
$V = (\begin{smallmatrix}2\sqrt{2}&6\sqrt{2}+5\sqrt{3}\\\ \sqrt{3}&4\sqrt{2}+3\sqrt{3}\end{smallmatrix})$.
Quite generally, the matrix $M = (\begin{smallmatrix}a&b\\c&d\end{smallmatrix})$ satisfies $MA = BM$ if and only if $b=3a+5c$ and $d = 2a+3c$, and then $\det M = 2a^2 - 5c^2$. We can't solve $2a^2 - 5c^2 = \pm 1$ in ${\mathbf Z}$ (look at it mod 5), but we can solve it in ${\mathbf Z}[\sqrt{2}]$ using $a = \sqrt{2}$ and $c = 1$. That is how I found $U$. We can solve $2a^2 - 5c^2 = 1$ using $a = 2\sqrt{2}$ and $c = \sqrt{3}$, which is how I found $V$.<|endoftext|>
TITLE: "psi-epistemic theories" in 3 or more dimensions
QUESTION [31 upvotes]: In their recent paper The Quantum State Can Be Interpreted Statistically, Lewis et al. end with a very nice mathematical question, one whose answer (either way) would have interesting implications for the foundations of quantum mechanics.  In the hopes of getting some non-quantum math folks interested in their question---and maybe even finding someone to say "the answer is trivial for the following reason..." :-)---I decided to state the question for the MO community, shorn of all the physics and philosophy.
Let Hd be the set of unit vectors in $\mathbb{C}^d$.  A ψ-epistemic theory in d dimensions consists of the following:

A measurable space Λ (called the "space of ontic states").
A function mapping each unit vector ψ∈Hd to a probability measure Dψ over Λ.
A function f(λ,M,i)∈[0,1], which takes as input an ontic state λ∈Λ, an ordered orthonormal basis M=(v1,...,vd) for $\mathbb{C}^d$, and an index i∈{1,...,d}.

f must satisfy the following two conditions:
(i) $\sum_{i=1}^{d}f(\lambda,M,i)=1$ for all λ and M.  (Intuitively, f must give rise to a probability distribution over the "measurement outcomes" v1,...,vd in M.)
(ii) $\int_{\lambda \sim D_{\psi}} f(\lambda,M,i) d\lambda = |v_{i}^{*}\psi|^{2}$ for all ψ, M, and i.  (Intuitively, the probability of the measurement outcome vi, averaged over all λ drawn from Dψ, must equal the squared projection of ψ onto vi.)
Note that we can trivially satisfy conditions (i) and (ii) as follows:

Λ=Hd
Dψ assigns probability 1 to λ=ψ, and probability 0 to all other states in Λ
f(ψ,M,i) = |vi*ψ|2

Thus, let Supp(D)⊆Λ be the support of D, and call a ψ-epistemic theory nontrivial if there exist ψ≠ϕ such that $Supp(D_{\psi})\cap Supp(D_{\phi}) \ne \emptyset$.
Observe that, if ψ and ϕ are orthogonal, then Supp(Dψ) and Supp(Dϕ) must be disjoint.  This is because, if we set v1=ψ and v2=ϕ, then $v_{1}^{*}\psi = v_{2}^{*}\phi = 1$ and $v_{1}^{*}\phi = v_{2}^{*}\psi = 0$, which is not possible if Dψ and Dϕ have any nonzero overlap.  Motivated by this observation, call a theory maximally nontrivial if $Supp(D_{\psi})\cap Supp(D_{\phi}) \ne \emptyset$ whenever ψ and ϕ are not orthogonal.
I can now state Lewis et al.'s open problem:
Does there exist a maximally-nontrivial ψ-epistemic theory in dimensions d≥3?
Update: See the comments for an extremely nice solution by George Lowther, plus my followup questions.
I know of two results directly relevant to this problem.
First, there exists a maximally-nontrivial theory in dimension d=2, which was found by Kochen and Specker in 1967.  See this paper by Rudolph for more details, including why the obvious generalizations to 3 or more dimensions seem to fail.  Briefly, the Kochen-Specker theory is defined as follows:

Λ=H2.
Dψ assigns probability measure $2 | \psi^{\*} \phi|^{2} - 1$ to ϕ if $| \psi^{\*} \phi|^{2} \geq 1/2$, and probability measure 0 to ϕ otherwise.
f(ψ,M,i) = 1 if $|v_{i}^{\*} \psi|^{2} \geq 1/2$, and f(ψ,M,i) = 0 otherwise.

(Warning: I converted from a different representation, and can't promise I didn't get a factor of 2 wrong or something like that.)
The second result is that, for all finite d, there exists a nontrivial ψ-epistemic theory (though it's far from being maximally nontrivial).  This is the main result of Lewis et al.
My own guess is that maximally-nontrivial theories don't exist for d≥3, but I'd only give it 60% confidence.
To anticipate some questions:

Yes, I'd also be interested in this problem with $\mathbb{R}$ in place of $\mathbb{C}$ (though I suspect the two cases are pretty similar).
Yes, I'd be interested in negative results for restricted classes of theories.  Here are a few examples of restrictions one could look at, in various combinations: Λ=Hd, f∈{0,1}, f is continuous, symmetry under unitary transformations, symmetry under relabeling of the vi's.
No, I don't know how to rule out that the answer could depend on the Axiom of Choice or something crazy like that (but I doubt it).

Update (March 20, 2013): Adam Bouland, Lynn Chua, George Lowther, and myself now have a paper on ψ-epistemic theories originating with this MO post.  The paper contains the construction below, but also proves impossibility results for ψ-epistemic theories when an additional symmetry condition is imposed.

REPLY [12 votes]: Since George Lowther seems to have a lot of late nights, I decided to express my gratitude to him by writing up his lovely answer myself and thereby saving him the trouble.
The answer to my (and Lewis et al.'s) question is that yes, maximally-nontrivial ψ-epistemic theories do exist for every finite dimension $d$.
The first realization is that we can "mix" small ε-balls around any two non-orthogonal vectors.
Lemma 1: Given any two non-orthogonal unit vectors $\psi,\phi \in \mathbb{C}^d$, there exists a ψ-epistemic theory $T = T(\psi,\phi)$ such that $Supp(D_{\psi})$ and $Supp(D_{\phi})$ have nonempty intersection.  Moreover, for this $T$, there exists an $\epsilon \gt 0$ (for example, $\epsilon = |\psi^{*} \phi|/2d$) such that $Supp(D_{\psi'})$ and $Supp(D_{\phi'})$ have nonempty intersection for all $\psi',\phi'$ such that $||\psi - \psi'||,||\phi - \phi'||\lt \epsilon$.
Proof: Our ontic state space will be $\Lambda = H_{d} \times [0,1]$.  Given an orthonormal basis $M=(v_1,...,v_d)$, first sort the $v_i$'s in decreasing order of $min(|v_{i}^{∗}\psi|,|v_{i}^{∗}\phi|)$. Then the outcome of measurement M on ontic state $(w,p)\in \Lambda$ will equal the smallest positive integer $i$ such that
$|v_1^* w|^2+...+|v_{i−1}^*w|^2 \le p \le |v_1^* w|^2+...+|v_i^*w|^2$.
In other words, $f((w,p),M,i)$ will equal $1$ if $i$ satisfies the above and no $j \lt i$ does, and $0$ otherwise.  For now, assume that $D_{w}$ is simply the uniform distribution over all states $(w,p)$ with $p \in [0,1]$.  It is easy to check that this yields a valid ψ-epistemic theory, albeit so far a trivial one.
Since $|\psi^{*} \phi|\gt 0$, I claim that there exists an $\epsilon \gt 0$ such that for all orthonormal bases $M=(v_1,...,v_d)$, there exists an $i$ such that $|v_{i}^{∗}\psi|\ge \epsilon$ and $|v_{i}^{∗}\phi|\ge \epsilon$.  Indeed, by the triangle inequality, setting $\epsilon := |\psi^{*} \phi| / d$ will work.
Now, the above means that, for all measurements $M$ and all $p \in [0,\epsilon]$, the outcome is always $i=1$ when $M$ is applied to either of the ontic states $(\psi,p)$ or $(\phi,p)$.  Following Lewis et al., this implies that we can "mix" the corresponding distributions $D_{\psi}$ and $D_{\phi}$―i.e., have them intersect each other in the region $p \in [0,\epsilon]$―without affecting any outcome of any measurement $M$.
Furthermore, suppose $\psi'$ and $\phi'$ are $\epsilon /2$-close to $\psi$ and $\phi$ respectively, in some standard metric such as trace distance.  Then by continuity, we can similarly mix the distributions $D_{\psi'}$ and $D_{\phi'}$―i.e., have them intersect each other in the region $p\in [0,\epsilon /2]$―without affecting any measurement outcome.   (One subtlety is that, as we vary $M$, the sorting procedure can make $v_1$ "jump" discontinuously from one basis vector of $M$ to another.  However, this jumping is not a problem, since it depends only on the fixed vectors $\psi$ and $\phi$, not on $\psi'$ or $\phi'$.  So it happens the same way everywhere in the $\epsilon /2$-balls.) QED
The second realization is that we can take "convex combinations" of ψ-epistemic theories.  Given two ψ-epistemic theories $T=(\Lambda,D,f)$ and $T'=(\Lambda',D',f')$ (where $D,D'$ are the functions that map vectors $\psi \in H^d$ onto ontic distributions), and a constant $c \in (0,1)$, define the new theory $c T + (1-c)T' =(\Lambda_c,D_c,f_c)$ as follows:

$\Lambda_c := \Lambda \cup \Lambda'$.
$D_c := c D + (1-c) D'$.
$f_c : \Lambda_c \rightarrow [0,1]$ equals $f$ on $\Lambda$ and $f'$ on $\Lambda'$.

Lemma 2: $c T + (1-c)T'$ is a ψ-epistemic theory.  Furthermore, if $T$ mixes the ontic distributions of two vectors $\psi$ and $\phi$, and $T'$ mixes the ontic distributions of two other vectors $\psi'$ and $\phi'$, then $c T + (1-c)T'$ mixes both pairs of distributions.
Proof: Immediate.
Using Lemmas 1 and 2, we now construct a maximally-nontrivial ψ-epistemic theory.  Let $T(\psi,\phi)$ be the theory returned by Lemma 1 given vectors $\psi,\phi\in H^d$.  Also, for all positive integers $n$, let $A_n$ be a $1/n$-net for $H^d$: that is, a finite subset $A_n \subseteq H^d$ such that for all $v\in H^d$, there exists a $w \in A_n$ satisfying $|| w - v || \lt 1/n$.  By making small perturbations, we can easily ensure the property that $u^{*}v \neq 0$ for all $u,v\in A_n$.  Then our theory $T$ is defined as follows:
$$T = \frac{6}{\pi^2} \sum_{n=1}^{\infty} \frac{1}{n^2} \left( \frac{1}{|A_n|^2} \sum_{u,v \in A_n} T(u,v) \right). $$
One can check that $T$ mixes the distributions $D_{\psi}$ and $D_{\phi}$ for all non-orthogonal $\psi$ and $\phi$.
Comment: As often in math, I'd say the true value of knowing this answer is that it points us toward the questions we (or rather Lewis et al.) "really" meant to ask!  In the above construction, the overlap between $D_{\psi}$ and $D_{\phi}$ is indeed nonzero for any non-orthogonal $\psi,\phi$, but the amount of overlap falls off (by my crude estimate) as $ ( |\psi^{*} \phi| / d) ^{ \Theta(d) } $ .
A skeptic of ψ-epistemic theories might argue that for large $d$ (and of course, $d$ can be huge in quantum mechanics), such an overlap is physically irrelevant.  So one obvious followup question is how large the overlap can be―for example, whether it can fall off only as $(|\psi^{*} \phi| / d)^{O(1)}$.  I'd better stop here, though, since I know MO is not for open-ended research discussions.  The question, as I stated it, has been answered.<|endoftext|>
TITLE: Reference request for the number of Sylow p-subgroups
QUESTION [5 upvotes]: Let $G$ be simple group of Lie type or Alternating group. I need reference for find the number of Sylow $p$-subgroup $G$ for every $p$. Thanks a lot.

REPLY [3 votes]: It's probably useful to compare the other questions recently asked here by Sara, which are sometimes stimulating but typically not well formulated or motivated:   for example here.   Like the other questions, this one seems to be asking a very broad question which isn't easily answered for all simple groups.  Here the number of Sylow subgroups is measured by the index of the normalizer, which itself requires a lot of case-by-case study.    There certainly aren't any convenient tables of results to consult, though the various references given here and in other answers are a good start.   
Leaving aside the alternating groups, there are two distinct questions for simple groups of Lie type: When the prime is the defining one, the internal structure of the group given by the BN-pair is a good guide to the normalizer of a Sylow subgroup (characterized in terms of algebraic groups via the unipotent radical of a Borel subgroup).   But for other primes there are more subtle questions raised, which many people have studied in the setting of block theory and Deligne-Lusztig character theory.   In the literature, it seems most natural to focus on the behavior of each prime according to which factor of the order polynomial for the group it divides.    This work aims mostly at comparing the ordinary and the modular representation theory for the given prime, but I'm not sure how explicitly it involves the normalizer order or index wanted here.   As Derek Holt points out, there may be extra detail available case-by-case in the case of finite classical groups. 
Anyway, it's better to start with a more fully developed question which might get a precise answer.<|endoftext|>
TITLE: Order of Ш (Sha)
QUESTION [8 upvotes]: To prove the BSD conjecture, one has to know about 'the finiteness of the Shafarevich Tate group'. But, an example of an elliptic curve of rank 2 (whose Sha group $Ш(E/\mathbb{Q})$ is finite) is not yet known.
Is there any example of an elliptic curve of rank 2 such that $p$-primary components of Ш are trivial for $p$ outside a finite set of primes?.
In particular, $Ш(E/\mathbb{Q})[p]$ is trivial for $p$ $\neq$ 2, 3, 5, 7.

REPLY [17 votes]: No, there are no such examples known. In fact, with the current technology, the two questions are more or less equally hard. That's because for any given prime $p$, you can, in principle, establish finiteness of $Ш(E/\mathbb{Q})[p^\infty]$ algorithmically by performing $p^n$-descent for higher and higher $n$, until the upper bound on the rank of $Ш(E/\mathbb{Q})[p^n]$ stabilises. Of course, we cannot prove a priori that this would ever happen, but in practice, if you knew finiteness of $p$-primary parts of sha outside a finite set of primes, you would run your computer to do $p^n$-descent for the remaining primes, until you establish finiteness for this finite set, too.<|endoftext|>
TITLE: OCR for handwritten mathematics
QUESTION [24 upvotes]: I am in the process of scanning a large collection of handwritten notes.  They consist of diagrams and formulae with a relatively small proportion of actual words.  Of course it would be hopeless to get an OCR program to digest the diagrams or formulae, but it would be useful if I could get one to find and transcribe enough of the words to build an index.  Has anyone tried this kind of thing?

REPLY [9 votes]: To supplement Bob Terrell's post, here is an example from InftyReader.
Snippet from an input image, scanned at 600dpi:
  
Snippet from output of the corresponding LaTeX produced by InftyReader:
  
It's not perfect—$\partial z \partial \bar{z}$ becomes $\partial z k$—but it's pretty impressive!<|endoftext|>
TITLE: Compact space in site -> compact object in topos
QUESTION [9 upvotes]: Given a site $C$, there are various standard notions for an object $X \in C$ being compact. For instance:

Every covering family $\lbrace U_i \to X \rbrace$ has a finite subfamily that is still covering.
The functor $C(X,-)$ commutes with filtered colimits.
After Yoneda-embedding, the functor $Sh_C(X, -)$ commutes with filtered colimits.
After $\infty$-Yoneda-embedding, the functor $\infty Sh_C(X, -)$ commutes with filtered $\infty$-colimits.

These notions are closely related but subtly different. For instance for $C = Top$ it is well known that the first two are not equivalent without further fine-tuning.
What can one say about the relation of 1. to 3. and 4. ?
It seems to me that one can say for instance: compactness in the first sense implies that $Sh_C(X,-)$ commutes with mono-filtered colimits, and this should generalize to the $\infty$-case in the suitable sense.
What else can one say?

REPLY [9 votes]: The question that you ask might be better phrased intrinsically without referring to sites. Fix an $0 \leq n \leq \infty$ and let $\mathbf{X}$ be an $n$-topos. Recall that an object $X \in \mathbf{X}$ is quasi-compact if for every effective epimorphism of the form $\coprod_{i \in I} U_i \to X$ there is a finite subset $I_0 \subseteq I$ such that $\coprod_{i \in I_0} U_i \to X$ is an effective epimorphism. On the other hand, let us say that $X \in \mathbf{X}$ is small if the functor $Map_{\mathbf{X}}(X,-)$ preserves filtered colimits. Your question can then be phrased as follows:
What is the relation of between the property of being quasi-compact and the property of being small?
Using the fact that colimits in an $n$-topos are universal one can easily verify that $X \in \mathbf{X}$ is quasi-compact or small if and only if the terminal object in $\mathbf{X}_{/X}$ is quasi-compact or small respectively. We can hence assume that $X$ itself was the terminal object $\ast \in \mathbf{X}$. Now it is not hard to prove that $\ast \in \mathbf{X}$ is quasi-compact if and only if $\ast$ is small when regarded as an object of the underlying $0$-topos of $\mathbf{X}$ (i.e. the full subcategory spanned by $(-1)$-truncated objects). Hence we see that $\ast$ being small implies $\ast$ being quasi-compact. However, when $n > 0$ the inverse implication is false in general.
Here is a counterexample that works for every $0 < n \leq \infty$. Let $G$ be a free group on an infinite set of generators and let $\mathbf{X}$ be the $n$-topos of $(n-1)$-types equipped with a $G$-action. Then the terminal object $\ast$ is quasi-compact but the functor represented by $\ast$ is the homotopy fixed point functor, which does not commute with filtered colimits for such $G$.<|endoftext|>
TITLE: Ideals of $C(X)$ with only finitely many number of zerosets
QUESTION [5 upvotes]: We denote the rings of all real valued continuous functions on compeletely regular Hausdorff space $X$ by $C(X)$.Let $I$ be a ring ideal of $C(X)$. define $$Z[I]:=\lbrace Z(f):\;f\in I\rbrace$$ where  $Z(f):=\lbrace x \in X:\;f(x)=0\rbrace$. We are interested in knowing about the relation  about algebraic properties of the ring $C(X)$ and topological properties of the space $X$. I have a question about the relation between the number of zero sets of $I$ in the finite case and it's relation with the topological space $X$. Let me pose my questions in the definite way.
Question1:  for which positive integer $n$ we can have a ring ideal $I$,  that $Z(I)$ contains only n elements?
Question2: If we have an ideal $I$, such that $|Z(I)|=n$ for some positive integer $n$, can we characterize the ideal $I$?
Question3: for which condition on the space $X$, we have an ideal $I$, with $|Z(I)|=n$, for some positive integer $n$?

REPLY [5 votes]: Edit: In the first version I had the additional assumption that each singleton in $X$ is a zero set. But as observed by AliReza Olfati the proof can be adapted to work in the general case. Therefore we have: 

$Z(I)$ is finite if and only if the following holds: 

There is $A \subseteq X$ closed and $x_1,...,x_m \in X\setminus A$ with $X=A\cup  \lbrace x_1,...,x_m\rbrace$.
$Z(I) = \lbrace A \cup T \mid T \subseteq \lbrace x_1,...,x_m\rbrace\;\rbrace$. In particular $|Z(I)|=2^m$ is a power of $2$.
$I=(1-\delta_{A \cup T} \mid T \subseteq \lbrace x_1,...,x_m\rbrace)$ 


Proof: $(\Rightarrow)$ Let $Z(I) = \lbrace Z(f_1),...,Z(f_n)\rbrace$. 
First note that $Z(f^2 + g^2) = Z(f) \cap Z(g)$. Hence $Z(I)$ is closed under $\cap$ and we have $$A := Z(f_1) \cap ... \cap Z(f_n) \in Z(I).$$
Let $A=Z(f_0)$.  
Next let's show that $X \setminus A$ is finite. Suppose $X \setminus A$ is infinite. Since $X$ is Hausdorff and $A$ is closed, then there is a sequence of closed subsets $A \cup \lbrace x_1,...,x_k\rbrace$. Hence it's enough to show that there are only finitely many closed subsets $C \supseteq A$ in $X$. Since $X$ is completely regular, such a $C$ is the intersection of zero sets (Gillman-Jerison: Rings of continuous functions, Theorem 3.2). Thus we have a surjection 
$$\lbrace Z \subseteq X \mid Z \supseteq A \text{ zero set }\;\rbrace\to \lbrace C \subseteq X \mid  C \supseteq A \text{ closed }\rbrace,\; \mathfrak{Z} \mapsto \bigcap_{Z \in \mathfrak{Z}}Z.$$
Let $Z=Z(h) \supseteq A$ be a zero set. Then $f_0h \in I$ with $Z(f_0h)=Z(f_0) \cup Z(h) = Z$, i.e. $Z \in Z(I)$. Hence the LHS of the map is just the finite set $Z(I)$ and consequently its image is also finite. Hence the finiteness of $X \setminus A$ and 1) are  shown. 
Let $X \setminus A = \lbrace x_1,...,x_m\rbrace$. Since $A$ is closed and $X$ is Hausdorff it follows that $A$, $\lbrace x_i\rbrace$ are both, open and closed. Hence a function $f: X \to \mathbb R$ is continuous iff $f|A$ is continuous. In particular we find continuous $h$ with $Z(h)=\lbrace x_i\rbrace$. 
To see 2) let $f \in I$. By definition $A \subseteq Z(f)$. Thus $T := Z(f) \setminus A \subseteq \lbrace x_1,...,x_m\rbrace$ and $Z(f)=A \cup T$ is in the RHS.  Conversely, if   $T = \lbrace x_{p_1},...,x_{p_k}\rbrace$  then choose $h_j \in C(X)$ with $Z(h_j) = \lbrace x_{p_j}\rbrace$. Now $f := f_0h_1\cdots h_p \in I$ and $Z(f) = A \cup T \in Z(I)$. 
3) Note that the functions used in the following are continuous by the remark preceding the proof of 2). Let $f \in I$ and let $X \setminus Z(f) = \lbrace x_{p_1},...,x_{p_k}\rbrace$. Then 
$f= \sum_j f(x_{p_j})(1-\delta_{X \setminus \lbrace x_{p_j} \rbrace})$ is contained in the RHS. Conversely, let $T \subseteq \lbrace x_1,...,x_m\rbrace$ be given. By 2) there is $f \in I$ with $Z(f) = A \cup T$. Define a continuous map $h$ by $h|Z(f) = 0$, $h(x) = 1/f(x)$ if $x \notin Z(f)$. Then $1-\delta_{A \cup T} = fh \in I$. 
$(\Leftarrow)$ If $X$ is given by 1) and $I$ by 3) then $Z(I) = \lbrace A \cup T \mid T \subseteq \lbrace x_1,...,x_m\rbrace\;\rbrace$ follows easily. q.e.d.<|endoftext|>
TITLE: Does this knot invariant distinguish trefoil chiralities?
QUESTION [27 upvotes]: Let $C_N$ denote the labelled configuration of $N^{th}$ roots of unity with $p_J = e^{\frac{2\pi iJ}{N}}$ for $J = 1\ldots N$.
As a corollary of something else I was playing around with, I recently proved the following:
Theorem: Every tame knot (or link) $K$ has a (not necessarily minimal) stick presentation which such that the sticks can be projected onto the set of chords of some $C_N$ with the following crossing condition: Whenever the projection has chords $p_{J_1}p_{J_3}$ and $p_{J_2}p_{J_4}$ such that $1\leq J_1 < J_2 < J_3 < J_4\leq N$ then $p_{J_1}p_{J_3}$ crosses in front of $p_{J_2}p_{J_4}$. (In other words for any two intersecting chords in the projection, the chord which has the lowest numbered endpoint passes in front of the other chord).
Germane to this question is the fact that the theorem leads to a knot invariant which, for lack of a better name, I will call the circular stick number of $K$ and which is defined to be the minimum $N$ needed to obtain a projection of $K$ with the above properties.
My proof of the above theorem was very non-constructive, so I wanted to see some concrete realizations of such projections. And after some work, I was able to find that for the left trefoil knot, the circular stick number is 7. After much more playing around on $C_7$ and not finding a projection of the right trefoil, I moved on to $C_8$ where I was able to find a projection of the right trefoil.

Main Question: Does this invariant actually distinguish chiralities of the trefoil, or can someone find a projection of the right trefoil with the above properties on $C_7$? If it doesn't distinguish chiralities in this case, is it possible it distinguishes chiralities of some other pair of knots, or does anyone see a slick way to see that it cannot distinguish chiralities?
Secondary Question: (Mainly for knot theorists, or anyone with a deeper knowledge of knots than I) Has this invariant been studied before, and if so, what is the terminology used for it?

I am fairly sure that the general answer to the main question is strongly related to the characterization (and in particular the disjointness) of the sets of forbidden minors of the following two sets (which are almost certainly not known in general as the Robertson-Seymour theorem is non-constructive):
$\bullet$ Graphs which are $L$-lessly embeddable for a given chiral knot $L$
$\bullet$ Graphs which are $L^+$-lessly and $L^-$-lessly embeddable for a given chiral knot $L$ with chiralities $L^+$ and $L^-$
Note that graphs in the second set may admit embeddings in $\mathbb{R}^3$ containing either $L^+$ or $L^-$, but have at least one embedding which does not contain both $L^+$ and $L^-$. As of yet, I have not been able to flesh out a proof using this approach however.
For those who don't want to search for the solutions I found, for the left trefoil the points of $C_7$ are connected in the following order:
$p_1 \rightarrow p_3 \rightarrow p_5 \rightarrow p_7 \rightarrow p_2 \rightarrow p_4 \rightarrow p_6 \rightarrow p_1$
For the right trefoil the points of $C_8$ are connected in the following order:
$p_1 \rightarrow p_3 \rightarrow p_7 \rightarrow p_5 \rightarrow p_2 \rightarrow p_8 \rightarrow p_4 \rightarrow p_6 \rightarrow p_1$
 EDIT: Here are pictures of these two projections to help clarify the situation:
https://docs.google.com/open?id=0B5BVGcL23IkoSTlNNGZKZnZtT1E
With regards to Dylan's question about how this projection arose, I was considering the parametric curve $S(t) = (t,t^2,t^3) \subset \mathbb{R}^3$. In a comment or answer on a past MO post which for the love of me I cannot find now, someone had shown in a very simple way that no two chords of $S$ intersect one another anywhere in $\mathbb{R}^3$ (unless they share an endpoint). Thus, for each $n$, $K_n$ can be embedded in $\mathbb{R}^3$ as the set of chords connecting the points $(1...n)$. Now for a fixed tame knot or link $K$, the Robertson-Seymour theorem implies there is a finite set of forbidden minors for graphs which are $K$-lessly embeddable. Hence, every embedding of $K_n$ for $n$ sufficiently large contains $K$, so one may ask (like I did) "what $n$ is sufficient for a given $K$, and what order must I connect the points $(1...n)$ in order to realize $K$?" The crossing condition comes from looking at the chords projected onto the $yz$-plane as viewed from the $-x$ direction (which I think is the direction I did my crossing calculations from). Finally, I moved the $yz$-projections of the integral $t$-valued points around so that they lay on the unit circle to make the pictures easier/clearer.
 UPDATE: I was just skimming through a paper on some results on knots which were unrelated to this one. That paper referenced the Ramsey number $r(L)$ of a link $L$, and upon following its references I found this paper which proves the existence of $r(L)$ using essentially the representation I layed out above:
http://www.ams.org/journals/tran/1991-324-02/S0002-9947-1991-1069741-9/S0002-9947-1991-1069741-9.pdf
So it would appear the answer to Question 2 is this is known (in at least one paper) as a plat representation (much to my surprise, Negami derives the existence of such representations using essentially the argument I gave above). So that would seem to settle this thread entirely...

REPLY [20 votes]: I'm very curious where this came up.  In any case, the answer to the first question is yes, it does distinguish these trefoils; you found the minimal representatives.
Let $a_0,\dots,a_{N-1}$ be the roots of unity that are visited along the knot, in (cyclic) order.  Suppose we have a minimal representative for some non-trivial knot.  Then we cannot have $|a_k - a_{k+1}| = 1$ for any $k$, as otherwise we could replace this pair $a_k, a_{k+1}$ by a single root of unity (for $N-1$), adjusting the other roots of unity as appropriate.  A little more subtly, we cannot have $|a_{k-1} - a_{k+1}| = 1$ either, as then we could again delete $a_k$ from the sequence to get a smaller representation.  With these simple constraints, the smallest possible sequence for a non-trivial knot is the one you found for one of the trefoils with $N=7$.  There are several possibilities for $N=8$, including the one you found for the other trefoil.  I've included a very short Haskell program below that computes this.  The possibilities for $N=8$ are
$$
(2,7,5,3,1,6,4,0)\quad
(2,5,7,3,1,6,4,0)\quad
(3,6,1,4,7,2,5,0)\quad
(2,6,4,1,7,3,5,0)
$$
$$
(3,1,6,4,2,7,5,0)\quad
(2,4,6,1,3,7,5,0)\quad
(3,5,1,7,4,2,6,0)\quad
(4,2,7,5,1,3,6,0)
$$
$$
(3,1,5,7,2,4,6,0)\quad
(5,3,1,6,4,2,7,0)\quad
(2,4,6,1,3,5,7,0)
$$
For the second question, I have never heard of this representation before.
Here is the code, for anyone interested.
 -- A (partial) circular stick representation is a list of integers,
 -- the order of the roots of unity to visit in order
 type CircStick = [Int]

 -- The next element ak after a partial representation a1, ..., a{k-1} 
 -- must satisfy
 --   (a) ak has not already been seen
 --   (b) |ak - a{k-1}| > 1
 --   (c) |ak - a{k-2}| > 1
 -- There are a few more "easy" constraint, eg the first and last entries
 -- cannot differ by one.  We do not impose those constraint here.
 nexts :: Int -> CircStick -> [Int]
 nexts n [] = [0]
 nexts n [a1] = filter (\a -> abs (a-a1) > 1) [0..n-1]
 nexts n (a1:a2:as) =
   filter (\a -> not (elem a as)) $
       filter (\a -> abs (a-a1) > 1) $
   filter (\a -> abs (a-a2) > 1) $
     [1..n-1]

 completions :: Int -> CircStick -> [CircStick]
 completions n as | length as >= n = [as]
 completions n as =
   concat [completions n (a:as) | a <- nexts n as]

 -- Impose final constraints:
 --    (a) Last entry cannot be 1
 --    (b) Take entry that is lexicographically less than its reverse
 --    (c) first and next-to-last entries cannot differ by one
 circSticks :: Int -> [CircStick]
 circSticks n = 
   filter (\as -> abs ((as!!0) - (as!!(n-2))) > 1) $
       filter (\as -> as < tail (reverse as)) $
   filter (\as -> head as /= 1) $
   (completions n [])

Edit: For those interested, here are the 108 possibilities for $N=9$.  I hope there's some way of checking what these are more efficiently than just going through them by hand.
[[2,7,5,3,8,1,6,4,0],[2,7,5,3,1,8,6,4,0],[2,5,7,3,1,8,6,4,0],[2,7,5,1,3,8,6,4,0],[2,5,7,1,3,8,6,4,0],[2,6,8,3,5,1,7,4,0],[2,7,5,3,1,6,8,4,0],[2,5,7,3,1,6,8,4,0],[2,7,5,1,3,6,8,4,0],[2,5,7,1,3,6,8,4,0],[3,8,6,1,4,7,2,5,0],[3,6,8,1,4,7,2,5,0],[3,7,1,4,6,8,2,5,0],[2,8,6,4,1,7,3,5,0],[2,6,8,4,1,7,3,5,0],[2,7,4,1,6,8,3,5,0],[2,4,8,6,3,1,7,5,0],[2,4,6,8,3,1,7,5,0],[3,8,1,6,4,2,7,5,0],[3,1,8,6,4,2,7,5,0],[3,6,1,8,4,2,7,5,0],[3,1,6,8,4,2,7,5,0],[2,8,4,6,1,3,7,5,0],[2,4,8,6,1,3,7,5,0],[2,6,4,8,1,3,7,5,0],[2,4,6,8,1,3,7,5,0],[3,7,1,4,6,2,8,5,0],[2,7,4,1,6,3,8,5,0],[3,8,5,1,7,4,2,6,0],[3,7,5,1,8,4,2,6,0],[3,5,7,1,4,8,2,6,0],[4,7,1,3,5,8,2,6,0],[4,1,7,3,5,8,2,6,0],[4,8,2,5,7,1,3,6,0],[4,7,2,5,8,1,3,6,0],[4,2,7,5,1,8,3,6,0],[2,4,7,1,5,8,3,6,0],[2,8,5,3,7,1,4,6,0],[2,7,5,3,8,1,4,6,0],[3,8,1,5,7,2,4,6,0],[3,7,1,5,8,2,4,6,0],[2,7,5,3,1,8,4,6,0],[2,5,7,3,1,8,4,6,0],[3,1,7,5,2,8,4,6,0],[4,2,7,5,3,1,8,6,0],[3,5,1,7,4,2,8,6,0],[4,2,7,5,1,3,8,6,0],[2,4,7,1,5,3,8,6,0],[3,1,5,7,2,4,8,6,0],[5,8,3,1,6,4,2,7,0],[5,3,8,1,6,4,2,7,0],[3,5,8,1,6,4,2,7,0],[5,3,1,8,6,4,2,7,0],[3,5,1,8,6,4,2,7,0],[5,1,3,8,6,4,2,7,0],[5,3,1,6,8,4,2,7,0],[5,1,3,6,8,4,2,7,0],[4,6,1,3,8,5,2,7,0],[4,1,6,3,8,5,2,7,0],[4,6,2,8,5,1,3,7,0],[5,8,2,4,6,1,3,7,0],[5,2,8,4,6,1,3,7,0],[2,6,4,8,1,5,3,7,0],[4,2,6,8,1,5,3,7,0],[2,4,6,8,1,5,3,7,0],[2,6,4,1,8,5,3,7,0],[2,4,6,1,8,5,3,7,0],[2,5,8,3,6,1,4,7,0],[5,3,1,8,6,2,4,7,0],[3,5,1,8,6,2,4,7,0],[5,1,3,8,6,2,4,7,0],[5,3,1,6,8,2,4,7,0],[5,1,3,6,8,2,4,7,0],[4,2,8,6,3,1,5,7,0],[2,4,8,6,3,1,5,7,0],[4,2,6,8,3,1,5,7,0],[2,4,6,8,3,1,5,7,0],[3,6,1,4,8,2,5,7,0],[3,1,6,4,8,2,5,7,0],[2,8,4,6,1,3,5,7,0],[4,2,8,6,1,3,5,7,0],[2,4,8,6,1,3,5,7,0],[2,6,4,8,1,3,5,7,0],[4,2,6,8,1,3,5,7,0],[2,4,6,8,1,3,5,7,0],[2,6,4,1,8,3,5,7,0],[2,4,6,1,8,3,5,7,0],[5,7,3,1,6,4,2,8,0],[4,6,1,3,7,5,2,8,0],[5,3,7,1,4,6,2,8,0],[3,5,7,1,4,6,2,8,0],[6,4,2,7,5,1,3,8,0],[5,7,2,4,6,1,3,8,0],[6,2,4,7,1,5,3,8,0],[2,6,4,1,7,5,3,8,0],[5,2,7,4,1,6,3,8,0],[6,3,1,5,7,2,4,8,0],[6,1,3,5,7,2,4,8,0],[2,7,5,3,1,6,4,8,0],[2,5,7,3,1,6,4,8,0],[3,6,1,4,7,2,5,8,0],[6,2,4,7,1,3,5,8,0],[2,6,4,1,7,3,5,8,0],[4,2,7,5,3,1,6,8,0],[5,3,1,7,4,2,6,8,0],[3,5,1,7,4,2,6,8,0],[4,2,7,5,1,3,6,8,0],[3,1,5,7,2,4,6,8,0]]<|endoftext|>
TITLE: Sheafification - Why does twice suffice?
QUESTION [15 upvotes]: Hi,
I'm currently reading through "Sheaves in Geometry and logic" by Mclane-Moerdijk and this one issue has been bugging me for a long time, which I hope you could help me resolve. 
It is known that for a general presheaf on a Grothendieck Topology, we must in general apply the plus construction twice to obtain a sheaf. The first application turns an arbitrary presheaf into a separated presheaf, and one more application gives a sheaf. So, exactly, intuitively, what obstructs us from getting a sheaf from just one application of the plus construction when the sheaf is non-separated? Let us take the example:
What is an example of a presheaf P where P^+ is not a sheaf, only a separated presheaf? 
given by Sherry. When we apply it once we get a separated presheaf, OK. But what exact component of that presheaf hindered us from getting the sheaf we wanted? I agree with what Sherry wrote in that case, namely that : "So in our example, 1 and 3, over ABC and BCD, in our original presheaf were compatible on a refinement of BC but not on BC"
Can we generalize this notion to become rigorous in the case for arbitrary non-separated presheaves?

REPLY [19 votes]: "Why" questions can often be answered in multiple ways; I'll give an answer that's different from the other good comments and answers that you've already had.  
Sheaves are the first rung on an infinite ladder of concepts.  The next rung is "stack".  A stack is something like a "sheaf of categories", but there are added complications.  A typical example of a stack on a topological space $X$ is the assignment $U \mapsto Sh(U)$, sending each open subset $U \subseteq X$ to the category $Sh(U)$ of sheaves on $U$.  Sheaves on subsets of $X$ can be patched together, not quite uniquely, but uniquely up to canonical isomorphism.  From this, you can work out what the definition of stack must be.  This appears as an exercise somewhere in Mac Lane and Moerdijk.  
After stacks, there are 2-stacks (something like sheaves of 2-categories), 3-stacks, and so on.  (According to this terminological scheme, stacks are 1-stacks and sheaves are 0-stacks.)  And just as sheaves are presheaves satisfying certain conditions, $n$-stacks are "$n$-prestacks" satisfying certain conditions.
The point now is that "twice suffices" fits into a larger pattern.  For stacks, you need to apply (a version of) the plus construction three times.  For 2-stacks, you need to apply it four times.  In general, for $n$-stacks, you need to apply it $n+2$ times.
(I confess that this is something I've only heard in conversation, I believe from Andrew Kresch. I don't know the details.)<|endoftext|>
TITLE: Books on logic for someone aiming to go to grad school in the field?
QUESTION [5 upvotes]: I have taken two introductory courses on logic. One was an undergraduate level and the second one was at the graduate level. Both used a set of notes written by the instructor. I'm thinking about graduate school in the field and would like to spend my next summer reading more about logic.
What I'm looking at is a good list of rigorous graduate level textbooks on the most important topics in logic that would get one started and which would provide a good foundation for graduate level studies. I have the whole summer and my senior year to prepare.
Which books would you recommend? My plan at this point would be to read a rigorous graduate level textbook on basic logic and then one on model theory. The ones that I've looked at are: 
Manin, A Course in Mathematical Logic for Mathematicians
Marker, Model Theory: An Introduction
Would this be a nice place to start or are there better options?

REPLY [6 votes]: For set theory, I recommend Drake's "Set theory: An introduction to large cardinals" (which also contains a good treatment of various other parts of set theory, before concentrating on large cardinals) and Kunen's "Set theory: An introduction to independence proofs."  The "bible" of set theory is Jech's "Set Theory" but I think it may be easier to learn from Drake and Kunen first.  [Do you get the impression that set theorists are not very creative when it comes to book titles?]
For logic in general, Shoenfield's "Mathematical Logic" covers an amazing amount of material in rather few pages.  Most of it is great (if you do the exercises, into which a lot of material has been squeezed) but I wouldn't recommend his approach to constructibility and forcing --- use Kunen instead.<|endoftext|>
TITLE: Finding lots of discrete vectors in fairly general position
QUESTION [15 upvotes]: How many vectors can there be in $\mathbb{F}_2^{2n}$ such that no $n$ of them form a linearly dependent set? The bounds I have so far are embarrassingly far apart, though that probably means I should have thought about the question for longer before posting it.
To get an upper bound, observe that you can partition $\mathbb{F}_2^{2n}$ into $2^{n+2}$ translates of an $(n-2)$-dimensional subspace. If you choose more than $(n-1)2^{n+2}$ vectors, then $n$ of them must lie in one of those translates, and therefore in an $(n-1)$-dimensional subspace. So you definitely can't choose more than $Cn2^n$ vectors with the required property.
In the other direction, if you choose $M$ vectors randomly, then the probability that some fixed set of $n$ of them lives in an $(n-1)$-dimensional subspace is at most $n2^{-n}$ (since one of them must lie in the linear span of the others). So the expected number of problematic sets of size $n$ is at most $\binom Mn n2^{-n}$. If this is at most $M/2$, then we can get rid of a vector from each problematic set and we end up with no such sets. But for $n\binom Mn$ to be less than $2^n$ we basically need $M$ to be proportional to $n$, so this gives a lower bound of something like $2n$, which is pathetic as we could have just taken $2n$ linearly independent vectors. 
I end up with a similarly pathetic bound if I try to pick vectors one by one, always avoiding the subspaces that the previous vectors require me to avoid.
I think I'm slightly more convinced by the lower bound, pathetic as it is. My rough reason is that the difficulty I run into feels pretty robust, and also that the result I prove in the upper bound is much stronger than it needs to be (since the subspace I obtain is essentially a translate of some fixed subspace). But basically I can't at the time of writing see even roughly what the bound should be.

REPLY [14 votes]: Update 2: Since there seems some to be some interest in small values, in addition to asymtotic results, at the end some data that can, using the interpretation detailed below, be obtained from tables of code parameters to be found at http://www.codetables.de/ 
Update:  the number of such vectors should be at most $(2 + \epsilon)n$ for any $\epsilon > 0$ if $n$ is large enough.
The approach is as detailed below, but instead of (or in some sense in addition to) the Hamming bound we use McEliece--Rodemich--Rumsey--Welch bound which gives $$R \le h (1/2  - \sqrt{\delta (1 - \delta)}) +o(1)$$ with notation as below. Now, since $R = 1 - 2 \delta$ and ignoring the $o(1)$ one gets that the only positive $\delta \le 0.5$ for which this holds is in fact $0.5$.
(I did not 'prove' this, but the functions seems nice enough to rely on computer aid and it should also be not too hard and is likely even written somewhere that this is so; for an illustration see for example the figure on page 2 in lecture notes of Atri Ruda; the Plotkin bound illustrates the $1 - 2 \delta$, the Elias--Bassalygo bound stays below it, and MRRW would at least where it is relevant be still better/smaller; one also sees nicely that the Hamming bound only works until $0.3...$; the GV is a lower bound so not relevant here).
Thus, the ratio $n/N$ needs to approach $0.5$ (we know already it cannot go to zero), and the claim follows. 

While there are already two nice answers, I thought I still sketch how one also could get the linear bound by the approach given in my comment. 
Suppose there is a collection of $N$ such vectors. We seek an upper bound for $N$, so one can assume they generate the space as otherwise on could take add some vectors.
Now let $H$ be the matrix whose columns are these vectors. The code with this check matrix will be an $[N,N-2n]$ binary lineary code with minimal distance greater $n$.
Write $R = (N-2n)/N$ and $\delta = n/N$.  
By the Hamming bound one knows that $R \le 1- h(\delta/2) + o(1)$, as $N$ tends to infinity, where $h$ is the binary entropy function. Since $R = 1 - 2 \delta $ one gets that
$h(\delta/2) \le 2 \delta + o(1)$. 
Since the derivative of $h$ at $0$ is $+\infty$ one gets that $\delta$ needs to stay away from $0$ showing the ratio $n/N$ stays bounded away from $0$. 
Moreover, 'solving' the equation for $\delta$ would yield an actual  constant.
Possibly using more sophisticated bounds from coding theory could yield a better constant; I will try when I have better access to a CAS and report if I find something.    

Some small values, up to 20 and then selcted others, the first two match those given in comments by Gerry Myerson and Robert Israel. I retrieved and typed them quickly so small errors are possible; format (n , max number of vectors); if the later is not just a number, it is to be interpreted as bounds in the obvious way. [In case somebody would now how to format this better, please let me know.]
(3,32), (4,17), (5,24), (6,24), (7,28), (8,23),  (9,28), (10,31), (11,34), (12,32), (13, 35-38), (14,37),, (15,40),  (16,39), (17,43-45) , (18,44-48), (19,48-50), (20,48)
(25,59-60), (30,70), (40,88-89), (50,109-110), (100, 209-211)

REPLY [12 votes]: It seems to me that Robert Israel's argument generalizes. The sum of any $\frac{n}{2}$ vectors in your collection has to be distinct, in order for any $n$ to be linearly independent. From this one gets the inequality 
$$\binom{M}{n/2}\le 2^{2n}-1$$
which in particular implies $M\le O(n)$.

REPLY [12 votes]: $\def\FF{\mathbb{F}_2}$ Indeed, the linear bound is close to right. I can show that we can't beat about $3.197 n$. 
For convenience, set $m=n/2$. Fix a constant $c$ and suppose that we can find $\geq cn$ vectors for infinitely many $n$. Notice that every $m$ element subset of our $cn$ vectors must have a different sum, or we could find a relation with at most $n$ terms. So
$$\binom{cn}{m} \leq 2^{2n}$$
or
$$(2cm)(2cm-1) \cdots (2cm-m+1) \leq 2^{2n} m! \leq 16^m m^m (e+o(1))^{-m} .$$
$$(2c)(2c-1/m)(2c-2/m) \cdots (2c-1+1/m) \leq (16 e^{-1} (1+o(1)))^m.$$
Taking logs
$$\frac{1}{m} \sum_{i=0}^{m-1} \log (2c-i/m) \leq \log 16 -1 + o(1).$$
or, sending $m \to \infty$. 
$$\int_{2c-1}^{2c} \log t dt \leq \log 16 - 1.$$
I get that this forces $c \leq 3.1965677$.<|endoftext|>
TITLE: Maximum singular value of a random $\pm 1$ matrix
QUESTION [11 upvotes]: Define a matrix $\mathbf{A} \in \mathbb{R}^{m \times n}$ such that each element is independently and randomly chosen with probability $\frac 12$ to be either $+1$, or $-1$. Do you know any result in the literature that talks about properties of this kind of matrices?  
I have seen that there are some results for other kind of random matrices (for example matrices whose entries are i.i.d. Gaussian.) but not for this simple matrix of $\pm 1$.
I would be interested for example on the distribution of the $\sigma_{\max}(A)$, but not in an asymptotic regime, as $m$, $n$ are finite numbers and usually small in my case.
Thank you very much for any pointer or any thoughts.

REPLY [3 votes]: In relation to the first question, a recent breakthrough result of Konstantin Tikhomirov answers affirmatively an old conjecture attributed to Von Neumann, namely that for $M_n$, an $n\times n$ random matrix with independent $\pm 1$ entries
$$\mathbb{P}(M_{n}\ \text{is singular})=\left(\frac{1}{2}+o(1)\right)^n.$$
Here is the pre-print and a relevant entry in Gil Kalai's blog can be found here.<|endoftext|>
TITLE: connected compact semisimple lie group finite fundamental group
QUESTION [7 upvotes]: I was told that the fundamental group of a connected, compact, semisimple Lie group is finite, with the outline of a possible way to prove this fact.  Is there any source however that fleshes this out in detail / are there several ways to prove this fact?
Thanks!
(The result is often known as Weyl's theorem, I think for his take on the proof, Knapp provides a fairly detailed exposition of his perspective.)

REPLY [4 votes]: Every connected Liegroup, which has a semisimple Liealgebra with a definite Killing form is compact.
The Liealgebra of a compact Liegroup is always the direct sum of an semisimple and abelian Liealgebra, where the killing form of the semisimple part is negative definite.
So we can conclude, that the universal cover of your liegroup is compact and the finiteness of the fundamentalgroup follows immediately.
The fundamentalgroup is even always abelian.
The first two things can be found nicely in Serre's "Lie Algebras and Lie Groups" in Theorem 6.2 and 6.3.<|endoftext|>
TITLE: what's the motivation of Weyl calculus ?
QUESTION [11 upvotes]: In the pseudo-differential operator theory, we can define a  pseudo-differential operator by $$a(x,D)u=(2\pi)^{-n}\int{a(x,\xi)e^{i\langle x-y,\xi \rangle}u(y)dyd\xi}$$ with $a(x,\xi)$ belong to some particular function space (denoted by $S^m$).In the Weyl calculus one adopts the symmetric compromise  $$a^{w}(x,D)u=(2\pi)^{-n}\int{a((\frac{x+y}{2}),\xi)e^{i\langle x-y,\xi \rangle}u(y)dyd\xi}$$ again defined in the weak sense. From this one can see that the adjoint of $a^w$ is equal to $\bar a^{w}$. In particular, $a^w$ is its own adjoint when a is real valued. Is this convenience making Weyl calculus more applicable for physics? In mathematics, are there other reasons to the motivation of Weyl calculus? Furthermore, Can anyone show some problems which are solved by using this tool?

REPLY [6 votes]: In the Weyl calculus there is closer agreement between operator and symbol composition than in the standard Kohn-Nirenberg calculus. For example, $(a^w)^2\equiv (a^2)^w$ holds modulo order zero if $a$ has order one and is real-valued. Similarly, $(f\circ a)^w$ approximates $f(a^w)$, which is defined by the functional calculus, to higher order than the corresponding construction using the standard calculus. This is related to the symplectic invariance properties of the Weyl calculus, more specifically to the appearance of the Poisson bracket on the subprincipal level of the composition formula. Note that the Poisson bracket $\{f\circ a,a\}$ vanishes for real-valued symbols $a$.<|endoftext|>
TITLE: Mandelbrot set and analytic functions such that $f(az)=f(z)^2+c$
QUESTION [6 upvotes]: It is well known that the function $f(z)=2\cos(\sqrt {-z})$ (or more accurately the entire function $f(z)=2\sum_{n=0}^\infty \frac{z^n}{(2n)!}$) satisfies such a functional equation, i.e. $f(4z)= f(z)^2-2$ ; it is not hard to show that this is the unique solution with $c=-2$ and $f'(0)=-1$. Patient calculations (see for instance  this note in french), or the classical results of Tan Lei, show that this implies that the small cardioids on the left main antenna of the Mandelbrot set (on the real axis near $-2$) are in position  $-2+3\pi^2/2^{2n+1}+o(4^{-n})$. It would be easy to generalise this result to other Misiurewicz points, if similar functions were known for other values of $c$, as  the small copies of the $M$-set lie similarly near the zeros of these functions (after appropriate rescaling)  ; it is not hard to show their existence and unicity, but have they been studied, and what is known on their zeros?

REPLY [8 votes]: Such functions were studied, since Poincare.
They are called Poincare functions, or "global linearizers".
Let $f$ be a rational function, and consider the functional equation
$F(kz)=f(F(z))$. Suppose $F$ is analytic at $0$. Then $a=F(0)$ must be a fixed point of $f$.
If $F'(0)$ not $0$, then $k=f'(0)$ by the chain rule.
If all this is so, and $|k|>1$, one can easily show that F exists and is meromorphic in the whole
plane. When $f$ is a polynomial, $F$ is entire. $F$ is uniquely defined by the condition
$F'(0)=0$, if the fixed point of $f$ is given.
The literature on Poincare functions is quite large.
Specially for quadratic polynomials, see the papers of P. Myrberg.
I have several papers on the general rational case.
(All my papers can be found on my web page).<|endoftext|>
TITLE: Transgressions commute with the Steenrod operations on the base and fiber in a central group extension?
QUESTION [8 upvotes]: The following sentence is quoted from the paper ON THE COHOMOLOGY OF SPLIT EXTENSIONS by D. J. BENSON AND M. FESHBACH:
In general, the differentials in the Lyndon-Hochschild-Serre spectral sequence
for a group extension are difficult to understand. In the case of a central extension,
the $E_2$ page is easy to calculate and the transgressions may be computed
using the fact that they commute with the Steenrod operations on the base and
fiber.
May I know what is the complete statement of the result "the transgressions commute with the Steenrod operations on the base and fiber" and which reference can I refer to for it? And why this makes calculation easy? Can anyone give some examples illustrating that?

REPLY [8 votes]: The statement in question refers to the Kudo-Serre transgression theorem: If $E_r$ is, for example, the LHS spectral sequence of a group extension and $x \in E_{2k+1}^{0,2k}$ is transgressive with $d_{2k+1}(x)=y$, then $$d_i(x^p)=0\;\;(i=2k+1,...,2kp)\quad \text{and }\;\; d_{2kp+1}(x^p)=P^ky$$ where $P^k$ denotes the Steenrod power operation  (a textbook reference is McCleary: A User's guide to spectral sequences, Theorem 6.14). 
I'll give an example that is related to the Benson-Feshbach paper: Let $p$ be a prime and let $$1 \to C \to G \to Q \to 1$$ be a central extension with $C \cong (\mathbb{Z}/p)^n$ elementray abelian. If $k$ is a field of char. $p$ (with trivial $G$-action), than $Q$ acts trivially on $H^\ast(C;k)$ and by the universal coefficient theorem, the LHS spectral sequence satisfies 
$$E_2^{ij} \cong H^i(Q;k) \otimes_k H^j(C;k).$$
Moreover, $E_2 \cong H^\ast(Q;k) \otimes_k H^\ast(C;k)$ is an isomorphism of $k$-algebras (up to a sign). This is what is ment by B-F when they say the $E_2$-page is easy to compute. 
The cohomology of $C$ is given by $H^\ast(C;k) = k[x_1,...,x_n,y_1,...,y_n]$ with $\deg(x_i)=1, \deg(y_j)=2$ and the relations $x_i^2=y_i$ if $p=2$ and $x_i^2=0$ for $p$ odd. Note that $y_i = \beta(x_i)$ where $\beta$ is the Bockstein homomorphism. 
In terms of the spectral sequence, we have $x_i \in E_2^{0,1}$. Hence $d_2(x_i) =: z_i \in E_2^{2,0}=H^2(Q;k)$, i.e. $x_i$ is transgressive. Now Kudo's theorem says that $d_2, \beta$ commute, i.e.
$$d_2(y_i) = d_2(\beta(x_i))=\beta(d_2(x_i))=\beta(z_i) \in H^3(Q;k).$$
Hence $y_i$ is again transgressive and by Kudo's theorem $d_5(y_i^2)=P^1\beta(z_i).$ Continuing this way, one obtains 
$$d_{2p^k+1}(y_i^{p^k})=P^{p^{k-1}}P^{p^{k-2}}\cdots P^1\beta(z_i).$$
As a result, the differentials on the fiber $E_\ast^{0,\ast}$ are completely determined by the values of $z_i \in H^2(Q;k)$ (the base) under the power operations.<|endoftext|>
TITLE: Non-commutator in simple group?
QUESTION [12 upvotes]: Hi,
For a group $G$, we say that $x\in G$ is a commutator if there exists $a,b \in G$ such that $x=a^{1}b^{-1}ab$, and we say that $x$ is a non-commutator if there is no $a,b \in G$ such that $x=a^{1}b^{-1}ab$.
Does there exist a non abelian simple group $G$ (finite or not) such that $G$ has at least one non-commutator?
I tried with $-I_n$ in $\mathrm{PSL}(n,q)$ but no luck.
Thanks.

REPLY [18 votes]: The proof of the Ore conjecture was recently completed by Liebeck, O'Brien, Shalev and Tiep. The proof depends heavily on the classification of simple groups.
If one weakens the condition that $G$ is nonabelian simple and assumes the much weaker condition that $G' = G$, then lots of examples exist where G contains noncommutators.
See, for example, my note in the MAA Monthly 84 (1977) 720-722.
Finally, I mention a character-theoretic condition that an element $x$ of $G$ is a noncommutator. It is that $\sum \chi(x)/\chi(1) = 0$, where the sum runs over all $\chi \in {\rm Irr}(G)$. This sum is always positive if $x$ is a commutator.<|endoftext|>
TITLE: iterating ultrapowers of C*-algebras
QUESTION [5 upvotes]: Let $A$ be something interesting like the C*-algebra of compact operators on a separable infinite-dimensional Hilbert space and let $A^1$ be an ultrapower of $A$. Then $A^1$ is a primitive C*-algebra strictly containing $A$ (Ge and Hadwin). Now let $A^2$ be an ultrapower of $A^1$. Then $A^2$ is a primitive C*-algebra containing $A^1$, presumably strictly. In the same way we may define $A^3$, $A^4$, etc. Since $A^n$ contains $A^{n-1}$ we may define an inductive limit $A^{\omega}$, and then continue transfinitely with $A^{{\omega} +1}$ etc.
My question is, Does the process ever stabilise? If so does it stabilise at some finite step, or at $\omega$ or at the first uncountable ordinal?

REPLY [2 votes]: In the C*-algebra ultrapower construction, instead of identifying two sequences if they agree on a set in the ultrafilter, you identify two sequences if their difference goes to zero along the ultrafilter.  You also throw out those sequences whose norm goes to infinity along the ultrafilter.
I think the simple answer to the question is that any infinite dimensional C*-algebra is properly contained in any ultrapower coming from a free (nonprincipal) ultrafilter.  In fact this will be true of any infinite dimensional Banach space, as a simple consequence of the fact that its unit ball is not precompact.  We can find a sequence of vectors in the unit ball such that the distance between any two of them is at least 1/2, and this will always give rise to a new element in the ultrapower (it is not identified with any constant sequence). So no, the process never stabilizes.<|endoftext|>
TITLE: The deep significance of the question of the Mandelbrot set's local connectedness?
QUESTION [30 upvotes]: I am given to understand that the celebrated open problem (MLC) of the Mandelbrot set's local connectness has broader and deeper significance deeper than some mere curiosity of point-set topology.  
From http://en.wikipedia.org/wiki/Mandelbrot_set I see that conjecture has implications concerning the structure of the Mandelbrot set itself, but I don't think I grasp its broadest implications for complex dynamics and matters beyond.
Request: Could someone explain the proper current context in which to view MLC and/or how the world would look it its full glory if MLC has a positive answer?  Alternatively, please give a pointer to somewhere in the literature that does the same.

REPLY [28 votes]: If a connected compact $K \subset C$ is locally connected then the Riemann map 
$h\colon C \setminus \Delta \to C \setminus K$ extends continuously to $\partial \Delta$. For each $z \in \partial K$, the boundary of the convex hull of $h^{-1}(\{z\})$ is the union of a set $\Lambda_z$ of chords; the union of these $\Lambda_z$ over all $z \in \partial K$ is a closed set $\Lambda_K$ of disjoint chords; it is called a lamination of $\Delta$. We can reconstruct the convex hulls of each $h^{-1}(\{z\})$ from $\Lambda_K$, and when we collapse every convex hull to a point, we obtain a topological model for $K$. 
In the case where $K$ is the Mandelbrot set $M$, the lamination $\Lambda_M$ can be described combinatorially, so MLC would mean that we know the topology of $M$. 
There is a second answer which is more subtle and more important. For each $c \in M$, the filled Julia set $K_c$ of $z \mapsto z^2 + c$ is compact and connected; if it is locally connected the resulting lamination $\Lambda_c \equiv \Lambda_{K_c}$ is, in the right sense, invariant under $z \mapsto z^2$ on $\partial \Delta$. Even if $K_c$ is not locally connected, there is a way of defining what the lamination would be if $K_c$ were locally connected. Every invariant lamination appears as $\Lambda_c$ for some c, and MLC is equivalent to the statement there is a unique $c$ with a given lamination. We think of $\Lambda_c$ as describing the combinatorics of $K_c$, and we think of this uniqueness conjecture as "combinatorial rigidity"---two maps of the form $z \mapsto z^2 + c$ are conformally conjugate (and hence equal) if they are "combinatorially equivalent". 
(Actually, if $z \mapsto z^2 + c$ has an attracting periodic cycle, then the set of combinatorial equivalent parameters form an open subset of $C$, so the statement of combinatorial rigidity must be suitably modified in that case. It is known that structural stability is open and dense in the family of maps $z \mapsto z^2 + c$, so combinatorial rigidity implies that every $z \mapsto z^2 + c$ in this open and dense set must have an attracting periodic cycle; this is the implication that Eremenko alluded to in his answer. )
In this sense MLC is closely analogous to Thurston's Ending Lamination Conjecture (proven by Brock, Canary, and Minsky), which says, broadly speaking, that a finitely generated Kleinian group is determined by the topology of its quotient and the ending laminations of its ends, which are also, when viewed appropriately, invariant laminations of the disk. 
There is a third answer which is more historical and empirical. We can prove MLC and combinatorial rigidity "pointwise" (or "laminationwise") by proving that for a given invariant lamination $\Lambda$, it appears as the lamination $\Lambda_c$ for a single $c$. This has been done in great many cases, first by Jean-Christophe Yoccoz, and then by Mikhail Lyubich, the author of this post, Genadi Levin, and Mitsuhira Shishikura. To prove this combinatorial rigidity for a given $c$ seems to require a detailed understanding of the geometry of the associated dynamical system, and this almost always leads to further results. So proving MLC would most likely mean having a thorough understanding of the geometry and dynamics of every map $z \mapsto z^2 + c$.<|endoftext|>
TITLE: Difference between Laver's and Mathias's forcing
QUESTION [9 upvotes]: Mathias forcing consists of conditions of the form $(s,A)$ where $s$ is a finite subset of $\omega$, $A$ is an infinite subset of $\omega$ such that $\max(s) < \min(A)$ and $(s,A)\leq (t,B)$ iff $s\supseteq t$, $A\subseteq B$ and $s\setminus t\subseteq B$.
Laver's forcing, on the other hand, consists of conditions $p$ that are subtrees of $\omega^{< \omega}$ (i.e. they are subsets closed under initial segments) such that there is a node $s\in p$, called the stem of $p$, that is comparable with every element of $p$ and moreover, for every $t\in p$, if $s\subseteq t$ then $\{n < \omega | t\frown n\in p\}$ is infinite.
In both forcings, the union of the "working" parts (i.e. the first coordinate in the case of Mathias, the stem in the case of Laver) of the elements of the generic filter give us a generic real. My question is whether there is an easy-to-grasp difference between the generic models that one can obtain by means of these two forcings. This is motivated by the (empirical) "fact" that, based on my experience, everything that can be done with Mathias forcing can usually also be done with Laver's (which is often much more intuitive). For example, both forcings "kill" uncountable strong measure zero sets from the ground model. Also, both forcings are proper, and the proofs for each of the forcings are fairly similar. I sort of understand why they are not forcing equivalent (well, at least I understand why my attempts to define a dense embedding of one into the other fail), and I know that if you iterate these forcings you can actually see some differences. For example, if you look at the table of cardinal characteristics of the continuum at the end of Blass's article in the Handbook of Set Theory, you can see that e.g. the splitting number $\mathfrak s$ attains a value of $\omega_1$ in Laver's model, as opposed to a value of $\mathfrak c$ (which in this case is $\omega_2$) in Mathias model. However, these models are the result of iterating these forcings $\omega_2$ many times with countable support, and I'm looking for something more immediate. So, my specific question is:
Suppose $V$ is a fixed model of ZFC (plus CH, or any other hypothesis you might need), and that $G$ is a Laver-generic filter over $V$, and that $H$ is a Mathias-generic filter over $V$. What would be a (or some) easy to grasp (as much as possible, but of course if it's too technical I don't mind going through it) difference between $V[G]$ and $V[H]$? (Thus, I'm not asking for a difference between Laver's and Mathias's model (that are the result of adjoining $\omega_2$ many Laver or Mathias reals, respectively), but for a difference between the models obtained by adding a single Laver or Mathias real).

REPLY [3 votes]: In this paper Alan Dow compares Laver and Mathias forcing regarding their effects on the algebra $\mathcal{P}(\omega)/\mathit{fin}$. He considers four iterations of length $\omega_2$: iterate $L$, iterate $M$, iterate $L*M$ or iterate $M*L$ and calculates the values of many familiar cardinal characteristics of the continuum in the resulting models.<|endoftext|>
TITLE: Are all free factors of finitely generated subgroups of free groups geometric?
QUESTION [6 upvotes]: Let $F$ be a fixed free group of finite rank. If $H$ is a finitely generated subgroup of $F$ and $A$ is a basis for $F$, then we can form the Stallings graph $\Gamma_A(H)$ for $H$. It is the unique smallest ($|A|$-labelled) subgraph of the covering space of a bouquet of $A$-circles corresponding to $H$ that contains the base point and carries the fundamental group. 
It is easy to see that if $\Gamma_A(H)$ embeds in $\Gamma_A(K)$, then $H$ is a free factor in $K$. If one fixes the basis $A$ then there are free factors of $F$ itself that are not induced by graph embeddings. 
Let us say that a finitely generated subgroup $H$ of $F$ is a geometric free factor of a subgroup $K$ if there exists some basis $B$ for $F$ such that $\Gamma_B(H)$ embeds as a labelled graph in $\Gamma_B(K)$. 


Question. Is every free factor $H$ of a finitely generated subgroup $K$ of a free group $F$ a geometric free factor?


History. I came upon this question in some joint work with Auinger on finite semigroups. To solve some problem we needed all geometric free factors of open subgroups to be closed in the pro-C topology in order for nice things to happen. We would have liked to have been able to drop the word "geometric." Enric Ventura has independently considered this question.

REPLY [2 votes]: Here is what I think will be an infinite index example, although I'm missing some details of proof. Take $F = \langle a,b \rangle$, identified with the fundamental group of a rose $R$ with two petals labelled $a$, $b$. Let $\Theta$ be the rank 2 graph with oriented edges $e_1,e_2,e_3$ having the same initial and terminal endpoints, labelled by three reduced words $w_1,w_2,w_3$ in the letters $a,b,a^{-1},b^{-1}$. Make sure the initial three letters of these words are all distinct, as are the terminal three letters, so the induced map $\Theta \to R$ is an immersion and the $\pi_1$-image of $\Theta$ is an infinite index, rank 2 subgroup $B$ in $F$. Make sure that the words $w_1,w_2,w_3$ are really long and random, for instance at the very least each of them should obtain every possible length 10 subword of $F$. Now start applying the standard generators of $Aut(F)$, for example $a \to a$, $b \to ab$, and you will see that the induced self-map on the graph $\Theta$ is homotopic to a homeomorphism. I believe that this property continues to be true as you apply any word in the generators of $Aut(F)$. As a consequence, the only geometric subgroups of $B$ are those represented by the three loops in $\Theta$. Roughly speaking the reason for this behavior is that any element of $Aut(F)$ can be represented by an equivariant Stallings fold sequence on the universal cover of $R$, and if two segments of $R$ are identified under this fold sequence then those two segments never have the property that they contain every length 10 word of $F$, so the only identifications induced on $\Theta$ are short initial segments of its three edges.<|endoftext|>
TITLE: (Approximately) bijective proof of $\zeta(2)=\pi^2/6$?
QUESTION [66 upvotes]: Given $A,B\in {\Bbb Z}^2$, write $A \leftrightarrows B$ if the
interior of the line segment $AB$ misses
${\Bbb Z}^2$.
For $r>0$, define
$S_r:=\{ \{A, B\} \mid  A,B\in {\Bbb Z}^2,\|A\|<r,\|B\|<r, \left| \|A\|-\|B\| \right|<1 \text{ and } A \leftrightarrows B \}\ .$
A little calculus gives the equivalence of $\zeta(2)=\pi^2/6$ and
$$\lim_{r\to\infty} \frac{|S_r|}{(2r)^3} = 1\ .$$
Of course $(2r)^3$ counts lattice points in a cube $C_r:=[-r,r)^3$.
Question: Does there exist an approximately bijective proof of this
limit (or some variant), one that matches most of $S_r$ with most of $C_r$?

REPLY [9 votes]: A proof of $\displaystyle \sum_{n=1}^\infty \frac 1{n^2} = \frac{\pi^2}6$ is described in a youtube video that I may find and post later today, after it appeared in a paper that I may cite here later today, so I'll let this present posting serve as a reminder to get back to this when I have more time. (As I thought might happen, someone beat me to the URL for the video. See the comment below.) (Another postscript: The paper is by Johan Wästlund. Here is the preprint.)
This involves a one-to-two correspondence (so just construe "bi-" in "bijective" as referring to that "two").
It goes like this. First observe that the proposition to be proved is easily seen to be equivalent to $\displaystyle \sum_{\text{odd } n\,\in\,\mathbb Z} \frac 1{n^2} = \frac{\pi^2}4 $.
Next, approximate this last sum by the sum of squares of the reciprocals of the distances from $0$ of certain points on the circle of circumference $2^n$ that touches the line $y=0$ in the plane $(x,y)\in\mathbb R^2$ at the point $(0,0)$. Those specified points are the those whose distance measured along the circle from $(0,0)$ is an odd integer.
The central lemma is that if $n$ increases by $1$, then that sum does not change. This reduces the problem to the the case $n=1$, and in that case the sum is $\pi^2/4$.
The proof of the lemma is from secondary-school geometry: Through each of the specified points draw the line through the very top of the circle. That line intersects the next circle, the case $n+1$, at two points. Show that those are two of the points at odd-integer arc lengths from $(0,0)$ and that the sum of the squares of the reciprocals of the distances from those two points to $(0,0)$ is equal to the reciprocal of the square of the distance to $(0,0)$ from the point you started with. $\quad\blacksquare$<|endoftext|>
TITLE: Mertens' function in time $O(\sqrt x)$
QUESTION [10 upvotes]: This MathOverflow question seems to indicate that the state of the art in computing
$$
M(x)=\sum_{n\le x}\mu(n)
$$
takes time $\Theta(n^{2/3}(\log\log n)^{1/3}),$ which matches my understanding. Recently I came across a paper [1] which gives, almost as an afterthought, an algorithm for computing $M(x)$ in $O(\sqrt x)$ time (in section 3.2).
The algorithm itself seems to be correct, being derived in a straightforward way from the identity
$$
M(x)=1-\sum_{d\ge2}M(x/d).
$$
But the claim that is runs in time $O(\sqrt x),$ or even $O(x^{1/2+\varepsilon}),$ seems unusual enough for me to ask for verification.
First, this would be a major breakthrough in computing $M(x),$ enough that I would think it would merit more mention than a substep of another algorithm. At the least, I would expect a mention that previous algorithms were slower.
Second, the algorithm is very simple, so that if the time is as indicated the implied constant should be low and the algorithm should be practical. In any case it is not hard to program.
Third, on coding the algorithm I found it to be very slow. In fact, it was slower for those values tested than sum(n=1, 10^7, moebius(n)) in GP which involves factoring each number up to $10^7.$ The time needed to factor numbers of those sizes is about $\sqrt x/\log x$ on average, so that's a $\Theta(n^{3/2}/\log n)$ algorithm (admittedly, well-optimized) beating a $\Theta(n^{1/2})$ algorithm. The constants would have to be worse by a factor of $5\cdot10^6$ for that to happen, and there's nothing in the algorithm to suggest anything that bad.
Of course I may have miscoded it (though I obtained the correct answer) or there may be reasons why the constant factors would be so high for this algorithm. But in any case this seemed suspicious enough to bring up here that I might be enlightened in any case.
Of course even if the result claimed is not correct this is no mark against the author, as the paper is only a preprint and the claim is peripheral in any case.
[1]: Jakub Pawlewicz, Counting square-free numbers (2011), arXiv:1107.4890.

REPLY [9 votes]: This is really a response to Dror Speiser's comment (but it is too long to give as a comment), and gives an analytic $O(n^{1/3+\epsilon})$ time and $O(n^{1/6+\epsilon})$ space algorithm to count the number of square free numbers less than $x$ (thus improving on the complexity of the algorithm given in the Jakub Pawlewicz paper cited in the question).  If we do exactly in the same way as in my other answer  for
$$\sum_{n\leq x} |\mu(n)|$$
instead of 
$$\sum_{n\leq x} \mu(n)$$
by using the fact that $$\frac{\zeta(s)}{\zeta(2s)} = \sum_{n=1}^\infty |\mu(n)| n^{-s}$$
it is true that the time complexity will be $O(x^{1/2+\epsilon})$. There is however a way to combine the analytic method (which is essentially the same as in Lagarias-Odlyzkos 1987 paper "Computing $\pi(x)$ an analytic method" http://www.dtc.umn.edu/~odlyzko/doc/arch/analytic.pi.of.x.pdf) with the elementary identity
$$
 S(n)=\sum_{d=1}^{\lfloor \sqrt n\rfloor} \mu(d) \lfloor \frac n {d^2} \rfloor 
$$
that is Pawleviec's starting point in obtaining his algorithm. Let $\Phi$ be defined as in my other answer, i.e.  let $\Phi(x)$ be a smooth test function such that $\Phi(x)=1$ for $x<0$ and $\Phi(x)=0$ for $x>1$. Consider the identity
$$ \sum_{n\leq x} |\mu(n)| =\sum_{n=1}^\infty |\mu(n)| \Phi \left( \frac {n-x} {x^{2/3}} \right )
- \sum_{x< n < x + x^{2/3} } |\mu(n)| \Phi \left(\frac {n-x} {x^{2/3}} \right)$$
The first sum can be estimated by a  complex integral of length $x^{1/3+\epsilon}$ and by the Odlyzko-Schönhage algorithm it can be calculated in $O(x^{1/3+\epsilon})$ time (and $O(x^{1/6+\epsilon})$ space). The remaining sum will have length $O(x^{2/3})$ so might at first glance not be so simple to treat. I claim however that in  fact it can be calculated in $O(x^{1/3+\epsilon})$ time. This is where using the elementary identity is handy. Let  $\Psi(x)=0$ for $x<0$ and $\Psi(x)=\Phi(x)$ for $x \geq 0$ (this will have a discontinuity at $x=0$) By a similar elementary identity we obtain
$$\sum_{x< n < x + x^{2/3} } |\mu(n)| \Phi \left(\frac {n-x} {x^{2/3}} \right) = \sum_{d=1}^{\lfloor \sqrt {x+x^{2/3}}\rfloor} \mu(d) \sum_{k=1}^\infty \Psi \left(\frac {d^2k-x} {x^{2/3}} \right).
$$
For any $d$ the inner sum can be calculated fast (certainly in $O(x^\epsilon)$ time), by either noticing that the sum is empty, just contains one element or the Poisson summation formula. The trick now is to see that there will only be $O(x^{1/3})$ integers $d$ where the inner sum (in $k$) is non empty, namely $d$ must either be $O(x^{1/3})$ or belong to an interval $(\sqrt{ x/k},\sqrt{(x+x^{2/3})/k})$ for $k \leq x^{1/3}$  and there are only $O(x^{1/3})$ such $d$.  Now if we can calculate the values of $\mu(d)$ fast (simple sieving seems more difficult in this case) we obtain the desired algorithm. Determining $\mu(d)$ basically comes down to factoring the number. However it is sufficient to determine the factors less than $d^{1/3}$ since if we know that all prime factors are greater than $d^{1/3}$, then either it has one or two prime factors.  We can determine if it is a prime fast by the Agrawal-Kayal-Saxena algorithm, then of course $\mu(d)=-1$. If it has two prime factors either $\mu(d)=0$, but then it has to be a square, and it can be easily checked, or $\mu(d)=1$. Now factoring the $O(x^{1/3})$ numbers of size $O(x^{1/2})$ into all prime factors less than $x^{1/6}$ will take $O(x^{1/3+\epsilon})$ time and $O(x^{1/6+\epsilon})$ space by Bernstein's algorithm http://cr.yp.to/papers/sf.pdf. I might possibly write up this answer as a paper proper. If anyone has seen these arguments anywhere else or have references I would be interested.<|endoftext|>
TITLE: Existence of elements in an Eichler order whose norm is minus one
QUESTION [6 upvotes]: Let $B$ be an indefinite quaternion algebra over $\mathbb{Q}$ of discriminant $D$, and $\mathcal{O}_N$ be an Eichler order of level $N$.  Is there an element $x\in \mathcal{O}_N$ such that its reduced norm $\mathrm{Nrd}(x)=-1$?
There exists an element $x\in B$ such that $\mathrm{Nrd}(x)=-1$. This follows from the norm theorem.  But I don't know if the existence stays to be true if $B$ is replaced with $\mathcal{O}_N$. 
The question is interesting in the following way.  According to the paper of S. Molina, section 2.2, Shimura curves $X_0(D,N)$ (defined as the upper half plane quotient by the group of norm one elements in $\mathcal{O}_N$) parametrizes abelian surfaces with QM by $\mathcal{O}_N$. 
Fix an isomorphism $i: B\otimes \mathbb{R} \to \mathrm{Mat}_2(\mathbb{R})$. 
Given $z$ in the upper half plane, the $\Gamma_0(D,N)$-orbit of $z$ corresponds to the isomorphic class of pairs $(\mathbb{C}^2/\Lambda_z, i)$, where $\Lambda_z$ is the lattice $i(\mathcal{O}_N)\begin{bmatrix} z \\ 1\end{bmatrix}$. 
Question: why all abelian surfaces with QM by $\mathcal{O}_N$ arises this way? 
There is nothing special in taking $i(\mathcal{O}_N)\begin{bmatrix} z \\ 1\end{bmatrix}$. We can just take any invertible left-ideal $I$ of $\mathcal{O}_N$ and look at the lattice   $i(I)\begin{bmatrix} z \\ 1\end{bmatrix}$.  Since the class number is one, any 
such ideal is equal to $\mathcal{O}_Ny$ for some $y\in B$.  So after multiplying a scalar, the lattice $i(I)\begin{bmatrix} z \\ 1\end{bmatrix}$ is equal to $i(\mathcal{O}_N)\begin{bmatrix} i(y)z \\ 1\end{bmatrix}$.  
Everything works well if $\mathrm{Nrd}(y)>0$. However, in the case $\mathrm{Nrd}(y)<0$, then $i(y)z$ is no longer on the upper half plane. This can be remedied if there is an element $x\in \mathcal{O}_N$ such that $\mathrm{Nrd}{x}=-1$, in which case 
$i(\mathcal{O}_N)\begin{bmatrix} i(y)z \\ 1\end{bmatrix}$ coincides with $i(\mathcal{O}_N)\begin{bmatrix} i(xy)z \\ 1\end{bmatrix}$ up to a scalar.   If such an element does not exists, I am afraid that we might have to throw in the lower half plane as well in defining the Shimura curve, and the quotient by $\Gamma_0(D,N)$ will not be connected.   
In fact, in section IX.5 of "Introduction to Algebraic and Abelian Functions" by S. Lang, he explicitly assumed that there is an element in the order whose reduced norm is -1.  Is there an theorem that says such an element will always exists in the indefinite case, and his assumption is just for exposition convenience rather than out of necessity?

REPLY [4 votes]: I can think of two ways to answer your question.
The first is to revisit what it is I think you want to conclude, which is that every right fractional O-ideal is (principal and) generated by an element with positive reduced norm.  It is in this way that the question most naturally arises when considering double coset spaces giving rise to Shimura curves.  I hate to plug my own papers, but see Section 2.2 in
  http://www.cems.uvm.edu/~voight/articles/classno-ants-fixed-errata.pdf.
In this context (and with this notation), what you want to know is that the map
$$
B_+ \backslash \widehat{B}^\times / \widehat{\mathcal{O}}^\times 
\to F_+ \backslash \widehat{F}^\times / \widehat{\mathbb{Z}}_F^\times = \mathrm{Cl}^+(\mathbb{Z}_F) $$
is a bijection; this is a consequence of strong approximation, which can be found in Vigneras "Arithmetique des Algebres de Quaternions", Theoreme III.4.3.  Or if you'll forgive their woefully incomplete nature, see Section 4.2 in 
  http://www.cems.uvm.edu/~voight/crmquat/book/quat-modforms-041310.pdf; 
here I also try to explain the link between the "usual" formulation of strong approximation (in terms of elements of $B^\times$ of reduced norm 1) and representatives of ideal classes.
In the case you are interested in (swapping right for left ideals), $F=\mathbb{Q}$ has trivial narrow class group $\mathrm{Cl}^+(\mathbf{Z})=\{(1)\}$; so every right fractional $\mathcal{O}$-ideal has a totally positive generator.  
The second way to answer your question is to prove independently that there is a unit of norm one.  This follows from the above reasoning by taking any element of negative norm $\alpha$, considering the right fractional $\mathcal{O}$-ideal $\alpha \mathcal{O}$ generated by it, and then noting that it is generated by a totally positive element $\beta$; then $\beta = \alpha\mu$ and by comparing norms we have $\mathrm{nrd}(\mu) = -1$.  But you might also find it to be a helpful exercise to derive this from the "usual" statement of strong approximation, involving the elements of reduced norm 1.<|endoftext|>
TITLE: What are supersingular varieties?
QUESTION [16 upvotes]: For varieties over a field of characteristic $p$, I saw people talking about supersingular varieties.
I wanted to ask "why are supersingular varieties interesting". However, as I don't want to ask an MO question with no background/context I thought I'd better define what a supersingular variety is. Unfortunately I can't. (And search engine doesn't help...) Can anyone here help me with this and explain why they are interesting?
A little bit more context: I have seen the definition of supersingular elliptic curves on textbooks by Hartshorne and Silverman. When I read about abelian varieties I saw "for abelian varieties of $\text{dim}>2$ being supersingular $\neq p$ rank being 0".
Illusie mentioned (in the "Motives" volume) that for $X$ a variety over a perfect field of characteristic $p$, $X$ is said to be ordinary if "$H_\text{cris}^*(X/W(k))$ has no torsion and $\text{Newt}_m(X)=\text{Hdg}_m(X)$ for all $m$". My guess is being supersingular should correspond to the other extreme, but what precisely is it? ($\text{Newt}_m(X)$ being a straight line for all $m$?)

REPLY [13 votes]: Not an answer, but some historical context (which I think is correct). An elliptic curve over $\mathbb{C}$ used to be called "singular" if its endomorphism ring was larger than $\mathbb{Z}$, i.e., what we now call having complex multiplication. Presumably this use of the word singular was to indicate that the curve was unusual. Then, when people looked at elliptic curves over finite fields, the found that some of them had endomorphism rings that were even larger than an order in a quadratic imaginary field, so those curves were "supersingular" in the sense of being even more unusual. Of course, it turns out that an alternative way to characterize those curves is as having no $p$-torsion (over the algebraic closure of their base field).<|endoftext|>
TITLE: Request for clarification of a point in the book Galois Theories by Borceux and Janelidze
QUESTION [6 upvotes]: In page 105 of the book Galois Theories, in Example 4.6.2, the authors state that a specific diagram is an internal groupoid, but I really cannot see why.

Let $\sigma \colon S \rightarrow R$ be a morphism in a category $\mathcal X$ with pullbacks. The diagram  is an internal groupoid, with $(p_0,p_1)$ the kernel pair of $\sigma$, $\Delta$ the diagonal of the pullback and $\tau$ the twisting isomorphism which interchanges factors. 

So, my main problem is, where does $p_4$ come from? Shouldn't it be $p_2$? And they also say $(p_0,p_1)$ is the kernel pair of $\sigma$. What is $p_0$? Shouldn't the kernel pair be $(p_1,p_2)$? In the beginning I thought it was just a typo, but in Section 5.1 they use the internal groupoid again, still with $p_4$ which makes me think that I must be doing something wrong. I have spent a lot of time trying to find out what happens, so please, if anyone could clarify this, I would appreciate it.

REPLY [10 votes]: They are thinking of the object $M=(S\times_RS)\times_S(S\times_RS)$ as a subobject of $S^4$, so there are four projections $p_1,p_2,p_3,p_4:M\to S$.  If $\mathcal{X}$ is some kind of category of sets with structure, then the composition map $(p_1,p_4)$ is just $((a_1,a_2),(a_3,a_4))\mapsto (a_1,a_4)$.  (Here $a_2$ is actually the same as $a_3$ because the middle pullback is over $S$.)  The other maps are $\tau(a_1,a_2)=(a_2,a_1)$ and $p_1(a_1,a_2)=a_1$ and $p_2(a_1,a_2)=a_2$ and $\Delta(a)=(a,a)$.  The kernel pair of $\sigma$ is indeed $(p_1,p_2)$ rather than $(p_0,p_1)$, which is just a typo.
If $\mathcal{X}$ is just the category of sets, you end up with the category whose object set is $S$, and there is a unique morphism from $a$ to $b$ if $\sigma(a)=\sigma(b)$, and no morphisms from $a$ to $b$ otherwise.<|endoftext|>
TITLE: Why could Mertens not prove the prime number theorem?
QUESTION [46 upvotes]: We know that 
$$
\sum_{n \le x}\frac{1}{n\ln n} = \ln\ln x + c_1 + O(1/x)
$$
where $c_1$ is a constant. Again Mertens' theorem says that the primes $p$ satisfy 
$$
\sum_{p \le x}\frac{1}{p} = \ln\ln x + M + O(1/\ln x).
$$
Thus both these divergent series grow at the same rate. Mertens' theorem was proved without using the prime number theorem, some 25 years before PNT was proved. However from these two examples, we cannot conclude that 
$$
\lim_{n \to \infty} \frac{p_n}{n\ln n} = 1
$$ 
otherwise Mertens' would have been the first to prove PNT. My question is - based on the above two series, what are the technical difficulties that prevent us from reaching the conclusion that $p_n/n\ln n = 1$. There may be counter examples with other series, so such conclusions may not be true in general. However I am not interested in the general case. Instead I am asking only in case of the sequence $1/n\ln n$ and $1/p_n$.

REPLY [10 votes]: Because the scale is too small in Mertens's theorem, and the prime number theorem as well as the Riemann hypothesis are hidden by the $O(1/\log{X})$ notation.
Indeed, the former amounts to strengthening this term to $o(1/\log{X})$; the latter - to $O(1/\sqrt{X})$.
[Incidentally, one could go for equivalent statements to a still smaller scale. Of course, an estimate like $\sum_{p < X} 1/p^2 = \mathrm{const} + O(1/X)$ does not say anything at all about the primes. But one could, if one wished, express the prime number theorem by elaborating on the $O(1/X)$ term. ]
To elaborate on this a bit, let me go to a slightly bigger scale where the prime number theorem begins to emerge outside of $o(1)$. This is also more natural; indeed, it was how Mertens's theorem was proved. 
By partial summation, Mertens's estimate is equivalent to $\sum_{p < X} (\log{p})/(p-1) = \log{X} + O(1)$; or, if one prefers, $\sum_{n < X} \Lambda(n)/n = \log{X} + O(1)$. The prime number theorem however is the statement that the $O(1)$ term converges to a constant: $\sum_{n < X} \Lambda(n)/n = \log{X} - \gamma + o(1)$. Indeed, the related bound $\sum_{n < X} \Big( \frac{1}{n} - \frac{1}{X} \Big) \Lambda(n) = \log{X} - 1 - \gamma + o(1)$, another form of the prime number theorem, is what de la Vallee Poussin actually obtained in his original paper. Here $\gamma = 0.57\ldots$ is Euler's constant, but this is of no importance for us, see the next paragraph. Also the $\log{X}$ logarithmic pole corresponds to the pole of $\zeta(s)$ at $s=1$, whereas the $o(1)$ term expresses there being no zeros with $\mathrm{Re}(s) = 1$. The Riemann hypothesis is the correspondingly stronger bound $O(1/\sqrt{X})$ on the $o(1)$ oscillating term. At this scale, in contrast to $\psi(X) = X + O\big( \sqrt{X} (\log{X})^2 \big)$ or $\pi(X) = \mathrm{Li}(X) + O(\sqrt{X}\log{X})$, a logarithmic factor in addition to the square root is not required, as $\sum_{\rho} 1 / |\rho|^2 < \infty$ over the zeros.
Here finally is how to deduce the more usual form $\psi(X) \sim X$ of the prime number theorem from the refinement $S(X) := \sum_{p < X} \Lambda(n)/n = \log{X} -\gamma + o(1)$ of Mertens's theorem: Summing by parts, $\psi(X) = XS(X) - \int_1^X S(t) \, dt = X(\log{X} - \gamma + o(1)) - \big( \int_1^X \log{t} \, dt - \gamma X + o(X) \big) = X + o(X)$.

Added (December, 2017). I came upon an observation giving also a 'trivial' proof of the reverse elementary implication of the two purely qualitative forms, multiplicative and logarithmic, of the prime number theorem: $\psi(X) \sim X \Leftrightarrow S(X) = \log{X} - \gamma + o(1)$. The following seems to have been missed in the literature on elementary methods which, at this point, seem all to quote a somewhat more involved Tauberian theorem of Axer; cf. section 8.1.1 of Montgomery and Vaughan's book (Multiplicative Number Theory: I) or, for a more general setting, chapter 14 of Diamond and Zhang's recent book on Beurling Generalized Numbers (really this paper of theirs). The simpler argument below also extends easily to number fields, supplying  a particularly easy proof of the 'elementary equivalence' of Landau's prime ideal theorem and number field sharp Mertens. Incidentally, as I happen to recall, this addresses a slightly curious point that had come up in the comments to this answer of Eric Naslund.   Remembering also my answer here, I figured it may be worth to record the following  observation as an addendum to it, sticking for simplicity to the rational case assumed in this question.
A proof of $\psi(X) \sim X \Rightarrow S(X) = \log{X} - \gamma + o(1)$. For simplicity, let me stick to $\mathbb{Q}$. The case of a number field $K$ has the same result with $\gamma$ generalized as the 'Euler-Kronecker invariant' $\gamma_K$. 
The key is to observe that the formula
$$
X^{-1} \log{X!} = \sum_{n \leq X/T} \frac{\Lambda(n)}{n} + \sum_{m \leq T} \frac{1}{X} \Big( \psi\Big( \frac{X}{m} \Big) -  \psi\Big( \frac{X}{T} \Big) \Big)  + O(1/T)
$$
holds uniformly in the two parameters $X, T \geq 1$, with an absolute implied coefficient. It interpolates between Mertens's estimate (case $T = 1$) and Chebyshev's convolution formula $\log{X!} = \sum_m \psi(X/m)$ (case $T = \infty$). But the general formula also follows, after a moment of reflection, from Chebyshev's argument with the prime factorization of $X!$. Divide the moduli into the ranges $n \leq X/T$ and $n > X/T$. The total contribution of the latter are exactly accounted for by the second sum. For a small modulus $n \leq X/T$, the contribution via the prime factorization is $X^{-1} \lfloor X/n \rfloor \Lambda(n) = \frac{\Lambda(n)}{n} + O\Big(\frac{\Lambda(n)}{X}\Big)$, neglecting the fractional part.  The $O(1/T)$ term then comes from summing these for $n \leq X/T$, and using Chebyshev's estimate $\sum_{n \leq Y} \Lambda(n) \ll Y$. (In the number field generalization, the latter estimates extend as lattice point counts.)
Now, by Stirling's asymptotic, the qualitative $\psi(X) \sim X \Rightarrow S(X) = \log{X} - \gamma + o(1)$ implication is immediate from the observed formula upon first letting $X \to \infty$ and then $T \to \infty$.<|endoftext|>
TITLE: Polar decomposition in C*-algebras
QUESTION [5 upvotes]: A very nice feature of W*-algebras is the following:
once you have an element $a$ of a W*-algebra $M$, and $a=u|a|$ (the polar decomposition), then $u\in M$.
It seems that it carries over to AW*-algebras without pain. This is simply because (A)W*-algebras have lots of projections, unlike general C*-algebras.
Can one give an abstract characterisation of C*-algebras with the above mentioned property?

REPLY [4 votes]: It appears that there is such a condition for so-called Rickart C*-algebras (to each element a in the algebra there is a selfadjoint idempotent generating the left annihilator of a). This condition is mentioned in the first paragraph of the paper "Polar decomposition in Rickart C*-algebras" by Dmitry Goldstein.
I don't expect there will be a trivial abstract characterization of the property in general.

REPLY [2 votes]: There exist further generalizations but they do not go very far from AW*-algebras. See
http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.pjm/1102650543
and
http://dmle.cindoc.csic.es/pdf/PUBLICACIONSMATEMATIQUES_1995_39_01_01.pdf<|endoftext|>
TITLE: A question about some  special compactifications of $\mathbb{R}$
QUESTION [6 upvotes]: We Know that the topological space $Y$ is a compactification of the topological space $X$, if the space $Y$ is compact and hausdorff and $X$ is dense in $Y$. If for a positive integer $n$ we have a compactification $Y$ that $Y-X$ has only $n$ elements, we say that $Y$ is the $n$ point compactification of $X$. I have some Questions about the existence of $n$-point compactification of a space $X$.
If the space $X$ has an $n$-point compactification for $n>1$ is it true that $X$ is locally compact?
We Know that $\mathbb{R}$ has a one point compactification, homeomorphic with the circle. and it has a two point compactification, homeomorphic with the closed interval $[0,1]$. is it true that $\mathbb{R}$ has a three point compactification? if it exists,it is homeomorphic with which well Known topological space? and if the answer is negative, why?

REPLY [9 votes]: Suppose a space $X$ has $k$ ends and an $n$ point compactification. We can show that $k\geq n$. Indeed, there are disjoint neighborhoods $A_1,\dots,A_n$ of each of these points at infinity. Now let $Y$ be the complement of $\cup A_i$. Then $Y$ is compact so under mild assumptions ($X$ is hemicompact) it is contained in a bigger compact subset of $X$, which we call $\bar{Y}$, whose complement in $X$ has exactly $k$ connected components. Then we form $\bar{A_i}$ as the intersection of $A_i$ and the complement of $\bar{Y}$. Since the $A_i$'s are disjoint any component can contain elements from at most one $A_i$ so we get $k\geq n$.
As an application, $\mathbb R$ has $2$ ends so it cannot have a $k$ point compactification for $k\geq 3$. Similarly $\mathbb R^m$ has one end for $m\geq 2$ so it cannot have a $k$ point compactification for $k\geq 2$.

REPLY [3 votes]: First of all, if $\gamma X$ is a compactification of $X$ and $\gamma X\setminus X$ is finite,
then $X$ is locally compact.
Namely, let $x\in X$.  In $\gamma X$, $x$ has a neighborhood $U$ that is disjoint from the finite set $\gamma X\setminus X$.
Since $\gamma X$ is normal, there is a neighborhood $V$ of $x$ whose closure is still a subset of $U$.  Since $\gamma X$ is compact, the closure of $V$ is a compact neighborhood of $x$.
Since $U$ is disjoint from $\gamma X\setminus X$, the closure of $V$ is a compact neighborhood of $x$ in $X$.
The definition of an "end" of a topological space can be found here: https://en.wikipedia.org/wiki/End_(topology)<|endoftext|>
TITLE: Are there any complete, first-order and unstable theories which have non-categorical second-order formulations?
QUESTION [9 upvotes]: Since it's not stable, $PA$ fails at being categorical in a power in the worst possible way, having $2^{\lambda}$ models in any uncountable $\lambda$. But $PA$ regains its categoricity in the move to second order logic, by which I mean, once we replace the induction schema with a single, second-order axiom. Are there natural examples of unstable first-order theories for which categoricity is not recoverable in this way? 
If the restriction to unstable $T$ is too strong, are there natural examples of complete, non-categorical first-order $T$ which are also non-categorical as second-order theories?
I'm struggling to think of such  examples because I don't know of any axiomatizable, semantically complete but non-categorical second-order theory. Also, of course, unlike $PA$, not every theory has a natural second-order extension. 
Thanks.

REPLY [10 votes]: Another spin on this question, that gets around the fact that there are only countably many second order sentences, is to look at categoricity in $L^2_{\kappa,\omega}$ where we allow conjunctions of size less that $\kappa$ and second order quantifiers. Hyttinen, Kangas and Vaananen (see http://www.math.helsinki.fi/logic/people/jouko.vaananen/HytKanVaa.pdf) show that, for suitable cardinals $\kappa$, all models of a complete first order theory $T$ of size $\kappa$ are characterized by and $L^2_{\kappa,\omega}$-sentence if and only if $T$ is
``classifiable" in the sense of Shelah's Main-Gap i.e., $T$ is stable, NDOP, shallow, and NOTOP.<|endoftext|>
TITLE: Results that are easy to prove with a computer, but hard to prove by hand
QUESTION [18 upvotes]: Consider the assertion:
There is no completely multiplicative function $f:\mathbb{N}\rightarrow \{\pm 1\}$ with $\left|\sum_{n\leq x}f(n)\right|\leq 2$ for all $x\geq 0$.
One can write a very short program to show that things go wrong for $x\geq 247.$ Doing this by hand with current technology requires an unmanagable amount of cases (cf. http://michaelnielsen.org/polymath1/index.php?title=Human_proof_that_completely_multiplicative_sequences_have_discrepancy_greater_than_2). I wonder whether you have encountered similar situations. I am aware that one can always construct artificial examples. That is not what I want. Rather that you consider some special case of a theorem you try to prove, such that you can prove this with a computer, but not by hand.
Btw. a short proof of the above would also count as an answer.

REPLY [7 votes]: I think this example fits the question nicely (but it concerns a paper I coauthored, so if there is some feeling that this is shameless self-promotion please say it frankly in the comments and I'll delete the answer). 
We were trying to classify all bilinear estimates in $\mathbb{R}^3$ for spaces of wave-Sobolev type. The precise definition of the spaces is pointless here, let's say they are some sort of Sobolev spaces with weights measuring the amount of mass close to the light cone in Fourier variables; they are characterized by two indices $H^{s,b}$ and are an essential tool in the theory of nonlinear waves. A bilinear estimate is a product estimate of the form
$$H^{s_1,b_1}\cdot H^{s_2,b_2}\subset H^{s_3,b_3}$$
and one is interested in finding all the 6-tuples $(s_1,s_2,s_3,b_1,b_2,b_3)$ for which the estimate is true. There is a long list of counterexamples (21 plus symmetries; some of the trickiest ones were found by Terry Tao when he was young :) which bound a polyhedron with a large number of faces in the space of 6-tuples $\mathbb{R}^6$. Now the problem is: if a 6-tuple is inside the polyhedron, is the corresponding bilinear estimate true? in other words, is the list of counterexamples exhaustive?
In order to solve the problem we had to compute the vertices of the conjectured polyhedron (and then prove the estimates at the vertices, or near the vertices). Usually, one manages to do this kind of computation by hand with some neat tricks. But in this case, just writing the complete set of inequalities by hand takes a good 20 minutes. We had to admit, reluctantly, that this was physically impossible to do with pencil and paper. Of course this is a trivial problem for a computer and a simple algorithm gave us the answer.<|endoftext|>
TITLE: Best Computational Knot Invariants
QUESTION [13 upvotes]: My apologies if this is too closely related to this closed post.
I have been collaborating with a physicist looking at long polymer chains. These chains form knots with 2D projections having on the order of 100s of crossings, but which may in some/most cases all cancel out to give knots of relatively few crossings. I realize that there is no absolute knot invariant to definitively classify the knots that arise. But what is the best or best combination of knot invariants that would be 1) straightforward to compute on static diagrams and 2) distinguish most of the knots?
(By "straightforward to compute" I mean something like skein relations and not something like the fundamental group of the complement space. I realize that this requirement might be somewhat ill-defined, so feel free to lead me in the direction of computational references.)
For instance, I seem to recall somewhere that the trio of the Alexander Polynomial along with the 2nd and 3rd Vassiliev invariants was pretty powerful. Can anyone substantiate that and/or provide a reference?

REPLY [3 votes]: If these knots can be put in the form of braids, either closed or plat, then you can compute their Jones polynomials quickly if the braid index is reasonably small, say \leq 12.  This is particularly useful if you have lots of twisting, which quickly increases crossing number but preserves braid index.  Snappy can also handle braids with hundreds of crossings, so you would have two effective and generally independent invariants.<|endoftext|>
TITLE: Existence of a compactification of $\mathbb{R}$ with $\aleph_0$ remainder
QUESTION [7 upvotes]: We know that the space $\mathbb{R}$ has compactifications with one point remainder, and two point remainder. but there is no compactification of $\mathbb{R}$ with three point remainder and the same holds for every finite natural number $n$ greater than 3.
We know that the Stone-Čech compactification of $\mathbb{R}$ has infinite remainder.
(i.e.$|\beta\mathbb{R}-\mathbb{R}|=2^\mathfrak{c}$
I have the same question for  other infinite cardinals less than $2^\mathfrak{c}$ as follows:
A. Is there any compactification $X$ of $\mathbb{R}$ with the property that $|X-\mathbb{R}|=\aleph_0$?
B. Can we improve our question to cardinals less than $2^{\aleph_0}$?

REPLY [11 votes]: I guess you are interested in Hausdorff compactifications. (It is easy to construct a non-Hausdorff compactification with 3-point remainder.)
Set $W=\beta\mathbb R\backslash \mathbb R$;
note that $W$ has two connected components.
If $Z$ is an other compactification of $\mathbb R$ then $V=Z\backslash \mathbb R$
is an image of $W$.
Therefore $V$ has at most two connected components.
It follows that cardinality of $V$ may be 1, 2, or something between $\mathfrak{c}$ and $|W|=2^{\mathfrak{c}}$.<|endoftext|>
TITLE: Conditions for non-triviality of Caratheodory measure
QUESTION [8 upvotes]: This may be too vague to end up being useful, but: 
Are there any (natural? reasonable?) conditions that can be imposed on an outer measure $\phi^*:{\mathcal P}(S)\to[0,\infty]$ to ensure that the $\sigma$-algebra of measurable sets (obtained through Caratheodory's construction) is non-trivial? 
By non-trivial, I mean that at least it contains some sets other than $\emptyset,S$; even better if the condition ensures the $\sigma$-algebra is infinite.

REPLY [6 votes]: One commonly used condition, even found in the original paper of Caratheodory, is what we now call a metric outer measure.  $(S,d)$ is a metric space.  A metric outer measure is an outer measure $\mu$ such that: If two sets $A, B \subseteq S$ have positive distance (there is $\delta > 0$ so that for all $a \in A$ and $b \in  B$ we have $d(a,b) \ge \delta$), then $\mu(A \cup B) = \mu(A) + \mu(B)$.  Then: if $\mu$ is a metric outer measure, then at least all Borel sets in $S$ are $\mu$-measurable sets.<|endoftext|>
TITLE: p-adic analysis of hypergeometric functions
QUESTION [6 upvotes]: Are there any p-adic techniques that can be applied to the 2F1 hypergeometric function?
For e.g. I'm interested in which values this function converges p-adically.

REPLY [8 votes]: The Gauss hypergeometric function is the main example in the theory of p-adic differential equations. See 
K. S. Kedlaya, p-Adic Differential Equations, Cambridge University Press, 2010,
for the general theory. There were also two books by Dwork, almost completely devoted to ${}_2F_1$ (its p-adic theory is much more complicated than the classical one):
B. Dwork, Generalized hypergeometric functions. Oxford: Clarendon Press, 1990. 
B. Dwork, Lectures on p-adic differential equations, Springer, 1982.
It is easy to check local p-adic convergence for the hypergeometric series, but to study and even correctly define its analytic continuation properties one needs subtle analytic and algebraic techniques.<|endoftext|>
TITLE: Rational points on smooth compactifications
QUESTION [6 upvotes]: Let $X$ be as smooth variety over a field $k$ of characteristic $0$.
Consider the following statements:

The variety $X$ has no $k((t))$-rational points.
No smooth compactification of $X$ has a $k$-rational point.

Are these equivalent? If not, what additional assumptions on $X$ would make them equivalent? I'm particularly interested in the case where $X$ is a homogenous space of a "nice" algebraic group over $k$.

REPLY [12 votes]: Yes, this is true.
One implication is immediate: if $X$ has a $k((t))$ point then by the valuative criterion of properness there is a map $Spec(k[[t]])$ to any compactification of $X$, so the image of the closed point gives a $k$-point of the compactification.
For the converse, if a smooth compactification has a $k$-point then choose a general curve $C$ through that point (which is smooth at that point). Since $C$ is general, it is not contained in the boundary. By completing the local ring of the curve at the smooth point you get a map $Spec(k[[t]])$ to the compactification. Since $C$ is not contained in the boundary, the map retricted to the generic point (which is sent to the generic point of the curve) gives a $k((t))$-point of $X$.<|endoftext|>
TITLE: Error to sum of Euler phi-functions
QUESTION [23 upvotes]: The number theory identity $\phi(1) + \phi(2) + \dots + \phi(n) \approx \frac{3n^2}{\pi^2}$ can be interpreted as counting relatively prime pairs of numbers $0 \leq \{ x,y \} \leq n$ .
      

Has anyone studied the distribution of error term?  $$\displaystyle \frac{1}{n} \left[\sum_{k=1}^n \phi(k) - \frac{3n^2}{\pi^2}\right]$$ It looks like white noise:
      

The histogram has a distinctive shape, maybe hard to prove.  I suspect it's the Gaussian Unitary Ensemble (a Hermite polynomial times a Gaussian).
      


Similar questions: 
Question concerning the arithmetic average of the Euler phi function:
averages of Euler-phi function and similar

REPLY [17 votes]: This is an interesting question. I don't think anyone has worked out what the distribution of the error term
$$\frac{E(x)}{x} = \frac{1}{x}\left(\sum_{n \leq x}{\phi(n)} - \frac{3x^2}{\pi^2}\right)$$
actually looks like in any useful sense; from what I can make out, it seems to essentially be the same as the distribution of
$$\sum_{n = 1}^{\infty}{\frac{\mu(n)}{n} \left\{\frac{x}{n}\right\}}$$
where $\mu(n)$ is the Möbius function and $\{x\}$ is the fractional part of $x$, but this doesn't really seem to tell you anything particularly useful.
Nevertheless, it certainly is known that $E(x)/x$ has a distribution function; this is proved on p.13 of "On the Existence of Limiting Distributions of Some Number-Theoretic Error Terms" by Yuk-Kam Lau. This also follows quite easily from the fact that
$$\frac{E(x)}{x} = H(x) + O\left((\log x)^{-4}\right)$$
where
$$H(x) = \sum_{n \leq x}{\frac{\phi(n)}{n}} - \frac{6 x}{\pi^2}$$
combined with the main result of the paper "The Existence of a Distribution Function for an Error Term Related to the Euler Function" by Erdős and Shapiro. What is better known is the average behaviour of $E(x)$, in the form of the asymptotics
$$\sum_{n \leq x}{E(x)} \sim \frac{3 x^2}{2 \pi^2}$$
and
$$\int^{x}_{0}{E(t)^2 dt} \sim \frac{x^3}{6 \pi^2}.$$
Probably the best reference for what is currently known about this error term is the paper "Oscillations of the remainder term related to the Euler totient function" by Kaczorowski and Wiertelak (but unfortunately this paper isn't available for free online).

EDIT: Very recently, Lemke Oliver and Soundararajan have reproven the existence of the limiting distribution of $E(x)/x$, as well as quantitative bounds for the tails of this distribution; see Theorem 1.3 of their paper.<|endoftext|>
TITLE: Examples of theorems with proofs that have dramatically improved over time
QUESTION [95 upvotes]: I am looking for examples of theorems that may have originally had a clunky, or rather technical, or in some way non-illuminating proof, but that eventually came to have a proof that people consider to be particularly nice.  In other words, I'm looking for examples of theorems for which have some early proof for which you'd say "ok that works but I'm sure this could be improved", and then some later proof for which you'd say "YES! That is exactly how you should do it!"
Thanks in advance.
A sister question: Examples of major theorems with very hard proofs that have NOT dramatically improved over time

REPLY [2 votes]: The four color theorem’s first proof by Appel and Haken, which actually contained mistakes, was proved again much more succinctly and successfully by Robertson, Sanders, Seymour, and Thomas (see for instance here).  As far as I know, there is no better proof as yet, and unlikely there will ever be one, but who knows.<|endoftext|>
TITLE: Fibrations with non-simply connected base and rational homology
QUESTION [5 upvotes]: Let $p\colon E\to B$ be a fibration with fibres simply connected and homotopy equivalent to a compact CW-complex.
Must $p_*\colon H_3(E;\mathbb{Q})\to H_3(B;\mathbb{Q})$ be surjective?
COMMENTS.
Yes if (EDIT) $B$ is simply connected, even if the fibre is not compact but just finite-dimensional.
In general, finite-dimensionality is not enough: consider the homotopy fibre sequence $\mathbb{R}^3\setminus\mathbb{Z}^3\to T^3\setminus\mathrm{point}\to T^3$.
MOTIVATION.
If $p_*\colon H_3(E;\mathbb{Q})\to H_3(B;\mathbb{Q})$ is surjective, then any bundle gerbe over $p\colon E\to B$ is rationally trivial,
cf. M. Murray, D. Stevenson, A note on bundle gerbes and infinite-dimensionality (http://arxiv.org/abs/1007.4922).

REPLY [2 votes]: I'm wondering the extent to which the assumptions can be tweaked. Let's assume 
$B$ is connected and with basepoint. Let $F$ be the fiber over the basepoint. 
However, I won't assume $F$ is homotopy finite (i.e., homotopy equivalent to a finite complex). Nor will I assume anything about the action of $\pi_1(B)$. Rather, I will
assume 

$F$ is $1$-connected (just as Semen does), and
$H_2(F;\Bbb Q)$ is trivial.

Assertion: 
With respect to these assumptions, $H_3(E;\Bbb Q) \to H_3(B;\Bbb Q)$ is surjective.
Proof:
By slight abuse of notation, let $E/F$ be the the mapping cone of the inclusion $F\to E$.
Then the Blakers-Massey theorem shows that
$E/F \to B$ is 3-connected.
We infer that $E\to B$ is $H_3({-};\Bbb Q)$-surjective if $E \to E/F$ is.
But the long exact homology sequence of $F \to E \to E/F$ and the assumption that $H_2(F;\Bbb Q)$ is trivial implies $H_3(E;\Bbb Q) \to H_3(E/F;\Bbb Q)$ is surjective.
$\square$
The above leads to the following question:  Is there a relationship between the hypotheses
(1) $F$ is homotopy finite, simply connected and $\pi_1(B)$ acts trivially on $H_*(F;\Bbb Q)$;
(2) $F$ is simply connected and $H_2(F;\Bbb Q)$ is trivial
?<|endoftext|>
TITLE: Second cohomology of group of $S_n$
QUESTION [10 upvotes]: Hello,
Let $k$ be a field of characteristic different from $2$.
Let $n\geq 1$ be an integer, and let $T$ be the maximal torus of the $k$-algebraic group $PGL_n$, namely the quotient of diagonal matrices by the diagonal action of $\mathbb{G}_{m,k}$.
Then let the symmetric group $S_n$ act on T by permuting the diagonal entries.
We know from general results that $H^2(S_n,T)$ is killed by $n!$, but can we do better ?
Is it killed by $n$, for example ? what is the exponent of $H^2(S_n,T)$?
Thanks in advance!

REPLY [3 votes]: This is an interesting question indeed. I can not offer a full answer,
only the following observation. 
First consider the case $n=3$, i.e., $T_3$ is the maximal torus of
$\operatorname{PGL}_3$ with the natural action of $S_3$. Let's use the
Hochschild-Serre spectral sequence to compute
$\operatorname{H}^\bullet(S_3,T_3)$.
First, let $G=\mathbb{Z}/3$ and use the standard resolution 
$$
\cdots\to \mathbb{Z}[G]\stackrel{\Delta}{\rightarrow}
\mathbb{Z}[G]\stackrel{t-1}{\rightarrow}\mathbb{Z}[G]\to\mathbb{Z}\to 0
$$
where $t$ is a generator of $G$ and $\Delta=1+t+t^2$. 
Now
$\operatorname{H}^\bullet(G,T_3)=\operatorname{H}^\bullet(\operatorname{Hom}_{\mathbb{Z}[G]}(\mathbb{Z}[G],T_3))$
is the cohomology of the complex 
$$
T_3\stackrel{(t-1)^\ast}{\rightarrow}
T_3\stackrel{\Delta^\ast}{\rightarrow} T_3\to \cdots
$$
To evaluate this, describe the action of $G$ on $T_3$. The maximal
torus $T_3$ in $\operatorname{PGL}_3$ is given by diagonal matrices
$\operatorname{diag}(a,b,c)$ modulo scalar matrices $\operatorname{diag}(a,a,a)$. Using
representatives $\operatorname{diag}(1,a,b)$, the action of $t$ on $T_3$ is
given by $\operatorname{diag}(1,a,b)\to \operatorname{diag}(b,1,a)\sim
\operatorname{diag}(1,b^{-1},ab^{-1})$. Consequently, the action of $t-1$ on
$T_3$ is $(a,b)\mapsto ((ab)^{-1},ab^{-2})$ and the action of $\Delta$
on $T_3$ is trivial. From this, we obtain 
$$
\operatorname{H}^i(G,T_3)=\left\{\begin{array}{ll}
\mu_3 & i \textrm{ even}\\
k^\times/(k^\times)^3 & i \textrm{ odd}
\end{array}\right.
$$
where $\mu_3$ is the group of third roots of unity in $k$. 
Next, we describe the action of $\mathbb{Z}/2$ on
$\operatorname{H}^i(G,T_3)$. On $T_3$, the non-trivial element $\sigma$ acts as
$\operatorname{diag}(1,a,b)\mapsto \operatorname{diag}(1,b,a)$. It follows, that $\sigma$
acts via $x\mapsto x^{-1}$ on $\operatorname{H}^i(G,T_3)$. Again, we use the
standard resolution to
compute the cohomology $\operatorname{H}^j(\mathbb{Z}/2,\operatorname{H}^i(G,T_3))$. The
element $\sigma-1$ acts as identity on $\mu_3$, and the element
$\sigma+1$ acts trivially on $\mu_3$. Similarly, $\sigma-1$ acts as
identity on $k^\times/(k^\times)^3$, and $\sigma+
1$ acts trivially. The result is that 
$$
\operatorname{H}^i(\mathbb{Z}/2,\operatorname{H}^j(\mathbb{Z}/3,T_3))=0
$$
for all $i,j$, so the Hochschild-Serre spectral sequence degenerates
and shows that $\operatorname{H}^i(S_3,T_3)=0$ for all $i$. 
I claim that the same argument shows that the  $p$-torsion in
$\operatorname{H}^i(S_p,T_p)$ is trivial. In fact, we only need to compute the
cohomology of the normalizer $N_p$ of the $p$-Sylow with
$T_p$-coefficients. This is done as above via the Hochschild-Serre
spectral sequence. In the first step, 
$$
\operatorname{H}^i(\mathbb{Z}/p,T_p)\cong\left\{\begin{array}{ll}
\mu_p & i \textrm{ even}\\
k^\times/(k^\times)^p & i \textrm{ odd}.
\end{array}\right.
$$
On these groups, the quotient
$N_p/(\mathbb{Z}/p)\cong\mathbb{Z}/(p-1)$ acts via cyclic permutation
of the powers. As above, the map $\tau-1$ is an isomorphism in the
Hom-complex, showing the vanishing of cohomology
$\operatorname{H}^\bullet(N_p,T_p)$ as above. 
So, as a partial answer to your question: the cohomology in case $n$ prime is only annihilated by $n$ if it is completely trivial.  
I would love to know the general statement, what is $\operatorname{H}^\bullet(S_n,T_n)$, and I would expect it to be known and written somewhere. A seemingly related question I would
also like to see answered: what
is a reference for cohomology of the normalizer of a maximal torus in
an algebraic group. The above cohomology groups are part of the
Hochschild-Serre spectral sequence computing the normalizer of the
maximal torus in $\operatorname{PGL}_n$.  Over algebraically closed fields, the
homology of the normalizer of the maximal torus (with finite
coefficients)  can be understood in terms of \'etale 
cohomology, see the discussion in Chapter 5 of K. Knudson ``Homology
of linear groups''. Can somebody take it from here?<|endoftext|>
TITLE: Examples of conjectures that were widely believed to be true but later proved false
QUESTION [45 upvotes]: It seems to me that almost all conjectures (hypotheses) that were widely believed by  mathematicians to be true were proved true later, if they ever got proved. Are there any notable exceptions?

REPLY [2 votes]: The longest-standing one of the sort is the "conjecture" that the parallel postulate can be proved using Euclid's first four postulates. I know that it is a far-fetched understanding of "conjecture". But, after all, it is something that people tried to prove by doing it for more than two thousand years, believing that it is true.<|endoftext|>
TITLE: Covering a Cube with a Square
QUESTION [25 upvotes]: Suppose you are given a single unit square, and you would like to completely cover the surface
of a cube by cutting up the square and pasting it onto the cube's surface.

Q1. What is the largest cube that can be covered by a $1 \times 1$ square when cut into at most $k$ pieces?

The case $k=1$ has been studied, probably earlier than this reference:
 "Problem 10716: A cubical gift," American Mathematical Monthly, 108(1):81-82, January 2001,
solution by Catalano-Johnson, Loeb, Beebee.
          

(This was discussed in an
MSE Question.)
The depicted solution results in a cube edge length of
$1/(2\sqrt{2}) \approx 0.35$.
As $k \to \infty$, there should be no wasted overlaps in the covering of the 6 faces,
and so the largest cube covered will have edge length
$1/\sqrt{6} \approx 0.41$.  What partition of the square leads to this optimal cover?

Q2. For which value of $k$ is this optimal reached?

I have not found literature on this problem for $k>1$, but it seems likely it has been explored.
Thanks for any pointers!

REPLY [16 votes]: Just illustrating Noam Elkies' 4-piece solution:



           


Bottom face is mostly yellow (except for a little green); two hidden back faces are mauve.<|endoftext|>
TITLE: Does the fundamental group of a surface have rigid subgroups?
QUESTION [5 upvotes]: Suppose $\Gamma$ is the fundamental group of a closed, oriented surface $S$. Let $B$ be a finitely generated, infinite index subgroup of $\Gamma$, and let $\Gamma_B$ be the compact core of the $B$-covering space of $S$ (well-defined up to isotopy in the $B$-covering space). Lifting hyperbolic structures gives a well-defined continuous function from $T(\Gamma)$, the Teichmuller space of hyperbolic structures $\Gamma$, to $T(\Gamma_B)$, the Teichmuller space of hyperbolic structures with totally geodesic boundary on $\Gamma_B$. 
The general question I would like to ask is: How rigid is the map $T(\Gamma) \to T(\Gamma_B)$? However, as it stands that question suffers from the problem that the boundary components of $T(\Gamma_B)$ can be arbitrarily long. So, in order to sweep that problem under the rug, I want to compose with the map from $T(\Gamma_B)$ to the Masur-Minsky marking complex $M(\Gamma_B)$, defined in their paper "Geometry of the curve complex II". 
Question 1: How rigid is the composed map $T(\Gamma) \to T(\Gamma_B) \to M(\Gamma_B)$?
EDIT: What I mean by "rigidity" here is, loosely speaking, some kind of measurement of the extent of the image of this map in the marking complex $M(\Gamma_B)$. Here are two examples which exhibit opposite extremes of rigidity.
For example, if $B$ is the fundamental group of an essential subsurface of $\Gamma$ then the composed map is essentially surjective. This is the "least rigid" possibility. 
For another example, if $B$ has rank $2$ and $\Gamma_B$ is a pair of pants then the composed map has bounded image in $M(\Gamma_B)$. This is the "most rigid" possibility (although this example is uninteresting because $M(\Gamma_B)$ is itself just a single point).
Question 2: Are there examples of subgroups $B$ where $\Gamma_B$ is not just a pair of pants but the image of the composed map $T(\Gamma) \to M(\Gamma_B)$ has finite diameter?
As a reminder of how the map $T(\Gamma_B) \to M(\Gamma_B)$ is defined, one fixes an appropriate Margulis constant $\epsilon$ for $\Gamma_B$, chosen so that for any hyperbolic structure in $T(\Gamma_B)$ the collection of closed geodesics and closed proper arcs of length $\le \epsilon$ fill $\Gamma_B$, and from that collection one uses surgery to construct a pants decomposition of $\Gamma_B$ and a transverse closed curve for each pants curve.
This question occurred to me while pondering a question of Benjamin Steinberg. I am not aware of any literature on this question.
It would also make sense to ask this question with regard to a natural "lifting" map $M(\Gamma) \to M(\Gamma_B)$, but the image of this map is probably coarsely the same as the image of the map $T(\Gamma) \to M(\Gamma_B)$.

REPLY [2 votes]: Regarding Question 2, you get lots of examples that are rigid for the lifting map $M(\Gamma) \to M(\Gamma_B)$. 
Let $B$ be finitely generated subgroup of $\Gamma$ (considered a fuchsian group) such that every nontrivial element of $B$ fills $S$ (there are lots of these).  Proposition 5.1 of [Kent, Leininger, and Schleimer. Trees and mapping class groups. J. Reine Angew. Math., 637:1–21, 2009] states that if you pull back all the essential simple closed curves on $S$ to the cover corresponding to $B$ and intersect with the convex core $\mathrm{core}(B)$, you get a finite set of arcs up to proper isotopy.
The idea of the proof is this:  Suppose that there are arcs obtained by pulling back simple closed curves to $\mathrm{core}(B)$ whose lengths tend to infinity.  These arcs limit (after passing to a subsequence) to a lamination $\widetilde \lambda$ on $\mathrm{core}(B)$ that covers a lamination $\lambda$ on $S$.  But then there is a loop in $\mathrm{core}(B)$ that is not filling in $S$ (since the boundary of the supporting subsurface of $\widetilde \lambda$ will project to a curve in $S$ that misses $\lambda$).  So there is a uniform bound to length of any of the arcs in our collection, and so the collection of isotopy classes is bounded.
So, if you lift any marking, you land in a finite set of markings for $\Gamma_B$.
It seems to me that, with some more work, this argument should give you a finite set of such arcs even if you allow $\Gamma$ to range over all of Teichmuller space (Dowdall and Leininger and I have been considering similar things, so I'll talk to them), which would answer Question 2 in its original form.  Though it's probably easier to just say that the lifting map is essentially the same as $T(\Gamma) \to M(\Gamma_B)$, as you suggest.<|endoftext|>
TITLE: Softwares for drawing hyperbolic surfaces , closed, with boundaries or with punctures ?
QUESTION [5 upvotes]: In a paper I am in the process of writing in LaTeX, I need to draw and incorporate some diagrams of hyperbolic surfaces in my LaTeX document. Is there any software I can use to draw hyperbolic surfaces, either closed, or with boundaries or punctures ? Are there any such freely downloadable softwares that I can use to draw such diagrams ? Please let me know if you know one. I apologize if this is off-topic. Thank you !

REPLY [6 votes]: Recently, Dani Calegargi put together a program called wireframe for this purpose. It is documented here and available here. Playing around with it, it seems to work pretty well.<|endoftext|>
TITLE: The gimbal lock shows up in my quaternions
QUESTION [31 upvotes]: I suspect this is a bit basic for mathoverflow, seeing I'm still just an undergraduate 
I've been playing around with quaternions as means to eliminate the gimbal lock. From what I understand, one place the gimbal lock occurs is when you rotate $\frac{\pi}{2}$ around the y-axis. If I create two rotation matrices, $R_{1}$ rotates first $\phi$ around x-axis and $\frac{\pi}{2}$ around the y-axis, while $R_{2}$ rotates first $\frac{\pi}{2}$ around the y-axis and then $\theta$ around the z-axis.
$\begin{equation}
R_{1} = R_{z}(0) R_{y}(\frac{\pi}{2}) R_{x}(\phi) \\
 = \begin{bmatrix}
   0 & 0 & 1 \\
   0 & 1 & 0 \\
   -1 & 0 & 0
\end{bmatrix} \begin{bmatrix}
   1 & 0 & 0 \\
   0 & \cos(\phi) & -\sin(\phi) \\
   0 & \sin(\phi) & \cos(\phi)
\end{bmatrix} \\
 = \begin{bmatrix}
   0 & \sin(\phi) & \cos(\phi) \\
   0 & \cos(\phi) & -\sin(\phi) \\
   -1 & 0 & 0
\end{bmatrix},
\end{equation}$
$\begin{equation}
R_{2} = R_{z}(\theta) R_{y}(\frac{\pi}{2}) R_{x}(0)  \\
 =  \begin{bmatrix}
   \cos(\theta) & -\sin(\theta) & 0 \\
   \sin(\theta) & \cos(\theta) & 0 \\
   0 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
   0 & 0 & 1 \\
   0 & 1 & 0 \\
   -1 & 0 & 0
\end{bmatrix}  \\
 = \begin{bmatrix}
   0 & -\sin(\theta) & \cos(\theta) \\
   0 & \cos(\theta) & \sin(\theta) \\
   -1 & 0 & 0
\end{bmatrix}.
\end{equation} $
Since $R_{1} = R_{2}^{-1} \Rightarrow R_{1}(\theta) = R_{2}(-\theta)$, we've lost a degree of freedom. Which is what I expect.
From what I understand, if I perform the same rotations using quaternions, I should be avoiding the gimbal lock?
$ Q_{1} = Q_{z}(0) \times Q_{y}(\frac{\pi}{2}) \times Q_{x}(\theta) = (1, 0, 0, 0) \times (\frac{1}{\sqrt(2)}, 0, \frac{1}{\sqrt(2)}, 0) \times (\cos\frac{\theta}{2}, \sin\frac{\theta}{2}, 0, 0)\\ = \frac{1}{\sqrt(2)}(\cos\frac{\theta}{2}, \sin\frac{\theta}{2}, \cos\frac{\theta}{2}, -\sin\frac{\theta}{2})$
$ Q_{2} = Q_{z}(\phi) \times Q_{y}(\frac{\pi}{2}) \times Q_{x}(0) = (\cos\frac{\phi}{2}, 0, 0, \sin\frac{\phi}{2}) \times (\frac{1}{\sqrt(2)}, 0, \frac{1}{\sqrt(2)}, 0) \times (1, 0, 0, 0) \\
= \frac{1}{\sqrt(2)}(\cos\frac{\phi}{2}, -\sin\frac{\phi}{2}, \cos\frac{\phi}{2}, \sin\frac{\phi}{2})$
By setting $\phi = -\theta$, $Q_{2}$ becomes
$ Q_{2} = \frac{1}{\sqrt(2)}(\cos\frac{-\theta}{2}, -\sin\frac{-\theta}{2}, \cos\frac{-\theta}{2}, \sin\frac{-\theta}{2})$ which due to trig properies becomes 
$ Q_{2} = \frac{1}{\sqrt(2)}(\cos\frac{\theta}{2}, \sin\frac{\theta}{2}, \cos\frac{\theta}{2}, -\sin\frac{\theta}{2})$
Which means that $Q_{1}$ and $Q_{2}$ rotates around the same axis only in the oppsite direction, and we've lost a degree of freedom (??). Am I missing something fundamental?

REPLY [66 votes]: There's no paradox here: you did the same calculation in two different ways and got the same answer, as you should.  The issue is how to think about gimbal lock.
How should you represent a rotation in three dimensions?  You can try using Euler angles to represent it using three rotation angles, but there's something fishy about this.  That naturally parametrizes a three-dimensional torus, but the rotation group is not a torus (rather, it's a projective space).  It doesn't even have a torus as a covering space, but rather a 3-sphere.  So the problem is that the naive coordinates just don't give the right topology, and therefore something must go wrong in degenerate cases to fix the topology.  Gimbal lock is essentially a name for what goes wrong.
When people say quaternions avoid gimbal lock, they mean the unit quaternions naturally form a 3-sphere, so there are no topology issues and they give a beautiful double cover of the rotation group (via a very simple map).  Keeping track of a unit quaternion is fundamentally a more natural way to describe a rotation than keeping track of three Euler angles.
On the other hand, if you describe your quaternion via Euler angles, then gimbal lock shows up again, not in the quaternions themselves but in your coordinate system for them.  That's what you are seeing in your calculations: you are doing a standard calculation to see the effects of gimbal lock, and then redoing the same calculation using quaternions.
Some explanations of gimbal lock don't distinguish clearly between the underlying geometry/topology and the choice of coordinates, which has always annoyed me, since that's essential for understanding what's going on mathematically.<|endoftext|>
TITLE: Classifying spaces of a profinite groups  
QUESTION [11 upvotes]: As a topological group, a profinite group $G$ has a classifying space $BG$. On the other hand, since $G = \underleftarrow{\lim}\; G/U\;\;(U \le G$ open$)$ is an inverse limit of finite groups, we also have an inverse system $B(G/U)$ of topological spaces and can form the space $\underleftarrow{\lim}B(G/U)$. 
Question 1: How do $BG$ and $\underleftarrow{\lim}B(G/U)$ relate ? Can we, for example say, that $\underleftarrow{\lim}B(G/U)$ is a classifying space for $G$ ? 
Question 2: Are there futher results about classifying spaces of profinite groups ?

REPLY [2 votes]: There's a lot of literature on classifying spaces of the $p$-adic integers. For instance, there are two 45-year old "computations" of the complex K-theory of the $p$-adic integers, resulting in two different answers. An explanation of why they are different, only found recently, is by considering a more refined notion of a classifying space, which is a uniform space well-defined up to uniform homotopy equivalence (see, for lack of a better account, subsection 1.4 here).<|endoftext|>
TITLE: Category of Uniform spaces
QUESTION [8 upvotes]: I suspect that the category of uniform spaces and uniformly continuous maps and the full subcategory of complete uniform spaces are both bicomplete and  cartesian closed.  Can anyone comfirm or deny, with reference if possible?

REPLY [2 votes]: There's a professor at my university (Mike Rice) who I sometimes talk to about uniform spaces. When completeness came up he pointed me to the following two papers by himself and Gloria Tashjian:

"Cartesian-closed Coreflective Subcategories of Uniform Spaces Generated by Classes of Metric Spaces", Topology and Applications 15 (1983), 301-312.
"Cartesian-closed Coreflective Subcategories of Uniform Spaces", Trans. Amer. Math. Soc., 276, 1 (1983), 289-300.

The second paper gives an example which shows Unif is not cartesian-closed. We're not sure if also works to show CompleteUnif is not cartesian-closed. Another interesting example is the coreflective hull of $[0, 1]$.
If you want a full subcategory of Unif which is cartesian closed you can of course work in compactly-generated spaces (in Unif the compactly generated spaces are k-spaces with the fine uniformity), but Professor Rice pointed out to me that in Unif this subcategory can act very different from k-spaces inside Top.<|endoftext|>
TITLE: What is the difference between holonomy and monodromy?
QUESTION [49 upvotes]: What is the difference between holonomy and monodromy?
And what is the simplest example in which one is trivial and the other is not?

REPLY [17 votes]: I'm not sure the literature is entirely consistent on the use of these terms. 
Here are some ideas I learned from Jean Pradines' explanations in 1981 in Toulouse of his note  

Pradines, J.,   Th\'eorie de Lie pour les
groupoides diff\'erentiables,  relation entre propri\'et\'es
locales et globales,  Comptes Rendus Acad. Sci. Paris, S\'er A,
263 (1966), 907-910.

the first of his 4 notes introducing the relation between Lie groupoids and Lie algebroids. The ideas for the first note were  written up in detail in various work by research students at Bangor, with the full knowledge of Pradines. (Kirill Mackenzie worked quite independently on the succeeding theory of Lie groupoids and Lie algebroids, published in his 1987 book.) 
Intuitively, non trivial holonomy may be explained  as an iteration of local procedures which return to the starting point with a change of phase. This idea is related to hysteresis, and shows the nice relation with physics. The problem is to define rigorously all the terms used in this explanation!
A Monodromy Principle is enunciated in Chevalley's famous book ``Lie groups"; the Principle may be explained as giving an extension of a restriction of a local morphism to a morphism on a simply connected cover.
In foliation theory, it is usual to define a monodromy groupoid as the disjoint union of the fundamental groupoids of the leaves, with a topology reflecting the local structure of the foliation; and to define the holonomy groupoid as a quotient of the monodromy groupoid. However this does not easily yield a Monodromy Principle.  
Pradines' idea for his Th\'eor`eme 2 in the Note was to use the Monodromy Principle as guiding the construction of a Monodromy Groupoid of a Lie groupoid, generalising the universal cover of a connected Lie group. So, given a neighbourhood $W$ of the identities of a Lie groupoid, one forms the groupoid $M(W)$ which is universal for all local morphisms of $W$ into groupoids. The problem is to define an appropriate topology on $M(W)$ and Pradines solves this using a notion of holonomy groupoid, although in this case the holonomy is trivial (!).
In the case of a Lie group $G$, the topology on $G$ is defined by a neighbourhood of the identity satisfying some reasonable conditions given in, for example, Bourbaki. Now one can define a local Lie groupoid to be a groupoid $G$ with a set $W$ containing the identities and satisfying a number of reasonable conditions. However it is no longer true that the topology of $W$ extends to a topology on $G$ making it a Lie groupoid. Instead there is, under reasonable conditions,  a Holonomy Groupoid $Hol(G,W)$ which projects to $G$ and which has a Lie groupoid structure locally like $W$. The construction of Pradines is written up in:  

Aof, M.E.-S.A.-F. and Brown, R.,  The holonomy groupoid of a locally topological groupoid,  Top. Appl.  47, 1992, 97-113

(with the agreement of Pradines). It really does use the idea of ``iteration of local procedures" where the local procedures here are given by Ehresmann's local admissible  sections of $G$ with values in $W$. The holonomy groupoid $Hol(G,W)$ has a universal property for maps of Lie groupoids into $G$.  
The application to  the monodromy groupoid is written up in: Brown, R. and Mucuk, O.,  The monodromy groupoid of a Lie groupoid, Cah.
Top. G\'eom. Diff. Cat. 36 (1995) 345-369. 
See also:   Mucuk, O., Kılı¸carslan, B., S¸ahan, T. and Alemdar N. Group-groupoid and monodromy groupoid, Topology and its Applications 158 (2011) 2034-2042.
So holonomy comes out as a kind of right adjoint, and monodromy as a kind of left adjoint, which explains one difference. But there seems  still work to do to explain everything stated in the two Theorems of Pradines' first Note, and to apply these ideas more widely. This is a reason for advertising Pradines' ideas here. 
May 29, 2019  In response to Arrow's request, here are some references,  The statements on adjoints are made in Pradines' cited paper, but very briefly, and with no proofs. 
One  idea behind holonomy is that of "an iteration of local procedures returning to the starting point but with a change of phase"; an example is moving a pencil at the North pole of a globe down a line of longitude, then along the equator, then back up to the pole, when it is pointing in a different direction. Pradines formalised this in terms of a "locally topological groupoid". An old theorem is that a locally topological group is extendible to a topological group (Bourbaki). For groupoids this is no longer true; the precise statement is in this edited version of a published paper; instead $(G,W)$ determines under good conditions a holonomy topological groupoid $H(G,W)$ which maps down to $G$ and is universal for maps of topological groupoids into $(G,W)$. 
The situation is different for monodromy, which relates to the old "monodromy principle", that a local morphism should be extendible to a universal cover. See papers [88,89] in my publication list, in which a "universal cover" is replaced by a "monodromy groupoid" of a topological groupoid.  Sorry to be so brief on this.<|endoftext|>
TITLE: How to construct a continuous finite additive measure on the natural numbers
QUESTION [13 upvotes]: I want to find some condition to construct a continuous finitely additive measure on the natural numbers, i.e. $f:P(\mathbb{N})\rightarrow [0,1]$ such that $f(\{n\})=0$, and $f$ is an additive measure.
I know in ZFC we can use an ultrafilter $U$ and define $f$ by $f(A)=1\Leftrightarrow A\in U$, but this is too trival.
How about ZF? or some other condition? like large cardinal.

REPLY [8 votes]: Eric van Douwen constructed a whole slew of finitely additive measures on the set of natural numbers, with various shifting and scaling properties in this paper. He discussed how much `choice' was needed for the constructions; often some version of the Hahn-Banach theorem would suffice.<|endoftext|>
TITLE: Proofs of the Stallings-Swan theorem
QUESTION [21 upvotes]: It is a well-known and deep${}^\ast$ theorem that if a group $G$ has cohomological dimension one then it must be free. This was proved in the late 60's by Stallings (for finitely generated groups) and Swan.
The proofs in the original articles are well-written and informative, bringing together a lot of ideas from topology, algebra and set theory. However there seems to be some scope for the proofs to be shortened, for instance by a clever choice of projective resolution, or the use of non-abelian cohomology.
My questions then is

Have any alternative (shorter) proofs of the Stallings-Swan theorem appeared since 1969?

${}^\ast$I am prepared to accept that the answer may be 'no', but in that case I wonder if someone could offer an explanation of why this theorem is so "deep", ie why there cannot exist some "quick trick" proof.

REPLY [33 votes]: The heart of the matter is the Stallings' "ends of groups" theorem: A finitely-generated group with infinitely many ends splits as graph of groups with finite edge groups. In addition, one also has to show that the decomposition process terminates for your group (this property is called accessibility). Neither one has a quick an dirty proof and for a good reason. 
a. Dunwoody has shown that accessibility fails for some finitely-generated groups with torsion, so something interesting (Grushko's theorem) is going on even in the easy part of the proof. 
b. The ends of groups deals with the key problem of geometric group theory: Relating geometric properties of a group and its algebraic structure. Each time one manages to recover algebraic structure of a group from geometric information about the group, some minor (or major) miracle has to happen and, to the best of my knowledge (with few trivial exceptions) there are no easy proofs of the results of this type. 
The "shortest" proof of the "Ends of groups theorem" is due to Gromov, see pages 228-230 of his essay on hyperbolic groups. The trouble with Gromov's proof is that it relies  on a compactness property  ("obvious" to Gromov) for a certain family of harmonic functions (in addition, a construction of the tree was missing in his proof, and this requires some trickery if one uses harmonic functions for finitely-generated groups). This compactness result (as far as I know) has no easy proof. I wrote a (somewhat long) proof in http://arxiv.org/pdf/0707.4231.pdf , Bruce Kleiner managed to shorten it to about 7 pages (this is not published), but his proof is still not quick. 
Dunwoody's proof (see John Klein's excellent comments) improved on the Stallings' proof, but his proof is still quite complicated. 
Niblo's proof in http://eprints.soton.ac.uk/29820/1/Stallingstheorem.pdf
provides another geometric argument using Sageev's complex, but Niblo's paper is still 20 pages long.    
Just to indicate how nontrivial Stallings' theorem is, consider the question: Is it true that every finitely generated group $G$ of homological dimension 1 is free? This is false for infinitely generated groups (like $G={\mathbb Q}$) and is true for finitely-presented groups (simply since in this case cohomological dimension is also 1). Otherwise, this problem is open since Stallings' theorem (and not for the lack of trying!).<|endoftext|>
TITLE: Type of 26-dimensional representation of different real forms of the complex simple Lie algebra $F_4$
QUESTION [12 upvotes]: The exceptional complex simple Lie algebra $F_4$ has an irreducible 26-dimensional representation $V$ with Dynkin label [0,0,0,1] in the usual ordering of the simple roots one can find, say, in Humphreys's book on Lie algebras and representation theory.  In fact, $F_4$ can be defined as the Lie subalgebra of $\mathfrak{sl}(V)$ which preserves a symmetric inner product and a certain cubic form on $V$.
Now there are three different real forms of $F_4$ and my question is about what happens to $V$ when restricting to these real forms.  The three real forms are the compact real form, the split real form and a third form.  They can be distinguished by the 'index' of the Killing form; i.e., if the Killing form $\kappa(X,Y) = \operatorname{Tr} \operatorname{ad}_X \operatorname{ad}_Y$ has signature $(p,q)$, its index is $p-q$.  I am most familiar with the compact real form, for which the Killing form is negative-definite, whence of index $-52$.  The split real form has index $4$ and the third real form has index $-20$, and are denoted $F_4^4$ and $F_4^{-20}$, respectively.
I would like to know the following (pointers to the literature would also be greatly appreciated):
Questions

What is the type of $V$ under the different real forms?  I know that for the compact real form it is real, but I would like to know also for $F_4^{-20}$ and $F_4^4$.
And if the type is real (as I suspect is the case), what is the signature of the invariant inner product on the underlying real representation $V_{\mathbb{R}}$?

Thanks in advance!
Edit
Based on Jim's answer below, the representations are of real type in all cases.  From Bruce's answer, it would seem that for the split case $F_4^4$ the signature is (14,12).
In a rather convoluted calculation, I seem to find that for $F_4^{-20}$ the signature is (16,10), but I would like confirmation since I have seen at least one claim in the Physics literature (see last equation in §4) that it is (25,1).

REPLY [4 votes]: All the forms of $F_4$ can be defined as automorphism groups of some Jordan algebra of three by three matrices with entries in octonions / split octonions / complexified octonions.  These algebras are all of dimension 27 over the appropriate field and the subspaces of trace-free matrices are the irreducible 26-dimensional representations of the various forms of $F_4$. The invariant quadratic form is $A\mapsto \mathrm{Tr}(A^2)$. (And the invariant cubic form is $A\mapsto \mathrm{det}(A)$.)
The group $F_4^{-20}$ is according to Yokota (but I guess that one can dig this up also out of the work of Veldkamp or Springer) the automorphism group of the real Jordan algebra $J(1,2,\mathbb{O}) = \{X\in \mathrm{M}(3,\mathbb{O}\otimes_\mathbb{R}\mathbb{C}) \, |\, I_1 \overline{X}^tI_1 = X \}$ where $I_1 = \mathrm{diag}(-1,1,1)$. Now the computation of the signature of $A\mapsto \mathrm{Tr}(A^2)$ is a matter of simple calculation.
The other two real cases $F_4^{-52}$, $F_4^4$ follow similarly since they are the automorphism groups of  $J(3,\mathbb{O}) = \{ X\in M(3,\mathbb{O})\,|\, X^t =X  \}$ and  $J(3,\mathbb{O}) = \{ X\in M(3,\mathbb{O}')\,|\,  X^t=X \}$ respectively.<|endoftext|>
TITLE: Representation theory of a finite p-group over a field of characteristic p: dim of invariants =1 => dim of coinvariants = 1? 
QUESTION [8 upvotes]: I am trying to understand the proof of Proposition 4 in
S. Ullom, Integral normal bases in Galois extensions of local fields, Nagoya Math. J. Volume 39 (1970), 141-148. The PDF is available here:
http://projecteuclid.org/euclid.nmj/1118798052
It appears that the following result is used, but I'm afraid that I don't quite see the proof.
Let $k$ be a field of characteristic $p$ and let $G$ be a finite $p$-group. Let $W$ be a left $k[G]$-module such that $\dim_k W = |G|$. Suppose that $\dim_k W^{G} = 1$. Then $\dim_k W_G = 1$.
Here $W^G$ denotes invariants and $W_G$ denotes coinvariants.
I have to admit that I know relatively little about representation theory in characteristic $p$. One idea would be to consider the dual representation $\hat{W}$, but I only got so far:
$\dim_k W^{G} = 1 \implies W$ is indecomposable $\implies \hat{W}$ is indecomposable. But maybe this is not the right approach. If I could show that the Tate cohomology groups of $W$ vanish, then of course the desired result drops out, but I think this is rather strong medicine.
Is anyone able to give a proof of the above claim? I suspect the solution is fairly easy, but I just don't see it at the moment.

REPLY [5 votes]: Maybe it's useful to add some general perspective to the efficient answer given.   The approach taken by modp is undoubtedly the most natural one, taking advantage of the extra assumption that $W$ has dimension equal to $|G|$.   This assumption is essential, since without it a $k[G]$-module may well have a space of covariants of higher dimension when the space of invariants is 1-dimensional.   (Concrete examples occur when you consider restrictions to a Sylow $p$-subgroup in a finite group of Lie type of various standard modules.)   
It's a standard fact that any finite $p$-group has only the trivial simple module over a field of characteristic $p$ (unlike the radically more complicated situation in characteristic 0).   In particular, any module with a 1-dimensional subspace of invariants is automatically indecomposable.    Though the indecomposable $k[G]$-modules can be arbitrarily difficult to study in general, the special assumption that $\dim_k W = |G| = \dim_k k[G]$  then yields an isomorphic embedding of $W$ onto $k[G]$:  For any finite group the group algebra over a field is both injective and projective as a left module.   In the special case of a $p$-group, the group algebra is itself the projective cover (= injective hull) of the unique simple module.   In particular, the socle (here the space of all invariants) is isomorphic to the head (here the space of all coinvariants).   Indeed, the module $W$ is self-dual.<|endoftext|>
TITLE: Stone-Weierstrass analogue for $L^p$
QUESTION [9 upvotes]: Let $A$ be a complex algebra of bounded measurable functions on the measure space $(X,\mu)$ (case of $[0,1]$ with Lebesgue measure is enough for me) closed under conjugation. Assume that $A$ separates points, i.e. there is no non-trivial measurable partition of $X$ such that each function in $A$ is constant on (almost every) part. Is it true that $A$ is dense in $L^p(X,\mu)$ for $1\leq p < \infty$?

REPLY [8 votes]: Yes (I assume that the measure is finite). Here is a proof that uses the von Neumann bicommutant theorem (or rather Kaplansky's density theorem).
See $A \subset L^\infty(X,\mu) \subset B(L^2(X,\mu))$ where $L^\infty$ acts on $L^2$ by pointwise multiplication. Then the assumption that $A$ separates points is exactly that the commutant of $A$ is $L^\infty(X,\mu)$, so that the bicommutant of $A$ is $L^\infty(X,\mu)$. Therefore, by Kaplansky's density theorem, any $f \in L^\infty$ with $\|f\|_\infty \leq 1$ belongs to the strong operator topology closure of $\{g \in A, \|g\|_\infty\leq 1\}$. Equivalently, there is a net $g_\alpha \in A$ with $\|g_\alpha\|_\infty \leq 1$ such that, for every $\xi \in L^2$, $\|g_\alpha \xi - f \xi\|_2\to 1$. In particular (using that the constant function $1$ belongs to $L^2$), $\|g_\alpha - f\|_2 \to 0$. But by this implies that for every $1\leq p < \infty$, $\|g_\alpha - f\|_p \to 0$~: if $p<2$ this is because the $L^p \subset L^2$ (the measure is finite), whereas if $p>2$ this is the inequality $\| \cdot \|_p \leq \|\cdot \|_\infty^\theta \|\cdot \|_2^{1-\theta}$ for $\theta=1-2/p>0$.
This proves that the $\| \cdot \|_p$-closure of $A$ contains $L^\infty$, and hence it is $L^p$.<|endoftext|>
TITLE: Reference Request: Relative De Rham Cohomology
QUESTION [14 upvotes]: I'm looking for a book, article, or lecture notes that does basic cohomology theory from a relative point of view (including the Thom isomorphism, the excision theorem, Lefschetz duality, the Gysin sequence, etc.) and uses the de Rham model for relative cohomology.  
Bott and Tu does most of basic cohomology theory using the de Rham model and even has a brief section on how to define the relative de Rham groups, but they mostly avoid the relative groups when formulating and proving the main results.  Hatcher uses relative cohomology groups all over the place but doesn't really do anything with de Rham cohomology.  I've been trying to build a dictionary between these two languages but I've run into some trouble at various points and I was hoping that somebody else has sorted all of this out.

REPLY [6 votes]: If $N \subset M$ is a closed (meaning a closed subset, not a compact submanifold without boundary) submanifold, then the restriction map $\Omega^{\ast}(M) \to \Omega^{\ast}(N)$ is surjective. You can see this using local adapted charts and partitions of unity. Thus you can define the relative cohomology as the cohomology of the complex that is the kernel of the restriction map. This is, as far as I remember, the viewpoint in Jost "Riemannian Geometry and Geometric Analysis".<|endoftext|>
TITLE: Eta Invariant of Spherical Space Form 
QUESTION [6 upvotes]: Is the eta invariant of spherical space form $\eta(S^3/\Gamma)$ always nonegative?
Can we calculate it with the information of $\Gamma\in SO(4)$ explicitly? 
In fact, i need a reference for the calculation of eta invariant. Can some one give me some advice or download the following paper for me? Thank you!
(1) Hitchin, N. J.(4-CAMB)
Einstein metrics and the eta-invariant. (Italian summary) 
Boll. Un. Mat. Ital. B (7) 11 (1997), no. 2, suppl., 95–105. 
(2) G.W. Gibbons, C.N. Pope
Index theorem boundary terms for gravitational instantons
Nuclear Physics B Volume 157, Issue 3, 1 October 1979, Pages 377–386.
Email: xuyiyan@math.pku.edu.cn

REPLY [3 votes]: The eta invariant of any locally symmetric space $\Gamma \backslash G/K$ is calculated by Moscovici and Stanton in Eta invariants of Dirac operators on locally symmetric manifolds. They related eta invariant with the conjugation class of $\Gamma$.<|endoftext|>
TITLE: Status of PL topology
QUESTION [56 upvotes]: I posted this question on math stackexchange but received no answers. Since I know there are more people knowledgeable in geometric and piecewise-linear (PL) topology here, I'm reposting the question. I'd really want to know the state of the question, since I'm self-studying the material for pleasure and I don't have anyone to talk about it. Please feel free to close this post if you think the topic is not appropriate for this site.    
I'm starting to learn about geometric topology and manifold theory. I know that there are three big important categories of manifolds: topological, smooth and PL. But I'm seeing that while topological and smooth manifolds are widely studied and there are tons of books about them, PL topology seems to be much less popular nowadays. Moreover, I saw in some place the assertion that PL topology is nowadays not nearly as useful as it used to be to study topological and smooth manifolds, due to new techniques developed in those categories, but I haven't seen it carefully explained.
My first question is: is this feeling about PL topology correct? If it is so, why is this? (If it is because of new techniques, I would like to know what these techniques are.)
My second question is: if I'm primarily interested in topological and smooth manifolds, is it worth to learn PL topology?
Also I would like to know some important open problems in the area, in what problems are mathematicians working in this field nowadays (if it is still an active field of research), and some recommended references (textbooks) for a begginer. I've seen that the most cited books on the area are from the '60's or 70's. Is there any more modern textbook on the subject?
Thanks in advance.

REPLY [7 votes]: These questions are ok, but it is important to understand 
as much as you can about manifolds.
 For each of the categories:
 a] Homotopy Types satisfying Poincaré Duality...
 b] Topological Manifolds... 
 c] Sobolev Manifolds eg Quasiconformal or  Lipschitz Manifolds...
 d] Piecewise Linear or Piecewise Differentiable Manifolds...
 e] C1, C2,...C-infiniy = Smooth Manifolds...
 f] Real Analytic Manifolds...
 g] Real Algebraic Manifolds...
    The types with canonical coordinates:::
 h] Poisson Manifolds...
 i] Symplectic Manifolds...
 j] Complex Manifolds...
 k] Generalized Complex-Symplectic manifolds...
 l] Geometrized Three-Manifolds... 
  One knows contexts where each of these categories are particularly useful.
  [Dennis Sullivan]<|endoftext|>
TITLE: General gluing theorem for adjunction spaces
QUESTION [9 upvotes]: Consider the following interesting theorem (7.5.7, p.294 in Topology and Groupoids by Ronald Brown):
Gluing theorem for adjunction spaces:
Suppose that we have the following commutative diagram of topological spaces and continuous maps:
    
where $\varphi_{A}$, $\varphi_{X}$ and $\varphi_{Y}$ are homotopy equivalences, and the inclusions $i$ and $i'$ are closed cofibrations. Then the map
$$
\varphi:X\cup_{f}Y\rightarrow X'\cup_{f\phantom{l}'}Y'
$$
induced by $\varphi_{A}$, $\varphi_{X}$ and $\varphi_{Y}$ is a homotopy equivalence. 
In this post I asked a question that would become a corollary if the gluing theorem were true for arbitrary cofibrations $i$ and $i'$. I would like to know if this is the case. Otherwise, does anybody know of a counterexample?
Perhaps, it is possible that although the map $\varphi$ may not in general be a homotopy equivalence when the cofibrations $i$ and $i'$ are arbitrary, the adjunction spaces $Y\cup_{f}X$ and $Y'\cup_{f'}X'$ will nonetheless be homotopy equivalent. A possible approach to this would be to try to find a third space containing both as deformation retracts. 
Note on notation: That an inclusion map $j:B\rightarrow Z$ is a closed cofibration means that it is a cofibration (i.e., the pair $(B,Z)$ has the homotopy extension property) and the set $B$ is closed in $X'$.
Thank you

REPLY [5 votes]: Yes, this is true even for not necessarily closed cofibrations. If you want a single source that gives a complete proof, then the only one that comes to my mind is this preprint.
Definition 1.1.1 introduces cofibration categories and then Lemma 1.4.1 says that the desired result holds in any cofibration category. Section 3.1 contains a detailed proof that the category of topological spaces equipped with Hurewicz cofibrations and homotopy equivalences is a cofibration category and thus the lemma applies. The crux of the matter is that acyclic cofibrations are closed under pushouts and this follows from a classical result of Dold (Lemma 3.1.9) that acyclic cofibrations admit deformation retractions, which doesn't depend on closedness.<|endoftext|>
TITLE: Are semi-direct products categorical (co)limits?
QUESTION [31 upvotes]: Products, are very elementary forms of categorical limits. My question is whether in the category of groups, semi-direct products are categorical limits.
As was pointed in:
http://unapologetic.wordpress.com/2007/03/08/split-exact-sequences-and-semidirect-products/
Bourbaki (General Topology, Prop. 27) gives a universal property:
Let $f \colon N \to G$, $g \colon H \to G$ be two homomorphisms into a group $G$, such that $f(\phi_h(n)) = g(h)f(n)g(h^{-1})$ for all $n \in N$, $h \in H$. Then there is a unique homomorphism $k \colon N \rtimes H \to G$ extending $f$ and $g$ in the usual sense.
However, I remain unsatisfied. The condition $f(\phi_h(n)) = g(h)f(n)g(h^{-1})$ is a condition on elements of groups, rather than a condition that says that some diagram is commutative.
So the question remains: are semi-direct products in the category of groups categorical limits?

REPLY [50 votes]: There is (another ?) description of the crossed product in categorical terms. 
Let ${\rm Mor}(Gp)$ be the category whose objects are homomorphisms of groups and morphisms are commutative diagrams. Let $C$ be the category of "groups acting on groups" whose objects are pairs of groups $(H,G)$ together with a homomorphism $H \to {\rm Aut}(G)$. Morphisms in this category are equivariant homomorphisms.
Now, there is a natural forgetful functor $T \colon {\rm Mor}(Gp) \to C$ which sends $H \to G$ to the pair $(H,G)$ with the homomorphism $H \to {\rm Aut}(G)$ given by conjugation. Now, almost by definition, the crossed product is the left-adjoint of this forgetful functor. Indeed, the left adjoint is easily seen to map $(H,G)$ with $H \to {\rm Aut}(G)$ to the inclusion $H \to G \rtimes H$.
Being a left-adjoint, the "crossed product" maps colimits to colimits.<|endoftext|>
TITLE: What makes a distance?
QUESTION [9 upvotes]: In the answers to my previous question, I learned that there are different concepts of distance, that is of distance-like functions with the usual metric being only the most popular and important one.
All distance-like functions seem to agree in being 

nonnegative: $d(x,y) \geq 0$
symmetric:  $d(x,y) = d(y,x)$
reflexive: $d(x,x)=0$ 

– let me call such functions proto-metrics – but differ in other respects:

a pseudo-metric fulfills the usual triangle inequality, but allows $d(x,y)=0$ for $x \neq y$.
an ultra-metric has $d(x,y)=0$ only for $x = y$, but fulfills a stronger triangle inequality, namely $d(x,z) \leq \text{max}(d(x,y),d(y,z))$ 
a dissimilarity function – conceived as some kind of distance, e.g. in a property space – is just any proto-metric, i.e. fulfills nothing but the core conditions above, esp. no triangle inequality at all.

One might ask whether such a dissimilarity function deserves its name, because it violates a strong intuition on dissimilarity (resp. distance) by allowing – without any restriction – two objects which are very similar (= close) to a third object to be arbitrarily dissimilar (= far away) from each other. 
Given that a distance-like function has to fulfill some kind of triangle inequality, my question boils down to:

Which conditions does a function $F$ have to
  fulfill such that
$$d(x,z) \leq F(d(x,y),d(y,z))\qquad(*)$$
can be seen as a reasonable triangle inequality,
  thus making the proto-metric $d$
  distance-like?

Some conditions on $F$ seem to spring to mind:

symmetry: $F(x,y)=F(y,x)$ 
monotonicity: $F(x,y) \leq F(x',y)$ for $x\leq x'$ 
scale-invariance: $\alpha F(x,y) \leq F(\alpha x, \alpha y)$ for $\alpha > 0$ 
associativity: $F(x,F(y,z)) = F(F(x,y),z)$ (Did I understand this correctly, Will Sawin? Where does this intution come from?) 
$F(x,0)=x$ (Thanks to David Feldman)

Is there hope to fix a set of conditions $\mathcal{C}$, such that one will be willing to accept any proto-metric $d$ to be distance-like if it fulfills $(*)$ for an $F$ that fulfills $\mathcal{C}$?

REPLY [3 votes]: Thinking about – and inspired by – an insightful example by Aaron Meyerowitz I found a convincing argument that there is probably no generally agreed upon concept of distance-likeness involving a sensibly generalized triangle inequality. The argument goes like this:
Consider a world of non-intersecting disks in the plane. Let the distance $d$ between two disks be the smallest Euclidean distance between any two points on their respective boundaries. This involves that any two touching disks have distance 0 and that the distance-law d(x,y)=0 iff x=y is violated. I assume that it is nevertheless generally agreed upon that this is a sensible distance. 
But any sensible triangle inequality is violated, too. That's because for every two non-touching disks $x,z$ that touch a common disk $y$ we have
$$d(x,z) > d(x,y) + d(y,z) = 0$$
and $d(x,z)$ can be arbitrarily large (depending on the radius of disk $y$) and thus greater than $F(0,0)$ for any function $F(u,v)$.
The question arises for what specific reasons we nevertheless do believe that $d$ is a
sensible distance?
Note, that for a given maximal disk radius $r$max we may get a sensible version of the triangle inequality (due to Aaron's example) and with $r$max $\rightarrow 0$ we get the usual triangle inequality.<|endoftext|>
TITLE: Axiomatic Set Theory
QUESTION [8 upvotes]: I am wondering if there exists a book that discuss different axiomatizations of set theory and compare them. Can you please give me any references?

REPLY [2 votes]: And if you happen to be (able to read) Czech try Sochor's Metamatematika Teorii Mnozin<|endoftext|>
TITLE: An exercise in Kunen. Getting Axiom of Replacement from set-like transitive closure.
QUESTION [8 upvotes]: I am studying Kunen's Set Theory (2011 edition) on my own. I got stuck at the excercise I.9.6 which is:
Excercise I.9.6. Derive the axiom of replacement from lemma I.9.5.
And the mentioned lemma is this:
Lemma I.9.5. For a relation R and a class A, if R is set-like on A, then R* is set-like on A.
Here R* is the transitive closure of R. Also, a relation R is set-like on a class A if { x\in A : xRy } is a set for all y\in A.
Help appreciated.

REPLY [15 votes]: Suppose that the lemma holds and that we are considering an instance of the replacement axiom, so we have a set $X$ and for some parameter $z$ and for every $x\in X$ there is a unique $y$ such that $\varphi(x,y,z)$. Fix any set $w$ not in $X$, and let $R$ be the class relation such that $R(x,w)$ for each $x\in X$, and such that $R(y,x)$ whenever $\varphi(x,y,z)$. That is, the children of $w$ are exactly the members of $X$, and the child of any $x\in X$ is precisely the corresponding $y$. Thus, the relation $R$ is set-like, since $X$ is a set, and $\{y\}$ is a set for each $y$ that arises. But the transitive closure of $R$ will relate all the $y$'s that arise from any $x\in X$ to $w$. And so if the transitive closure of $R$ is set-like, then the set $\{y\mid \exists x\in X\, \varphi(x,y,z)\}$ will be a set, thereby verifying this instance of the replacement axiom.<|endoftext|>
TITLE: Analogues of the Knuth and Forgotten equivalences on permutations: have they been studied?
QUESTION [10 upvotes]: Consider a totally ordered alphabet $A$ of $n$ letters. Let $W$ be the set of all words over $A$ which have no two letters equal. Then, for example, we can define the Knuth equivalence on $W$ as the smallest equivalence relation $\equiv$ which satisfies the following two properties:

If a word $w$ is obtained from another word $w^{\prime}$ by finding in $w^{\prime}$ a subword of the form $bca$ with $a < b < c$ and replacing it by $bac$, then $w \equiv w^{\prime}$.
If a word $w$ is obtained from another word $w^{\prime}$ by finding in $w^{\prime}$ a subword of the form $acb$ with $a < b < c$ and replacing it by $cab$, then $w \equiv w^{\prime}$.

In short, we say that the Knuth equivalence is "the equivalence relation generated by $bca\equiv bac$ and $acb\equiv cab$".
The Knuth equivalence has been studied a lot (suitably extended to words with possibly equal letters, it gives rise to the so-called plactic monoid and has a significant role in the modern theory of the symmetric group and Young tableaux).
A similar relation, namely the one generated by $acb\equiv bac$ and $bca\equiv cab$, has been studied by Novelli and Schilling in arXiv:0706.2996v3 and been called the "forgotten equivalence".
William Kuszmaul, a student I am mentoring in the MIT Primes project, has been working on systematically analyzing equivalence relations like this (continuing the work started in Linton, Propp, Roby, West, arXiv:1111.3920 and Pierrot, Rossin, West, FPSAC 2011), and was able to, e. g., compute the number of equivalence classes for many of them.
What we would like to know is how many such equivalence relations have already been studied. We are particularly interested in those of the form "equivalence relation generated by $...\equiv ...$ and $...\equiv ...$", since both the Knuth and the forgotten equivalence are of that type (as opposed to, say, the Chinese one), and it is these relations that, for some reason, turn out in algebraic contexts (symmetric functions, in particular).
We are interested in three-letter relations only, for the time being; so the hypoplactic monoid is not what we care about.

REPLY [3 votes]: The sylvester monoid (with the "s" of "sylvester" in lowercase) is a very good example of a plactic-like monoid. It was introduced by Novelli, Hivert, and Thibon (see The Algebra of Binary Search Trees). It is defined in the following way: Let $(A, \leq)$ be a totally ordered alphabet and consider the monoid congruence $\equiv$ generated by
\begin{equation}
    ac-b \equiv ca-b
\end{equation}
for any $a, b, c \in A$ such that $a \leq b < c$ (here, the symbol $-$ denotes an arbitrary factor). Then, the sylvester monoid is $A^*/_\equiv$.
Since $\equiv$-equivalence classes of words the have same commutative image, we can consider $\equiv$-equivalence classes of permutations. Equivalence classes of permutations of $\mathfrak{S}_n$ are in bijection with standard binary search trees (that are binary trees with $n$ internal nodes bijectively labeled on $[n]$).
The analog of the Robinson-Schensted insertion in the context of the sylvester monoid is simply the usual algorithm of insertion of a word (from right to left) in a binary search tree. Furthermore, this monoid is the main ingredient of a very interesting construction of ${\bf PBT}$, a Hopf algebra involving binary search trees which provides a generalization of symmetric functions.
There are a lot of other interesting plactic-like monoids like the Bell monoid (see Algebraic constructions on set partitions). It is the quotient of the free monoid generated by $A$ by the monoid congruence $\equiv$ generated by
\begin{equation}
    ac \; u \; b \equiv ca \; u \; b,
\end{equation}
where $a, b, c \in A$, $u \in A^*$, $a \leq b < c$, and any letter $d$ of $u$ satisfies $d \geq c$. Maybe this particular definition can give you some ideas to define new interesting monoids.
Besides, the Baxter monoid (see Algebraic and combinatorial structures on pairs of twin binary trees and The Hopf algebra of diagonal rectangulations) is another plactic-like monoid which involves four letters instead of three.<|endoftext|>
TITLE: Positively curved metrics on $S^2\times S^2$
QUESTION [6 upvotes]: As you know, the Hopf conjecture is about the existence of positively curved metric on $S^2\times S^2$. Hsiang-Kleiner have shown that there exists no positively curved metric admitting $S^1$-action on $S^2\times S^2$. 
My question is simple. If $(S_1=S^2,g)$ and $(S_2=S^2,h)$ are positively curved, 
 then for any positive function $f: S_1 \rightarrow \mathbb{R}$, is a warped metric $g+ fh$ not positively curved ? Or is this statement not proved ?

REPLY [5 votes]: An alternative proof of this result is to realize that the injection into the first factor $i:S^2\to S^2\times S^2$ gives you a "soul" of the warped product. By that I mean that the composition of the Riemannian submersion (for the warped metric) $S^2\times S^2\to S^2, (p,q)\to p$ with $i$ provides you with an analog of the Sharafutdinov map for open nonnegatively curved manifolds. Then you can adapt the proof of Perelman's proof of the soul conjecture to rule out positive curvature, as well as gain some rigidity for nonnegative curvature. The details appear in the last section of (sorry for the shameless self-promotion)
V. Berestovskii and L. Guijarro, A metric characterization of Riemannian submersions, Annals of Global Analysis and Geometry 18 (2000), pp.577-588<|endoftext|>
TITLE: Categorical description of the restricted product (Adeles)
QUESTION [18 upvotes]: Background on the Adèles
The Adèles $\mathbb{A}_K$ of a number field or function field $K$ are defined as a restricted product of the complete local fields $K_\nu$, where $\nu$ ranges over all places of $K$. The restricted product is usually defined as the subset of $\prod_\nu K_\nu$ given by
$\mathbb{A}_K := \prod_\nu' K_\nu := \{ (x_\nu)_\nu \in \prod_\nu K_\nu\ |\ \text{ all but finitely many } x_\nu \in \mathcal{O}_\nu\}$
where $\mathcal{O}_\nu$ is the ring of integers in $K_\nu$.
Tensor product description
An alternative description, for the sake of concreteness given for the rationals $K=\mathbb{Q}$ can be made by using the tensor product:
$\mathbb{A}_\mathbb{Q} = \left(\left(\prod_p \mathbb{Z}_p\right) \otimes_\mathbb{Z} \mathbb{Q}\right) \times \mathbb{R}.$
This is the same, because $\mathbb{Z}_p \otimes \mathbb{Q} = \mathbb{Q}_p$ and the tensor product captures the finiteness condition. As there are always only finitely many infinite places, this description can be given for any number field as well (and of course function fields, since they don't have infinite places at all).
The topology on the Adèles
The restricted product comes with a restricted product topology, which is not the subspace topology from the ordinary product (despite its name), but the topology whose subbasis sets are
$V_{\eta,U_\eta} := \{(x_\nu)_\nu \in \prod_\nu K_\nu\ |\ x_\nu \in \mathcal{O}_\nu \text{ for } \nu \neq \eta, \text{ and } x_\eta \in U_\eta\}$
with $\eta$ a place and $U_\eta \subseteq K_\eta$ any open subset.
The subspace topology from the product differs from this by requiring only $x_\nu \in \mathcal{O}_\nu$ for all but finitely many places, which are not fixed uniformly for a subbasis set.
Given a subset $U$ of $\mathbb{A}_K$ which is open in the ordinary subspace topology from the ordinary product, for every place $\nu$ there might be an $x \in U$ such that $x_\nu \notin \mathcal{O}_\nu$. If instead $U$ is open in the restricted product topology, there is a fixed finite set of places $\{\nu_1,...,\nu_m\}$ such that for every $x \in U$ and every other place $\nu \neq \nu_i$ we have $x_\nu \in \mathcal{O}_\nu$.
Nice properties of this topology are: You get again a locally compact group with compact open subgroup $\prod_\nu \mathcal{O}_\nu$ and that the Haar measure on $\mathbb{A}_K$ gives the quotient $\mathbb{A}_K/K$ a finite measure (with $K$ embedded diagonally by the maps $K \to K_\nu$).
The question: how to describe the Adèles categorically?
More specifically, I'd like to understand the restricted topology as well.
The ordinary product is a limit, and as such it carries the initial topology. Any subspace carries the initial topology as well, but this gives the wrong topology, not the restricted product topology but the topology restricted from the product.

Is it impossible to give a categorical description?
Would it even be useful to have a categorical description?
Does one have to apply a limit-colimit procedure or might a single limit or colimit suffice?
There are some similarities with ultraproducts, which are classically not defined in a categorical way, but it is possible. The restricted product is somewhat dual to an ultraproduct. Could that help?
Is there a good canonical way to topologize the tensor product of topological algebras over a topological ring? Would that solve my problem?
Which (universal) properties do the Adèles satisfy?


(there was a section with my (non-working) ideas on this, which I removed after the answers came in.)

REPLY [2 votes]: Let $P$ the pullbak (in the topological groups category) of the natural maps $\prod_{\nu\in N} K_\nu\to \prod_\nu K_\nu/\mathcal{O_\nu}$ and  $\coprod_\nu K_\nu/\mathcal{O_\nu}\to \prod_\nu K_\nu/\mathcal{O_\nu}$, then $P$ as set is the restricted product, but has the subspace topology of the product $\prod_\nu K_\nu$. Now $\coprod_\nu K_\nu/\mathcal{O_\nu}$ is the (filtrant) colimit of $(\coprod_{\nu\in F} K_\nu)_{F\subset N finite}$ 
with coproiections $\coprod_{\nu\in F}K_\nu\to \coprod_\nu K_\nu\to \coprod_\nu K_\nu/\mathcal{O_\nu}$.  And by pullback we have that the set $P$ is a (filtrant) colimits $(P_F\to P)_{F\subset N\ finite}$ and $P_F$ is the produt of the $K_\nu\ \nu\in F$ and the $\mathcal{O_\nu},\ \nu\not\in F$, and observe that $P_{F_1}\cap P_{F_2}=P_{F_1\cap F_2}$. Give on $P_F$ the topology inducted by the projection $P_F\to \prod_{\nu\in F} K_\nu$ (the codomain by the product topology). 
If we take the colimit  topology  on $P$ we get the restricted topology: 
If $U$ is open on the colimit topology then each $U\cap P_F$ is open, then is open in the restricted topology (where the family of the $P_F$'s is a open covering), viceversa, let $U$ a unions of a family of $P_F$, is enought show that each $P_F$ is open in the colimit topology, i.e. that the intersection with any $P_{F'}$ is open in $P_{F'}$, but this follow from the observation above (and observe that any $\mathcal{O}\subset K$ is supposed open, see http://modular.math.washington.edu/129/ant/html/node82.html).<|endoftext|>
TITLE: Diffeomorphisms of finite order not in the image of a circle action
QUESTION [25 upvotes]: Does there exist a closed smooth manifold $M$ and a diffeomorphism $f\colon M \to M$ such that:

$f$ is isotopic to the identity,
$f$ is of finite order, $f^n=ID$, and 
$f$ is not contained in the image of any circle action, $S^1\to \operatorname{Diff}(M)$?

If the answer is yes, what is an example?

REPLY [19 votes]: Such examples exist in dimension 5, they are contained in the paper by Cameron Gordon "On the higher-dimensional Smith conjecture", Proc. London Math. Soc. (3) 29 (1974), 98–110. Namely, Gordon  proves in Theorem 5 of this paper that (for every $n\ge 5$) there are infinitely many smooth knots $K=S^{n-2}\subset S^n$ so that $K$ is the fixed-point set of a ${\mathbb Z}_p$-action $\alpha_p$ for every prime $p$. He also proves (Theorem 4) that, given $K$, if every action $\alpha_p$ extends to a circle action on $S^n$ then $\pi_1(S^n\setminus K)\cong {\mathbb Z}$. He then notes (a theorem by Levine) that, for $n\ge 5$, if $\pi_1(S^n\setminus K)\cong {\mathbb Z}$ then $K$ is smoothly unknotted in $S^n$. Since Gordon's theorem 5 yields infinitely many smooth isotopy classes of knots $K$, it then follows for every such (nontrivial) knot, at least for some prime $p$, one of Gordon's actions $\alpha_p$ does not extend to a smooth circle action. 
Finally, every diffeomorphism of $S^5$ is PL isotopic to the identity (by the Alexander's trick). Since in dimensions $<7$, PL=DIFF, we conclude that the generator of $\alpha_p({\mathbb Z}_p)$ is smoothly isotopic to the identity. 
Lastly, note that in the topological category, the examples exists already for $M=S^3$: Bing 
("Inequivalent Families of Periodic Homeomorphisms of $E^3$",  Ann. of Math.  80  (1964) 78-93.) constructed finite order homeomorphisms whose fixed-point sets are wild knots, while F.Raymond (Classification of the actions of the circle on $3$-manifolds.  Trans. Amer. Math. Soc.  131  (1968) 51-78.) proved that the every  $S^1$-action on $S^3$ is topologically conjugate to an orthogonal action.<|endoftext|>
TITLE: Extending birational isomorphisms between planar curves to the P^2
QUESTION [10 upvotes]: Let $k$ be a field and let $C,D$ be two integral curves in $\mathbb{P}^2_k$. Now let $f:C \to D$ be a birational isomorphism. Can $f$ be extended to $\mathbb{P}^2_k$. To be precise, does there exist a birational isomorphism $F:\mathbb{P}^2_k \to \mathbb{P}^2_k$ such that $F$ and $f$ agree on a (non empty) open subset of $C$?
I am mainly interested in the case when $k=\mathbb Q$. 
[Edit] Since it is apparently easy to read over I will state it here explicitly. The map $F$ is allowed to be a birational isomorphism. It is clear that my statement is false when you want $F$ to be an isomorphism since it has to send curves of the same degree to curves of the same degree.
If we take for example $C$ to be the curve $x=0$ and $D$ to be the curve $y^2z=x^3+x^2z$ then $C$ and $D$ are birationally equivalent but clearly no automorphism of $\mathbb{P}^2_k$ sends C to D. However it is possible with a birational isomorphism.

REPLY [4 votes]: Another way to see that a nodal rational curve of degree $\ge 6$ cannot be sent to a line by a birational map of the plane is to see that it cannot be contracted by a birational map of the plane. Each birational map of the plane decomposes into blow-ups and contractions of smooth $(-1)$-curves (curves isomorphic to $\mathbb{P}^1$ and of self-intersection $-1$). If you want to contract the curve, you have to blow-up all nodes. The number of nodes is $(d-2)\cdot (d-1)/2$, and the self-intersection of the curve on $\mathbb{P}^2$ is $d^2$. After blowing-up, the self-intersection decreases by $4$ for each double point, so becomes $d^2-2\cdot (d-1)(d-2)=6d-d^2+4$. If $d\ge 6$, this number is $\le -2$, so the curve is not contractible. If $d\le 5$, the curve is contractible and in fact one can check easily that it can be sent onto a line by a birational map of the plane.
Similarly, one sees that a nodal rational curve of degree $d\ge 6$ cannot be sent onto a nodal rational curve of degree $d'\ge 6$ when $d\not=d'$.
Similar arguments work with curves of genus $1$. See for example  Proposition 3.3.3 of "On birational transformations of pairs in the complex plane", J. Blanc, I. Pan, T. Vust, Geom. Dedicata 139 (2009), 57-73.<|endoftext|>
TITLE: About the quantum spectrum of a certain potential. 
QUESTION [5 upvotes]: Intuitively one understands that if one is solving the Schroedinger's equation for energies $E$ such that $\{ x \vert  U(x)\leq E \}$ is compact (..is there a weaker criteria?..) then the spectrum will turn out to be discrete and the wave-functions will decay exponentially for large values of $x$. What is the most rigorous statement and proof of this? 

I want to focus on one potential, 
$U = \vert G(s_i)\vert ^2 + \vert p \vert ^2 \sum _ {i} \vert \frac{\partial G}{\partial s_i} \vert ^2 + \frac{e^2}{2}(\sum _i \vert s_i \vert ^2 - n^2 \vert p \vert^2 - r  )  + 2\vert \sigma \vert ^2 (\sum _i \vert s_i \vert ^2 + n^2 \vert p \vert^2) $
where $e$ is a real constant, $s_i$, $p$ and $\sigma$ are complex and $r$ is a real field and $G$ is a degree $n$ transverse homogeneous function in $s_i$. 
Now apparently the following claims are true, 

If $r = 0$ then for any value of $\sigma$, the range of $s_i$, $p$ where $U(s_i,p) \leq E$ is true is compact and hence the spectrum is discrete. 
If $\sigma \neq 0$ then for any fixed non-zero value of $r$, the region of $s_i$ and p where $U(s_i,p;r) \leq E$ is true is compact for small enough values of $E$ and hence the spectrum is discrete for low-lying values of $E$ below some $E_{critical}$. 

(..the above is apparently motivated by the fact that at $s_i=p=0$, $U$ becomes constant and equal to $\frac{e^2 r^2}{2}$ and hence independent of $\sigma$..so apparently if one goes above some critical value of $E$ the spectrum is continuous thanks to field configurations with large $\vert \sigma \vert$ but at $s_i = p =0$... )

My reading of the literature is that the above two claims are true independent of the topology of the space on which the fields are valued though in a case of interest one wants the theory to be on a circle and hence I guess one wants to think of $s_i$, $p$, $\sigma$ to be maps from $S^1$ to $\mathbb{C}$ or $\mathbb{R}$. If the theory is on a circle then semiclassically apparently the estimate for $E_{critical}$ is $\frac{e^2 r^2}{2} 2\pi R$ where $R$ is the radius of the circle.  

I would be glad if someone can help justify the above three claims.
This paper is the reference for this question

REPLY [2 votes]: This question (and a partial converse with counterexamples) is answered in 
B. Simon, Some quantum operators with discrete spectrum but classically continuous spectrum, Annals of physics 46 (1983), 209-220. http://www.math.caltech.edu/SimonPapers/158.pdf<|endoftext|>
TITLE: What is the homotopy type of a free simplicial ring? 
QUESTION [7 upvotes]: Is there a good description of the homotopy type of a free simplicial ring (or simplicial $R$-algebra) on a given simplicial set, in terms of the homotopy type of that simplicial set?
(This is mostly an idle question, but also motivated by the fact that it is a theorem of Milnor that a similar construction with the free group gives a model for the $\Omega \Sigma X$, perhaps believable in view of the fact that $\Omega \Sigma$ is supposed to be the left adjoint from spaces into grouplike $A_\infty $-(i.e., with a coherently associative multiplication law) spaces.)

REPLY [9 votes]: Let $R[-]$ be the free $R$-module functor, from sets to $R$-modules, and $T_R$ the free (tensor) $R$-algebra functor, from $R$-modules to $R$-algebras. The free $R$ algebra functor from sets to $R$-algebras is the compoisite $T_RR[-]$.
Given a simplicial set $X$, the homotopy type of $R[X]$ is well understood. It is a product of Eilenberg-MacLane spaces, and its homotopy groups are the homology groups of $X$ with coefficients in $R$,
$$R[X]\simeq \prod_{n\geq 0}K(H_n(X,R),n).$$
Since 
$$T_RM=\bigoplus_{m\geq 0}M^{\otimes m}$$ 
we have 
$$T_RR[X]=\bigoplus_{m\geq 0}R[X\times\stackrel{m}\cdots\times X]$$ 
therefore
$$T_RR[X]\simeq \prod_{n\geq 0}K\left(\bigoplus_{m\geq 0}H_n(X\times\stackrel{m}\cdots\times X,R),n\right).$$<|endoftext|>
TITLE: Which formulae of Euler is Fröhlich referring to?
QUESTION [9 upvotes]: In A. Fröhlich's article Local Fields in Algebraic Number Theory, the following claim is made: if $R$ is a Dedekind domain with field of fractions $K$, $L$ is a finite separable extension of $K$ and $S$ is the integral closure of $R$ in $L$, and $x$ is an element of $S$ with minimal polynomial $g$, then, "by Euler's formulae",
$$\text{tr}_{L/K}(x^i/g'(x)) \in R$$
for each $0 \leq i \leq n-1$, where $n=\text{deg } g$.
Which formulae of Euler are being referred to? The claim can be proven by the Lagrange interpolation formula; in fact the given quantity is $1$ if $i=n-1$, and $0$ for $0 \leq i < n-1$. However, I have no idea what proof Fröhlich has in mind. I also cannot resist pointing out the humor in appealing to Euler's "formulae" without further precision. Perhaps the formulae in question are well-known, and I am the only one who has not been invited to the party? In any case, more details would be greatly appreciated!
Thank you.

REPLY [5 votes]: I believe the reference is to this formula of Euler (see here): If $P(x)/Q(x)$ is a rational function and $ax+b$ is a simple factor of $Q(x)$, then the coefficient of $1/(ax+b)$ in the partial fraction decomposition of $P/Q$ is given by
$$ \lim_{x\to \frac{-b}{a}} \frac{a P(x)}{Q'(x)}. $$

To see how this applies here, proceed as Serre does in Local Fields. That is, write (in a suitable extension of $L$)
$$ \frac{1}{g(X)} = \sum_{k=1}^n \frac{a_k}{X-x_k} \qquad (*) $$
where the $x_k$ are the conjugates of $x$. Then a formal application of Euler's formula gives
$$ a_k = \lim_{X \to x_k} \frac{1}{g'(X)} = \frac{1}{g'(x_k)}. $$
Now expand both sides of (*) as power series in $1/X$ and compare coefficients.<|endoftext|>
TITLE: What is the number field analogue of the Narasimhan-Seshadri theorem ?
QUESTION [23 upvotes]: In his famous 1940 letter from prison in Rouen to his sister Simone, André Weil talks about the analogy between number fields and functions fields (in one variable) over finite fields, and the analogy between these functions fields and functions fields over $\mathbf{C}$ (or equivalently compact connected curves over $\mathbf{C}$).  This letter is reproduced in his Scientific papers and has been recently translated into English (Notices of the AMS 52(3) 2005).
Question What is the number field analogue of the Narasimhan-Seshadri theorem (Stable and unitary vector bundles on a compact Riemann surface. Ann. of Math. (2) 82 1965 540–567) ?
Addendum (in response to Felipe's comment)  The original paper of Narasimhan and Seshadri is available on JSTOR.  An excerpt from their introduction : D. Mumford has defined the notion of a stable vector bundle on a compact 
Riemann surface $X$ and proved that the set of equivalence  classes  of stable 
bundles (of fixed rank and degree) has a natural structure of a non-singular, 
quasi-projective, algebraic variety [13].  We prove in this paper that, if $X$ has 
genus $\ge2$, the stable vector bundles are precisely the holomorphic vector bundles 
on $X$  which arise from certain irreducible  unitary  representations of suitably defined fuchsian groups acting on the unit disc and having $X$ as quotient 
(Theorem 2, $\S12$).
[...]
A  particular case  of our result is  that a  holomorphic vector bundle of 
degree zero on $X$  is stable if and only if it arises from an irreducible unitary 
representation of the fundamental group of $X$.  As a consequence one sees that 
a holomorphic vector bundle on $X$  arises from a unitary representation of the 
fundamental group of $X$ if and only if each of its indecomposable components 
is of degree zero and stable.
Their main result is summarised by Atiyah (MR0170350) and Le Potier (Séminaire Bourbaki Exposé 737) as follows :
Atiyah: Let $X$ be a compact Riemann surface. If $W$ is a (holomorphic) vector bundle of rank $n$ over $X$ we define $d(W)$ to be the degree of the associated line bundle $\bigwedge^n W$. A bundle $W$ is stable, in the sense of Mumford, if $(\mathrm{rank}W)d(V)<(\mathrm{rank}V)d(W)$ for all proper sub-bundles $V$ of $W$. According to Mumford [Proc. Internat. Congr. Mathematicians (Stockholm, 1962), pp. 526--530, Inst. Mittag-Leffler, Djursholm, 1963], the set of isomorphism classes of stable bundles of rank $n$ and degree $q$ over $X$ has a natural structure of an algebraic variety. In this paper the authors give a complete characterization of stable bundles in terms of unitary representations of a certain discrete group (provided genus $X$ $≥2$). 
Their main theorem runs as follows. Given integers n and q, with $-n< q \le0$, we can choose (i) a discrete group $\pi$ acting on a simply connected Riemann surface $Y$ with $Y/\pi=X$ and with the map $p:Y\to X$ being ramified over only one point $x_0\in X$; (ii) a representation $\tau:\pi_{y_0}→\mathrm{GL}(n,\mathbf{C})$ of the isotropy group of $\pi$ at a point $y_0\in p^{−1}(x_0)$ by scalars such that the following holds.  A vector bundle over $X$ of rank $n$ and degree $q$ is stable if and only if the corresponding sheaf is isomorphic to a sheaf of the form $p_∗^\pi(\mathbf{V})$, where $\mathbf{V}$ denotes the $\pi$-sheaf of holomorphic mappings $Y\to V$, $V$ is an irreducible unitary representation of $\pi$ coinciding with $\tau$ when restricted to $\pi_{y_0}$, $p_∗$ is the direct image functor and $p_∗^\pi$ denotes the subsheaf invariant under $\pi$. Moreover, two such stable bundles are isomorphic on $X$ if and only if the corresponding unitary representations of $\pi$ are equivalent. 
It should be observed that the inequality $-n< q \le0$ presents no essential restriction since it can always be realized by tensoring with a line bundle $L$ and, on the other hand, the definition of stable bundle shows that $W$ is stable if and only if $W\otimes L$ is stable. 
Le Potier: En 1965 Narasimhan et Seshadri établissaient une correspondence bijective entre l’ensemble des classes d’équivalence de représentations unitaires irréductibles du groupe fondamentale $\pi$ d’une surface de Riemann compacte $X$, et l’ensemble des classes d’isomorphisme de fibrés vectoriels stables de degré $0$ sur $X$ : ils associent à une representation $\rho:\pi\to\mathbf{U}(r)$ le fibré vectoriel holomorphe $E_\rho$ défini par
$$
E_\rho=\tilde X\times_\pi\mathbf{C}^r
$$
où $\tilde X$ est le revêtement universel de $X$, et où le produit ci-dessus est le quotient de $\tilde X\times\mathbf{C}^r$ par l’action de $\pi$ définie par 
$(\gamma,(x,v))\mapsto(x\gamma^{-1},\gamma v)$ pour $\gamma\in\pi$ et $(x,v)\in \tilde X\times\mathbf{C}^r$.

REPLY [8 votes]: Theorem of Narasimhan and Seshadri is a special case of what Carlos Simpson calls nonabelian Hodge theory developed by Hitchin and Simpson. This theory was generalized to the characteristic $p$ case in the paper of Ogus and Vologodsky, Nonabelian Hodge Theory in Characteristic p. Hope this helps. 
Update: Below, on Alexander's request, is a brief explanation of relation between nonabelian Hodge theory and NS theorem. Consider vector bundles with vanishing 1st and 2nd Chern classes (I will call this condition ($\star$)), then the story in higher dimensions is exactly the same as for complex curves. The detailed explanation is in the pages 12-19 of Simpson's 1992 paper [S1992]. From there, it follows that flat unitary connections correspond exactly to vanishing Higgs fields (subject to ($\star$)). Briefly, every semistable Higgs bundle $E=(V, \bar\partial, \theta)$ has a hermitian YM metric $K$. Define the connection $D_K$ (as in [S1992], page 13), then then $D_K$ is flat (subject to ($\star$), page 17 of [S1992]).
If Higgs field $\theta$ vanishes then $D_K=\partial_K+ \bar\partial$ and, hence, by definition of $\partial_K$, connection $D_K$ preserves the metric $K$. Thus, our bundle reduces to a flat unitary bundle. Conversely, if bundle is flat unitary (with unitary metric denoted $K$) then the associated (multivalued) map $\Phi_K$ defined on page 16 of [S1992], is constant, so it has zero derivative. But its derivative is $0=d\Phi_K=\theta+ \bar\theta$  (here $\theta$ is the Higgs field determined by $K$). Since $\theta, \bar\theta$ have different types, the only way we can have $\theta+ \bar\theta=0$ is that $\theta=0$.<|endoftext|>
TITLE: Cramer's rule for eigenvectors
QUESTION [7 upvotes]: I know that the above doesn't exist, but do bear with me. I need estimates/formulas for entries of certain eigenvectors and Cramer's rule keeps popping up in my mind. So, what can play an anlogous role in this case?

REPLY [17 votes]: If $u$ is a unit eigenvector with eigenvalue $\lambda$ of a Hermitian matrix
$$ A_n = \begin{pmatrix} a & X^* \\\\ X & A_{n-1} \end{pmatrix}$$
with $a$ a real number, $X$ an $n-1$-dimensional row vector, and $A_{n-1}$ an $n-1 \times n-1$ Hermitian matrix, then (provided that $\lambda$ is not an eigenvalue of $A_{n-1}$) the magnitude of the first coefficient $u_1$ of $u$ is given by the formula 
$$ |u_1|^2 = \frac{1}{1 + \| (A_{n-1} - \lambda)^{-1} X \|^2};$$
see e.g. Lemma 41 of this paper of mine with Van Vu (and we learned of this formula from this previous paper of Erdos, Schlein, and Yau).  Similarly for other components of $u$.  The proof proceeds by expanding out the bottom $n-1$ components of the eigenvalue equation $A_n u = \lambda u$ and using the resulting equation and the formula $\|u\|^2=1$ to solve for $u_1$.  
The expression $\| (A_{n-1} - \lambda)^{-1} X \|$ can be expanded further by a number of formulae (e.g. Cramer's rule); in our applications, it turns out that the spectral theorem applied to $A_{n-1}$ is useful.  The above identity is particularly useful for establishing delocalisation of eigenvectors for random matrices (i.e. that the energy of the coefficients of a unit vector are spread out almost uniformly).  It may seem a bit strange that the top left coefficient $a$ plays no explicit role in this formula, but it is implicitly present due to its influence on the eigenvalue $\lambda$.
The phase $u_1$ of the unit eigenvector does not have a clean formula, because eigenvectors are only determined up to phase rotations unless one somehow selects an artificial normalisation.<|endoftext|>
TITLE: Operator Theoretical Models for $K(\mathbb{Z}, 3)$
QUESTION [5 upvotes]: I  am  looking  for  a  reference  concerning  operator  theoretical  Models  of  $K(\mathbb{Z},3)$. Stolz-Teichner   briefly say  in "what  is  an  elliptic  object" that  a  certain hyperfinite Type  III-factor, called "local  fermions  on the  circle"  should    be  the  right thing. Thanks

REPLY [7 votes]: Here is a $C^*$-algebraic version of the model described in Andre Henriques' answer (the latter was linked by David Corfield in the accepted answer above): 
Let $\mathcal{O}_2$ be the Cuntz algebra generated by two partial isometries $s_1$ and $s_2$ subject to the relations $s_i^*s_j = \delta_{i,j}$ and $s_1s_1^* + s_2s_2^* = 1$. This algebra has vanishing $K$-theory as was calculated by Cuntz. By the universal coefficient theorem and Bott periodicity, $KK(\mathcal{O}_2, S^n\mathcal{O}_2)$ should vanish as well, where $S^n\mathcal{O}_2$ denotes the $n$-fold suspension of $\mathcal{O}_2$. The automorphism group of the stabilized algebra $\mathcal{O}_2 \otimes \mathbb{K}$ (where $\mathbb{K}$ denote the compact operators on a separable Hilbert space) fits into a short-ish exact sequence
$$1 \to U(1) \to U(M(\mathcal{O}_2 \otimes \mathbb{K})) \to Aut(\mathcal{O}_2 \otimes \mathbb{K}) \to Out(\mathcal{O}_2 \otimes \mathbb{K}) \to 1$$ 
where $M(\mathcal{O}_2 \otimes \mathbb{K})$ is the multiplier algebra. The homotopy groups of $Aut(A \otimes \mathbb{K})$ for so-called Kirchberg algebras have been calculated (yeah, I was surprised too :-). You can find them in a paper by Dadarlat called "The homotopy groups of the automorphism groups of Kirchberg algebras". The result is 
$$\pi_n(Aut(A \otimes \mathbb{K})) \cong KK(A,S^nA).$$
Now, $\mathcal{O}_2$ fits into that class and therefore has weakly contractible automorphism groups, but - by a theorem of Mingo - $U(M(\mathcal{O}_2 \otimes \mathbb{K}))$ is contractible as well. Analyzing the above sequence, we see that $Out(\mathcal{O}_2 \otimes \mathbb{K})$ has the weak homotopy type of a $K(\mathbb{Z},3)$... at least if
$$1 \to PU(M(\mathcal{O}_2 \otimes \mathbb{K})) \to Aut(\mathcal{O}_2 \otimes \mathbb{K}) \to Out(\mathcal{O}_2 \otimes \mathbb{K}) \to 1$$ 
is a fibration. In fact, it could very well be that the topology on the quotient $Out(\mathcal{O}_2 \otimes \mathbb{K})$ is quite horrible.<|endoftext|>
TITLE: Open problems/questions in representation theory and around?
QUESTION [43 upvotes]: What are open problems in representation theory?
What are the sources (books/papers/sites) discussing this?
Any kinds of problems/questions are welcome - big/small, vague/concrete.
Some estimation of difficulty and importance, as well as, small description, prerequisites and relevant references, ... are welcome.

To the best of my knowledge, there are NO good lists of representation theory problems on the web. E.g. the sites below contain lots of unsolved problem in other areas, but not in representation theory:
http://en.wikipedia.org/wiki/Unsolved_problems_in_mathematics
http://garden.irmacs.sfu.ca/
http://maven.smith.edu/~orourke/TOPP/
MO questions also discuss other fields, but not representation theory:
What are the big problems in probability theory?
What are some open problems in algebraic geometry?
What are some open problems in toric varieties?
More open problems
Open problems with monetary rewards
Open problems in Euclidean geometry?
Open Questions in Riemannian Geometry
What are some of the big open problems in 3-manifold theory?
Open problems in continued fractions theory

REPLY [12 votes]: There are many open problems on modular representation theory of finite groups. There is a well-known list of problems of R. Brauer which date to around 1960, about ordinary and modular representations of finite groups. There are other major conjectures ( Broue's Abelian defect group conjecture has already been mentioned in comments). There are Alperin's Weight conjecture, Dade's Projective Conjecture, the Isaacs-Navarro conjecture, and several related conjectures. This is only the tip of the iceberg (and only in one corner of representation theory).<|endoftext|>
TITLE: A simple looking problem in partitions that became increasingly complex
QUESTION [6 upvotes]: I began with problem which looked simple in the beginning but became increasingly complex as I dug deeper. 
Main questions: Find the number of solutions $s(n)$ of the equation
$$
n = \frac{k_1}{1} + \frac{k_2}{2} + \ldots + \frac{k_n}{n}
$$
where $k_i \ge 0$ is a non-negative integer. This is my main questions. After tying different approaches, the one that I found most optimistic is as follows. But soon even this turned out to be devil (as we shall see why). 
Let $l_n$ be the LCM of the first $n$ natural numbers We know that $\log l_n  =\psi(n)$. Multiplying both sides by $l_n$ we obtain
$$
n l_n = \frac{k_1 l_n}{1} + \frac{k_2 l_n}{2} + \ldots + \frac{k_n l_n}{n}
$$
Each term on the RHS is a positive integer thus our question is equivalent to finding the number of partitions of $nl_n$ in which each part satisfy some criteria.
Criteria 1: How small can a part be? Assume that there is a solution with $k_n = 1$ then the smallest term in the above sum will be the $n$-th term which is $l_n / n$. Hence each term in our partition is $\ge l_n/n$.
Criteria 2: How many prime factors can each part contain? If my calculation is correct then for $n \ge 2, 2 \le r \le n$, the minimum number of prime factors that $l_n /r$ can contain is $\pi(n)-1$. With these two selection criterion we have:
$s(n) \le $ No. of partitions of $n l_n$ into at most $n$ parts such that each part is greater than $l_n / n$ and has at least $\pi(n) - 1$ different prime factors.
May be we can narrow down further by adding sharper selection criterions but I thought it was already complicated enough for the time being. The asymptotics of the number of partitions of $n$ into $k$ parts $p(n,k)$ is well known, but I have not found in literature any asymptotics for the number of partitions of $n$ into $k$ parts such that each part is at least $m$, let alone the case when each part has a certain minimum number of prime factors. I am looking for any suggestions, reference materials that would help in these intermediate questions that would ultimately help in answering the main question.

REPLY [2 votes]: I am impressed by the counts found by Max. Here are some comments which are perhaps already included in his dynamic program. For any non-negative integer $m$ let $f(m,n)$ be the number of solutions to $m = \frac{k_1}{1} + \frac{k_2}{2} + \ldots + \frac{k_n}{n}$ with the $k_i$ non-negative integers. We could actually consider $f(u,n)$ for rational $u$ but won't pay much attention to that general case. The numbers requested are the diagonal of the table of $f(m,n)$ for $m,n$ integers. The obvious generating function procedure for $\sum f(u,n)x^u$ is effective, at least for a while; To calculate the values of $f(u,n)$ $u \le U$ form the product $$\prod_{d=1}^n\frac{1}{1-x^{\frac{1}{d}}}$$  and truncate at $x^U.$ In practice this would be done one factor at a time (computing all the $f(u,s)$ for $s \lt n$ along the way.) If desired, all terms with $x$ to an exponent greater than $U$ can be truncated before going on. The coefficient of $x^u$ is $f(u,n)$. Of course $f(u,n)$ for fixed $n$ is given by some polynomial function depending on the denominator of $u.$ 
A more efficient modification is to treat seperately all groups of $j$ fractions $\frac{1}{j}.$ Call the sum of these the $w$-part and the remainder the $v$-part.If we have a given expansion $u = \frac{k_1}{1} + \frac{k_2}{2} + \ldots + \frac{k_n}{n}$ let $k_d=dq_d+r_d$ with $0 \le r_d \lt d$ then $u=w+v$ where $v = \frac{r_2}{2} + \ldots + \frac{r_n}{n}$ will be a number less than $n$ with the same fractional part as $u$ while $w=\frac{q_1}{1} + \frac{2q_2}{2} + \ldots + \frac{nq_n}{n}$ will be an integer expressed as a sum of units.  The number of ways to get a fixed integer $w$ is $\binom{w+n-1}{n-1}$ because this is just the number of ways to put $w$ identical balls into $n$ boxes (a ball in box $j$ denotes a pack of $j$ fractions $\frac{1}{j}$.) Here is an analysis of this process carried out for a few steps.
For an integer $n \ge 0$, 

$f(n+\frac{y}{2},2)=n+1$ for $y=0,1$. 
$f(n+\frac{y}{6},3)=\binom{n+2}{2}$ for $y=0,2,3,4,5$ but $f(n+\frac{1}{6},3)=f((n-1)+\frac{7}{6})=\binom{n+1}{2}$ This is because the $v$ part is less than $1$ with the exception of $\frac{1}{2}+\frac{2}{3}=\frac{7}{6}$ 
$f(n+\frac{y}{12},4)=\binom{n+2}{3}+\binom{n+3}{3}=\frac{(n+1)(n+2)(2n+3)}{6}$ for $y=0,3,4,7,8,11$ This is the sum of the squares up to $(n+1)^2$ so a square-pyramidal number. The other possibilities are $2\binom{n+2}{3}$ for $y=1,2,5$ and $2\binom{n+3}{3}$ for $y=6,9,10$.
$f(n,5)=\binom{n+3}{4}+\binom{n+4}{4}=\frac{(n+2)^2((n+2)^2-1)}{12}$ these are four-dimensional pyramidal numbers . Note that the expansion looks like the previous case. This is because the $v$ part can not use anything of the form $\frac{r}{5}$ . This only becomes possible at $n=10$ with $\frac{2}{5}+\frac{1}{10}=\frac{1}{5}+\frac{3}{10}=\frac{5}{10}=\frac{1}{2}$ as well as four ways to get $1$ and two ways to get $\frac{3}{2}$ 
$f(n,6)=4\binom{n+3}{5}+7\binom{n+4}{5}+\binom{n+5}{5}.$ The $7$ in the middle comes from the five cases $\frac{1}{2}+\frac{2}{4}=\frac{1}{2}+\frac{3}{6}=\frac{2}{4}+\frac{3}{6}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{2}{4}+\frac{1}{3}+\frac{1}{6}=1$ along with $\frac{1}{3}+\frac{4}{6}=\frac{2}{3}+\frac{2}{6}=1$ . This appears  without much comment in OEIS.
$f(n,7)=4\binom{n+4}{6}+7\binom{n+5}{6}+\binom{n+6}{6}$ The coefficients are as in the previous case because there can be no contribution of $\frac{r}{7}$ to the $v$ part until $n=14.$ This sequence of numbers $1,14,81,308,910,2268,4998,10032,\cdots$ does not appear in the OEIS at this moment.<|endoftext|>
TITLE: Membership problem for cyclic subgroups
QUESTION [8 upvotes]: Question: is there any example of a finitely presented  (or at least finitely generated) group $G$ with an infinite cyclic subgroup $C \leqslant G$ such that the word problem in $G$ is solvable but the membership problem for $C$ in $G$ is unsolvable?
I would guess that such examples should exist, since otherwise amalgams and HNN-extensions of f.p. groups with solvable word problem along cyclic subgroups would again have solvable word problem. But using such free constructions one could (possibly?) produce a group with unsolvable word problem, starting with a group for which the word problem is solvable...

REPLY [8 votes]: @Ashot: Finitely presented examples exist. It is in our paper with Olshanskii, Olshanskii, A. Yu.; Sapir, M. V., Length functions on subgroups in finitely presented groups. Groups—Korea '98 (Pusan), 297–304, de Gruyter, Berlin, 2000.  
An easier construction is this. Take the free Abelian group $A$ with generators $x_i, i=0,1,...$. Consider an injective recursive function $\mathbb{N}\to \mathbb{N}$ with non-recursive image $f(n)$. 
Impose the following relations on $A$:
$x_{f(n)}=x_0^{n!}, n=1,2,...$. The word problem in the resulting (infinitely generated) group $A_f$ is solvable. Indeed, consider any word $w=x_{i_1}^{k_1}...x_{i_m}^{k_m}$, $i_1\lt i_2\lt...$. That word is equal to 1 if and only if each $i_s$ is of the form $f(n_s)$ for some $n$ and $k_1n_1!+k_2n_2!+...=0$. That means all $n_i$ cannot be much bigger than $\max |k_i|$. Thus checking $w=1$ requires finite number of calculations of the function $f(n)$, so it is decidable. On the other hand the membership problem of $x_m$ into $\langle x_0\rangle$ is clearly undecidable since $f$ has non-recursive image. 
Now use Clapham's version of Higman's embedding and embed $A_f$ into a finitely presented group $G$ with decidable word problem. Given $i$, one can recursively find a word in $G$ representing $x_i$. Thus the membership problem in the cyclic subgroup $\langle x_0\rangle$ of $G$ is undecidable. 
 Update  Here is how to construct an injective recursive function with non-recursive image. Using Matiyasevich's theorem, find a polynomial $p(x_1,...,x_s)$ such that the solvability of the equation
$a=p(x_1,...,x_s)$ is undecidable. Now take any Turing machine $M$ that computes $p(x_1,...,x_n)$ in binary. Add a new tape to  $M$ and during the computation of $M$ write down the history of computation (i.e. numbers of the executed commands in binary, separated by number 3) on the new tape. When the machine stops, interpret the word written on the extra tape and the value of $p(x_1,...,x_s)$ as one number $f(x_1,...,x_s)$ (say, write first the value of $p$, then number 4, then the history, view the whole number as written in the 5-ary system). The function $n$ is clearly recursive and injective because the history determines the values of $x_1,...,x_s$ (make the machine scan the input before it computes $p$). It is also clear that the image of $n$ is not recursive. To make $n$ a function in one variable, precompose it with a (Goedel) function that enumerates ${\mathbb Z}^s$.<|endoftext|>
TITLE: Do you use the Mathematics Subject Classification (MSC) when searching for literature?
QUESTION [9 upvotes]: I suppose most of you are familiar with the Mathematics Subject Classification (MSC). Particularly, when submitting an article for publication one has to 
choose appropriate classification codes.
But I am wondering if it does play a significant role in searching for
literature on a specific subject, e.g. through MathSciNet or ZBMATH. 
Or do you prefer searching by subject terms?
Thank you very much!

REPLY [5 votes]: MSC codes are a nice way to keep an eye on what's appearing in your area. Back in the "bad old day" before online MathSciNet and the ArXiv, I used to go to the library every month or two and skim all of the Math Reviews in 11D and 11G, and I reasonably frequently found articles of interest that I would then go read. Now I'm a bit lazy, but at least a few times a year I'll bring up all the 11G and 37P reviews and skim the titles, then look at the reviews of the ones that seem interesting. (Of course, these days often I already know about them from the ArXiv.) So I guess I'm echoing some of the other responses in saying that if you're researching a particular topic, searching on key words is much better, but if you want a general overview of what sorts of things people in your area are working on, it can be quite instructive to look at titles (and reviews) in an MSC category. BTW, there's a feature on MathSciNet that lets you bring up articles whose reviews have appeared in the past month(s). This is useful, since otherwise there will be a lot of articles that appear in the list that are waiting to be reviewed.<|endoftext|>
TITLE: Intersection of an uncountable number of sets.
QUESTION [7 upvotes]: Let $\mathcal{I}$ be an uncountable set. Let $(\Omega, \mathcal{F},\mathbb{P})$ be a probability space, and $E_i, i\in \mathcal{I}$ be a measurable set such that $\mathbb{P}(E_i)=1$. What can we say about $\mathbb{P}(\cap_{i\in \mathcal{I}} E_i)?$. 
I know that an uncountable intersection of measurable sets is not necessarily measurable. But are there in general conditions that allow such infinite intersections to be measurable? 
Apologies if the question is too elementary or unclear.

REPLY [3 votes]: There is a quite general case when such an uncountable intersection is measurable:  Let $I$  be the set of real numbers, and assume that not only every $E_i$ is measurable, but that these sets are Borel, and that moreover the set 
$$E:= \{ (x,i):  x \in E_i \}$$
is a Borel set.  Then the intersection of all $E_i$ is co-analytic (also called $\bf \Pi^1_1$) and hence measurable by a classical theorem. 
("Borel" can be replaced  by "co-analytic", but not by "measurable".) 
It has already been pointed out that if $I$ is has sufficiently small cardinality then the intersection still has  full measure, where "sufficiently small" depends on the underlying set theory.  
But there is an interesting case where $I$ has size continuum.  Assume that $I= \omega^\omega $ is the set of all functions from the natural numbers to the natural numbers. Assume (as above) that $E $ is a Borel or $\Pi^1_1$ set. Assume also that the map $i \mapsto E_i$ is decreasing in the following sense: 
Whenever $i,j\in \omega^\omega$ and $i(n)\le j(n)$ for all $n$, then $E_i \supseteq E_j$.
Then the intersection of all $E_i$  is not only measurable but has full measure.<|endoftext|>
TITLE: Identity of the Weyl-Tensor
QUESTION [6 upvotes]: Let $(M^n,g)$ be a Riemannian manifold and let $W$ be its Weyl tensor. For a given ONB, does the identity
$$W_{ijkl}W_{ijkm}=\frac{1}{n}|W|^2g_{lm}$$
hold? I think I've seen it somewhere but I'm not sure whether this is valid only in dimension four (in this case, this is certainly true).

REPLY [8 votes]: This does not hold for $n>4$.  To see this, start with a Ricci-flat 4-manifold $N^4$ that is not flat, and let $M^{n}= N^4 \times \mathbb{R}^{n-4}$, endowed with the product metric.  Then the metric on $M$ is Ricci-flat, so its Riemann curvature tensor is its Weyl tensor and $W_{ijkl}=0$ when any index is greater than $4$.  However it is now clear that we can't have your equation when $l = m > 4$. 
(This argument doesn't start to work by $n=4$ because the Weyl curvature is identically zero until you get to dimension $4$.)<|endoftext|>
TITLE: Sliding blocks puzzle
QUESTION [12 upvotes]: Consider a 'game' played on a subset $S$ of an $n^2$ square grid as follows. There are 3 types of pieces, each occupying a square of $S$, 1 green, some red and the rest are blue, a move consists of shuffling the green piece with any of its 4 adjacent pieces (if they are within $S$). $S$ consists of squares, squares not in $S$ are static, $S$ can be any subset of squares of the $n^2$ square.
If two board configurations are reachable from eachother, is it possible to obtain an upper bound on the number of moves needed, given only the board size $n$, is it polynomial in $n$?

REPLY [3 votes]: Not a solution, just three observations that might trigger other ideas.

The pieces filling $S$ continue to fill $S$ at all times.  In other words, the shape of $S$ is
fixed, and lattice cells exterior to $S$ are irrelevant as they can never be used.
In order to move a particular tile from one square $a$ to another square $b$, 
there must be a simple
cycle in $S$ on which they both lie.  For example, in the first illustration below
(where $X$ plays the role of the empty/green square, and $0$ and $1$ are red and blue tiles),
the 1 in the lower-left corner cannot reach the upper-right corner because a connecting
cycle pinches in the middle and so is not simple.
In some sense the $2 \times n$ example illustrated feels like the worst case for moving one tile
from end to end of $S$, and that requires $2 n (n-1)$ moves, if I've counted correctly.

      
Update1. Zack Wolske's sequence of moves is more efficient and shows the $2 \times n$
example only needs a linear number of moves.  Gerhard Paseman's width-1 ring, however,
clearly needs a quadratic number of moves.
Update2. I think the key parameter is, not the OP's $n$, but rather $m=|S|$, the number
of cells in $S$. We have examples that require $\Omega(m^2)$ tile moves. Can anyone think of an pair
of configurations that requires more than a quadratic number of moves in $m$?<|endoftext|>
TITLE: Reference wanted for generalized Dirichlet's Theorem
QUESTION [5 upvotes]: I'm looking for a reference for the following result, which is a generalization of the classical theorem of Dirichlet on the approximability of real irrationals by rational numbers:
Let $k$ be a number field, $O$ its ring of integers, $v$ an infinite place of $k$, $\alpha$ any element of the completion $k_v$.  Let $\|\cdot\|_v$ be the usual absolute value (or its square, if $v$ is a complex place).  Let $H$ denote the multiplicative height function relative to $k$ -- that is, for any element $x\in k$, let $H(x)=\prod_w \max(1,\|x\|_w)$, where the product is over all places $w$ of $k$.  Then there is a positive real constant $C$ depending only on $k$ such that
$$\|\alpha-x\|_v < \frac{C}{H(x)^2}$$
for infinitely many $x\in k$.
I think I can prove this, but I am surely not the first.  If anyone can tell me a good place to point to for this result, I'd be very grateful -- thanks!

REPLY [4 votes]: It seems like this book has the reference you need.
Schmidt, Wolfgang M. Diophantine approximation. Lecture Notes in Mathematics, 785. Springer, Berlin, 1980.
Chapter VIII, Theorem 2A.
It's available on the Springlink website.
Hope it helps.<|endoftext|>
TITLE: Continuity of barycentre in Hausdorff metric
QUESTION [8 upvotes]: Let $K_1$, $K_2$ be two convex compact sets in $\mathbb{R}^d$, and $p_1,p_2$ be their barycenters. Is it true that the distance between $p_1$ and $p_2$ does not exceed a Hausdorff distance between $K_1$ and $K_2$? If not, maybe there is some weaker estimate, say, uniform continuity of the map (convex body)$\rightarrow$ (its barycenter) for bodies inside, say, unit ball? 
UPDATE. As Anton pointed out, the answer is obviously no, just take rectangle $\varepsilon\times 1$ and divide it by diagonal onto two triangles. Let me ask another question: is the barycentre of a closed $\varepsilon$-neighborhood close to the barycentre of initial body uniformly for all bodies inside unit ball?

REPLY [3 votes]: This question has been first discussed in the paper [ABB] below. They show that, in the plane, the barycenter of the boundary has the desired property: It is Lipschitz-continuous with respect to the Hausdorff distance. (However, the Lipschitz constant is larger than 1.)
Other such "reference points" which are Lipschitz-continuous w.r.t. the Hausdorff-distance, or w.r.t. some other metric (like the area of the symmetric difference) have been investigated. The objective is to get a fast heuristic or initial solution for matching two shapes under translation. A good starting point into the literature might be Oliver Klein's Ph.D. thesis "Shape Matching With Reference Points" from 2008.
The best Lipschitz constant (for the Hausdorff distance) is achieved by the so-called Steiner point (or Steiner center). In functional analysis, such "reference points" are known under the name "continuous selectors". It has been proved (by non-constructive methods, however) that the Steiner point has the smallest possible Lipschitz constant w.r.t. the Hausdorff distance.
One can show by an elementary example that a reference point (a mapping from compact convex sets to points that is equvariant under, say, isometries) cannot have Lipschitz-constant 1 in dimension 2 or larger. (Thus the answer to the first, strong, part of the original question is NO, even if you think of other points than the barycenter.)

[ABB] Helmut Alt, Bernd Behrends, Johannes Blömer. Approximate matching of polygonal shapes.
Ann. Math. Artif. Intell., Volume 13, Pages 251-266, 1995. The conference version may be easier to access: 
Proceedings of the seventh annual symposium on computational geometry, pp.186-193, June 10-12, 1991, North Conway, New Hampshire. ACM Press. doi>10.1145/109648.109669.<|endoftext|>
TITLE: Where to find Astérisque online?
QUESTION [29 upvotes]: Is there a place online where one can download papers that have appeared in the Astérisque if "your institution subscribes"? I am especially interested in back issues: Astérisque series published a lot of interesting papers which I would like to read, yet it seems to me the only way to access them is to take a trip to your library and pray that it has them in print. This being ridiculous in 21st century, I wonder if there is a place online with *.pdf versions of the issues...

REPLY [33 votes]: The first 180 volumes of Astérisque are now electronically available for free 
http://www.numdam.org/journals/AST/
EDIT: now 300+ volumes<|endoftext|>
TITLE: What are the benefits of viewing a sheaf from the "espace étalé" perspective?
QUESTION [53 upvotes]: I learned the definition of a sheaf from Hartshorne---that is, as a (co-)functor from the category of open sets of a topological space (with morphisms given by inclusions) to, say, the category of sets. While fairly abstract at the outset, this seems to be (to me) an intuitive view; in particular, all of the manipulations and constructions with sheaves fit nicely into this schema.
I know that the older view of a sheaf on $X$ was to consider it as a triple
$$
(E, X, \pi)
$$
where $\pi : E \to X$ is a local homeomorphism, and so that the "sheaf of sections" of this map $\pi$ is the sheaf in the functorial sense described above. 
This view makes much less sense to me, but I have to wonder if that is simply due to my having learned it second. However, it also makes me wonder if I am missing something, and so my question is as follows.

What are some (edit:) specific benefits of viewing a sheaf in this sense? What is gained by considering a sheaf as the espace étalé over $X$?

REPLY [39 votes]: To me the obvious answer involves sheafification of a presheaf.  If you look at the construction of the associated sheaf to a presheaf  in, say, Hartshorne it goes through the étalé space construction without specifically telling you, and to me it makes the construction somewhat unmotivated.
Namely, if $P$ is a presheaf on $X$, then taking the stalk $P_x$ at each point of $X$ gives you an $X$-indexed set, or as Tom would say above, a set over $X$.  One can then define a topology on $\biguplus_{x\in X}P_x$ so that the natural projection $\biguplus_{x\in X}P_x\to X$ is a local homeomorphism in the obvious way namely if $U$ is a neighborhood in $X$ and $s\in P(U)$, then $(s,U) = \lbrace germ_x(s)\mid x\in U\rbrace$ is a basic neighborhood.  This topology immediately makes $(s,U)$ homeomorphic to $U$ and makes $s$ a section over $U$ via $x\mapsto germ_x(s)$ for $x\in U$.  The sheaf of sections of $p$ is the associated sheaf of $P$.  I find this construction completely unmotivated without going through étalé spaces.
Added.  Another good reason is it is convenient for defining actions of a topological groupoid on a sheaf.  If $G=(G_0,G_1)$ is a groupoid, a $G$-sheaf is an étalé space $p:X\to G_0$ over $G_0$ together with an action map $G_1\times_{d,p} X\to X$ satisfying obvious axioms.  This is more difficult to phrase in the sheaf as a functor language.  A theorem of Joyal and Tierny says that every Grothendieck topos is equivalent to the topos of sheaves on a localic groupoid.
 Additional additions  From the étalé space point of view it is clear that covering spaces are indeed elements of the topos $Sh(X)$ of sheaves on $X$ and that the fundamental group of $Sh(X)$ (in the sense of Barr and Diaconescu) is the usual fundamental group of $X$ if $X$ is locally simply connected.  
Of course it is not hard to see that covering spaces correspond to locally constant sheaves but I don't think this is the way people think about covering spaces.<|endoftext|>
TITLE: Square roots of the Laplace operator
QUESTION [27 upvotes]: In several places in the literature (e.g. this paper of Caffarelli and Silvestre), I've seen an integral formula for fractional Laplacians.  I'd like to understand it.  In this question, I'll stick to the case of the square root.  
The formula I've seen is this:
$$((-\triangle)^{1/2}f)(x)=
C_n \int_{\mathbb{R}^n}\frac{f(x) - f(y)}{\|x - y\|^{n + 1}}\ dy.
$$
Here $x \in \mathbb{R}^n$ and $C_n$ is a constant.  Also, $f$ is a function $\mathbb{R}^n \to \mathbb{R}$, but I'm not sure what regularity assumptions it's supposed to satisfy.  
For this notation to be justified, it must surely be the case that
$$
(-\triangle)^{1/2} \bigl((-\triangle)^{1/2} f\bigr) = -\triangle f
$$
for all nice enough $f$.  My question is: why?  I haven't been able to prove this identity even in the case $n = 1$.
Comments

It's clearly the case that the Laplace operator has a square root defined by
$$
\widehat{((-\triangle)^{1/2}f)}(\xi) = \|\xi\| \hat{f}(\xi).
$$
The paper linked to says that this operator $(-\triangle)^{1/2}$ is the same as the operator $E$ defined by the integral formula.  If I'm understanding correctly, proving this is equivalent to proving (i) that $E$ really is a square root of the Laplacian, and (ii) that $E$ is a positive operator on functions of compact support.  
I've seen a couple of references to Landkof's 1972 book Foundations of Modern Potential Theory.  Unfortunately, those citing Landkof's book don't say which part of the book they're referring to, and I've been unable to find the relevant part myself.  I'd be happy for someone to simply tell me where in that book to look.
I can see that the integral formula has something to do with Laplacians.  Switching to spherical coordinates, the formula is
$$
((-\triangle)^{1/2}f)(x) = 
\text{const}\cdot\int_0^\infty \frac{\int_{S^{n-1}} f(x + ru)\ du - f(x)}{r^2}\ dr
$$
where $du$ is surface area measure on $S^{n-1}$ normalized to a probability measure.  The integrand converges to $(\triangle f)(x)$ as $r \to 0$ (up to a constant factor).  Also, the integrand is identically zero if $f$ is harmonic, which is promising.

REPLY [3 votes]: This answer is a modified version of Tom's edit on Michael's answer just to make it a bit more rigorous.
Take $b(\xi)$ to be a smooth non negative radially symmetric function supported in $1< |\xi| <2$. Then we see that $|\xi|^\alpha = c \int_0^\infty t^{\alpha-1} b(\xi/t) dt$.
Now, let us say $b$ is the fourier transform of some kernel $k$, which will also be smooth, radially symmetric, and its integral is zero because $b(0)=0$. Then $b \cdot \hat u = \widehat{k \ast u}$. Moreoever, since k has integral zero and is symmetric
$$ b \cdot \hat u = \left(\int k(y) (u(x+y)-u(x)) dy\right)^\wedge. $$
Let us assume that $\alpha \in (0,2)$.
I want to use the identity above for $|\xi|^\alpha$ as an integral of scaled versions of $b$. For that recall that $b(\cdot/t)^{\wedge} = t k(t\cdot)$. So, we have
$$ |\xi|^\alpha \cdot \hat u = \left( c \int_0^\infty t^{\alpha-1} \int t k(ty) (u(x+y)-u(x)) \ dy dt \right)^\wedge$$
Exchanging the order of integration we have that
$$ |\xi|^\alpha \cdot \hat u = \left( \int (u(x+y)-u(x)) K(y) dy \right)^\wedge$$
for
$$ K(y) = c \int_0^\infty t^{\alpha} k(ty) dt = c_0 |y|^{-n-\alpha}.$$
The last identity is obtained from the change of variables $s = t|y|$.
There are still a few things that should be clarified, like the application of Fubini, and that the integrals may be principal values when $\alpha \geq 1$. But everything should be fine if $u$ is smooth enough.<|endoftext|>
TITLE: Average rank of elliptic curves over $\mathbb{Q}$
QUESTION [9 upvotes]: So it was conjectured that if all elliptic curves over $\mathbb{Q}$ are ordered by their heights, then the average rank is $\frac{1}{2}$. 
Brummer initially showed assuming BSD and GRH that the average rank is bounded by 2.3. Since then many improvements have been made. In my search, I found the slides for a talk by Manjul Bhargava (linked here: http://www.dpmms.cam.ac.uk/research/BSD2011/bsd2011-Bhargava.pdf), where he talks about his result showing that the average rank is bounded by 1.5 unconditionally. 
My question is has there been any improvement on his result since then? A reference to such a paper would be appreciated as well.

REPLY [4 votes]: Thank you everyone for the great references. 
I also recently found this really good summary by Alice Silverberg about all things related to the rank of an elliptic curve if anyone is interested.
http://math.uci.edu/~asilverb/connectionstalk.pdf<|endoftext|>
TITLE: Extensions of the modularity theorem
QUESTION [12 upvotes]: In 1995 (if I'm not mistaken) Taylor and Wiles proved that all semistable elliptic curves over $\mathbb{Q}$ are modular. This result was extended to all elliptic curves in 2001 by Breuil, Conrad, Diamond, and Taylor.
I'm asking this as a matter of interest. Are there any other fields over which elliptic curves are known to be modular? Are there any known fields for which this is not true for? 
Also, is much research being conducted on this matter?

REPLY [15 votes]: Yes, this is a very active area -- one of the major themes of current research in number theory. 
Much of the recent work has focussed on proving something slightly weaker, but easier to get at, than modularity. An elliptic curve $E$ over a number field $K$ is said to be potentially modular if there is a finite extension $L / K$ such that $E$ becomes modular over $L$. This notion of potential modularity has been much studied by Richard Taylor and his coauthors, and turns out to be almost as good for most purposes as knowing modularity over $K$. 
It's now known, for instance, that any elliptic curve over a totally real number field $K$ becomes modular over some totally real extension $L / K$; a bit of googling turns up http://www2.math.kyushu-u.ac.jp/~virdol/basechange2.pdf (which shows that one can choose $L$ in a rather specific way, using work of Taylor and Skinner-Wiles to do the heavy lifting).
I'm not an expert in the area, but my impression from talking to genuine experts is that current methods are very much limited to the case where the elliptic curve is defined over a field which is either totally real or CM -- outside these situations modularity is much less well understood.
(EDIT: I should add that there are some totally real fields for which one can show modularity, rather than just potential modularity; Jarvis and Manoharmayum have shown, for instance, that every semistable elliptic curve over $\mathbb{Q}(\sqrt{2})$ is modular.)<|endoftext|>
TITLE: Weil reciprocity vs Artin reciprocity
QUESTION [11 upvotes]: This is probably an easy question for the experts:
Given two rational functions $f$, $g$ on a non-singular projective algebraic curve X (over an algebraically closed field $k$) and $p \in X$, one defines the Weil symbol $(f, g)_p$ as 
the value of $(-1)^{ab} f^a g^{-b}$ at $p$ where $a = v_p(g)$ and $b = v_p(f)$. (Here $v_p$ means order of zero/pole at $p$.)
Weil reciprocity claims that product of $(f, g)_p$ for all $p \in X$ is equal to $1$.
My question is whether the Weil symbol can be realized as a special case of the Artin symbol (for an extension of fields of rational functions)? 
(Note that the ground field $k$ is not assumed to be of positive characteristic.)

REPLY [8 votes]: Weil reciprocity actually holds for arbitrary fields $k$, not necessarily algebraically closed:  just let $x$ run over all closed points of $X$, and replace your expression for $(f,g)_x$ by its norm from $k(x)^* $ down to $k^*$.
Then the connection with Artin reciprocity occurs when $k$ is a finite field (of size say $q$), as Chandan suggests.  More precisely, if $K$ denotes the function field of $X$, then Weil reciprocity for $X$ is equivalent to Artin reciprocity for the Galois $K$-algebra $L_f = K[t]/(t^{q-1}-f)$, with Galois group $k^*$ acting by multiplication on $t$.
Even more precisely, we mean that for all $x$ in $X$ we have $(f,g)_x = Art(L_f/K)_x(g)$.  To verify this, note that since both sides have a product formula and finite modulus, by weak approximation it suffices to consider the case where $x$ is not in the support of $f$.  Then the left-hand side is the norm of $f(x)^{v_x(g)}$ and the right hand side is $Frob_x^{v_x(g)}$, so it suffices to show that $Frob_x$ is the norm of $f(x)$.  But the residue field extension is $k(x)[t]/(t^{q-1}-f(x))$, so letting $d$ denote the degree of $k(x)$ over $k$ we can calculate the Frobenius as sending $t$ to $t^{q^d} = t^{q^d-1} \cdot t = (t^{q-1})^{1+q+...+q^{d-1}} \cdot t = f(x)^{1+q+...+q^{d-1}} \cdot t = Norm(f(x)) \cdot t$, as desired.<|endoftext|>
TITLE: A hard diophantine equation
QUESTION [10 upvotes]: Hello !
I would like prove that the following diophantine equation is unsolvable: $m!+27=n^3$.
Thanks in advance.

REPLY [30 votes]: I am happy to report that the equation has no solution. I kept my original response, and put the remaining arguments in the "EDIT" section below.
Here is a quick proof that there are only finitely many solutions.
We use $m!=(n-3)(n^2+3n+9)$. Here $n$ is divisible by $3$, hence $n^2+3n+9$ is not divisible by any prime $p\equiv 2\pmod{3}$. In other words, all the prime divisors $p\equiv 2\pmod{3}$ of $m!$ are contained in $n-3$ with multiplicity. It follows, with the usual notations, that
$$ \frac{\log m!}{3}>\log(n-3)\geq\sum_{p\equiv 2 \ (3), \ p\leq m}v_p(m!)\log p> \sum_{p\equiv 2\ (3), \ p\leq m} \left(\frac{m}{p}-1\right)\log p.$$
The left hand side is $\sim (m\log m)/3$, while the right hand side is $\sim (m\log m)/2$ by Dirichlet's theorem. Hence for large $m$ the inequality must fail.
EDIT. Assume that $m\geq 1000$, and denote by $\chi$ the nontrivial character modulo $3$. Then
$$ \sum_{p\leq m}\frac{\chi(p)\log p}{p}<\sum_{n\leq m}\frac{\chi(n)\Lambda(n)}{n}-\sum_{p\leq m,\ p\neq 3}\frac{\log p}{p^2+p}. $$
This implies, in combination with some ideas of Bordelles (cf. the proof of (4.2) here), that
$$ \sum_{p\leq m}\frac{\chi(p)\log p}{p}<3\left|\frac{L'(1,\chi)}{L(1,\chi)}\right|+1.53<2.64\ .$$
By including the contribution of the prime $p=3$ to $n-3$ in the original inequality, and using also some classical bounds by Rosser and Schoenfeld (cf. (3.15) and (3.21) here), it follows that
$$\frac{m(\log m-0.9)}{3}>\frac{m(\log m-6.1)}{2}\quad\text{for}\quad m>e^{16.5}.$$
Hence $m < e^{16.5}$. I checked with SAGE that in fact
$$\sum_{p\leq m}\frac{\chi(p)\log p}{p}<-0.63\quad\text{for}\quad e^{16.5}>m>e^{7}.$$ 
This can be used to improve the previous bound to
$$\frac{m(\log m-0.99)}{3}>\frac{m(\log m-2.92)}{2}\quad\text{for}\quad e^{16.5}>m>e^{7},$$
which in turn forces $m < 1000$. The above shows that all solutions of the original equation satisfy $m < 1000$. However, I checked with SAGE that in this range the equation has no solution.<|endoftext|>
TITLE: Adelic formulations of complex multiplication and modular curves
QUESTION [8 upvotes]: In modular curves and modular forms, there is an adelic formulation, in which smaller open subgroups of some adelic group relate to higher level structure. As we know, higher level structure corresponds to higher torsion points (when we say "higher" we mean, e.g., $E[n]$ is "higher" than $E[m]$ if $m \mid n$). In complex multiplication, higher torsion points correspond to higher ray class fields, which correspond to smaller open subgroups of the ideles of the field of complex multiplication in question! (*Since someone asked, I sketch this in more detail below)
Thus I would like to ask whether there is a formulation that combines the adelic formulation of complex multiplication with the adelic formulation of modular curves.
To simplify and thus explain the principles**, it would be best to work with elliptic curves with complex multiplication. Let $F$ be a quadratic imaginary field. 
Then I have the following guess for an idea: We have an injection from the ideles of $F$ into the rational adelic points of $\mathrm{GL}_2$, and after quotienting by the rational (i.e. non-adelic) points, we still have a nice inclusion? I would guess that the resulting points at those corresponding to CM elliptic curves and their level structures, but I don't entirely understand how to prove this.
Furthermore, assuming this is true, what does this say about the Galois action? Does this allow us to construct the Galois action (i.e. rational structure) on modular curves?
(Also, to be even more precise, we should let the ideles of $F$ be represented by the adelic points of the Weil restriction of $G_m$ from $F$ to $\mathbb{Q}$).
*In the adelic formulation of modular curves, we view modular curves as quotients of the adelic points of $GL_2$. We get different modular curves, e.g. $\Gamma(N)$ instead of $\Gamma(M)$ (for $M,N$ some positive integers) by considering different open subgroups of the adeles. As is well-known, $\Gamma(N)$ parametrizes elliptic curves with "level N" structure, meaning pairs consisting of elliptic curves and $N$-torsion points. In complex multiplication, the $N$-torsion points correspond to the ray class field of level $N$, and this is, in particular, done through an adelic formulation of the main theorem of complex multiplication.
**Though I assume everything will extend to higher-dimensional Shimura varieties in an analogous way.

REPLY [10 votes]: This is an expansion of my comment above: You can view the modular curve of level $K$ over $\mathbb{C}$ as the double coset space
$$GL_2(\mathbb{Q})\backslash\mathbb{H}^{\pm}\times GL_2(\mathbb{A}_f)/K.$$
Here, $\mathbb{H}^{\pm}$ is the union of the half planes. This set is in bijection with isogeny classes of elliptic curves over $\mathbb{C}$ with $K$-level structure, and the bijection is obtained as follows: A point $\tau\in\mathbb{H}^{\pm}$ gives us an isomorphism of vector spaces $\mathbb{R}^2\xrightarrow{\simeq}\mathbb{C}$, which in turn equips $\mathbb{Q}^2$ with a Hodge structure of weights $(0,-1),(-1,0)$. This allows us to interpret it as the $\mathbb{Q}$-homology of an elliptic curve $E_{\tau}$ (This is just a fancier way of saying that $E_{\tau}=\mathbb{C}/(\mathbb{Z}+\mathbb{Z}\tau)$). So, given a point $[(\tau,g)]$ in the double coset space, we attach to it the elliptic curve (up to isogeny) $E_{\tau}$, with the $K$-level structure given by the $K$-orbit of the isomorphism
$$\mathbb{A}_f^2\xrightarrow{g}\mathbb{A}_f^2= H_1(E_{\tau},\mathbb{A}_f).$$
Let $\alpha$ be some generator of $F$ over $\mathbb{Q}$: it allows us to identify $F$ with $\mathbb{Q}^2$. Choose an embedding $F$ in $\mathbb{C}$, so that $\alpha$ can be viewed as an element of $\mathbb{H}$. Let $H$ be the rank $2$ torus over $\mathbb{Q}$ attached to $F$: we can view it as the sub-group of $GL_2$ that commutes with the action of $F$. We then get a map 
$$\eta:H(\mathbb{Q})\backslash H(\mathbb{A}_f)\rightarrow GL_2(\mathbb{Q})\backslash\mathbb{H}^{\pm}\times GL_2(\mathbb{A}_f)/K$$
$$\;\;\;\;[h]\mapsto [(\alpha,h)].$$
The image of this map consists exactly of those elliptic curves admitting CM by $F$, whose level structures admit an $F$-equivariant representative. (One subtlety is that, if $E$ is an elliptic curve in the image, then you can also look at the curve $\overline{E}$, where you twist the action of $F$ by complex conjugation. This won't show up in the image of $\eta$).
Now, we have the global reciprocity isomorphism
$$Gal(\overline{F}/F)^{ab}\xrightarrow{\simeq}H(\mathbb{Q}\backslash H(\mathbb{A}_f),$$
which equips $H(\mathbb{Q})\backslash H(\mathbb{A}_f)$ with the structure of a pro-finite $Gal(\overline{F}/F)$-set. The main theorem of CM for elliptic curves says that, if we identify the right hand side of $\eta$ with the set of $\overline{\mathbb{Q}}$-points of the modular curve, then $\eta$ is in fact Galois-equivariant. 
There is a slight sign issue here, since there are two possible choices for the reciprocity map (arithmetic or geometric Frobenius), but this is the general shape.<|endoftext|>
TITLE: complex manifold with corner
QUESTION [5 upvotes]: I was reading Dominic Joycee article on Manifold with corner.  He talk about manifold with corner modeled over $[0,\infty)^k\times \mathbb R^{n-k}$ for some $k\leq n$.  From here i moved to Melrose unpublished book on Manifold with corner.   
Is the theory for Complex manifold with boundary and corner is developed.  I mean is there some literature available where complex manifold with corner has been discussed. 
"Complex manifold with corner"  is a vague word.  But i mean, i want to see as $[0,1]\times [0,1]$ as complex manifold with corner where boundary is CR manifold.  
I think "main problem" is the extension of holomorphic function defined in the interior of domain.  In real case, we have whitney extension theorem.... There may be many other issues...  Can i have reference, suggested reading along these lines.  Thanks

REPLY [2 votes]: There are some related results about compact Stein 4-manifolds with boundary as Lefschetz fibrations over the disk (whose fiber has non-empty boundary). Corners in this case arise naturally on the total space. References includes Loi-Piergallini's theorem, and subequent works of Akbulut-Ozbagci (simply google with these keywords).<|endoftext|>
TITLE: the left hand side of the  Ricci flow equation at the initial value
QUESTION [6 upvotes]: I just started to learn about the Ricci flow and try to understand the Ricci flow evolution equation. It states that a one-parameter family $g_t$, $t\in[0,T)$ of Riemannian metrics on a smooth closed manifold $M$ is a solution of the equation
$$ \frac{\partial g_t}{\partial t}=-2 Ric_{g_t}. $$
But, in order to be able to write down the expression
$$ \frac{\partial g_t}{\partial t} $$ 
I need a little bit more than the fact that $g_t$ is a one-parameter family of Riemannian metrics. 
There are several ways to understand this expression, which probably are equivalent to each other. I think the most naive idea is to consider $g_t$ as a smooth curve in the infinite-dimensional vector space of all Riemannian metrics on $M$ and $\frac{\partial g_t}{\partial t}$ as tangent vectors to this curve.
Now, my question is: How do I have to understand the left hand side of the Ricci flow equation at the initial value $t=0$?
Does one consider $g_t$ as a smooth family at the open interval $(0,T)$ and requires continuity at the point $t=0$ or has one to deal with differentiability from the right? 
Thank you very much!

REPLY [3 votes]: Pardon, this is not an answer, but it was too long to put in as a comment, and I wasn't sure how to include this random thought otherwise. Let $g$ be a Riemannian metric on a closed manifold $M$. We can define its temperature $\tau(g)\in\lbrack0,\infty]$ to be the supremum of
all $T$ such that there exists a solution to the Ricci flow on $M\times
\lbrack0,T]$ with $g(T)=g$. By Brett Kotschwar's backward uniqueness result,
there exists a unique solution to the Ricci flow on $M\times(0,\tau(g)]$ with
$g\left(  \tau\left(  g\right)  \right)  =g$. This solution may or may not be
extendable to $t=0$. The temperature of an Einstein metric with nonnegative
scalar curvature is infinity (when $R>0$, more generally, a shrinking gradient
Ricci soliton). The temperature of an Einstein metric with negative scalar
represents its scale. More generally, $\tau (cg) =c\tau (g)$. By Shigetoshi Bando's result, if $\tau(g)>0$, then $g$ is real analytic (as Otis Chodosh wrote above). I'm not sure what this invariant, nontrivial only for real analytic metrics, is good for, though.<|endoftext|>
TITLE: Albert classification of rational endomorphism rings of simple Abelian varieties over finite fields
QUESTION [5 upvotes]: Recall the Albert classification of rational endomorphism rings with involution of simple Abelian varieties over arbitrary fields:

Type I: totally real, trivial involution
Type II and III: quaternion algebras over totally real number fields
Type IV: center is a CM field for which the restriction of the involution is complex conjugation

Now my question is: Which of these division algebras can actually occur if the base field is a finite field?

REPLY [2 votes]: See http://www.math.nyu.edu/~tschinke/books/finite-fields/final/05_oort.pdf, Section 15:
Only Type III and IV can occur, and III only for dimension $1$ or $2$.<|endoftext|>
TITLE: Knight tour prime (conjecture)
QUESTION [9 upvotes]: Hello,
I have the following conjecture:
Write all numbers from $1$ to $n^2$ over an $n\times n$ board as usually. There not exists $n$ such that we can find a hamiltonian path on primes numbers with a knight.
Andres Sanchez Perez (Ecole Polytechnique, Paris, France) verified this conjecture for $3\leq n \leq 100$.
Any comments or reference will be appreciated.

REPLY [4 votes]: An argument for even $n$: Suppose that $n\geq16$ is even, and consider the first two rows, consisting of $1,\ldots n$ and $n+1,\ldots,2n$, respectively. Since $3$, $7$, and $11$ are primes, in order for them to be reachable, at least $3$ of $n+1$, $n+5$, $n+9$, and $n+13$ must be primes. Modulo $3$ these are $n+1$, $n+2$, $n$, and $n+1$, so there is at least one composite number among these $4$ numbers. Hence one has to choose one of the following two triples: $(n+1,n+5,n+13)$ or $(n+1,n+9,n+13)$. In the first case, it leaves each of $7$ and $11$ reachable from only one square in the second row ($n+5$ and $n+13$), and similarly in the second case, each of $3$ and $7$ would be reachable from only one second-row square ($n+1$ and $n+9$). In any case, this means that each of the remaining primes in the first row must be reachable from two second-row squares, in order for a Hamiltonian path to exist. Now we look at the primes $5$ and $13$, which are in the first row, and conclude that $n+3$, $n+7$, $n+11$, and $n+15$ must all be primes. Modulo $3$, these are $n$, $n+1$, $n+2$, and $n$, making it clear that there is at least one composite number among them.
A different argument for even $n$, assuming that $2$ is not a prime:
Color the odd squares white, and the even squares black, as in Eric's answer. Since $n$ is even, this would mean simply that the odd columns are white, and the even columns are black. We do not want the knight falling on a black square, so we can remove the black columns, and assume that the knight moves diagonally. If we colour this board in checkerboard pattern, then a knight started on a white square must stay on white squares, and a knight started on a black square must stay on black squares. So essentially half of the odd numbers are forbidden, no matter where the knight's initial position is. Now, we note that $3$ and $5$ are in different groups (if $n\geq6$), so at least one of them is unreachable.
When we relax the assumption that $2$ is not a prime, there is a way to switch colour for the knight by going through $2$. But this is the only way, and it requires $2n+1$ and $2n+3$ be both primes.<|endoftext|>
TITLE: Can the holomorphic image of $(\mathbb{C}^*)^n$ be open but not dense
QUESTION [29 upvotes]: Let $M$ be a compact complex connected [but not necessarily kähler] $n$-manifold, and suppose we have a holomorphic map $$(\mathbb{C}^*)^n \to M$$ such that the image is open. Is the image necessarily dense in $M$?

Motivation: My intuition (which comes from the algebraic world) says that the answer ought to be "yes." On the other hand, I know that many properties of smooth algebraic varieties do not hold for complex manifolds in general. Knowing whether this statement has a counterexample would improve my intuition about the complex world.

REPLY [11 votes]: Fatou-Bieberbach domains were first constructed by Fatou in 1927; the construction was improved by Bieberbach in 1932.  The idea is linearizing coordinates at attracting cycles.
Let $f:\bf C^2 \to C^2$ be an analytic automorphism, such that $(0,0)$ is an attracting fixed point, for instance the Hénon mapping
$$
f: (x,y) \mapsto (x^2-y,x/2).
$$
Denote by $L$ the derivative of $f$ at the origin.
There is then (usually, in particular in the case above) a unique analytic mapping $\phi:\bf C^2\to C^2$ with $\phi(0,0)=(0,0)$, $D\phi(0,0)=id$ and $L\circ \phi= \phi\circ f$.
In this case (and in many others), $\phi^{-1}$ is defined on the basin of attraction of the origin by the formula
$$
\phi(x,y)= \lim_{m \to \infty} L^{-\circ m} \circ f^{\circ m} (x,y).
$$
The map $\phi$ is injective, but its image is certainly not dense. The image is the basin of attraction of the origin. There are lots of points not attracted to the origin, for instance  the set
$$
U=\{(x,y) |\  |x|\ge 2, |y|\le x\}.
$$
it is easy to see that $f(U)\subset U$, and that if you iterate $f$ in $U$ all orbits tend to infinity.
Showing that the limit above exists and defines an analytic map is not hard; I will write a proof if you are interested.
There is an enormous literature about linearization at periodic points.<|endoftext|>
TITLE: "Purely local" proof of local Langlands
QUESTION [12 upvotes]: As from this website
http://math.uchicago.edu/~lxiao/workshop_site/
My question is: What does it mean by "purely local"? 
Also, I heard about this phrase "purely local" in other problems as well, mostly with the phrase "a purely local proof". 
The other question is, for GL(1) and GL(2), are there already a "purely local proof"?
Thanks.

REPLY [17 votes]: The short answer to the question is that all currently known proofs of the local Langlands correspondence (and I'm just referring to GL(n) here) are "global" in the sense that they involve embedding the local problem into a global one.  That is, the local field in question is realized as the completion of a global field at one of its places.  Then the theory of automorphic forms over the global field may be applied.  In particular, under certain circumstances, we know that Galois representations may be attached to automorphic representations.  A purely local proof would not make reference to global fields at all.
Kevin commented that a purely local characterization (he uses the word statement) of the correspondences is a prerequisite for a purely local proof.  The established characterization for GL(n) (and indeed, the one used in the proofs of Henniart and Harris-Taylor) is, as Kevin points out, through epsilon factors of pairs, and the existence of these is only defined through global means.  (Rob is correct that Langlands has unpublished notes on the subject, but these are so complicated as to be unsatisfactory, and in any case it is truly unclear what the right characterization is for groups other than GL(n).)  
Now to Alexander Chervov's important comment:  what is the right characterization in the case of $n = 1$?  Sure, you can make some quantitative conditions involving ramification.  But let's recall that the most elegant path to local CFT is unquestionably through Lubin-Tate theory:  the maximal totally ramified abelian extensions of a nonarchimedean local field are obtained by adjoining the torsion of a one-dimensional formal module of height one.  Let us declare that Lubin-Tate theory itself provides the correct characterization of the local Langlands correspondence in the $n=1$ case (and to hell with conductors, Gauss sums, etc.).
This point of view suggests that variations on the theme of formal modules ought to provide the right purely local characterization of local Langlands (and also a hope for a purely local proof).  Now already by 1990, Carayol conjectured ("Nonabelian Lubin-Tate theory") that certain deformation spaces of formal modules ("Lubin-Tate spaces") exhibit the local Langlands correspondence in their cohomology, at least for some classes of representations of GL(n).  Harris and Taylor prove Carayol's conjecture for supercuspidal representations, which is enough to prove the existence of the correspondence in general.  Here the characterization is still through epsilon factors of pairs, and therefore still global in nature. 
The next big development along these lines is Peter Scholze's new proof of the correspondences for GL(n).  While still global in nature, Scholze gives a purely local characterization of the correspondences, which satisfies Kevin's requirements for a "natural bijection", and which is compatible with the global theory.   Suppose $\pi$ is a smooth irreducible representation of $\text{GL}_n(F)$ ($F$ a $p$-adic field).  Scholze characterizes the corresponding (semisimplified) Weil representation $\sigma$ by giving an actual formula for the trace of $\sigma(\tau)$, for any element $\tau$ in the Weil group of $F$!  Alas, the other side of Scholze's formula is too involved to describe here, but it involves deformation spaces of $p$-divisible groups in an ingenious way.  When $n=1$, the formula reduces to the statement that local class field theory is realized in the torsion of Lubin-Tate formal modules. In my mind, purely local attacks on the local Langlands correspondence ought to start here.
(Not that any of the preceding is going to be mentioned in my talks tomorrow.  My own meager contributions to this story don't yet connect to Scholze's work, but only to the theory of types, which figure prominently in the Bushnell-Henniart book mentioned by Keerthi.)<|endoftext|>
TITLE: Strong Atiyah conjecture
QUESTION [11 upvotes]: Who introduced the Strong Atiyah Conjecture?  
Recall that the conjecture says the following. Let $G$ be a group, $A$ a $n\times n$-matrix over ${\mathbb Z}G$. We view $A$ as a bounded operator $l^2(G)^n\to l^2(G)^n$. Let $K$ be the kernel of that operator and $p$ be the orthogonal projection onto $K$. Let $\delta$ be the function of $l^2(G)$ which is 1 on $1\in G$, and 0 everywere else. Let $e_i$ be the vector from $l^2(G)^n$ with $i$-th coordinate $\delta$, all other coordinates $0$, and $t_A$ be the sum of dot products $\sum_i \langle p(e_i), e_i\rangle$. The conjecture says that $t_A$ always belongs to $\frac1g\mathbb{Z}$ where $g$ is the least common multiple of the orders of finite subgroups of $G$ (so if $G$ is torsion-free, $t_A$ must be an integer).

REPLY [9 votes]: I received a message from Thomas Schick, which answers my question. The strong 
Atiyah conjecture was introduced jointly by Lueck and Schick (this was also one of the alternatives in Wolfgang Lueck's email to me), and Thomas Schick is responsible for the super-strong one. He was a postdoc working with Wolfgang Lueck at that time.  Peter Linnell came to the $l_2$-Betti theory via Kaplansky's zero divisors conjecture (these are related) and learned about various forms of Atiyah conjectures from Wolfgang Lueck. In any case, the various forms of Atiyah's conjecture have generated a lot of interesting mathematics already and probably will generate more. 
If there are no other answers, I will then accept this one.<|endoftext|>
TITLE: Finiteness of Tate-Shafarevich
QUESTION [12 upvotes]: Does anyone happen to know who conjectured the finiteness of the Tate-Shafarevich group?
We recall the conjecture. Let $E/K$ be an elliptic curve where $K$ is a number field. Then $Ш(E/K)$ is finite.

REPLY [12 votes]: In Cassels's 1962 ICM paper (available here), he says the following: "Indeed, Tate and Šafarevič have, I believe, independently conjectured(5) that Ш itself is always finite", and the (5) is a footnote stating: "In his lecture, Tate denied paternity but adopted the conjecture. In conversation during the Congress Šafarevič expressed strong doubts."
So, maybe no one knows!
EDIT: As an added bonus, I found the following quote in Cassels's review of Silverman's book: "Without doubt the reviewer's most lasting contribution to the theory is the introduction of the Cyrillic letter Ш ("sha") to denote this group, a usage which has become universal."<|endoftext|>
TITLE: References/surveys concerning characteristic classes of flat vector bundles
QUESTION [8 upvotes]: I'm looking for good surveys about characteristic classes of flat real vector bundles.  Letting $G$ be $\text{SL}_n(\mathbb{R})$ with the discrete topology, orientable flat $n$-dimensional real vector bundles are classified by $BG$, so the characteristic classes I'm looking for are elements of the cohomology of $BG$ with respect to various systems of coefficients.
The one source I know of is Morita's book "The geometry of characteristic classes", but this is not very comprehensive or up-to-date.  I also have found many references about Chern-Simons invariants, but they all seem to be written from the perspective of algebraic geometry or mathematical physics or differential geometry.  I'd really like a source that is as topological as possible; for instance, for ordinary characteristic classes I prefer the approach taken in Milnor-Stashef to the approach via Chern-Weil theory.  But any sources at all are welcome.

REPLY [6 votes]: I am not aware of much recent activity (after 2001, when Morita wrote his book) in this field, so I think it is still very valuable. Chern-Simons theory means something rather different today. The useful information on characteristic classes of flat vector bundles (CCFVB) seems to be scattered throughout the literature; one of the reasons might be that there is no complete picture known. Googling gives you a lot of sources, but I completely agree that a modern detailed and comprehensive exposition is missed.
Also, as far as I know, Chern-Weil theory is an indispensible ingredient of the theory. Purely topological methods are not very sensitive to flat bundles while Chern-Weil theory allows you to import methods from Lie algebra cohomology etc.
Here is a list of texts from which I learnt something on the subject:
Karoubi: Homologie cyclique et K-Theorie algebrique.
Jones, Westbury: Homology spheres, eta-invariants and algebraic K-theory. The first few sections form an informative survey which is worth reading even if you are willing to ignore the main problem they consider.
Dupont: Curvature and characteristic classes (the last section)
Kamber, Tondeur: Foliated bundles and characteristic classes 
A modern approach to CCFVB is to use differential cohomology (or differential characters). If you google, you find more information on that.<|endoftext|>
TITLE: Applications and motivation of $\eta$-quotient generators and algorithms
QUESTION [8 upvotes]: So initially Dummit, Kisilevsky and McKay found all Dedekind $\eta$-products which are newforms. This result was subsequently generalized by Martin. Further, Ono and Martin provide an exhaustive list of modular elliptic curves whose associated modular forms is expressible as an $\eta$-quotient.
Also, it is known that 
$$
E_4(z) = \dfrac{\eta(z)^{16}}{\eta(2z)^8} + 2^8 \cdot \dfrac{\eta(2z)^{16}}{\eta(z)^8}
$$
and
$$
E_6(z) = \dfrac{\eta(z)^{24}}{\eta(2z)^{12}} - 2^5 \cdot 3 \cdot 5
\cdot \eta(2z)^{12} - 2^9 \cdot 3 \cdot 11 \cdot \dfrac{\eta(2z)^{12}\eta(4z)^8}{\eta(z)^8} + 2^{13} \cdot \dfrac{\eta(4z)^{24}}{\eta(2z)^{12}},
$$
leading Ono to pose the question in his book "Web of Modularity", which spaces of modular forms are generated by eta-quotients. This question was addressed somewhat in this paper (http://arxiv.org/pdf/math/0701478v1.pdf) by Kilford. 
My question is, what is the motivation for this question by Ono as most of these expressions are rather "messy". Is this just a question of interest in itself or are there further consequences to this open question? Why are spaces generated by $\eta$-quotients of interest as opposed to some other modular form such as the Eisenstein series?
Also is there an algorithm that given a modular form that can be expressed as a linear combination (or even rational function), can compute the corresponding expression?

REPLY [2 votes]: I am not a specialist of that subject, but one of the interest of the $\eta$-function is its
direct relation with the partition function $p(n)$, the number of ways to write $n$ as a sum of positive integers disregarding order. There are a long line of works since Ramanujan
on the arithmetic property of $p(n)$ with still many beautiful open questions.
The relation of $\eta$ with $p(n)$ is simply, as you probably know well, that
$$\eta(z) = q^{1/24} \prod_{n \geq 1} (1-q^n) = q^{1/24} (\sum_n p(n) q^n)^{-1}$$
with $q=e^{2 i \pi z}$.
Relating $\eta$ with eigenforms of integral weights (basic example: $\eta^{24}=\Delta$), a kind of object for which we have a lot of tools, in particular the existence of a Galois representation attached to it, may if cleverly used give some information about the partition function (and I know that some results of Ahlgren, Nicolas, Serre, Boylan, Ono, and many others, that use this method. I am just quoting here at random papers that I happen to have looked at recently).
Now, this is not to say that this is the only and even main motivation to look for such kind of identities, nor that all such identities are useful in the study of the partition function,
but at least that is one motivation I can understand.<|endoftext|>
TITLE: iterating ultrapowers of C*-algebras: the Calkin algebra
QUESTION [6 upvotes]: Elsewhere I asked about ultrapowers of the C*-algebra $A$ of compact operators on separable infinite-dimensional Hilbert space. My question was whether the process of taking ultrapowers of ultrapowers ever stabilizes. 
It was pointed out that the definition of an ultrapower of a C*-algebra is slightly different from the usual definition of an ultrapower; that the process never stabilizes in the sense that the canonical embedding is never an isomorphism; but that nevertheless (assuming CH) that if $A^1$ is an ultrapower of $A$ then the ultrapowers of $A^1$ are isomorphic as C*-algebras to $A^1$.
This last statement depends on the fact $A$ is separable. Here I would like to ask about the corresponding question for the C*-algebra $B$ of bounded operators on Hilbert space and also for the Calkin algebra $C=B/A$. Does the process of taking ultrapowers of ultrapowers of $B$ ever stabilize in the sense of producing isomorphic C*-algebras? And the same question for $C$?

REPLY [11 votes]: The statement does not depend on separability of A. Actually in the Ge-Hadwin paper they explicitly state the result for C*-algebras of cardinality continuum. So under CH you have
that the ultrapowers of C^1 are isomorphic to C^1. 
It is not completely trivial that C is not isomorphic to C^1. 
This was proved in 
    arXiv:1112.3898 Countable saturation of corona algebras. Ilijas Farah, Bradd Hart.
Also, if CH fails then every C*-algebra has 2^c nonisomorphic ultrapowers, so the 
sequence does not stabilize. This is in 
    arXiv:0908.2790 Model theory of operator algebras I: Stability. Ilijas Farah, Bradd Hart, David Sherman.<|endoftext|>
TITLE: Finite groups admitting free isometric actions on round spheres
QUESTION [5 upvotes]: Is there some sort of classification of finite groups $G$ such that for at least one $n$ the group $G$ admit a free isometric action on the standard  sphere $S^n $of curvature 1? Are there some simple criteria that permit to check (in some particular cases) if a given group has such an action (for at least on $n$) or not?

REPLY [3 votes]: After all these comments, a possible answer to your question goes as follows.
If $n$ is even, then the only group that can act freely and isometrically on $S^n$ is $\mathbb{Z}/2$. One way to see this is as in Max's answer above, i.e. by looking at the behavior of eingenvalues of orthogonal matrices. Here is another reason: an action of $G$ on $S^n$ means that there is a group homomorphism $G\rightarrow\mbox{Homeo}(S^n)$. But any homeomorphsim has degree $\pm 1$. So we get a homomorphism $G\rightarrow\mathbb{Z}/2$. But the action has no fixed points, then the degree of the image of every non trivial element in $G$ is $(-1)^{n+1}=-1$, therefore the map has a trivial kernel and therefore it is an isomorphism. 
As macbeth already pointed out, when $n$ is odd, the problem is not so easy. In that case, we consider $S^n$ as the universal cover of a complete Riemannian manifold $M$ with constant sectional curvature. A simple argument using lifting properties of covering spaces shows that there is an isometry between $M$ and $S^n/G$ (whenever the action is free and properly discontinuous). So the problem of finding the subgrups with that particular action on the sphere is the same as the classification of complete Riemannian manifolds with constant sectional curvature (=1). That is done in Wolf's book Spaces of Constant Curvature.<|endoftext|>
TITLE: Does the classifying space of monoids commute with wedge sum up to weak equivalence?
QUESTION [5 upvotes]: If $G$ and $H$ are groups, then the map $BG\vee BH\to B(G\ast H)$ is a weak equivalence by van Kampen's theorem. However the classifying space $BM$ of a monoid can have arbitrary homotopy type so higher homotopy groups are involved.
If $M$ and $N$ are monoids, is the map $BM\vee BN\to B(M\ast N)$ still a weak equivalence? If there are counterexamples, is there still some criterion that insures such a weak equivalence?

REPLY [3 votes]: Charles has given a very good answer to the question. 
The following is not meant to be an answer, but just a heuristic argument which I cannot 
make into a proof. 
There should be an operation, "free product," denoted $\sharp$, in the category of associative topological monoids. If $X,Y$ are based spaces, then 
$$\Omega (X \vee Y)$$ (Moore loops), should decompose (at least up to homotopy) in this category as
$$(\Omega X) \sharp (\Omega Y) .$$ The reason  I find this to be plausible is that a loop in $X \vee Y$ is clearly a word of loops of $X$ and $Y$, and a word is supposed to represent an element of the free product.
Supposing this to be the case, we could take $X = BM$ and $Y = BN$, then we would have
$$
\Omega (BM \vee BN) \simeq (\Omega BM) \sharp (\Omega BN)
$$
It should also be the case that the inclusion 
$$M \ast N \to (\Omega BM) \sharp (\Omega BN)$$ 
is group completion, since $ (\Omega BM)$ and $ (\Omega BN)$ are group-like and the operation $\sharp$ should preserve grouplike monoids (furtheremore, we also should have
$M \ast N \simeq M\sharp N$). If this is true, then $(\Omega BM) \sharp (\Omega BN) \simeq \Omega B (M\ast N)$.
If the above works, then the homomorphism
$$
\Omega (BM \vee BN) \to (\Omega BM) \sharp (\Omega BN)
$$
is an equivalence. Now apply the classifying space to get the desired equivalence
$BM \vee BN \simeq B(\Omega BM) \sharp (\Omega BN) \simeq B(M \ast N)$.
Question: Can this heuristic sketch be made into a proof?<|endoftext|>
TITLE: Trig functions based on convex curves
QUESTION [7 upvotes]: Pardon my naivety, but I wonder if
much use has been found for
trigonometric functions
defined in terms of a centrally symmetric convex curve $K$ replacing
the circle $C$.
For example, here is the equivalent of the sine function
defined on a diamond, and on a quadratic curve
(with the true $\sin \theta$ function superimposed for comparison):

          


Perhaps for certain $K$ nice properties are retained for the
corresponding trig functions: trig identities, orthogonality,
Fourier series, etc.?
In some sense I am seeking to understand why the standard trig functions
are so ubiquitous and useful, by imagining a variant.
Thanks for any insights!

REPLY [2 votes]: One of the ways to define the usual sine function is to consider
the boundary value problem of the second order ODE $y''+ cy =0, y(0)=
y(\pi)=0.$  Generalizing this definition, P. Lindqvist studied generalized trigonometric functions [L]. Numerous other authors
have worked on this topic thereafter. Some of the latest 
papers include [BV],[BE],[EGL], [T].
[BV] B. A. Bhayo and M. Vuorinen:
On generalized trigonometric functions with two parameters. J. Approx. Theory 164 (2012) 1415--1426, doi: 10.1016/j.jat.2012.06.003
[BE] P. J. Bushell and D. E. Edmunds:
Remarks on generalised trigonometric functions.
 Rocky Mountain J. Math.  42 (2012), Number 1, 25--57.
[EGL] D. E. Edmunds, P. Gurka, and J. Lang:
 Properties of generalized trigonometric functions.
J. Approx. Theory 164 (2012) 47--56, doi:10.1016/j.jat.2011.09.004.
[L] P. Lindqvist: Some remarkable sine and cosine functions. 
Ricerche di Matematica, Vol. XLIV (1995), 269--290.
[T] S. Takeuchi: Generalized Jacobian elliptic functions and their application to bifurcation problems associated with p-Laplacian. J. Math. Anal. Appl. 385 (2012), 24--35, doi:10.1016/j.jmaa.2011.06.063.<|endoftext|>
TITLE: Can the pushforward of a very ample invertible sheaf under a birational morphism be reflexive?
QUESTION [5 upvotes]: Let $f: Y \to X$ be a birational  morphism of  projective varieties.  Let $\mathcal{M}$ be a very ample invertible sheaf on $Y$.  Suppose also that:

$f^{-1}$ is defined away from a single point $x \in X$.
$f_* \mathcal{O}_Y = \mathcal{O}_X$.

Two questions:
(1) If $V$ is a set of global sections of $\mathcal{M}$ that generate $\mathcal{M}$, consider the induced evaluation map
$ V \otimes \mathcal{O}_X \to f_* \mathcal{M}$. 
Let $\mathcal{N}$ be the image of this map.  Is it possible for $\mathcal{N}$ to be reflexive?  (Full disclosure:  I would like the answer to be no.)
If we know that $X$ is smooth at $x$, or more generally that $(f_* \mathcal{M})^{\vee\vee}$ is invertible, then the answer   is no.  Let $E$ be the exceptional locus of $f$.  Then $E$ is positive-dimensional, so any hyperplane section meets $E$ nontrivially, and thus any section of $f_* \mathcal{M}$ must vanish at $x$.  But if $(f_* \mathcal{M})^{\vee\vee}$ is only reflexive, then I don't see how to generalise this argument.  
(2) 
Is it possible for $f_* \mathcal{M}$ to be reflexive?  If so, what is the weakest possible condition that will guarantee that $f_* \mathcal{M}$ is not reflexive?
My chief interest in (2) is that a negative answer to (2) would be a cheap way to get a negative answer to (1).

REPLY [3 votes]: Hi Sue, I think it can be reflexive.
Let's consider $X = \text{Proj} k[x,y,u,v,t]/\langle xy - uv \rangle$.  This has only an isolated singularity at $x=y=u=v=0$ (it's the simplest non-Q-factorial singularity I know of).  Fix $U$ to be the regular locus of $X$.
(Blow up a divisor:) Set $\pi: Y \to X$ to be the blowup of the prime divisor $D = V(x,u)$.  This is a small resolution of $X$ (and also contains $U$ as an open set).  Lets define $F$ on $Y$ to be such that $O_Y(-F) = O_X(-D) \cdot O_Y$.  In other words, $F$ is the inverse image of $D$.  Notice that $\pi_* O_Y(-F) = O_X(-D)$ since $\pi$ is a small resolution.  Let me explain this point.
Choose $V \subseteq X$ an open set.  Then 
$$
\Gamma(V, \pi_* O_Y(-F)) = \Gamma(\pi^{-1} V, O_Y(-F)) = \Gamma((\pi^{-1} V) \cap U, O_Y(-F))
$$
since $\pi^{-1}V \setminus U$ is codimension 2 (see for example THIS ANSWER by Sándor Kovács).
But $\Gamma((\pi^{-1} V) \cap U, O_Y(-F)) = \Gamma(V \cap U, \pi_* O_Y(-F))$.  It follows that $\pi_* O_Y(-F)$ is determined away from the singularity, and is therefore S2 / reflexive.
Since $Y$ is smooth, $F$ is a Cartier divisor and we know that $-F$ is $\pi$-ample by construction.  Set $A = V(t)$ to be an ample Cartier divisor on $X$ (the choice of $A$ doesn't matter so much here).
(Define $\mathcal{M}$): It follows that $\mathcal{M} = O_Y(-F + n \pi^*A)$ is ample on $Y$ for $n \gg 0$.  
Then we know
$$
\pi_* O_Y(-F + n \pi^* A) = (\pi_* O_Y(-F) ) \otimes O_X(nA) = O_X(-D + nA).
$$
which is reflexive.  Note that you can also consider $-mF + mn \pi^* A$ to make  $\mathcal{M}^m$ as ample as you'd like.  The same sort of computation still holds.
Ok, so this gives a negative answer for (2).  
(Question 1.): Let's now consider (1).  Note $\pi_* \mathcal{M}$ is globally generated (since $n \gg 0$).  But on the other hand,
$$H^0(X, \pi_* \mathcal{M}) = H^0(U, \mathcal{M}) = H^0(Y, \mathcal{M})$$ again since $X \setminus U$ and $Y \setminus U$ are codimension 2.
Thus the global section of $H^0(Y, \mathcal{M})$ already globally generate $\pi_* \mathcal{M}$, so I don't think (1) works either.
Obviously if $X$ is Factorial then you won't run into this problem as you already pointed out.<|endoftext|>
TITLE: Levy-Gromov Isoperimetric Inequality
QUESTION [7 upvotes]: In his paper "Paul Levy's Isoperimetric Inequality", Gromov gives the following isoperimetric inequality:

Let $V$ be a closed $(n+1)$-dimensional Riemannian Manifold with $\mathrm{Ric}(V) \geq n \space (= \mathrm{Ric}(S^{n+1}))$.  Let $V_0 \subset V$ be a domain with smooth boundary and let $B$ be a round ball in $S^{n+1}$ such that
  $$ \frac{ Vol(V_0)}{Vol(V)}= \frac{Vol(B)}{Vol(S^{n+1})}.$$
  Then it follows that
  $$ \frac{Vol( \partial V_0)}{Vol(V)} \geq \frac{Vol(\partial B)}{Vol(S^{n+1})}. $$

Now my question: in a (slightly earlier) article 'Isoperimetric Inequalities In Riemannian Manifolds', Gromov states that the above inequality will still be true even if $V$ only admits a negative lower bound on its Ricci curvature.  Does anyone have a reference for a proof of this, or is the statement obvious? It just seems to me that the hypothesis compares the curvature of $V$ to that of $S^{n+1}$, so allowing $\mathrm{Ric}(V)$ to be negative will obscure this.

REPLY [2 votes]: See Cavaletti-Mondino's paper for the most general version that also works for metric measure spaces satisfying Riemannian weal lower Ricci bounds!<|endoftext|>
TITLE: Matrices generating non-discrete subgroups of SL(2,R)
QUESTION [5 upvotes]: Jorgensen's inequality $\mid \left(Tr\left(A\right)\right)^2-4\mid+\mid Tr\left[A,B\right]-2\mid\ge 1$ gives a necessary condition for two matrices A,B to generate a discrete subgroup of SL(2,R). Are there examples showing that this inequality is not a sufficient condition? Are there methods to prove that two given elements do NOT generate a discrete subgroup?
Background: A recently published paper http://xm10402.reportworld.co.kr/data/paper/view/2882/P2881838.html claims (in Theorem 4.5. in the appendix) that a group generated by two hyperbolic matrices A,B of positive trace is discrete if and only if their axes intersect and the trace of the commutator [A,B] is smaller than -2. This theorem seems a bit strong to me but it is not obvious to me how to construct counterexamples. In fact the condition on the trace of the commutator automatically guarantees that Jorgensen's inequality holds, so the first check for non-discreteness must fail.

REPLY [2 votes]: The answer is negative. If $G$ is a discrete subgroup of $SL(2, {\mathbb R})$ then  it cannot contain elliptic elements of infinite order. Thus, let $A$ be an elliptic of infinite order with fixed trace. In order for Jorgensen's inequality to be satisfied, we want $|tr([A,B])|$ be large. Note that $ABA^{-1}B^{-1}=AC$, where $C$ is also elliptic whose trace is the same as the one of $A$. Consider the character variety  $X$ of representations of $F_2$ to $SL(2,R)$, where we send free generators of $F_2$ to $A$ and $C$. Then $X$ is parameterized by traces of $A, C, AC$. Thus, if trace of $AC$ is bounded, then be get a bounded domain in $X$. However, if you take a sequence $B_n=g^n$, where $g$ is a fixed hyperbolic translation, then the distance between the fixed points of $A$ and $C_n= B_n g B^{-n}$ tends to infinity as $n\to \infty$. Therefore, the resulting sequence of elements of  $X$ diverges to infinity. Thus, we obtain a sequence of pairs $(A, B_n)$ so that $|tr([A,B_n])|$ diverges to infinity. Hence, for all but finitely many $n$'s Jorgensen's inequality holds but the group generated by $A, B$ is nondiscrete. 
On the other hand, the result that you mention does seem plausible in view of Goldman's thesis, see here and here. Goldman proves that representations of surface groups to $SL(2, {\mathbb R})$ with maximal Euler number are Fuchsian. (Namely, the representation you mentioned looks like a representation of the fundamental group of the 1-holed torus with maximal relative Euler number.)<|endoftext|>
TITLE: Actions of finite groups on exotic smooth manifolds of dimension >4
QUESTION [5 upvotes]: Let $M_1^n$ and $M_2^n$, $n>4$ be two smooth compact manifolds that are homeomorphic but not diffeomorphic. Suppose that a finite group is $G$ acting faithfully on $M_1^n$ by diffeomorphisms. Is it true that $G$ admits as well a faithful action on $M_2^n$ by diffeomorphisms?
If no, what would be the a (relatively) simple example?
For example, can one differentiate exotic structures on $S^7$ this way?

REPLY [2 votes]: I think the answer is no. But I don't know the reason. I take the fact that exotic spheres in dimension 7 do not admit orientation reversing diffeomorphisms (but the standard sphere does) as a clue to say that one shouldn't expect an affirmative answer to your question. 
EDIT: I gave a wrong answer and I'm not sure if I can delete the post. I decided to change my answer by a comment and make it CW. My apologies to those who read this expecting to see a right answer...<|endoftext|>
TITLE: Edge Colorings of Directed Graphs which Respect an Involution 
QUESTION [6 upvotes]: Let G be a graph and let C be a set of coloring. Suppose that there is an involution $\phi$ from C to C. We can think about the element of C as the nonzero elements of some Abelian group and $\phi(x)=-x$. (For concreteness we can consider especially the special case  that C corresponds to the cyclic group, so that $\phi$ has no fixed points when |C| is even and one fixed point when |C| is odd.) 
Consider a colorings where for every edge {u,v} in G we color the directed edge from u to v with some color c and the directed edge from v to u with -c. We will also require that two edges with the same tail and two edges with the same head must have different colors. If $\phi$ is the identity this is a usual edge coloring of $G$. 
My question is if such colorings were considered and what is known about them. For example, is there an analog of Vizing theorem?  (Vizing theorem asserts that the edges of every graph with maximum degree d can be colored by d+1 colors.)

REPLY [2 votes]: It seems to me that such colorings are closely related to 2-factorizations of graphs. Indeed every color class represents a subgraph which is a disjoint union of directed paths and cycles. Let's denote by $C(x)$ the set of edges colored $x\in G$. We can see that the underlying graphs of $C(x)$ and $C(-x)$ coincide, but they have opposite orientation. When $2x\neq 0$ then the underlying subgraph must have maximum degree 2. When $2x=0$ then $C(x)$ is a matching.
It's possible to show that a graph with $n$ vertices and maximum degree $\Delta$ can be written as a union of at most $\lfloor \frac{\Delta+1}{2}\rfloor$ subgraphs of maximum degree 2. Each of these subgraphs is a union of paths and cycles, so we can assign two colors to the corresponding directed edges. This shows that there is always a coloring satisfying the properties in the question coming from $\mathbb Z/(\Delta+1)\mathbb Z$. This would be the most straightforward analog of Vizing's theorem. Notice that $\Delta+1$ cannot be improved.<|endoftext|>
TITLE: Algorithm for Brauer lifting via Brauer tree?
QUESTION [6 upvotes]: Background: Given a finite group $G$ and a prime $p$ dividing its order, Brauer theory compares the ordinary characters of $G$ with the Brauer characters arising from $p$-modular representations.   On the character level there is a well-defined "reduction mod $p$", which Brauer showed to be surjective in the sense that each irreducible Brauer character lifts to a virtual character ($\mathbb{Z}$-linear combination of ordinary irreducible characters).   Part III of Serre's textbook works this out in the framework of Grothendieck groups.   The lifting is not unique, though J.A. Green's 1955 work here provides a simple character formula for one lifting.   (This is developed in a more sophisticated way by Lusztig, Ann. of Math. Studies No. 81, 1974.) 
An important subtheme, initiated about 70 years ago, concerns blocks (indecomposable 2-sided ideals of the modular group algebra) which have cyclic defect groups: here a defect group is a certain $p$-subgroup of $G$ determined up to conjugacy.   To such a block is associated a Brauer tree, a graph with edges labelled by irreducible Brauer characters and vertices labelled by one (or exceptionally several) irreducible ordinary characters.   Chapter VII of Feit's 1982 book has an extensive treatment, while his 1984 paper studies the possible Brauer trees using the classification of finite simple groups.    
For finite groups of Lie type with defining characteristic $p$ (simple or close to simple), the only family giving rise to Brauer trees consists of the groups $\mathrm{SL}(2,p)$ (say for $p>3$).   Leaving aside the Steinberg character (with trivial defect group), there are two blocks with defect group of order $p$.  Consider just the block containing the trivial ordinary (and modular) character, or the corresponding block of $\mathrm{PSL}(2,p)$.   Here the tree is an open polygon with $1_G$ at one end and two exceptional characters at the other end.  The edges correspond to even highest weights $0, p-3. 2, p-5, \dots$.  It's easy to specify Brauer liftings here by following vertices to the left or right with alternating signs.

For an arbitrary $G$ having a block with a nontrivial Brauer tree, is there a similar algorithm for Brauer lifting (in particular, an algorithm for Green's special lifting)?

I don't know all the literature well enough to sort this out, but the idea would be somewhat in the spirit of Green's "walk around the Brauer tree" (J. Austral. Math. Soc. 17, 1974) for producing a projective resolution (necessarily infinite, but periodic).  For this purpose the Brauer tree simultaneously encodes the projective covers needed, via Brauer reciprocity.

REPLY [5 votes]: I know this is a very old question, but for anyone else who finds this, here is a simple way to write an edge (=simple module) of the Brauer tree as an alternating sum of the vertices (=ordinary characters).
Pick your edge $e$. Removing this edge from the tree yields two connected components, so we choose the one that does not contain the exceptional node, for simplicity, say $X$. (Choose either if there is no exceptional node.)
We assign value +1 to the unique vertex $v$ in $X\cap e$, and value $(-1)^n$ to every vertex of $X$ of distance $n$ from $v$. The reduction modulo $p$ of this alternating sum yields the Brauer character labelling $e$.
To see this, note that the ordinary character of $v$ reduces to $e$ plus the other incident edges. These are subtracted off by removing, for each other edge incident to $v$ (hence in $X$) a copy of the other vertex incident to it (they have distance $1$), which gives us $e$ minus those edges in $X$ that are separated from $e$ by a single edge. The rest is an induction or standard inclusion-exclusion argument.
The only difference between this and fherzig's algorithm is I've used the fact that there are two different ways to write $e$ as an alternating sum of the vertices (one for each component of the tree once the edge is deleted) to avoid the exceptional node.<|endoftext|>
TITLE: When have we lost a body of mathematics because errors were found?
QUESTION [69 upvotes]: The history of mathematics over the last 200 years has many occasions when the fundamental assumptions of an area have been shown to be flawed, or even wrong. Yet I cannot think of any examples where, as the result the mathematics itself had to be thrown out. Old results might need a new assumption or two. Certainly the rewritten assumptions often allow wonderful new results, but have we actually lost anything?
Note I would like to rule out the case where an area has been rendered unimportant by the development of different techniques. In that case the results still hold, but are no longer as interesting.
I wrote up a longer version of this question with a look at a little of the history:
http://maxwelldemon.com/2012/05/09/have-we-ever-lost-mathematics/
Edit in response to comments
My thinking was about results that have been undermined from below. @J.J Green's example in the comments of Italian algebraic geometry seems like the best example I have seen. The trisection and individually wrong results do not seem to grow into areas, but certainly I would find interesting any example where a flawed result had built a small industry before it was found to be wrong. I am fascinated by mathematics that has been overlooked and rediscovered (ancient and modern) but that is perhaps a different question.

REPLY [2 votes]: I'm not sure that Drach's 1898 thesis on differential Galois theory "built a small industry", but it was certainly accepted and praised by his examiners before Vessiot pointed out a very serious flaw.  However, there was no public acknowledgement of this at the time (or later) by any of the parties involved.
It wasn't until the 1983 publication of Pommaret's Differential Galois Theory that the story came to light. In his 1988 book Lie Pseudogroups and Mechanics, Pommaret reproduced and translated into English the original examiners' reports, and the key correspondence describing the error.
For more context and details about Drach's work, see T. Archibald, "Differential equations and algebraic transcendents: French efforts at the creation of a Galois theory of differential equations 1880 - 1910", Revue d'histoire des mathématiques, 17 (2011) 373- 401.<|endoftext|>
TITLE: Is $\mathbb{Z}[2\cos(\frac{\pi}{k})]$ a Euclidean domain?
QUESTION [7 upvotes]: The ring $\mathbb{Z}[2\cos(\frac{\pi}{k})]$ is known to be a Euclidean domain for $k=3,4,5$ and $6$, because in those cases $2\cos(\frac{\pi}{k}) = 1, \sqrt{2},$ the golden ratio $\phi$, and $\sqrt{3}$ respectively, and $\mathbb{Z}, \mathbb{Z}[\sqrt{2}], \mathbb{Z}[\phi]$, and $\mathbb{Z}[\sqrt{3}]$ are all known to be Euclidean domains.
I haven't been able to find a reference that will tell me whether $\mathbb{Z}[2\cos(\frac{\pi}{k})]$ is (or isn't) a Euclidean domain for integers $k$ higher than $6$, though. Is this an open problem?

REPLY [9 votes]: There's certainly no reason to expect that it will always be a Euclidean domain, since (as Voloch points out) to be a Euclidean domain it must necessarily be a PID. On the other hand, the converse is (essentially) true in this case. Namely, a theorem of Weinberger implies that if $K/\mathbb{Q}$ is Galois, totally real, and has degree $\ge 4$, then $\mathcal{O}_K$ is a Euclidean domain if and only if it is a PID. So you are really asking about the invariant known as $h^{+}$, that is, the class numbers of totally real subfields of cyclotomic fields.  If
one supposes that $p$ is prime, then the smallest example which is provably not a PID is
$p = 163$, since there exists a totally real $A_4$ extension $L/\mathbb{Q}$ which is unramified over the degree three extension $L \cap \mathbb{Q}(\zeta_{163})$. (Indeed, assuming GRH, this is even the first example. There's a paper of Schoof where he talks about computing these numbers.)
The problem of proving that there are infinitely many such fields with class number one seems as  least as hard as proving the result for real quadratic fields of prime conductor, which is totally open.<|endoftext|>
TITLE: Request: Mazur' "Arithmetic in the geometry of symmetric spaces"
QUESTION [9 upvotes]: Does anyone have a copy of Mazur's unpublished notes:
Arithmetic in the geometry of symmetric spaces
that they are willing to post? Thank you!

REPLY [9 votes]: Dear modular1,
I sent the link to your question to Barry Mazur, and he has very kindly sent the paper to me.  I've got it scanned and sent it back to him, so if you ask him gently, he'll send the pdf file to you.  
I'm happy to have helped you in a small way, because there is a very kind soul who helps me in a big way.  His name is Dino Lorenzini, and he has sent me some papers which cannot be found anywhere in the closed disk of radius 3000 km centered at Allahabad.  Merci, Dino.
Best,
Chandan
Addendum.  With the permission of Barry Mazur, here is the typescript.  He'll also put it up on his website.<|endoftext|>
TITLE: the inverse for the trace theorem
QUESTION [7 upvotes]: The trace theorem says that the restriction of a  $W^{1,p}(\Omega)$ function $u$, $Tu$ belongs to $W^{1-1/p,p}(\partial\Omega)$ if $\Omega$ satisfies some smooth condition, for example, $\Omega$ is convex. 
Now my question is the inverse of the Trace Theorem. Suppose $\Omega$ is convex, and $\phi\in W^{1-1/p,p}(\partial\Omega)$, is there exists a fucntion $\Phi\in W^{1,p}(\Omega)$ with
$\|\Phi\|\leq C\|\phi\|?$
Dees the extension theorem is related to this question? But usually the extension Theorem talks about the extension from a domain to the whole space.

REPLY [5 votes]: I think your question is answered in 
Article (JonWal1978) Jonsson, A. & Wallin, H. A Whitney extension theorem in $L_p$ and Besov spaces Ann. Inst. Fourier (Grenoble), 1978, 28, vi, 139-192
and 
Article (Marsch1987) Marschall, J. The trace of Sobolev-Slobodeckij spaces on Lipschitz domains Manuscripta Math., 1987, 58, 47-65
Theorem 2 of the latter paper states that if $\Omega$ is a Lipschitz domain, with $s,p$ satisfying some inequalities (as usual), then $W^{s,p}(\Omega)$ traces to (as a surjection) some Besov space on the boundary $\partial\Omega$ and that the trace operator has bounded linear right inverse if $s-1/p$ is not an integer.<|endoftext|>
TITLE: Tamagawa Number of Elliptic Curves over $\mathbb{Q}$
QUESTION [6 upvotes]: I am currently reading a paper by De Weger and one theorem in it proves a bound for the Tamagawa number of any elliptic curve defined over $\mathbb{Q}$. 
I was wondering if anyone has any good references/texts that provide an exposition on the Tamagawa number of an elliptic curve as I was unable to find one in the Arithmetic of Elliptic Curves. 
I know the definition of the Tamagawa number from this reference, but not really much more than that (http://math.uci.edu/~asilverb/connectionstalk.pdf).
EDIT: I am looking for something a little more in depth than the intuition behind the definition. For instance useful recent applications of Tamagawa numbers or what is known about Tamagawa numbers for elliptic curves over $\mathbb{Q}$.

REPLY [4 votes]: I think I would start with Weil "Adeles and algebraic groups", but if you are looking for something more specifically associated to elliptic curves, maybe this survey of Guido Kings:
http://epub.uni-regensburg.de/13613/1/MP6.pdf is a good starting point to see the connection between the Equivariant Tamagawa Number conjecture and the BSD conjecture, and consider the references therein.
And here is anothor one of M.Flach: http://www.math.caltech.edu/papers/baltimore-final.pdf<|endoftext|>
TITLE: Lucky chance or combinatorial cause?
QUESTION [12 upvotes]: Consider an $n \times 1 - $rectangle where the $n$ squares are numbered $1$ to $n$. Cover this rectangle with white squares, black squares,  and dominoes. To each covering of the rectangle  associate the following weight: Each white square has weight 1, each black square at position $i$  and each domino at position ${(i,i + 1)}$ have weight $q^i.$ As usual the weight of a covering is the product of its components and the weight of a set of coverings is the sum of their weights. 
Then it is easy to verify that the weight $u(n,k)$ of all coverings with precisely $k$ dominoes is the product  $u(n,k)=a(n,k)b(n,k)$ with  $a(n,k)= {q^{k^2}}{n-k\brack k} $ and $b(n,k) = (1 + {q^{k + 1}})(1 + {q^{k + 2}}) \cdots (1 + {q^{n - k}}).$
Here ${n\brack k}=[1][2]\dots[n]/(([1]\dots[k])\([1]\dots[n-k]))$
with $[n]=(1-q^n)/(1-q)$ denotes a $q-$binomial coefficient.
It is often claimed that there are no accidents in mathematics. Therefore my question is: 
Is there a simple combinatorial reason for the fact that $u(n,k)$ is the product of two terms with simple combinatorial interpretations or is it an accident after all?
Since this seems to be rather elementary I have posted this question in https://math.stackexchange.com/questions/142158/lucky-chance-or-combinatorial-cause  but did not get an answer.
Edit
Some days ago Ilse Fischer has shown me a simple bijection. Associate with a tiling of an $n - $board with $k$ dominoes , $\ell $ black squares and $n - 2k - \ell $ white squares the word  ${c_1}{c_2} \cdots {c_{n - k}}$ in the letters $w,b,d,$  where $d$ occurs $k$ times, $b$ occurs $\ell $ times and $w$ occurs  $n - 2k - \ell $ times. Let $W({c_1}{c_2} \cdots {c_{n - k}})$ be the weight of the tiling. 
First reverse in ${c_1}{c_2} \cdots {c_{n - k}}$ the order of the letters $b,d$ and obtain a word ${C_1}{C_2} \cdots {C_{n - k}}.$ 
Let e.g. $(n,k,\ell ) = (12,3,2)$ and ${c_1}{c_2} \cdots {c_9} = wbdwwdbwd.$ Then ${C_1}{C_2} \cdots {C_9} = wdbwwddwb.$ 
Then replace in ${C_1}{C_2} \cdots {C_{n - k}}$ all  $b$ by $w.$ This gives a word $A$ with $k$ letters $d$ and $n - 2k$ letters $w.$ In our example we get $A = wdwwwddww.$ 
Then delete in ${C_1}{C_2} \cdots {C_{n - k}}$ all letters $d$ and get a word $B$ with $n - 2k$ letters $w,b.$ In our example $B = wbwwwb.$
Then $W({c_1}{c_2} \cdots {c_{n - k}}) = {q^{k\ell }}W(A)W(B).$
In our example we have $W({c_1}{c_2} \cdots {c_9}) = W(wbdwwdbwd) = {q^{2 + 3 + 7 + 9 + 11}} = {q^{32}},$ $W(A) = W(wdwwwddww) = {q^{2 + 7 + 9}} = {q^{18}},$ $W(B) = W(wbwwwb) = {q^{2 + 6}} = {q^8}.$
If $u(n,k,\ell )$ denotes the weighted enumeration of all tilings this implies
$u(n,k,\ell ) = {q^{k\ell }}u(n,k,0)u(n - 2k,0,\ell ).$

REPLY [5 votes]: Yes, there is a simple reason why this happens, but instead of giving the "magic" bijection, I will describe how to construct it. I will need to express this in the language of partitions, where I'm more familiar, although it shouldn't be hard to switch back to the dominoes interpretation. Below, I assume the weight of a partition of $n$ is $q^n$, and the weight of a pair of partitions is the product of their weights.
Proposition A: There is a weight preserving bijection between tilings of an $n\times 1$ rectangle with $k$ dominoes and $a$ black squares and pairs of partitions $(P,Q)$ where $P$ has $k+a$ parts each of size $\le n-2k-a$, and $Q$ has $k+a$ parts $q_1\geq q_2\geq \cdots \geq q_{k+a}=1$ and $q_{i}-q_{i+1}\in \lbrace 1,2\rbrace$, where these differences are $=2$ exactly $k$ or $k-1$ times.
Proof: Let $p_{k+a}$ denote the number of white squares before the first black square or domino, $p_{k+a-1}$ denote the number of white squares before the second black square or domino etc. We will let $P$ be the partition $p_1,p_{2},\dots$. We let $q_{k+a-i}-q_{k+a-i+1}=2$ if the $i$'th black square or domino is a domiono, and $q_{k+a-i}-q_{k+a-i+1}=1$ otherwise. It should be clear that this is a bijection. Proving that it is weight preserving is also easy and is left as an exercise.
Proposition B: The partitions $Q$ from the previous theorem are in a weight preserving bijection (up to a power of $q$ which depends only on $k$ and $a$) with partitions which have $k$ parts which are $\le a$.
Proof: Take the partition $Q$ and subtract the partition $(k+a,k+a-1,\dots,1)$, so we obtain $Q'=(q_1-k-a,q_2-k-a+1,\dots,q_{k+a}-1)$. Now let $\hat{Q}$ be the transpose partition of $Q'$. Check that $Q'$ has $k$ distinct parts which are $\in [0,k+a-1]$. Therefore we make a fourth partition $R=\hat{Q}-(k-1,\dots,1)$. Check that $R$ has $k$ parts which are $\le a$.
Now let's denote by $B(x,y)$ the set of partitions contained in an $x\times y$ rectangle (so partitions that have at most $x$ parts and each part is at most $y$). It is well known that the generating function for these is ${x+y\brack x}$.
Proposition C There is a weight preserving bijection between pairs of partitions $B(k+a,n-2k-a)\times B(k,a)$ and pairs of partitions $B(n-2k,k)\times B(n-2k-a,a)$.
Proof: Look up your favourite bijective proof of
$${n-k \brack k+a}{k+a\brack a}={n-k\brack k}{n-2k\brack a}.$$
(Note that this is the part which messes things up a bit, so that the resulting final bijection won't be "obvious".)
So propositions A,B give a bijection between tilings with $k$ dominos and $a$ black squares and $B(k+a,n-2k-a)\times B(k,a)$. Applying proposition C, we get a bijection with $B(n-2k,k)\times B(n-2k-a,a)$. Finally since $B(n-2k,k)$ is in a weight preserving bijection (up to an inconsequential power of $q$) with domino tilings (no black squares) and $B(n-2k-a,a)$ is up to a power of $q$ the coefficient of $z^a$ in $(1+zq^{k+1})\cdots(1+zq^{n-k})$, we get the desired result. $\square$<|endoftext|>
TITLE: On elimination of Noetherian conditions in the geometry of schemes
QUESTION [7 upvotes]: Grothendieck seemed to try to eliminate Noetherian conditions as much as possible in EGA.
For example, he developed the cohomology theory of schemes without Noetherian conditions.
On the other hand, the Hartshorne's book assumes Noetherian conditions to do it.
If one's main interest is in the geometry of varieties or schemes of finite type over Dedekind domains, does one need non-Noetherian theory like Grothendieck's?

REPLY [19 votes]: For the most part, the answer is "no"; most people who work with varieties will never need to worry about non-Noetherian rings.  But there are reasons to be open to the non-Noetherian setting.  First, they can actually come up (as pointed out in the comments).  As just one example, the normalization of a Noetherian ring can be non-Noetherian.  You can either work hard to show that the rings you care about don't have this pathology, or you can just relax and not worry about it.  Second, there is often no harm in working with schemes in general.  (One caution with Hartshorne:  often the Noetherian conditions are unnecessary, and not used in the proofs; and in some cases, the arguments are more complicated because of the desire to work in the Noetherian setting.)
My suggestion is this:  don't worry about reading how to remove Noetherian conditions until you have a need to remove them.  But don't gratuitously add Noetherian conditions for no good reason.  (Note also:  sometimes Noetherian conditions can be removed with little pain, see for example my answer Flatness for family of hypersurfaces.)<|endoftext|>
TITLE: Origin of the sign convention in the Tensor product of graded vector spaces
QUESTION [7 upvotes]: Suppose $V := \bigoplus_{i \in \mathbb{N}}V_i$ and $W := \bigoplus_{i \in \mathbb{N}}W_i$ are $\mathbb{N}$-graded vector spaces. Then their graded tensor power is defined by 
$V \bigotimes W := \oplus_{n \in \mathbb{N}} \oplus_{i+j=n} V_i \otimes W_j$ and there is a natural isomorphism $\sigma_{V,W}: V \bigotimes W \to W \bigotimes V$ defined on homogeneous elements
by the 'sign convention'
$$\sigma_{V,W}(v \otimes w) = (-1)^{\deg(v)*\deg(w)}w \otimes v$$ 
What is the origin of this natural isomorphism? Is this the only one or are there others defined by other sign conventions?

REPLY [7 votes]: Is this the only one or are there others defined by other sign conventions?
As has already been mentioned - there are essentially two but the Koszul one is amongst other things, the only one which makes the tensor product of complexes a complex.
Since I have been interested in this topic of late, I would point you towards the following internal (within MathOverflow) references:

For a recent discussion see:
Tensor product of linear mappings versus chain complexes
In particular Simon Letner and Qiaochu Yauns' answers provide a nice conceptual backdrop.
I also found Theo Johnson-Freyd's answer here quite illuminating:
References for sign conventions in homological algebra
Theo Johnson-Freyd's answer in this thread is also instructive:
Sign convention for derivations in CDGAs

REPLY [3 votes]: If you want to find another $\sigma'_{V,W}\colon V\otimes W\to W\otimes V$ so that $\sigma'_{U,V\otimes W}=\sigma'_{U,V}\sigma'_{U,W}$ and assume that on homogeneous elements $\sigma'_{V,W}(v\otimes w)=f(\deg(v),\deg(w))w\otimes v$ for some elements $f(\cdot,\cdot)\colon\mathbb{N}\times\mathbb{N}\to F$, then it is easy to deduce that $f(a,b)=f(1,1)^{ab}$. This, together with the requirement that $\sigma'_{V,W}\sigma'_{W,V}=Id_{V\otimes W}$, instantly shows that $f(1,1)=\pm1$. But of course there were many assumptions made along the way. 
As for the origin, I think that in addition to what Mark Grant says, when you have the product on the cohomology $H^*(M)$ (over a field), you then can apply the Künneth formula and say $H^*(M\times N)=H^*(M)\otimes H^*(N)$. In what sense these two are isomorphic as algebras? To define aa product on the tensor product $H^*(M)\otimes H^*(N)$, you need those isomorphisms $\sigma$, so that you can do $H^*(M)\otimes H^*(N)\otimes H^*(M)\otimes H^*(N)\to H^*(M)\otimes H^*(M)\otimes H^*(N)\otimes H^*(N)$ and then compute the product in $H^*(M)$ and in $H^*(N)$. The Koszul convention is precisely the choice for which you obtain an algebra isomorphism.<|endoftext|>
TITLE: Equidecomposable graphs, unimodality and asymptotics
QUESTION [6 upvotes]: I will call two graphs $G$ and $H$, $r$-equidecomposable (in analogy with Hilbert's third problem) if they can be written as unions of disjoint subgraphs
$$G\cong \bigsqcup_{i=1}^r G_i\quad  ,\quad  H\cong \bigsqcup_{i=1}^r H_i$$
(disjoint here means they have no common edges), and $G_i\cong H_i $ for $1\le i \le r$. Let $\epsilon(G,H)$ be the smallest $r$ for which $G$ and $H$ are $r$-equidecomposable. Let
$$f(n,m)=\max_{G,H\in V(n,m) } \epsilon(G,H) $$
where $V(n,m)$ is the collection of all graphs with $n$ vertices and $m$ edges.
Is the sequence $f(n,m)$ unimodal for fixed $n$? Can one find asymptotics for values such as $f\left(n,\frac{n(n-1)}{4}\right)$ when $n$ is large?

REPLY [2 votes]: If Joseph is going to take this as an opportunity to show some nice and illustrative pictures, I am going to do similarly, but using words instead.  Gjergji probably already knows what I am going to say, but others might find the remarks a useful stepping stone to the subject.
My initial thought was using a star graph, which on m nodes has a degree sequence of (m-1) 1's and 1 entry of (m-1).  Embedding this into n=2m nodes and choosing H to be a graph of m disconnected edges gives that
n/2 is a lower bound for f(n,n/2) and n even, and an analogous bound for n odd.  Gjergji hints at a better construction in the comments which asymptotically gives a lower bound of 2n/3 .
It so happens that every graph on n nodes is r-decomposable into star graphs for some r less than n.  A nice argument on degree sequences allows one to remove a star graph from
each of G and H leaving at least one vertex in one of the graphs with no edges, and then
a second star graph can be removed from each to guarantee that a vertex in the other
graph has no edges.  This allows us to look at the situation on n-1 nodes and gives
an upper bound of 2n (which can be tightened to 2n-6 for n>3) for f(n,m) for any m.
If there is a relation between decomposability for a pair of graphs and decomposability for their complements, I have not found it yet.  However having min (m, floor(2n/3)) and
2n as upper and lower bounds on parts of f suggests to me that f is unimodal, that
there are nice asymptotics and that for m between n and ((n choose 2) - n), the
asymptotic value for f(n,m) will be Cn for some constant C less than 2, most likely C=1.
Using star graphs for the pieces provides nice results.  It might be good to
consider decomposition into certain graph classes like paths or trees and see what
asymptotics can be found using such restrictions.
Gerhard "Ask Me About System Design" Paseman, 2012.05.11<|endoftext|>
TITLE: Is it true that all real projective space $RP^n$ can not be smoothly embedded in $R^{n+1}$ for n >1
QUESTION [15 upvotes]: So first for n even, $RP^n$ is not orientable, hence can not be embedded in $\mathbb{R}^{n+1}$.
For odd n, $RP^{n}$ is orientable, hence the normal bundle is trivial. Now using stiefel-Whitney classes, one can prove when $n$ not of the form $2^k - 1$, it can not be embedded in $\mathbb{R}^{n+1}$. I would appreciate if someone can give an more elementary proof in this case.
Then for the left cases. $RP^3$ can not be embedded proving by homology thoery (Alexander sphere duality, lefschetz duality and a long exact sequence).
For the other cases, I do not know how to prove. I realized Don Davis has a table for the immersion and embedding of $RP^n$ (http://www.lehigh.edu/~dmd1/immtable). But the question I am asking is easier, hence there may be an answer and a proof I could follow.

REPLY [28 votes]: I'll use cohomology with coefficients $\mathbb{Z}/2$ everywhere.
Suppose that the space $P=\mathbb{R}P^{n-1}$ embeds in $S^{n}$ (where $n>2$).  Recall that 
$$ H^*(P)=(\mathbb{Z}/2)[x]/x^{n} = (\mathbb{Z}/2)\{1,x,\dotsc,x^{n-1}\} $$ 
By examining the top end of the long exact sequence of the pair $(S^{n},P)$ we find that $H^{n}(S^{n},P)$ has rank two.   Lefschetz duality says that this group is isomorphic to $H_0(S^{n}\setminus P)$, so we see that $S^{n}\setminus P$ has two connected components.  (I don't need any orientation conditions here as I am working mod 2.) Let $A$ and $B$ be the closures of these components, so $A\cap B=P$ and $A\cup B=S^{2n}$.  Lefschetz duality also gives $H^{n}(A)\times H^{n}(B)=H^{n}(S^{n}\setminus P)=H_0(S^{n},P)=0$.
We now have a Mayer-Vietoris sequence relating the cohomology groups of $A$, $B$, $P$ and $S^{n}$.  As $H^1(S^{n})=H^2(S^{n})=0$ this gives an isomorphism $H^1(A)\times H^1(B)\to H^1(P)=\{0,x\}$.  After exchanging $A$ and $B$ if necessary, we can assume that $H^1(B)=0$ and that there is an element $a\in H^1(A)$ that maps to $x$ in $H^1(P)$.  It follows that $a^{n-1}$ maps to $x^{n-1}$, which generates $H^{n-1}(P)$, so the Mayer-Vietoris connecting map $H^{n-1}(P)\to H^{n}(S^{n})=\mathbb{Z}/2$ must be zero.  This contradicts exactness at the next stage, because $H^{n}(A)\times H^{n}(B)=0$.<|endoftext|>
TITLE: Ternary "Lie structure"
QUESTION [15 upvotes]: One of the motivation of the theory of Lie Algebras is that every associative algebra $A$ is a LA when the bracket is defined by $[a,b]=ab-ba$ : this is skew-symmetric and satisfies the Jacobi identity $[[a,b],c]+[[b,c],a]+[[c,a],b]=0$. Conversely, every abstract LA can be embedded into an associative algebra (its envelopping algebra). And for some good reason, one is really interested in sub-LAs rather than sub-algebras. A similar attitude, with different motivation lead to the notion of Jordan algebras.

If $A$ is an associative algebra, one may consider instead the ternary bracket $$[a,b,c]_3=abc+bca+cab-acb-cba-bac.$$ 
  Does $[.,.,.]_3$ satisfy non-trivial identities, besides skew-symmetry? Is there any theory of abstract objects, vector spaces endowed which a ternary skew-symmetric product satisfying these identities?

More generally, we may consider a $d$-bracket, which bears the name of standard non-commutative polynomial in $d$ non-commuting variables. For $d=2$, it is nothing but the standard bracket. When $d=2p$, the $d$-bracket does satisfy non-trivial identies, for instance
$$\sum_{i\in\frak A_7}[[a_{i_1},a_{i_2},a_{i_3},a_{i_4}],a_{i_5},a_{i_6},a_{i_7}]=0,\qquad\forall a_1,\ldots,a_7\in A.$$
I don't know if something non-trivial exists when $d$ is odd.

REPLY [3 votes]: You might want to look at the paper "On Lie k-Algebras" by P. Hanlon by M. Wachs (http://www.sciencedirect.com/science/article/pii/S0001870885710389). They consider algebras satisfying the generalized Jacobi identity you specify. I wanted to leave this as a comment but I don't have enough reputation.<|endoftext|>
TITLE: Do Baumslag-Solitar Group von Neumann algebras have Property $\Gamma$?
QUESTION [13 upvotes]: A type $II_{1}$ factor $M$ with trace $\tau$ has Property $\Gamma$ if for every finite subset $\{ x_{1}, x_{2},..., x_{n} \} \subseteq M$ and each $\epsilon >0$, there is a unitary element $u$ in $M$ with $\tau (u)=0$ and $||ux_{j}-x_{j}u||_{2}<\epsilon$ for all $1 \leq j \leq n$. (Here $||T||_2=(\tau(T^{*}T))^{1/2}$ for $T\in M$.) 
A countable discrete group $G$ is inner amenable if there is a finitely additive measure $m$ on the subsets of $G \backslash${$e$} with total mass 1 and satisfying $m(gXg^{-1})=mX$ for all $X\subseteq G \backslash${$e$} and all $g \in G.$
I should mention that if the left group von Neumann algebra of an i.c.c. group has property $\Gamma$ then the group is inner amenable, however there exist i.c.c. inner amenable groups whose group von Neumann algebras don't have $\Gamma$, as recently shown by Stefaan Vaes.

Given a non-residually finite
  Baumslag-Solitar group $$BS(m,n)
 = \langle b,s\mid s^{-1}b^ms = b^n\rangle$$ does its group von
  Neumann algebra have property
  $\Gamma$?

It is known that all such groups are inner amenable, and it recently has been shown that the associated group factors have no Cartan subalgebra, are prime and yet are not solid.

REPLY [11 votes]: Yes, it has property $(\Gamma)$. This follows from Stalder's proof of inner amenability plus a fact that the semigroup $\langle T_m, T_n \rangle$ admits an approximately invariant subsets having proportional measures. Here, $T_m$ is the $m$-times map on $[0,1)$, $T_m x = mx \mod 1$. As Stalder proves, $\sum_{1\le i,j \le n} z^{m^i n^j}$ is approximately invariant under $T_m$ and $T_n$. By a standard procedure, it gives rise a subset $E \subset [0,1)$ which is approximately invariant under $T_m$ and $T_n$, i.e., $| T_l^{-1}(E) \bigtriangleup E | < \epsilon |E|$ for $l=m,n$. Then, Abert--Nikolov's argument (see Chifan--Ioana's paper
arXiv:0802.2353 Lemma 10) allows one to widen $E$.<|endoftext|>
TITLE: Exploding primes
QUESTION [48 upvotes]: Suppose every prime $n$ could "explode" once.
An explosion results in $\lfloor \alpha \ln n \rfloor$ particles being
uniformly distributed over the integers in a range $n \pm \lfloor \beta \ln n \rfloor$.
If a particle hits a composite or a previously exploded prime, nothing happens.
If a particle hits a new prime $p$, then $p$ explodes under the same rules.
Here is a little simulation starting at $n=23$,
with $\alpha=10$ and $\beta=5$:

          


So, on the first step, $n=23$ explodes into $\lfloor 10 \ln 23 \rfloor = 31$
particles, uniformly spreading over $23 \pm 15$. In the run depicted, these particles hit four primes:$11, 13, 29, 31$.  Then each of those explodes; and so on. In the last frame, $199$ is hit.

Q1. For which $\alpha$ and $\beta$, if any, will this process almost
  surely explode an infinite number of primes?

I am hoping the answer is independent of the starting $n$.
Being largely ignorant of number theory, I am wondering if current
knowledge of the distribution of the primes is sufficient to
answer this question.
Thanks for insights or even speculations!
Q1 Answered. quid shows that a ballistic range of $\pm \beta \log n$ is not enough
to bridge the prime gaps:
$\pm \beta \log^2 n$ conjecturally suffices, but a range of more than $\pm \beta \sqrt{n}$
is all that current technology can prove.
Addendum. Both quid and joro confirm that even assuming RH
(the Riemann Hypothesis),
the range needed for my simulation to provably explode all primes is at least $\pm \beta \sqrt n \log n$.  May I add this question: 

Q2. What is the minimum ballistic $\pm$-range that would suffice to provably
  (under "current technology") explode (with sufficent $\alpha$) all primes
  without any RH or otherwise conjectural assumptions?

REPLY [37 votes]: This process will never explode an infinite number of primes (independent of $n$, $\alpha$, $\beta$). 
Let $g_n$ denote the gap between the $n+1$-th and the $n$-th prime. 
It is known that $g_n/\log n$ is unbounded; indeed by a result of Rankin there is an infinitude such that this is larger than $$\frac{c \log \log n \log \log \log \log n }{ \log \log \log n}$$ for an explicty $c$; see the wikipedia page on prime gaps for additional details.
Yet the largest number one can reach from the $n$-th prime (or any prime before) with the given rules is 
$p_n + \beta \log p_n$. Since $p_n$ is asymptotically $n \log n$ its logarithm is of order $\log n$. So the largest number one can reach is bounded above by $p_n + \gamma \log n$ for some $\gamma$ independent of $n$. 
Yet, by the above mentioned result on prime gaps there are primes for which the next largest prime will be graeter than $p_n + \delta \log n$ for any $\delta$. Since one can never bridge a gap with a $\delta$ greater than the $\gamma$; the process can at most explode all the primes before such a gap, appearing after the starting value. That is, at most finitely many. 
Addition for added Q2: First a partial summary of comments, in view of the above a minimal requirement for such a process to have the potential to go on indefinitely is that the range will exceed the maximal (relative) size of prime gaps. While conjecturally their size should (only) be of order $(\log x)^2$ (for numbers of size $x$), the best known upper bounds are much weaker, namely asymptotically $x^{6/11 + \epsilon}$ for every positive $\epsilon$. 
Now, for the actual Q2, I am not completely sure what is asked regarding 'probabilty', and I am not well placed to comment on this 'probabilty' aspect either. If the range is large enough to bridge all the gaps and the number of newly created explosions is not too small (but even more than one seems sufficient) it is possible that some instance of the process explodes all primes; always explode just the next largest prime (and also the next smallest if one started in the middle) but nothing else. Yet no matter the choices at each step there is a positive probabilty that the process stops at that step. Yet, perhaps, the question when there is a positive porobability that everything explodes is intended. As said I am not well placed to say something on this.<|endoftext|>
TITLE: Neat results from algebraic graph theory
QUESTION [5 upvotes]: Studying algebraic graph theory, I've stumbled across a wide range of results that I found pretty stunning and also useful. I'd like to share them here and ask for your favorite result from this area?
Let $G$  be a simple graph  $A$ and $L$ its adjacency and Laplacian matrix $\lambda_1\leq \cdots \leq\lambda_n$ and $\mu_1 \leq \cdots \leq \mu_n$ the respective eigenvalues of $A$ and $L$.

(Wilf, Hoffman) For a nontrivial graph $G$  $$1+\frac{\lambda_n}{-\lambda_1} \leq \chi(G) \leq 1+\lambda_n$$
(Sachs,Harary) Let $G$ be a graph of odd girth $2r+1.$ Let $p(x) = \sum_{i=0}^n c_{n-i} x^i$ be the characteristic polynomial of $A.$ Then $$c_3 = \cdots = c_{2r-1} = 0$$ and $\frac{-c_{2r+1}}{2}$ is the number of $(2r+1)$-cycles in G.
(Folklore) The number of triangles in $G$ equals $tr(A^3)/6.$ Since matrix multiplication can be done in $O(n^k)$ for $k \leq 2.37$ this presents an improvement over the straightforward approach for counting triangles in graphs that is currently the fastest way to compute the number of triangles in simple graphs.
(Kirchhoff) The number of spanning trees in G is $\frac{1}{n}\mu_2 \cdots \mu_n$
(McKay) $diam(G) \geq  \frac{4}{n\mu_2}$

Do you have any neat results like this to share?

REPLY [6 votes]: I have incorporated some results I like in algebraic graph theory into my notes http://math.mit.edu/~rstan/algcomb.pdf. See for instance Example 9.12 (number of spanning trees of the $n$-cube), Corollary 10.11 (enumeration of binary de Bruijn sequences), Section 11.5 (squaring the square), Section 12.3 (complete bipartite partitions of $K_n$), and Section 12.5 (odd neighborhood covers). Many other nice examples are in Matoušek's Thirty-three Miniatures.<|endoftext|>
TITLE: Determine if you can build a polygon from segments
QUESTION [6 upvotes]: Is there a way to determine whether it is possible to build a polygon from given n segments?
Maybe triangle inequality generalized?

REPLY [12 votes]: The answer is: Given a collection of positive numbers $r_1,...,r_n$, there exists a polygon in $R^2$ with the side-lengths $r_1,...,r_n$ if and only if for every $i$, 
$$
r_i\le \frac{1}{2}(r_1+...+r_n)
$$ 
which is just a form of triangle inequalities (equivalently, every side is at most the sum of the rest of the sides). If you do not accept degenerate polygons as legitimate polygons then in the above answer you replace non-strict inequalities with the strict ones. 
There are different ways to prove this, one is to use elementary Euclidean geometry, I will explain a better solution, which uses hyperbolic geometry, since it teaches you something interesting. Suppose that $r_1,...,r_n$ satisfy the strict triangle inequalities above. Now, think of the numbers $r_i$ was masses and place them at distinct points $x_i$ on the unit circle $S^1$. The result is a finite measure $\mu$ on $S^1$. The triangle inequalities ensure that this measure is stable, therefore, the measure $\mu$ has (unique) conformal barycenter $c(\mu)$ in the open unit disk $D^2$, which you regard as a model of the hyperbolic plane. Since conformal barycenter is preserved by hyperbolic isometries, there exits a hyperbolic isometry $g$ of $D^2$, so that $g_*(\mu)$ has conformal center at the center $0$ of the disk. Let $v_i:=g(x_i)$, $i=1,...,n$. By the properties of the conformal barycenter, $c(\mu)=0$ if and only if the Euclidean barycenter of the measure $\mu$ is also at zero. Thus, 
$$
\sum_{i=1}^n r_i v_i=0. 
$$ 
This means that you get a closed polygon $P$ in the Euclidean plane whose edges are represented by the vectors $r_i v_i, i=1,...,n$. The  polygon $P$ you get need not be embedded. However, you can triangulate $P$ from a single vertex and then inductively straighten it to get an embedded polygon if you wish. Alternatively, if you choose points $x_i$ on the circle in their natural cyclic order $1,...,n$, then the resulting polygon $P$ will be convex.  
See here for the details and 
here for generalizations. 
This argument above has interesting generalizations, with hyperbolic plane replaced by the symmetric space of the group $GL(n)$ (and other symmetric spaces and buildings), which allow one to solve the problem about eigenvalues of sums of symmetric matrices and other problems in algebra.<|endoftext|>
TITLE: Shimura-Taniyama-Weil VS Grothendieck's dessins
QUESTION [33 upvotes]: When listening to the beautiful lectures by Gilles Schaeffer at
the SLC68, the following (perhaps crazy) question occurred to me:
did anyone attempt (succeed?) to combinatorially prove modularity of elliptic curves using dessins d'enfant?
Of course I am not talking about a combinatorial proof of the general result due to Wiles, Taylor, Breuil, Conrad and Diamond. If such a thing existed, everyone and their dog would have heard about it. I am interested in learning about combinatorial proofs, if any, even for very modest examples. As I do not know anything about the subject,
references to the relevant literature would be appreciated.
This question can be broken down into the following three:
1) Can one tell `by looking at a dessin' if the corresponding curve is defined over $\mathbb{Q}$? If this is too hard, can one construct an explicit collection of dessins which catches all elliptic curves defined over $\mathbb{Q}$?
2) Does one know explicit dessins for all modular curves?
3) Let $X\rightarrow\mathbb{P}^1$ and  $Y\rightarrow\mathbb{P}^1$ be two coverings given by dessins. Is there some sufficient criteria for the existence of a cover $X\rightarrow Y$?

Crazy addendum to a crazy question:
Can one `count' $H_{X,Y}$ the number of covers in question 3)? Again, I am talking about examples.

REPLY [5 votes]: This does not answer your question. But it was a bit too long to put as a comment.
Firstly, it seems that the following old question is of some relevance.
Families of curves for which the Belyi degree can be easily bounded
In fact, dessins $X\to \mathbf{P}^1$ are also called Belyi maps/morphisms/functions on $X$.  I wanted to know of curves for which one has explicit bounds on the Belyi degree, i.e., the minimal degree of a dessin $X\to \mathbf{P}^1$. Here are the examples 

Fermat curves
Modular curves (congruence or non-congruence)
Hurwitz spaces (see JSE's answer to the above question)
Galois Belyi curves = Wolfart-curves = Galois three-point covers
Elkies' curves (see his answer to the above question).

Let me elaborate on 2. If $\Gamma\subset \mathrm{SL}_2(\mathbf{Z})$  is a finite index subgroup, you can consider the quotient $Y_\Gamma = \Gamma\backslash \mathbf{H}$, where $\mathbf{H}$ is the complex upper half-plane and $\mathrm{SL}_2(\mathbf{Z})$ acts on $\mathbf{H}$ by Mobius transformations. The curve $Y_\Gamma$ naturally inherits the structure of a connected Riemann surface from $\mathbf{H}$. We compactify $Y_\Gamma$ by adding "cusps". The compactification of $Y_\Gamma$ is usually denoted by $X_\Gamma$. Note that there is a natural map $Y_\Gamma \to Y_{\mathrm{SL}_2(\mathbf{Z})} = Y(1)$ induced by the inclusion $\Gamma\subset \mathrm{SL}_2(\mathbf{Z})$.  This morphism extends to the compactifications $X_\Gamma \to X(1)$ and induces a dessin $X_\Gamma \to \mathbf{P}^1(\mathbf{C})$ after you compose with the isomorphism given by the $j$-invariant $j:X(1)\to\mathbf{P}^1(\mathbf{C}$. (The branch points are the elliptic points $0$, $1728$ and the cusp $\infty$ of $X(1)$.)
Let me adress your third question. The above is about your second question. I don't have much to say about your first question, unfortunately. What do you mean by a dessin which "captures" all elliptic curves over $\mathbf{Q}$? 
Firstly, assume that $X\to \mathbf{P}^1$ is a dessin of prime degree. It's clear that this morphism will not factor.
I get the feeling (but I might be wrong) that  you are interested in modular parametrizations of elliptic curves in the following sense. You want to know whether the above explicit dessins on $X_0(n)$ can be shown to factor through some elliptic curve. If this is the case, the answer is likely to be no for $n$ big.
Now, you can bound  the number of dessins on a curve $X$ of given degree $d$ by the number of dessins of degree $d$, i.e., the number of topological covers of $\mathbf{P}^1-\{0,1,\infty\}$.
But your $H_{X,Y}$ will be zero or infinite. 
In fact, if it not zero then there exists a dessin $X\to \mathbf{P}^1$ which factors through a dessin $Y\to \mathbf{P}^1$. But Belyi proved that for any finite set $B\subset \mathbf{P}^1(\overline{\mathbf{Q}})$ there exists a dessin $R:\mathbf{P}^1_{\mathbf{Q}}\to\mathbf{P}^1_{\mathbf{Q}}$ (defined over $\mathbf{Q}$ even!) such that $R$ sends $B$ to the set $\{0,1,\infty\}$. So from a given factorization $X\to Y\to \mathbf{P}^1$ you can construct an infinite number of really different dessins (and associated factorizations).
The former paragraph is just applying the fact that given a dessin  $f:X\to \mathbf{P}^1$ you can construct an infinite number of dessins $g :X\to \mathbf{P}^1$ by composing $f$ with an arbitrary dessin on $\mathbf{P}^1$. (Belyi actually gave an algorithm to compute a dessin $R$ on $\mathbf{P}^1$ associated to $B$ as above.)
So to make sense of your last "crazy" question, you might want to fix a dessin $X\to \mathbf{P}^1$ on $X$  and try to look at possible factorizations, where $Y\to \mathbf{P}^1$ is a dessin and $Y$ is not fixed. Thus, let $H_{\pi}$ be the number of pairs $(Y,f)$ up to isomorphism, where $f:Y\to \mathbf{P}^1$ is a dessin and there exists a factorization $g:X\to Y$ such that  $\pi = fg$.
I don't think it is possible to give a precise formula for $H_\pi$ easily, but it is certainly possible to bound this number in terms of  the degree of your dessin.<|endoftext|>
TITLE: What is / are the softwares to use to draw surfaces of the form of a two or three-holed torus , or torus, or torus with cusps attached to it?
QUESTION [7 upvotes]: I am trying to draw surfaces with complete hyperbolic structures and surfaces which are topologically tori. The hyperbolic surfaces I need to draw are torus with one or two holes on it, or torus with punctures on it, or torus with a cusp attached to it . They could also be torus torus with one or more than one handles attached to it.
For example, see the diagrams on : http://www.maths.bris.ac.uk/~mazag/hyperbolic/index.html
Or see the diagrams on : http://lamington.wordpress.com/2010/04/18/hyperbolic-geometry-notes-4-fenchel-nielsen-coordinates/
to get ideas about what surfaces I am talking about. They are not given by any easy equations.
Is there a software I can use to draw them ? People who study Riemann Surfaces or Hyperbolic Geometry or Teichmmuller Theory would definitely know exactly what surfaces I am talking about.Please let me know if you use such a software. Thanks a lot in advance !!

REPLY [4 votes]: If you go to the Algebraic Surface page of The Virtual Math Museum  at
http://virtualmathmuseum.org/Surface/gallery_o.html#AlgebraicSurfaces
you will see many nice examples. These were created using the program 3D-XplorMath which you can download at  http://3D-XplorMath.org  .   If you install that and go to the Implicit Surface category you will see many examples with documentation.<|endoftext|>
TITLE: Decomposing $\mathbf{\Pi}^1_1$ sets into closed sets
QUESTION [14 upvotes]: It is well known that every $\mathbf{\Pi}^1_1$-set is a union of $\aleph_1$-many Borel sets. I wonder whether it can be improved under certain reasonable set theory axioms assumption.
For example, assuming  $ZFC+CH$, then it is trivially true that every set is a union of $\aleph_1$-many closed sets. But this seems heavily depends on $CH$ since  if $ZFC+\neg CH+MA$, then there is a lightface $\Pi^0_2$-set which cannot be a union of $\aleph_1$-many closed sets.
So my question is: is it consistent with $ZFC+\neg CH$ that every  $\mathbf{\Pi}^1_1$-set is a union of $\aleph_1$-many closed sets?

REPLY [9 votes]: Every ${\bf\Pi}_1^1$ set is the union of $\aleph_1$ Borel sets, so we only have to make sure that every Borel set is the union of $\aleph_1$ closed sets.
But every Borel set is analytic and thus a continuous image of the Baire space $\omega^\omega$.
Now the Baire space is the union of $\mathfrak d$ (the dominating number) compact sets.  Continuous images of compact sets are again compact and hence every analytic set is the union of $\mathfrak d$ compact (and hence closed) sets.
It is well known that $\mathfrak d=\aleph_1$ is consistent with the failure of CH.
This happens for instance in the so called Sacks model.

Edit:  I forgot to mention that it was Paul Larson who turned my attention to this question.  Paul conjectured (or should I say, was convinced) that in the model obtained by forcing with a large countable support product of Sacks forcing over a model of CH every ${\bf\Pi}_1^1$ set is the union of $\aleph_1$ closed sets, 
which is true, since $\mathfrak d$ is $\aleph_1$ in that model.<|endoftext|>
TITLE: Current Status of the Riemann Hypothesis
QUESTION [12 upvotes]: Does anyone know the current progress in showing the Riemann hypothesis? I was only able to find this paper of Conrey that says at least 40% of the zeros of the Riemann Zeta function lie on the critical line.

REPLY [7 votes]: It is unsolved as of today.
However, some latest researches are:

A paper from 2002 talks by Freeman J. Dyson on Random Matrix Theory, Quasicrystals and zeta function
Fractal Geography of the Riemann Zeta Function
Andrew Odlyzko's collection of papers on such topic

Edit: 
FWIW, do note that the shortest "proof" of RH by Mark Colyvan as mentioned in IEP here using paraconsistent logic:


As the founders of relevant logic,
  Anderson and Belnap, urge in their
  canonical book Entailment, a ‘proof’
  submitted to a mathematics journal in
  which the essential steps fail to
  provide a reason to believe the
  conclusion, e.g. a proof by explosion,
  would be rejected out of hand. Mark
  Colyvan (2008) illustrates the point
  by noting that no one has laid claim
  to a startlingly simple proof of the
  Riemann hypothesis:
Riemann’s Hypothesis: All the zeros of the zeta function have real part equal to > 1/2. 
Proof: Let R stand for the Russell set, the set of all sets that are not
  members of themselves. It is
  straightforward to show that this set
  is both a member of itself and not a
  member of itself. Therefore, all the
  zeros of Riemann’s zeta function have
  real part equal to 1/2.
Needless to say, the Riemann
  hypothesis remains an open problem at
  time of writing.

The cited 2008 article by Colyvan however does not use this for RH but Fermat's Last Theorem.
Speaking of non-classical approach, Douglas S. Bridges further writes about the self-referential nature in Reality and Virtual Reality in Mathematics (2006):

There is an even more dramatic example of a proof which might cause the
  same unease. A famous conjecture of Riemann in the nineteenth century, the
  Riemann Hypothesis, remains unsolved today despite the efforts of some of the
  greatest mathematicians in the intervening 150 years. Early last century, the
  English mathematician J.E. Littlewood produced a theorem whose difficult proof
  was split into two cases. In the first case, Littlewood assumed that the Riemann
  Hypothesis was true, and in the second that it was false. Writing R to denote the
  Riemann Hypothesis, and P to denote the conclusion of Littlewoods theorem,
  we can express his proof in the schematic form
  $(R\bigvee\neg R)\Rightarrow P.$ (1)
  Here I have introduced the standard logical symbols $\bigvee$ (or), $\neg$ (not), and $\Rightarrow$ (implies)
  What is the meaning of Littlewoods proof? Since we are unable at this
  date to decide whether or not the Riemann Hypothesis is true, we cannot say
  which of the two cases of his proof actually applies. If, as most mathematicians
  expect, the Riemann Hypothesis turns out to be provable, then that part of
  Littlewoods proof that is based on the assumption that the Riemann Hypothesis
  is false is worthless and can be thrown away. Moreover, in such a proof, if $P$
  is an existential statement one that asserts the existence of a certain object
  $x$ with certain properties then the two cases of a proof of $P$ that follows the
  Littlewoods schematic form (1) may produce di¤erent objects $x$ with the desired
  properties (as in our earlier proof involving $\sqrt{2}^{\sqrt{2}}$
  ); under such circumstances, we
  might be unable to tell which of the two possibilities for x was the desired one
  until we could prove the truth or falsity of the Riemann Hypothesis.
  The formalist might attempt to remove our unease about Littlewoods proof
  as follows. Suppose that the desired conclusion $P$ of Littlewoods theorem is false.
  Then Littlewoods arguments, schematised in (1), show that neither the Riemann
  Hypothesis nor its negation can hold (since each of these alternatives leads us
  to a proof of $P$): In other words, if $P$ is false, then the Riemann Hypothesis is
  false and it is false that the Riemann Hypothesis is false! This is plainly absurd.
  Hence we conclude that $P$ cannot be false and is therefore true.<|endoftext|>
TITLE: Cosets and conjugacy classes
QUESTION [13 upvotes]: I'm interested in the following situation:

$G$ is a finite group;
$C$ is a conjugacy class in $G$;
$H$ is the centralizer of an element $h$ of $C$.

I want information on $|C\cap Hg|$ as $g$ varies across $G$. In particular I'd
like to prove that there exists $k<1$ such that for all $g\in G$ we have
$$|C\cap Hg| \leq k|H|.$$
Unfortunately for me such a bound does not exist in complete generality: consider $C_p\rtimes C_{p-1}$ for a prime $p$ (semidirect product of two cyclic groups). Let $C$
be the conjugacy class of elements of order $p$, all of which have the same
centralizer $H$. Then $C$ is a subset of $H$ and we have
$$|C\cap H| = (p-1/p)|H|.$$
So as $p$ goes to infinity we have $(p-1/p)\to 1$.
So we can only prove a bound of the given form for particular cases. With this in mind here are some questions:

Is it true that $|C\cap Hg|\leq |C\cap H|$? Edit: No it is not true. Mark Wildon has provided counter-examples in his answer below.
If we assume that $G$ is simple does a bound of the given form with $k<1$ exist?
Does anyone know if this problem appears in the literature in an
alternative formulation? I'm interested even in particular cases, e.g.
taking G to be a particular family of simple groups and C a particular
family of conjugacy classes.
Edit: As discussed in comments below, the case when $|C\cap Hg|=1$ for all $g\in G$ corresponds precisely to the situation $G=HC$. An example of this phenomenon is given below when $G=C_p\rtimes C_{p-1}$, a Frobenius group. Does this ever happen for $G$ simple? Has the problem of decomposing a group $G$ into the product of a centralizer and conjugacy class been studied in the literature?

REPLY [2 votes]: Here are  a couple of character theoretic observations which do not require CFSG.
In the situation of the question (where $C$ denotes the conjugacy class of $g$), we have $|C \cap Hx| = 1$ for every $x \in G$ if and only if $\langle {\rm Res}^{G}_{H}(\chi), 1 \rangle = 0$ whenever $\chi \in {\rm Irr(G)}$ is a non-trivial character with $\chi(g) \neq 0.$ More generally, $\sum_{ t \in T} |C \cap Ht|^{2} = [G:H] \sum_{\chi \in {\rm Irr}(G)} \frac{|\chi(g)|^{2}}{\chi(1)} \langle {\rm Res}^{G}_{H}(\chi,1 \rangle,$ where $T$ is a transversal to $H$ in $G.$ 
This formula is derived by considering the product (in the group algebra $\mathbb{Z}G$), of class sums ${\tilde C}{\tilde C^{-1}},$ where $C^{-1}$ denotes the class of $g^{-1}$ and we use ${\tilde C}$ to denote the class sum of the class of $g.$ The coefficient of $x \in G$ in this product is well-known to be $\frac{|G|}{|H|^{2}} \sum_{ \chi \in {\rm Irr}(G)} \frac{|\chi(g)|^{2} \chi(x^{-1})}{\chi(1)}.$ This is always non-negative, and if we sum these quantities over $x \in H,$ the claims follow easily ( in the first case, the trivial character already contributes $[G:H]$ to the RHS, and all other terms on the RHS are non-negative. In the second case, the conjugates of $g$ in $Ht$ contribute $|C \cap Ht|^{2}$ elements of $H$ to the given product of class sums, including multiplicities).
Note that we easily obtain $\sum_{t \in T}(|C \cap Ht|-1)^{2} \leq d(|G|-[G:H]),$ where $d$ is the maximum value over non-trivial irreducible characters $\chi$ with $\chi(g) \neq 0$ of $\frac{\langle {\rm Res}^{G}_{H}(\chi), 1 \rangle}{\chi(1)}$.
Later edit: In fact, this problem is quite closely related to an earlier one on MO about doubly transitive action on a conjugacy class: if $p$ is a prime, and $G$ is a (putative) doubly transitive permutation group whose point stabilizer $H$ has a non-trivial center (and with $F(G) =1$), then it can be shown that for an element $z \in Z(H)$ of prime order, there is one conjugate of $z$ in each coset of $H,$ ie the conjugates of $z$ form a transversal to $H.$ 
Also, I mention (without proof, but in case it is useful to anyone else), the following facts which may be proved using block theory: if $G$ is a finite group and $z \neq 1$ is an element of order a power of a prime $p$ whose conjugates form a transversal to $H = C_{G}(z),$ the following hold:
Whenever $y$ is a $p$-regular element of $H,$ we have ${\tilde C}_{zy} {\tilde H} = [H:C_{H}(y)]{\tilde G}$ (where, as before, for $S$ a subset of $G,$ we let ${\tilde S}$ denote the sum of the elements of $S$ in the group algebra $\mathbb{Z}G,$ and where $C_{u}$ denotes the conjugacy class of $u.$ 
Whenever $x$ is a $p$ -element (possibly the identity element)of $H,$ we have $|S_{p}^{G}(x)| = [G:H]|S_{p}^{H}(x)|$
and ${\tilde S}_{p}^{G}(x){\tilde H} = |S_{p}^{H}(x)|{\tilde G},$ where $S_{p}^{H}(x)$ denotes the $p$-section of $x$ in $G$ ( that is, the set of elements of $G$ whose $p$-part is conjugate to $x$).
By way of explanation, these last facts follow because the trivial character is the only constituent in the principal $p$-block of the character ${\rm Ind}_{H}^{G}(1),$ and central characters associated to irreducible characters outside the principal $p$-block annihilate $p$-section sums by Brauer's Second Main Theorem. For the first, we also have the more precise fact that the trivial character is the only constituent of ${\rm Ind}_{H}^{G}(1)$ in a $p$-block with a defect group containing $z,$ so all non-trivial constituents of ${\rm Ind}_{H}^{G}(1)$ vanish on the $p$-section of $z.$<|endoftext|>
TITLE: Transversal Matroid
QUESTION [8 upvotes]: I am trying to figure out why the transversal matroid is a matroid.
Specifically, if $L$ and $M$ are two independent sets s.t. $|L| < |M|$, why is there an $i$ in $M \setminus L$ s.t. $L \cup \{ i \} $ is independent?
thank you in advance!

REPLY [9 votes]: Let me check that we have the same definition of transversal matroids. For me, the input is a bipartite graph $\Gamma$, with white vertex set $W$ and black vertex set $B$. The ground set of the matroid is $B$, with a subset of $B$ being independent if it can be matched to a subset of $W$.
Direct Proof that this is a matroid: Suppose that $M$ and $L$ are subsets of $B$ can be matched, with $|L|<|M|$. Superimpose the matchings. 
This gives a graph $G$ where the elements each vertex has degree $\leq 2$ -- so a union of paths and cycles (including possibly some $2$ cycles.) We'll call an edge of $G$ mauve or lavender acording to whether it comes from $M$ or $L$. In each path or cycle of $G$, the edges alternate color.
Each element of $M \setminus L$ gives a degree $1$ in $B$ at the end of a mauve edge, and each element of $L \setminus M$ gives a degree one element of $B$ at the end of a lavender edge. Since $|M| > |L|$, there must be a path one of whose ends is a mauve edge ending in $W$ and whose other end is not a lavender edge ending in $B$. The only other possibility for the end of that path is a mauve edge ending in $W$. In other words, this path has vertices $b_1$, $w_1$, $b_2$, $w_2$, ..., $b_r$, $w_r$ where $b_i$ is matched to $w_i$ in $M$,  $w_i$ is matched to $b_{i+1}$ in $L$ and $b_1$ and $w_r$ are not matched in $L$. Now modify $L$ to match $b_i$ to $w_i$, leaving all other elements of $L$ matched as before. This is a matching of $L \cup \{ b_1 \}$.
Conceptual proof Take a matrix $A$ with rows labeled by $W$ and columns labeled by $B$. Place a $0$ in $A_{bw}$ if $\Gamma$ has no edge between $b$ and $w$, and place algebraically independent elements in the positions where $\Gamma$ does have an edge. The transversal matroid is realized by the columns of this matrix.<|endoftext|>
TITLE: Classification of surfaces and the TOP, DIFF and PL categories for manifolds
QUESTION [33 upvotes]: A surface is simply a 2-manifold. The classification theorem for compact connected surfaces (with boundary) is commonly regarded in the categories TOP, DIFF and PL. Well known proofs (e.g. via triangulations, or Morse theory) yield the same classification because of results that connect these categories for surfaces. Informally speaking, here is what I know to be true for compact connected surfaces

[TOP & PL]. Topological surfaces always admit a triangulation, and any two triangulations of a surface are piecewise-linear equivalent (Hauptvermutung for surfaces)

[DIFF & PL (without using 1.)]. Every smooth surface admits a PL-structure, as every smooth manifold does (See the paper "On $C^{1}$ Complexes", by J.H.C. Whitehead).


Next is where I seek to be enlightened:

[TOP & DIFF, (without using either 1. or 2.)]. Two smooth surfaces are diffeomorphic iff they are homeomorphic, and a topological surface always admits a smoothing. (Proved by J.R. Munkres- see the last paragraph of the answer below)

Where can I find a formal statement, and a complete proof of 3.?
Finally, consider non-compact connceted surfaces (with boundary). There seems to be a complete classification of non-compact connected triangulable surfaces with boundary (See the paper "Classification of Noncompact Surfaces with Boundary", by A.O. Prishlyak and K.I. Mischenko).What about the TOP and DIFF categories? That is, do the results 1-3 above hold for non-compact surfaces?
NOTE: I want to mention the post Classification problem for non-compact manifolds for a related, yet different discussion. The paper: "On the Classification of Noncompact Surfaces", by Ian Richards is mentioned there in a comment. This paper considers the case of non-compact triangulable surfaces without boundary.
Thank you!

REPLY [44 votes]: I have come to believe that answering the questions I posted would be more enlightening if I try to provide an overview of the larger context that they are part of.
The literature treating and generalizing the topics mentioned in the post for surfaces is as extensive as it is interesting. The 1960's and 70's were times of very active research in this part of topology, and it still is today. Three wonderful resources are
Kirby & Siebenmann's book Foundational Essays on Topological Manifolds, Smoothings and Triangulations, Milnor's paper Differential Topology Forty-six Years Later, and A. Ranicki's slides. I will refer to these as [KS], [Mil2011] and [Ran], respectively. Also, a word on notation: uniqueness will mean up to PL, DIFF or TOP homeomorphism, depending on the category at hand. Unless otherwise stated, words like manifold or surface will have general meaning (i.e. possibly with boundary and possibly non-compact). Finally, a list of references is included at the bottom.

[DIFF & PL] (Strictly speaking PL and DIFF are not comparable. One uses the category PDIFF, which is equivalent to PL. However, this distinction is not normally made unless technicalities may require so.) Differentiable manifolds admit canonical PL structures. A differentiable manifold can be triangulated uniquely up to PL equivalence. S.S. Cairns first proved this result for compact $C^{1}$ manifolds, including those having a finite number of boundary components (See [Cai1934], [Cai1936]), although he generalized these results later (see [Cai1961]). J.H.C. Whitehead proved it for $C^{1}$ manifolds without boundary (see [Whi1940]), and J. Munkres finally included $C^{r}$ manifolds with boundary, $1\le r\le\infty$ (see [Mun1966] or Theorem 3.10.2 in [TL]).

A given PL structure on a topological manifold may have compatible differentiable structures that are inequivalent. That is, $$\mathrm{DIFF}\rightarrow \mathrm{PL}$$ is not injective. In [Mil1956] J. Milnor gave an example of a manifold PL-homeomorphic to the usual 7-dimensional sphere $S^{7}$, but not diffeomorphic to it. In fact, it is known that for $n\neq 4$ a topological $n$-sphere admits a unique PL structure (For $n\le 3$ see [Moi1977] or [TL], for $n\ge 5$ is due to Smale and can be found in [Sma1962]. The case $n=4$ is an open question). Therefore, the inequivalent differentiable structures that Milnor constructed in [Mil1956] are all compatible with the usual PL structure on $S^{7}$.
Even More, on the topological manifold $\mathbb{R}^{4}$ it is possible to define uncountably many inequivalent PL or differentiable structures. An excellent account of this exotic $\mathbb{R}^{4}$'s can be found in Chapter XIV of [Kir1989].
The functor above is also not surjective. That is, there are PL manifolds that do not admit a compatible differentiable structure. M. Kervaire gave such an example in [Ker1960]. Later, J. Ells and N.H. Kuiper (see [EK1961]), and I. Tamura (see [Tam1961]) gave examples in dimension 8, the lowest possible.
In dimensions 7 or less, PL manifolds always admit compatible differentiable structure, and in dimensions 6 or less this happens in an a unique way (See Theorem 2 in [Mil2011] and Theorem 3.10.8 and Problems 3.10.19-20 in [TL] for dimension up to three). In this sense, DIFF=PL for manifolds of dimension $n\leq 6$, which means that the number of inequivalent differentiable structures on a topological 4-sphere is also unknown.
The obstruction to finding a differentiable structure on a given PL manifold is called the Munkres-Hirsch-Mazur obstruction (see the last paragraph on [Mil2011]).

[TOP & PL] $$\mathrm{PL}\rightarrow \mathrm{TOP}$$ is neither surjective nor injective. Indeed, there are topological manifolds, such as Freedman's E8 manifold, that do not admit a PL structure, or are even triangulable even if we allow non-PL triangulations. (A proof of this now follows from the proof of the 3-dimensional Poincaré conjecture, which implies that any triangulation of a 4-dimensional manifold is necessarily a PL-triangulation).

The exotic $\mathbb{R}^{4}$'s mentioned above provide an example of a topological manifold having uncountably many inequivalent PL structures. This disproves the manifold version of the Haupvermutung. The non-manifold version of the Haupvermutung was disproven by J. Milnor ([Mil1961]), who found two homeomorphic compact simplicial complexes that are not PL homeomorphic.
In dimension $3$ or less, the Haupvermutung is true (see Chapters 35 & 36 in [Moi1977] or Thurston/Levy's book). In this sense PL=TOP for manifolds of dimension $n\le 3$. Moreover, as mentioned earlier, except possibly for $n=4$ there is only one $n$-dimensional PL sphere.
The obstruction to finding a PL structure on a given topological manifold culminated with the resuts of Kirby and Siebenmann (the Kirby–Siebenmann class). (see [KS] and Theorem 1 in [Mil2011]).

[TOP & DIFF] As John Klein points out in the comments, smoothing a topological manifold is in general a question formulated by first putting a combinatorial structure on the manifold (normally a handlebody structure or a PL structure). The examples of Kervaire, Ells & Kuiper and Tamura mentioned above yield topological manifolds having no differentiable structure. However, these are still PL manifolds.

More striking is the E8 manifold which, not being triangulable, cannot have a differentiable structure. It provides and example of a topolgical manifold of dimension four that admits neither PL nor differentiable structures.
The exotic $\mathbb{R}^{4}$'s mentioned earlier give an example of a topological manifold having uncountably many inequivalent differentiable structures.
In dimension 3 or less the results above yield DIFF=PL=TOP.
Coming back to surfaces I want to point out that Theorem 8.3 in [Moi1977] shows that every surface is triangulable. At the begining of the proof it is shown that triangulations and PL structures are equivalent structures on a surface. Moreover, Theorem 8.5 is the Haupvermutung for surfaces. Therefore, a complete classification of non-compact surfaces (with boundary) seems to have been achieved by the results contained and mentioned in  Prishlyak and Mischenko's paper.
Finally, I want to point out that the result that two smooth surfaces are diffeomorphic iff they are homeomorphic is due to J. Munkre's and can be found in his dissertation Some Applications of Triangulation Theorems, U. of Michigan, 1955. The proof uses the triangulation theorems proven by E.E. Moise, who was Munkres' advisor.
REFERENCES:
The references [KS], [Mil2011] and [Ran] are listed (with links) right at the beginning, in the second paragraph above.
[Cai1934] Cairns, S. S., On the triangulation of regular loci, Ann. Math. (2) 35, 579-587 (1934). ZBL0012.03605.
[Cai1936] Cairns, S. S., Polyhedral approximations to regular loci., Ann. Math., Princeton, (2) 37, 409-415 (1936). ZBL62.0806.04.
[Cai1961] Cairns, S. S., A simple triangulation method for smooth manifolds, Bull. Am. Math. Soc. 67, 389-390 (1961). ZBL0192.29901.
[Ker1960] Kervaire, M.A., A manifold which does not admit any differentiable structure, Comment. Math. Helv. 34, 257-270 (1960). ZBL0145.20304.
[Kir1989] Cairns, S. S., A simple triangulation method for smooth manifolds, Bull. Am. Math. Soc. 67, 389-390 (1961). ZBL0192.29901.
[EK1961] Eells, J.; Kuiper, N.H., Manifolds which are like projective planes, Publ. Math., Inst. Hautes 'Etud. Sci. 14, 181-222 (1962). ZBL0109.15701.
[Mil1956] _Milnor, J. W., On Manifolds Homeomorphic to the $7$-Sphere, Ann. Math. (2) 64, 399-405 (1956). ZBL0102.38103.
[Mil1961] _Milnor, J. W., Two complexes which are homeomorphic but combinatorially distinct, Ann. Math. (3) 74, 575-590 (1961). ZBL0102.38103.
[Moi1977] _Moise, E. E., Geometric topology in dimensions 2 and 3, Graduate Texts in Mathematics. 47. New York - Heidelberg - Berlin: Springer-Verlag. X, 262 p. (1977). ZBL0349.57001.
[Mun1960] _Munkres, J.R., Obstructions to the smoothing of piecewise-differentiable homeomorphisms, Ann. Math. (2) 72, 521-554 (1960). ZBL0108.18101.
[Mun1966] _Munkres, J.R., Elementary differential topology. Lectures given at Massachusetts Institute of Technology, Fall, 1961. Revised ed, Annals of Mathematics Studies. 54. Princeton, N.J.: Princeton University Press. XI, 112 p. (1966). ZBL0161.20201.
[Sma1962] _Smale, S., On the structure of manifolds, Am. J. Math. 84, 387-399 (1962). ZBL0109.41103.
[Tam1961] _Tamura, I., 8-manifolds admitting no differentiable structure, J. Math. Soc. Japan 13, 377-382 (1962). ZBL0109.16302.
[TL] _Thurston, W. P., Three-dimensional geometry and topology. Vol. 1. Ed. by Silvio Levy, Princeton Mathematical Series. 35. Princeton, NJ: Princeton University Press. x, 311 p. (1997). ZBL0873.57001.
[Whi1940] _Whitehead, J.H.C., On $C^1$-complexes, Ann. Math. (2) 41, 809-824 (1940). ZBL0025.09203.<|endoftext|>
TITLE: How to quantify noncommutativity?
QUESTION [40 upvotes]: If I have two operators or finite-dimensional matrices $A$ and $B$, how can I quantify the amount to which they commute or don't commute? (I would consider it a big plus if it is computable easily for finite complex-valued matrices $A, B \in \mathbb C^{n\times n}$.) 
Let me try the obvious thing here: by definition if $A$ and $B$ commute, then the commutator $[A, B] = AB-BA = 0$. Naively would use some sort of functional like an operator norm to reduce this to a number that could potentially behave like a metric. The first thing I thought of was the trace, but clearly that doesn't work since $\mathrm{tr } [A, B] =\mathrm{tr } (AB-BA) = \mathrm{tr }AB - \mathrm{tr }AB = 0$ always. One could then turn to, say, the Frobenius norm of $[A, B]$. What is known about the maximal (or supremal) value of such norms?
Are there quantifiers of noncommutativity that can also account for higher-order effects, e.g.  cases where $[A, B] \ne 0$ but $[A, [A,B]] = 0$? This should be "less" non-commuting than if $[A, B] \ne 0$ and $[A, [A,B]] \ne 0$ and $[B, [A,B]] \ne 0$ but, say, $[A, [B, [A, B]]] = 0$.
For those who prefer a free algebraic setting, the question can be framed as: how free is a non-free algebra? Is there a sensible way to measure proximity to a free algebra? What if I had an algebra where $AB=BA$ is the only one relation that makes it not a free algebra; is there a sense it is "less free" or "more free" than an algebra where $ABABAB=BAA$ is the only such relation, or example.
Motivation: it is sometimes said that free probability is the study of "maximally" non-commuting objects. I would like to know if this statement can be made precise in the sense of how one can define "maximally non-commuting" in a sensible fashion.

REPLY [2 votes]: A good way to quantify the non-commutativity of $A,B$ is to compare their flows, e.g. to compare the solutions of
$
\dot u=Au,\quad\dot v=Bv
$
with same initial datum. A complete answer is given by the Campbell-Hausdorff formula
$$
\exp{tA}\  \exp{tB}=\exp{\bigl(t(A+B)+\frac{t^2}{2}[A,B]+\frac{t^3}{12}([A,[A,B]]+[B,[B,A]])+\dots\bigr)}.
$$
In the exponent are appearing successively the first bracket and the iterated brackets.

REPLY [2 votes]: One very algebraic way to approach it (alas not that easily approached computationally) - somewhat related to David Jordan's answer - would be to look at the Lie subalgebra generated by $A$ and $B$ in the algebra of matrices (that is, the linear span of all commutator words in $A$ and $B$) and impose one or another condition relaxing commutativity, e.g. say that it is nilpotent of degree $k$ or solvable of degree $k$. In that case, $k$ will of course somewhat quantify the magnitude of noncommutativity $A$ and $B$ share.<|endoftext|>
TITLE: Surface Eversions: Generalizing from Sphere and Torus Eversions
QUESTION [10 upvotes]: In 1958, Smale proved that a $2$-sphere can be "turned inside out", and throughout the 60s, 70s, and 80s, explicit constructions such as Thurston Corrugations, and Minimax eversions were developed to visualize this sphere eversion. I just encountered a recent youtube video: Torus Eversion, which depicts the eversion of a torus. My general question is then this,

Is there a characterization, by Euler characteristic, genus, or some other topological invariant, for which surfaces have eversions?

If not a characterization of all such surfaces which have eversions, is there a finite list of surfaces which can be everted?

REPLY [10 votes]: I'll expand on my comment above. Consider an eversion of a sphere, namely a path of immersions $e_t: S^2\to \mathbb{R}^3$ such that $e_0$ and $e_1$ are embeddings, and which exchanges the two sides. By possibly reparameterizing the maps $e_t$, we may assume that there exists $D^2\subset S^2$ such that  $e_{t|D^2}: D^2\to \mathbb{R}^3$ is an embedding for all $t$. The map $e_{t|D^2}$ may be extended to an ambient isotopy $f_t$ of $\mathbb{R}^3$, which we may also assume takes a  ball $B_0$ intersecting $e_0(S^2)$ in $e_0(D^2)$ to a ball $B_1$ intersecting $e_1(S^2)$ in $e_1(D^2)$. Add some small trivial handles to $D_0$ inside $B_0$, to make a (smooth) embedded surface $\Sigma_g\subset \mathbb{R}^3$ of genus $g$. Then one may perform the regular homotopy $\sigma_t: \Sigma_g \to \mathbb{R}^3$ where $\sigma_{t | S^2-D^2} = e_{t | S^2-D^2}$, and $\sigma_{t| \Sigma_g-(S^2-D^2)}=f_t(\Sigma_g\cap B_0)$. Then $\sigma_1$ will be an embedding which exchanges the two sides of $\Sigma_g$. 
One may also use the torus eversion shown in the link to evert all surfaces of genus $g>0$ in a simpler fashion. 
For the 2-sphere, I like the eversion given a couple of years ago by Aitchison.<|endoftext|>
TITLE: Finite time hitting probabilities for Brownian motion in the plane
QUESTION [8 upvotes]: Consider a Brownian particle in the plane with a circular trap at the origin.  If we give the particle enough time it falls into the trap (since Brownian motion is space filling in 2D).  However, suppose we only let the particle evolve for a finite interval of time.  What is the probability it is trapped?  
Scaling time and space appropriately we may assume that the trap is the unit disk.  Formulated precisely the problem is. Start a standard Brownian motion in the plane at some distance $r>1$ from the origin.  Let it run up to time $T$.   What is the probability $p(r,T)$ that it hits the unit disk at some time $t\in [0,T]$?  
This can, of course, be rephrased entirely in terms of a boundary value problem:
$$\frac{\partial p}{\partial T}= \frac{1}{2r} \frac{\partial }{\partial r} r \frac{\partial p}{\partial r},$$
$$p(1,T)=1, \quad p(r,0)=0, \ 1 \lt r \lt \infty, \quad \ \lim_{r \rightarrow \infty} p(r,T)=0.$$ 
If we take $T\rightarrow \infty$ then $p(r,T)\rightarrow 1$ for every $r$, but for any fixed $T<\infty$, $p(r,T)<1$ and decays as $r\rightarrow \infty$ at least like a Gaussian. 
Surely this has been studied somewhere in the literature. My question is, "Where?" Maybe even some sort of "exact" formula exists for $p(r,T)$ involving special functions -- probably Bessel functions.  I am particularly interested in understanding the "effective area of the trap", i.e., the area of the disk over which $p(r,T)$ is "almost 1", say $>1-\epsilon$ for some fixed $\epsilon$, but any reference to a study of this type of finite time hitting problem would be appreciated.

REPLY [9 votes]: Instead of a concrete answer, I will give what appears to be the most useful reference. I quote the first paragraph of Wendel, J. G. "Hitting spheres with Brownian motion". Ann. Prob. 8, 164 (1980).

Let $X_t$ be a standard $d$-dimensional Brownian motion with nonrandom starting point $X_0$. When $d \ge 2$ we seek explicit formulas which will determine the joint distributions of the first time $T \le \infty$ and place $X_T$ (which is only defined when $T$ is finite) where $X_T$ hits a sphere centered at the origin, either from the inside or from the outside, or exits from the region bounded by concentric spheres.

See also Betz, C. and Gzyl, H. "Hitting spheres from the exterior". Ann. Prob. 22, 177 (1994) and various cites of these papers.<|endoftext|>
TITLE: Computer package for representation theory of the symmetric group
QUESTION [24 upvotes]: Is there a computer algebra package in which I can compute the following for representations of a specific symmetric group (e.g. $S_7$):
(1) $V \otimes W$
(2) $S_\lambda V$, where $S_\lambda$ is a Schur functor, or even just $\wedge^s V$,
where $V$ and $W$ are input as sums of irreducible representations, i.e. by partitions with coefficients, and output in the same format?

REPLY [16 votes]: In the years since I left the answer below, Sage has improved dramatically, especially its symmetric function theory.  In order to compute the third exterior power of the $S_7$-irrep corresponding to the partition $3+3+1$, one need only type the following:
s = SymmetricFunctions(QQ).schur()
s[1,1,1].inner_plethysm(s[3,3,1])

which returns
s[2, 1, 1, 1, 1, 1] + 2*s[2, 2, 1, 1, 1] + 4*s[2, 2, 2, 1] + 6*s[3, 1, 1, 1, 1] + 9*s[3, 2, 1, 1] + 7*s[3, 2, 2] + 4*s[3, 3, 1] + 7*s[4, 1, 1, 1] + 9*s[4, 2, 1] + 4*s[4, 3] + 2*s[5, 1, 1] + 4*s[5, 2] + s[6, 1] + s[7]

the same answer discovered by the GAP code below.  Sage is also faster.  From what I understand, GAP still wins when you want to study representations over a finite field.  (But it's not really a contest since GAP is included in Sage!)  From Sage, you may start a GAP console with
gap_console()

The answer from 2012 appears below.

Here is the GAP code I use to do these computations:
SchurFunctorOfCharacter:=function(char,p)
  local n,t,c;
  if p=[] then
       return TrivialCharacter(UnderlyingCharacterTable(char));
  fi;
  n:=Sum(p);
  t:=CharacterTable("Symmetric",n);
  c:=List(CharacterParameters(t),u->u[2]);
  return Symmetrizations([char],n)[Position(c,p)];
  end;;
CharacterFromPartition:=function(table,p)
  local c;
  c:=List(CharacterParameters(table),u->u[2]);
  return Irr(table)[Position(c,p)];
  end;;
DecomposeCharacter:=char->List(Irr(UnderlyingCharacterTable(char)),x->ScalarProduct(x,char));;
t:=CharacterTable("Symmetric",7);;
chi:=CharacterFromPartition(t,[3,3,1]);;
DecomposeCharacter(SchurFunctorOfCharacter(chi,[1,1,1]));

This code computes the character table of $S_7$, finds the character corresponding to the partition $3+3+1$, and applies the Schur functor corresponding to the partition $1+1+1$ (otherwise known as $\wedge^3$).  Here is the result:
[ 0, 1, 2, 4, 6, 9, 7, 4, 7, 9, 4, 2, 4, 1, 1 ]

These are the multiplicities of the irreducible constituents of our character.  The ordering on partitions is lexicographic.  For example, to determine the meaning of the $6$, just take the fifth partition of $7$:
Partitions(7)[5];

The output shows that the coefficient $6$ appears before the L-shaped partition $3+1+1+1+1$:
[3,1,1,1,1]<|endoftext|>
TITLE: Does the minima of a sequence of convex convergent functions converge?
QUESTION [5 upvotes]: Suppose $f_1,f_2,\ldots $ is a sequence of convex functions that converges to a continuous convex $f$. Let $x_1^*,x_2^*$ be their respective (not necessarily unique) minima, and let y be a minima of $f$ (once again need not be unique). Can we prove that there exists a version of $x_1^*,x_2^*,\ldots$ such that $x_n^*\rightarrow y$ ?

REPLY [10 votes]: No; here's a counterexample: let $f = 0$ and consider the minimizer $y = 0.$ Then you can construct convex functions which converge to $0$ pointwise but whose minima are always moving away from $y =0,$ e.g. $f_n(x) = (x - n)^2/n^n.$

REPLY [6 votes]: No. Let $f_n=x^2/n$ for $n$ odd and $(x-1)^2/n$ for $n$ even. Then $x_n^*$ is an alternating sequence of $1$s and $0$s, which does not converge to anything. But $f_n$ converges pointwise to $f=0$.
We can modify this to make the convergence uniform, by using an absolute value instead of a square, or to make $f$ nonconstant, by adding $\max(|x-1/2|,1)$ to $f_n$.
If $f$ has a unique minimum, the statement is true. Let $a$ be the $\lim\inf$ of $x_n^*$ and $b$ be the $\lim\sup$. Let $y$ be the unique minimum of $f$. Assume $a< y$. Clearly $f( (a+y)/2) > f(y)$. By convexity, $f(a)> f((a+y)/2)$. So for $n$ sufficiently large, $f_n(a)> f_n((a+y)/2)> f_n(y)$. But this implies that the minimum of $f_n$ is greater than $(a+y)/2$. So the $\lim \inf$ of $x_n^*$ is greater than or equal to $(a+y)/2$, which is greater than $a$. This is a contradiction so $a\geq y$. By symmetry $b\leq y$. Hence because $a \leq b$, $a=b=y$ and  $y$ is the limit.<|endoftext|>
TITLE: Non-trivial consequences of Lob's theorem
QUESTION [6 upvotes]: Informally, Löb's theorem (Wikipedia, PlanetMath) shows that:

a mathematical system cannot assert its own soundness without becoming inconsistent [Yudkowsky]

In symbols:

if $PA\vdash$ $Bew$(#P) $\rightarrow P)$, then $PA\vdash P$ 

where $Bew$(#P) means that the formula $P$ with Gödel number #P is provable.
Other than Leon Henkin's application to show that Santa Claus exists (also see here), Michael Detlefsen wrote about limitations of mechanism which I do not have a copy but I did read through a refutation of it here. (Note: The first page of Detlefsen's paper can be accessed here). 
Additionally, Drucker mentions Kripke's 1967 "new proof" of the theorem here.
My question is has there been any other interesting non-trivial applications of the theorem other than the cited ones?

REPLY [8 votes]: Löb's theorem can be used to show that there exist equilibria in games like prisoner's dilemma when the participants are computer programs that can read each other's source code.
If player 1 can show that player 2 will cooperate when given player 1's source code, and vice versa, then we can have an equilibrium. The catch is that player 2 needs to predict what will happen when its own source code is fed to player 1 and vice versa. So implicitly each player must reason about itself. Löb's theorem shows there is a consistent way to to this. See Robust Cooperation in the Prisoner's Dilemma: Program Equilibrium via Provability Logic.
It's not entirely practical but it's interesting anyway.<|endoftext|>
TITLE: Sums of unitaries with small norm in full group $C^*$-algebras
QUESTION [7 upvotes]: Suppose $G$ is a finitely generated group, with given generating set $S={g_1, \dots, g_n}$.  (Assume that if $g\in S$, then $g^{-1}\notin S$.  (EDIT: Also assume that $S$ is minimal in the sense that no proper subset of $S$ is generating.))  Given complex numbers $a_1, \dots , a_n$, we can form the element $\sum a_j g_j$ of the group ring $\mathbb C[G]$ and consider its norm as an element of the full group C*-algebra $C^\ast(G)$:
$$
\|\sum a_j g_j \| =\sup_\pi \|\sum a_j \pi(g_j)\|
$$
where the supremum is over all unitary representation $\pi$ of $G$ on Hilbert space.  I am interested in the quantity
$$
\alpha(S) = \inf_{\|a\|_1=1} \|\sum a_j g_j\|.
$$
(Here $\|a\|_1 =\sum |a_j|$). 
Question:  Are there known examples of groups $G$ and generating sets $S$ for which $\alpha(S)<1$?  
(I am mostly interested in the case where all the generators have infinite order, so I haven't spent much time looking at finite groups. So, in case there are finite examples, a follow-up question would be for examples with all $g$ of infinite order.)  (EDIT: Since the C*-norm will be strictly smaller than $\ell^1$ norm, there should be lots of examples, see answers & comments below. We can still ask for examples when the generating set is minimal, and then there is the harder question of just how small $\alpha(S)$ can be.)
Some background and discussion: The question arose from an attempt to analyze the convex set of all points in $\mathbb C^n$ of the form
$$
(\langle \pi(g_1)x,y\rangle, \dots \langle \pi(g_n)x,y\rangle )
$$
where $\pi$ ranges over all representations of $G$ on Hilbert spaces $\mathcal H$ and $x,y$ range over all vectors of the unit ball of $\mathcal H$. I am interested in how large a polydisk centered at the origin such a set can contain; estimating the norm of $\sum a_j g_j$ is essentially the dual problem to this one. In this setting, cases where $\alpha(S)$ is small are the "enemy", so I'm trying to understand more about when this happens.  
It is not hard to prove the bounds
$$
\frac{1}{\sqrt{n}} \leq \alpha(S) \leq 1
$$
(indeed the upper bound is trivial, and the lower bound comes from considering the regular representation of $G$ on $\ell^2(G)$. )  As simple examples, it can be shown that the free abelian group $\mathbb Z^n$ and the free group $\mathbb F^n$ (with their standard $n$-element generating sets) both have $\alpha(S)=1$.  Now, one can try to get better lower bounds on $\alpha(S)$ in particular cases by considering particular representations $\pi$ (such as the regular representation), since then one is dealing with particular concrete unitaries, but it seems much harder to improve the upper bound (when it can be improved) unless one has a handle on a faithful representation of $C^\ast(G)$.  Of course this is the case when $G$ is amenable, though at the same time it's not clear that amenability is relevant, given the free (non-amenable) and free abelian (amenable) examples just mentioned.

REPLY [6 votes]: Here is an algebraic characterization of $\alpha(S)=1$: if $S=\{g_1,\dots,g_n\}$,  $\alpha(S)=1$ if and only if $\{g_1^{-1}g_k,k=2\dots n\}$ generate a free abelian group in the abelianization $G^{ab}$ of $G$.
In fact, since $\alpha(S)$ is unchanged under the action of $G$ by left (or right) multiplication, it is natural to request for example that $1$ belongs to $S$. We therefore consider, for a subset $S \subset G$, $\widetilde \alpha(S) = \alpha (S \cup \{1\})$. Studying $\alpha$ and $\widetilde \alpha$ are clearly the same problem, since for example if $S=\{g_1,\dots,g_n\}$, $\alpha(S)=\widetilde \alpha(g_1^{-1} g_2,\dots,g_1^{-1} g_n)$. 
The condition $\widetilde \alpha(S)=1$ can be expressed in terms of the image $q(S)$ of $S$ in the abelianization $G^{ab}$ of $G$. Namely the claim is that $\alpha(S)=1$ if and only if the group generated by $q(S)$ in $G^{ab}$ is $\mathbf Z^S$ (i.e. it is free abelian).
One direction is clear: $\|a_0+\sum_{g \in S} a_g g\|_{C^*(G)} \geq \|a_0+\sum_{g \in S} a_g g\|_{C^*(G^{ab})}$, so that if $q(S)$ generates a free abelian group, $\widetilde \alpha(S)=1$.
For the other direction, assume that $q(S)$ does not generate a free abelian group. Equivalently there exists a word $g_{1}^{\varepsilon_1} \dots g_{k}^{\varepsilon_k}$ with $g_i \in S$, $\varepsilon_i \in \{-1,1\}$ that represents $1$ in $G$ but such that $\sum_{i, g_i=g} \varepsilon_i \neq 0$ for at least one $g \in S$ (in other words, this word is an element of the free group on $S$ generators $F_S$ that is trivial in $G$ but nontrivial in $F_S^{ab}=\mathbf Z^S$). In particular, one can choose $a_g \in \mathbf C$ for $g \in S$ such that $\prod a_{g_i}^{(\varepsilon_i)} <0$ where I use the notation $z^{(1)}=z$ and $z^{(-1)}=\bar z$.
Consider now the element $X=1+\sum_{g \in S} a_{g} g$. I claim that $\|X\|_{C^*(G)}< 1+\sum |a_g|$. Indeed, $\|X\|^{2k} = \|(XX^*)^k\|$, and formally $(XX^*)^k$ is a sum of terms of the form $\sum_w c_w w$ where the sum is over all $|S+1|^{2k}$ words in $S\cup{1}$ and $S^{-1} \cup 1$ alternatively, with $\sum |c_w| = (1+\sum |a_g|)^{2k}$. But in this sum, there are two particular elements: $w_0$, the one corresponding to the word with all $1's$, for which $c_w=1$, and (at least) one word, $w_1$ that corresponds to $g_{1}^{\varepsilon_1} \dots g_{k}^{\varepsilon_k}$ once we remove all the $1$'s, for which $c_w=\prod a_{g_i}^{(\varepsilon_i)} <0$. But both words correspond to $1$ in $G$, so that by the triangle inequality we can write 
$$\|(XX^*)^k\|_{C^*(G)} \leq |1+c_{w_1}| + \sum_{w \neq w_0,w_1} |c_w|< \sum_w |c_w|=(1+\sum_{g \in S} |a_g|)^{2k}.$$
QED

Some additional remark. The fact that $\widetilde \alpha(S)=\widetilde \alpha(S')$ if $S$ and $S'$ are the generators of $F_n$ and $\mathbf Z^n$ says that the linear space space spanned by $S$ and $S'$ in $C^*(G)$ can be isometrically isomorphic when the groups generated are not. Such phenomenon cannot happen if one considers the operator space version of the question. It is a theorem by Pisier that if a map $T:span(1,u_1,\dots,u_n) \to span(1,v_1,\dots,v_n)$ sending $1$ to $1$ and the unitary $u_i$ to the unitary $v_i$ is completely isometric, then $T$ extends to $*$-isomorphism of the $C^*$-algebras generated. When $u_i=g_i$ (resp $v_i=g_i'$) in $C^*(G)$ (resp. $C^*(G')$), this implies that $G$ and $G'$ are isomorphic.<|endoftext|>
TITLE: Lie groups vs. algebraic groups and deformations
QUESTION [13 upvotes]: I am interested in deformations of (discrete subgroups of) Lie groups. But, as I understand it, deformation theory, as a theory, prefers to speak schemes.
At least the classical Lie groups can be turned into group schemes, allowing for a standard treatment with deformation theory.
Are there any results from (the deformation theory of) algebraic groups/group schemes, which can not be translated back to the language of Lie groups?
Does studying groups like $\mathrm{SO}_n$ or $\mathrm{U}_n$ as algebraic groups yield equivalent results for $\mathrm{SO}_n$ or $\mathrm{U}_n$ as Lie groups? Can one translate results between differential geometry and algebraic geometry, or are there major stumbling blocks that prevent algebraic results from having any meaning for the differential side?
In particular, in the case of moduli spaces/stacks, the topology is of some interest. But the groups (Lie vs. algebraic) have quite different topologies; the Zariski topology is not Hausdorff and I have no idea what the fppf topology would look like...

REPLY [15 votes]: If you take, say, set of real points of the group-scheme $O(n)$, i.e., $O(n, {\mathbb R})$, then you recover the usual orthogonal (real Lie) group, which you know as $O(n)$. Same applies to $SL(n)$, etc. There is one case when this does not work well, namely when you deal with character varieties. For instance, take $\pi$, say, the free group on two generators, and try to form the quotient $Q=Hom(\pi, SU(2))/SU(2)$. The standard way to do this is to consider the corresponding character variety (or, rather, affine scheme) $X$ and take its set of real points. However, the result will contain both equivalence classes of representations of $\pi$ to $SU(2)$ (as you expected), but also equivalence classes of representations to $SL(2, {\mathbb R})$! The easiest way to see this is to realize that the coordinate ring of $X$ is generated by traces of the elements $A, B, C=AB$ of $\pi$ (where $A, B$ are the free generators). To get the set of real points, you need to use points with real traces, so you end up with the elements of both real Lie groups $SU(2)$ and $SL(2, {\mathbb R})$. This is rather annoying, but one can learn to live with this problem. Namely, in order to isolate $Q=Hom(\pi, SU(2))/SU(2)$ inside $X({\mathbb R})$, you impose also some inequalities, so $Q$ becomes a real semi-algebraic subset. Same problem appears if you consider $Hom(\pi, SL(2, {\mathbb R}))$: Character variety will give you unitary representations as well. The standard way to deal with this problem (in Teichmuller theory) is to consider not all representations to $SL(2, {\mathbb R})$, but only discrete and faithful ones, so that the commutator $[A,B]$ maps to elements of the fixed trace. 
Then you can form the (topological) quotient by $SL(2, {\mathbb R})$ by taking slice, i.e., restricting to representations $\rho$ so that the (attractive, repulsive) fixed points of $\rho(A)$ are $0, \infty$ and the attractive fixed point of $\rho(B)$ is $1$.  
Addendum: More generally, in all "interesting" case I know, the desired quotient can be constructed without algebraic geometry. Suppose you are interested in $R:=Hom(\pi, O(n,1))$ and the group $\pi$ is "nonelementary", i.e., is not virtually abelian. Then, in $R$ consider the open subset $R'$ consisting of representations $\rho$ so that $\rho(\pi)$ does not fix a point in $S^{n-1}$. For instance, $R'$ will contain all discrete and faithful representations. Then you can just take the "naive" quotient $Q=R'/O(n,1)$ instead of the (set of real points of) character variety $X({\mathbb R})$. Then $Q$ will embed in $X({\mathbb R})$. 
Concerning existence of a slice: The same argument I described works for representations to $SL(2, {\mathbb C})$. However, if you consider representations to $O(n,1), n\ge 4$, there is no (in general) global slice. However, locally, it does exist, see e.g. 
Slice Theorem for the general information about slices for group actions. 
Recommended reading: D. Johnson and J. Millson, Deformation spaces, associated to compact hyperbolic manifolds, in "Discrete Groups in Geometry and Analysis" (Papers in honor of G. D. Mostow on his sixtieth birthday), 1984. I read this paper as a graduate student and still find it useful.  
By the way, here is where algebraic viewpoint is definitely superior to the Lie theoretic. Suppose that you are interested in understanding local structure of the analytic variety $Hom(\pi, G)$, where $G$ is a real Lie group, i.e., what singularity it has at a representation $\rho: \pi\to G$. In general it is a rather difficult problem as singularities could be "arbitrarily complicated." See M.Kapovich, J.Millson, "On representation varieties of Artin groups, projective arrangements and the fundamental groups of smooth complex-algebraic varieties", Math. Publications of IHES, vol. 88 (1999), p. 5-95, one of the main results there is that singularities of character varieties could be arbitrary (defined over ${\mathbb Q}$). 
So, assume that $G=\underline{G}({\mathbb R})$, where $\underline{G}$ is an algebraic group (scheme) over reals. Let $A$ be an Artin local ${\mathbb R}$-ring with the projection to the residue field $\nu: A\to {\mathbb R}$. Then the set of $A$-points $G_A:=\underline{G}(A)$ is a certain nilponent extension of the Lie group $G$ with the quotient $\nu_G: G_A\to G$ induced by $\nu$. Then, instead of analyzing the scheme $Hom(\pi, \underline{G})$ at $\rho$, you consider the collection of real-algebraic sets $Hom_{\rho}(\pi, G_A)\cong  Hom_{\rho}(\pi, \underline{G})(A)$, consisting or representations $\tilde\rho: \pi\to G_A$ which project to $\rho$ under $\nu_G$. The point is that the collection of real-algebraic sets $Hom_{\rho}(\pi, G_A)$ "knows everything" (and even more!) about the singularity of  $Hom(\pi, G)$. For instance, to recover the (Zariski) tangent space $T_\rho Hom(\pi, G)$, you just take $A$ to be the "dual numbers", which is the quotient ${\mathbb R}[t]/(t^2)$. Then 
$$
T_\rho Hom(\pi, G)\cong Hom_{\rho}(\pi, G_A)$$
for this choice of $A$.
This staff is explained in the paper 
W.Goldman, J.Millson, The Deformation Theory of Representations of Fundamental Groups of Compact Kahler Manifolds, Publ. Math. I.H.E.S.; 67 (1988).<|endoftext|>
TITLE: Existence of a large leaf in a foliation of the ball
QUESTION [7 upvotes]: Consider a smooth $k$-dimensional foliation of the unit ball $B$ of $\mathbb{R}^n$, all of whose leaves are diffeomorphic to $k$-disks. 
Question: Is there a leaf whose $k$-volume is at least $\omega_k$?
Here $\omega_k$ denotes the volume of the unit ball of $\mathbb{R}^k$.
Particular cases. If $k=1$, the leaf through the origin (in this case a curve) has the desired property. If $k=n-1$, the leaf which divides the ball into two regions of equal volume has the desired property, by a relative isoperimetric inequality. 
A possible argument. Replace each leaf $F$ by a $k$-submanifold which minimizes the $k$-volume among those with boundary $F\cap \partial B$. Among the minimal submanifolds obtained in this way, consider one which passes through the origin: its $k$-volume is at least $\omega_k$ by the monotonicity formula, and the $k$-volume of the leaf with the same boundary is even larger. 
I know how to make this argument rigorous when the foliation is close enough to the foliation by parallel affine subspaces (but a reference or a more elementary proof also for this perturbative case would be very useful). In general, besides for possible singularities of the minimizers (which should not disturb), the problem I see is how to guarantee that at least one of them passes through the origin.

REPLY [7 votes]: I think one might be able to prove this  without using the heavy duty machinery from the geometric measure theory that Anton mentioned by modifying the proof of the main result of the following paper of Gromov "Isoperimetry of Waists and Concentration of Maps".
He proves a stronger version of your statement but for the sphere instead a disk:
for any continuous map $f: \mathbb S^n\to\mathbb R^{k}$ there is a fiber $F$ such that $vol(U_\epsilon(F))\ge vol (U_\epsilon(\mathbb S^{n-k}))$ for any $\epsilon>0$. 
This of course immediately gives some lower bound on the volume of the maximal leaf in your case because the double of a disk is bilipschitz to a sphere but I think it's likely that one can  get a sharp bound too.
There is a very readable explanation of Gromov's proof by one of his students posted on the arxiv. That's what I looked at as reading Gromov's papers can be tough.
The construction is roughly as follows. He looks at convex polyhedral partitions of $\mathbb S^n$ into $2^N$  (with $N\to\infty$) subsets of equal volume. Given a map  $f: \mathbb S^n\to\mathbb R^{k}$ he defines a section of a certain vector bundle over the space of partitions and checks that this bundle has a nonzero top Stiefel-Whitney class so that this section must have a zero. by construction a zero of the section gives a fiber of $f$ passing through the center of mass of every convex set in the partition. The argument is very similar to (and is in fact a generalization of) the proof of the Borsuk-Ulam theorem.
Furthermore one can make sure that the convex sets are $\delta(N)$ close to being k-dimensional with $\delta(N)\to 0$ as $N\to\infty$. One then passes to the limit to get a "partition" into convex sets of dimension $\le k$ and the fiber in question is the one that passes through the center of mass of every convex set  with respect to the limit of the normalized volume measure. The key geometric part is to prove that for every $k$-dimensional convex set $C$ in the limit and the normalized limit measure $\mu$ on it one has that $$\mu(B_\epsilon(p)\cap C)\ge \frac {vol (U_\epsilon(\mathbb S^{n-k}))}{vol (\mathbb S^n)}$$ where $p$ is the center of mass of $C$.
Since the elements of the partitions along the sequence had equal volume this yields the result.
The proof of the inequality in the displayed formula is a convexity argument similar to the proof of the Bishop-Gromov volume comparison.
I haven't checked the details but I think the whole construction can be adapted to a disk instead of a sphere by using convex partitions of the disk obtained by coning off the elements of the partitions of the sphere to give  a sharp bound for the disk. This may even be discussed somewhere in Gromov's paper.<|endoftext|>
TITLE: non-isomorphic stably isomorphic fields
QUESTION [20 upvotes]: Q1: What is the simplest example of two non-isomorphic fields $L$ and $K$ of characteristic $0$ such that $L(x)\simeq K(x)$ (here $x$ is an indeterminate)? 
Q2: Do we have a sufficient criterion for a general field $K$ of characteristic $0$ which guarantees that if $K(x_1,\ldots,x_n)\simeq L(x_1,\ldots, x_n)$ (here $L$ is a field and the
$x_i$'s are indeterminates) then $K\simeq L$?

REPLY [8 votes]: An answer to Q2, generalizing Ralph's comment: "$K$ is algebraically closed" is a sufficient condition. Indeed, you can characterize $K$ inside $K(x_1,\dots,x_n)$ as the set of elements having $m$-th roots for infinitely many integers $m$. More generally, it is enough to assume that for some $m>1$, the $m$-th power map on $K$ is onto. Examples: $K$ perfect of positive characteristic, or $K=\mathbb{R}$.<|endoftext|>
TITLE: Manifolds are paracompact
QUESTION [7 upvotes]: By Definition, smooth manifolds are assumed to be Hausdorff and to satisfy the second countability axiom.
I have heard (but never seen written) that these assumptions imply paracompactness (and thus the existence of a Riemannian metric by the well-known construction using Partition of unity). 
Does anybody know a reference or Proof for paracompactness?

REPLY [27 votes]: Theorem: A countable atlas of charts for a Hausdorff $n$-manifold $M$ can be refined to a locally finite atlas.  In fact, each chart only needs to be trimmed.
Proof: Let $U_1,U_2,\ldots$ be the charts.  Each $U_i$, as a subset of $\mathbb{R}^n$, is the limit of a nested sequence of compact subsets $K_{i,1} \subseteq K_{i,2} \subseteq \ldots$.   Since $M$ is Hausdorff, each $K_{i,j}$ is closed in $M$.  So it suffices to delete
$K_{1,i} \cup \cdots \cup K_{i-1,i}$ from $U_i$ to make a new chart $V_i$.  Some of the $V_i$ might be empty, but this is no problem.<|endoftext|>
TITLE: vector balancing problem
QUESTION [49 upvotes]: I believe the solution posted to the arXiv on June 17 by Marcus, Spielman, and Srivastava is correct.

This problem may be hard, so I don't expect an off-the-cuff solution. But can anyone suggest possible proof strategies?
I have vectors $v_1, \ldots, v_k$ in ${\bf R}^n$. Each of them has euclidean length at most $.01$, and for every unit vector $u \in {\bf R}^n$ they satisfy
$$\sum_{i=1}^k |\langle u,v_i\rangle|^2 = 1.$$
Is it possible to find a set of indices $S \subset \{1, \ldots, k\}$ such that
$$.0001 < \sum_{i \in S} |\langle u,v_i\rangle|^2 < .9999$$
for every unit vector $u$? This will imply the same bounds when summing over the complement of $S$.
The $.01$ and $.0001$ aren't important; I just need the result for some positive $\delta$ and $\epsilon$. But they have to be independent of $k$ and $n$. (This may seem unlikely, until you try to construct a counterexample.)
The motivation is that this is a (very slightly simplified) equivalent version of the famous Kadison-Singer problem. A solution would have important consequences in operator theory, harmonic analysis, and C*-algebra. Many people have worked on this problem, but perhaps not in the above form, which I feel exposes the combinatorial difficulty which is the real root of the problem.

REPLY [16 votes]: This paper contains a proposed solution to the problem. The acknowledgements suggest that MO might have facilitated the solution, should it be correct. (I'm speculating that Gil Kalai found out about Nik Weaver's formulation of the problem from this MO question.)<|endoftext|>
TITLE: Embedding a Riemann surface in the sphere
QUESTION [9 upvotes]: Assume we have a Riemann surface, the underlying topological surface of which is a sphere with (possibly uncountably many) points removed. Can we always conformally embed this Riemann surface in the Riemann sphere? If not, can someone suggest a counter example?

REPLY [10 votes]: See e.g. here: 
Theorem 3.2.7. Any planar connected Riemann surface is biholomorphic to an open subset of $S^2$.
The proof is very straightforward: Exhaust a genus $0$ surface $S$ by relatively compact domains $D_n$ each of which necessarily has genus $0$. For each $D_n$ find a conformal embedding $f_n$ to $S^2$. Now, normalize the family of mappings $f_n$ to to send a point $x\in D_1$ to a fixed point $z\in {\mathbb C}$ and to have unit derivative (in a chart) at $x$. Then use normality of the family of maps $f_n$ to get the limit (for a subsequence). Lastly, check that the limit is injective. This is the same argument Caratheodory used in his proof of uniformization theorem.<|endoftext|>
TITLE: Combinatorial interpretation of the power of a series
QUESTION [14 upvotes]: I am trying to understand a result involving the power of a series that occurs in Gradstein and Ryzhik's Table of Integrals, Series, and Products. Result 0.314 (p.17, 7th ed.) is:
$$\left(\sum_{k=0}^\infty a_k x^k\right)^n=\sum_{k=0}^\infty c_k x^k$$
where $$c_0 = a_0^n, c_m=\frac 1 {ma_0} \sum_{k=1}^m (kn-m+k) a_k  c_{m-k}$$ for $m\ge1$ and $n\in\mathbb N$.
What is an appropriate combinatorial interpretation of this result?

One way I am trying to understand it is to see how it arises from the multinomial expansion
$$
\left(\sum_{k=0}^\infty b_k \right)^n = \sum_{\kappa\vdash k} \binom k \kappa b^\kappa
$$
which has the usual nice combinatorial interpretation of how to put objects in bins. This is suggested in the reference given in Gradstein and Ryzhik, which is an even older book: Smithsonian mathematical formulae and tables of elliptic functions, p.118. However, the additional structure provided by regrouping powers of $x$ after substituting $b_k = a_k x^k$ must surely have some significant, nontrivial and well-known combinatorial implications that I am simply unaware of. (I hope this is clear; the multiindex notation is new to me and I don't know a nice way to write the result of this last step.)

REPLY [7 votes]: Ralph gave a nice proof but I don't think it counts as a combinatorial interpretation.  However, we can work directly with his rearrangement (1).  If $a_k$ is the number of "objects" of "size" $k$, then $c_k$ is the number of vectors of $n$ objects with total size $k$. (This is the standard interpretation of the power of an ordinary generating function.)
Now consider an object of size $k$ to be a sack of $k$ distinguishable "atoms".  Let $N$ be the number of $(n+1)$-tuples of objects, of total size $m$, with one atom (out of the $m$ atoms altogether) distinguished.
We can start with one object, say of size $k$, append $n$ more objects of total size $c_{m-k}$  to make the total size up to $m$, then distinguish one of the $m$ atoms. So $N$ is the left side of (1).
Alternatively, start with one object, say of size $k$, distinguish one of its atoms, then extend this in either or both directions in $(n+1)c_{m-k}$ ways to make $n+1$ objects of total size $m$. (The factor of $n+1$ is the number of positions that the originally chosen object might end up in.) So $N$ is the right size of (1).

REPLY [3 votes]: A train is composed putting in series $n$ carriages. Carriages of size $k=0,1,\dots$ (number of places) are available in $a_k$ different colors. So, there are $c_k$ possible trains of $n$ carriages and $k$ places.
Now suppose a further carriage is attached, reaching an amount of $m$ places. Clearly, there are $n$ times as many ways of having your seat in the old $n$ carriages of such trains than in the new one (just imagine to switch the new carriage with one of the $n$ old ones), and this is your identity, written
$$n\sum_{k=0}^m ka_k c_{m-k}=\sum_{k=0}^m kc_k a_{m-k}\, .$$
(and, of course, this is the identity  $n\alpha'\alpha^n=\alpha(\alpha^n)'$ between g.f.'s)
Note that one may state this in a probability language in terms of an independence property, coming from a symmetry argument.

REPLY [2 votes]: Nice question! You are looking for a combinatorial interpretation of nth-fold self-convolution of a sequence. A quick google serach gave me this result on Catalan sequence. In this paper, a similar result is worked out on a series with Catalan coefficients. May be you can use similar idea for generic sequences.<|endoftext|>
TITLE: Set of permanents over binary square matrices
QUESTION [5 upvotes]: I am interested in the collection of possible values for permanents over square binary matrices. Consider $n \times n$ 0-1 matrices. The possible permanents for $n=1$ are $\{0,1\}$. For $n=2$ the possible permanents are $\{0,1,2\}$, and for $n=3$ the permanents are $\{0,1,2,3,4,6\}$. Note that $5$ is missing for $n=3$.
Let $p(n)$ denote the set of permanents for $n \times n$ binary matrices, and let $$p(\mathbb{N}) = \bigcup_{n \in \mathbb{N}} p(n)$$ We can show at least that $n! \in p(\mathbb{N})$ for all $n$ and that $p(\mathbb{N})$ is closed under products.    
Is this set $p(\mathbb{N})$ well understood? A pointer would be welcome. If $\mathbb{N} \subset p(\mathbb{N})$, a construction proof would be appreciated. It is not clear how to extend the trivial rectangular matrix with permanent $k$ to a square matrix.
The following related integer sequence counts distinct permanents: http://oeis.org/A087983

REPLY [8 votes]: That set consists of all natural numbers. Consider the $n\times n$-matrix $A=(a_{ij})$ where all numbers on the diagonal, first row and first column are equal to 1, all the other entries are 0. Then the permanent of that matrix is $n$. Indeed, each non-zero summand in the definition of permanent should start with some $a_{1,i}$. Then the $i$-th factor of that product, $i > 1$,  cannot be $a_{i,i}$. It must then be $a_{i,1}$. Hence all the other factors in that product are from the diagonal. Therefore exactly one non-zero summand in the permanent contains $a_{1,i}$, for every $i=1,...,n$. So the permanent contains exactly $n$ non-zero summands, and is equal to $n$.<|endoftext|>
TITLE: New arXiv procedures?
QUESTION [99 upvotes]: Recently I encountered a new phenomenon when I tried to submit a paper to arXiv. The paper was an erratum to another, already published, paper and will be published separately. I got a message from arXiv saying that I need to join the erratum with the original file. I was a little surprised receiving a reply from, obviously, a human being. Although I thought the request was a bit silly, I did what was requested, submitted the joint paper (the original union the errata), and forgot about it. But today I got a call from another mathematician. She tried to submit a paper with a title  "... II". The paper "... I" was already in the arXiv and submitted to a (very good) journal. Both papers solve similar but different problems. One of these problems is at least 40 years old. Her submission was denied:  she got a request from the arXiv to submit a union of that new paper and the old paper instead. This is quite silly. Is there now a special person in the arXiv who is making these decisions? It looks like there has been a change in how arXiv is managed. I understand that this is not a research question, and I make it a community Wiki. I post it here because several frequent MO users are affiliated with arXiv.

REPLY [51 votes]: I talked to the arXiv staff about Olga Kharlampovich's submissions and I now have some answers.  The letter that Olga posted here is a form letter that doesn't fit the facts.  The text overlap tool reported that the new submission substantially overlapped with the old submission.  After that, as far as I know, no moderator and no advisory committee was ever contacted.  Instead, an arXiv employee sent this stock response just to keep things moving.  After that, I was told, her case was added to the to-do list.  I was assured that as of last week, before this question was posted to MathOverflow, her submission was already slated to be reverted in her favor on Monday.
Obviously this is not satisfactory.  I am one of the moderators (and not the only one) who should have seen the appeal.   The e-mail said that someone like me had seen it and rejected her appeal, but apparently no such thing happened.  It seems that the submitted version (which I think is now version 3) had something like 75% text overlap with the previous version (version 2) of arXiv:1111.0577.  It's not so unreasonable to flag such a submission.  After that it wasn't handled properly.  I do not want to name names and lead people to pour opprobrium on the overworked arXiv staff.  (There are only two of them who handle daily submissions.)  But I want to make this story sound accountable, so I can say that some of my information came directly from Paul Ginsparg.
To go back to the title question, no there has not been any great change in arXiv management.  You could certainly argue that there is insufficient management, but that's not the same thing.
People are also asking about the policy by which papers are labelled as having text overlap with other papers.  A clearer statement of that policy would be useful, but that is a separate question from Olga's case.

According to e-mail that I just saw, this morning Olga was given the option of reverting the previous arXiv paper to Part I and submitting Part II separately.  Her answer, according to what I saw, was that she elected to keep it as a replacement after all.  I am mentioning this so that readers who see arXiv postings this week won't think that injustice continues.
I stand by my explanation that the stock e-mail that she was sent didn't fit the facts, and that her appeal should not have been stonewalled.  (In fact her appeal was soon seriously considered internally, but that was not explained.)  However, in the original posting, Olga's name was withheld supposedly to protect her interests.  Although I understand that anonymity is sometimes vital even in a public accusation, in this case I don't see how it helped matters.<|endoftext|>
TITLE: injectivity of the pull-back via a finite map
QUESTION [5 upvotes]: Let $f:X \to Y$ be a finite map of (smooth, compact) complex algebraic varieties.
Then a map $f^*$ is defined at the level of Chow and cohomology rings. Say that for simplicity we work with rational coefficients.
Question: is $f^*$ injective? I know this is true when $f$ is the natural map onto the quotient via the action of a finite group. Indeed, it is simply the inclusion of the $G$-invariants.
In the case when $X$ and $Y$ are smooth and projective, is $f^*$ injective at the level of each individual $H^{p,q}$?
(in the case of cohomology, Proposition 2.2 of http://www.maths.ed.ac.uk/~aar/papers/smith2.pdf answers affirmatively my question, but I would anyway be curious of seeing a simpler proof in my generality)

REPLY [13 votes]: Let $f:X\to Y$ be a finite surjective map of degree $d$ between smooth projective varieties of dimension $n$. Then $f$ is flat (apply EGA IV 2 Prop 6.1.5), so Example 1.7.4 of Fulton's Intersection Theory shows that $f_{\ast}\circ f^{\ast}$ is multiplication by $d$ on the Chow groups $CH^*(Y)$.
Hence the kernel of $f^{\ast}$ on $CH^{\ast}(Y)$ is of $d$-torsion. 
It remains true that $f_{\ast}\circ f^{\ast}$ is multiplication by $d$ on the integral cohomology of $Y$, hence that the kernel of $f^{\ast}$ on $H^{\ast}(Y,\mathbb{Z})$ is of $d$-torsion. It may be seen as follows.
Recall that if $\alpha\in H^{k}(Y)$ and $\beta\in H^{k}(X)$, their Poincaré-dual classes are $\alpha\cap [Y]\in H_{n-k}(Y)$ and $\beta\cap [X]\in H_{n-k}(X)$  and that $f_*:H^{k}(X)\to H^{k}(Y)$ is defined to be the push-forward at the level of Poincaré-dual classes. Combining the projection formula $f_{\ast}(f^{\ast}\alpha\cap[X])=\alpha\cap f_{\ast}[X]$ and the fact that $f_{\ast}[X]=d[Y]$ because $f$ is a degree $d$ map, we get the formula  $f_{\ast}\circ f^{\ast}(\alpha)= d\alpha$ we wanted to prove.
This shows that $f^{\ast}$ is injective on Chow groups modulo torsion, on integral cohomology modulo torsion, on rational cohomology, on cohomology with complex coefficients... This last point moreover implies that $f^{\ast}$ is injective on the pieces $H^{p,q}(Y)$ of the Hodge decomposition.
However, $f^{\ast}$ need not be injective on the integral cohomology. If $Y$ is an Enriques surface and $X$ is its universal cover (a K3 surface), $H^2(Y,\mathbb{Z})_{tors}=\mathbb{Z}/2\mathbb{Z}$ and $H^2(X,\mathbb{Z})$ has no torsion so that $f^{\ast}$ is not injective on $H^2$.
Torsion in cohomology is not the only reason why $f^{\ast}$ may fail to be injective on $CH^{\ast}(Y)$. For example, let $X$ be an elliptic curve. Suppose that its $2$-torsion is generated by $\sigma$ and $\tau$, and let $Y=X/\sigma$. We still denote by $\tau$ the image of $\tau$ in $Y$. Then, working in $CH^1=Pic$, we get $f^*(\tau-0)=\tau-0+\tau+\sigma-0-\sigma=2\tau-2.0=0$, so that $f^{\ast}$ is not injective on $CH^1$.
When dealing with $0$-cycles, it is possible to apply the theorem of Roitman. For example, if $H^1(Y,\mathbb{C})=0$, $f^*$ is injective on $CH_0$. Indeed, the Albanese variety of $Y$ is then trivial, and, by Roitman, $CH_0$ has no torsion.<|endoftext|>
TITLE: Enveloping algebras which are solvable as Lie algebras
QUESTION [8 upvotes]: Let $L$ be a Lie algebra over a field $F$ and denote by $U(L)$ the universal enveloping algebra of $L$. Regard $U(L)$ as a Lie algebra with respect to the Lie bracket $[a,b]=ab-ba$ for any $a,b\in U(L)$. If $U(L)$ is solvable as a Lie algebra, is $L$ necessarily abelian?

REPLY [12 votes]: If $F$ has characteristic different from 2 the answer is yes. This follows from Corollary 6.1 in the paper by D. Riley - A.Shalev: The Lie structure of enveloping algebras, J. Algebra 162, 46-61 (1993).
On the other hand, in characteristic 2 this conclusion is false. For instance, if $L$ is a 2-dimensional nonabelian Lie algebra or a 3-dimensional Heisenberg algebra, then one can see by explicite calculations that $U(L)$ is Lie solvable of derived length 3.<|endoftext|>
TITLE: Variational formulation for bilaplacian
QUESTION [8 upvotes]: I am trying to derive a variational formulation for the following problem $$\left\{ \begin{array}{ll} \Delta^2u=f, & \Omega \\ \Delta u+\rho \partial_{\nu}u=0, & \partial \Omega \end{array}\right.$$
where $\rho>0$ is constant. I intend to show that the right functional setting is $H^2(\Omega)\cap H^1_0(\Omega)$ and to prove that the resulting problem is well posed.
I am confused as to how to establish the right functional setting, so for a start I choose $C_0^2(\Omega)$ as a space of test functions (functions in $C^2(\Omega)$ compactly supported in $\Omega$) so that the boundary condition makes sense.
Multiplying the equation by $v\in C_0^2(\Omega)$ and integrating over $\Omega$ we obtain $$\int_{\Omega} \Delta^2u\cdot v\,dx=\int_{\Omega} fv\,dx$$
and now integrating by parts (Green's identity) twice on the left hand side we obtain 
\begin{eqnarray*}
\int_{\Omega} \Delta^2u\cdot v\,dx &=& \int_{\Omega} div (\nabla \Delta u)\,dx \stackrel{Green}{=} \int_{\partial \Omega} \partial_{\nu}(\Delta u)v\,d\sigma- \int_{\Omega} \nabla \Delta u\cdot \nabla v \,dx
\\
&\stackrel{Green}{=}& \int_{\partial \Omega} \partial_{\nu}(\Delta u)v\,d\sigma - \int_{\partial \Omega} \Delta u\cdot \partial_{\nu}v\,d\sigma + \int_{\Omega} \Delta u\cdot \Delta v\,dx
\\
&=& \int_{\partial \Omega} \partial_{\nu}(\Delta u)v\,d\sigma + \int_{\partial \Omega} \rho \partial_{\nu}u \cdot \partial_{\nu}v\,d\sigma + \int_{\Omega} \Delta u\cdot \Delta v\,dx
\end{eqnarray*}
where in the last equality I use the boundary condition. Now, enlarging the space of test functions by taking the closure of $C_0^2(\Omega)$ in $H^2(\Omega)$, namely $H_0^2(\Omega)\subset H_0^1(\Omega)\cap H^2(\Omega)$ the first integral vanishes ($v$ has zero trace) so our variational formulation is $$\int_{\partial \Omega} \rho \partial_{\nu}u \cdot \partial_{\nu}v\,d\sigma + \int_{\Omega} \Delta u\cdot \Delta v\,dx=\int_{\Omega} fv\,dx$$
How can I rigorously conclude that I need to take the whole $H_0^1(\Omega)\cap H^2(\Omega)$ as my space of test functions?
In order to prove that the problem is well posed I intend to use Lax-Milgram theorem as usual, but I am confused as to how to tackle the integral over $\partial \Omega$. I define a bilinear form $$B(u,v)=\int_{\partial \Omega} \rho \partial_{\nu}u \cdot \partial_{\nu}v\,d\sigma + \int_{\Omega} \Delta u\cdot \Delta v\,dx$$ and I want to check continuity and coercitivity. For the first one I have $$|B(u,v)|\leq \int_{\partial \Omega} \rho |\partial_{\nu}u \cdot \partial_{\nu}v|\,d\sigma + \int_{\Omega} |\Delta u\cdot \Delta v|\,dx$$
For the second integral we have $$\int_{\Omega} |\Delta u\cdot \Delta v|\,dx\leq ||\Delta u||_0||\Delta v||_0\leq ||u||_{H^2(\Omega)} ||v||_{H^2(\Omega)}$$ but what about the first one?
Thanks in advance for any insight.

REPLY [9 votes]: To begn with, your Boundary-Value Problem (BVP) is under-determined, because it lacks one boundary condition: because the PDE is elliptic and fourth-order, you need two boundary conditions, not only one. Because you insist on working with $H^2\cap H^1_0$, it seems that the hidden BC is
$$u=0\qquad\hbox{on }\partial\Omega.$$
So let us assume that your BVP includes this condition. Then your calculation works whenever $v\in C^2(\overline\Omega)$ (you should avoid $C^2_0$, because its elements satisfy $\partial_\nu v=0$ as well).If $v$ vanishes on the boundary (but not necessarily its normal derivative), a solution $u$ of the BVP does satisfy
$$B(u,v)=0.$$
If $\Omega$ is bounded with a smooth boundary, then 
$$|\partial_\nu w|_{L^2(\partial\Omega)}\le C|w|_{H^2}\le C'|\Delta u|_{L^2},$$
where the second inequality is a kind of Poincar\'e inequality. Therefore $B$ is continuous over $H^2(\Omega)$ and $B(u,v)=0$ extends to all elements $v\in H^2\cap H^1_0$ by density.
The existence follows from Lax--Milgram, using the Poincar\'e inequality to show that $B$ is coercive over $H^2\cap H^1_0$.
Conversely, suppose that $u\in H^2\cap H^1_0$ and $B(u,v)=0$ for every $v\in H^2\cap H^1_0$. Then making the calculations backward, you find
$$\langle \Delta^2 u,v\rangle_\Omega+\int_{\partial\Omega}(\Delta u+\rho\partial_\nu u)\partial_\nu vds(x)=0,$$
where the first term is duality between $H^{-2}(\Omega)$ and its dual. Taking all $v\in{\mathcal D}(\Omega)$, you first obtain $\Delta^2 u=0$ in the sense of distributions. There remains therefore
$$\int_{\partial\Omega}(\Delta u+\rho\partial_\nu u)\partial_\nu vds(x)=0,$$
for every $v\in H^2\cap H^1_0$. Because $v\mapsto\partial_\nu v$ is onto over $H^{1/2}(\partial\Omega)$, this gives you $\Delta u+\rho\partial_\nu u=0$. Finally, $u=0$ on the boundary is ensured because $u\in H^1_0$.<|endoftext|>
TITLE: Identifying lattices
QUESTION [10 upvotes]: I wrote a program that numerically searches for lattices in $\mathbb{R}^d$ with high sphere packing densities. As I have been running the program, it has been able to find, in addition to well-known lattices such as the laminated lattice $\Lambda_d$ and the Coxeter Todd-related lattices $K_d$, a few interesting looking lattices, which I have been unable to identify. The lattices I have found so far are not as dense as $\Lambda_d$ or $K_d$, but are reasonably dense, and are nice integral lattices. Since I found them through a sort of a local optimization, I suppose they are probably perfect. I looked through the lattices listed on the Sloane-Nebe Catalog of Lattices and did not find any matches, but there do not seem to be many lattices listed there.
Here is an example of one of the lattices I find in $\mathbb{R}^{11}$. The Gram matrix is given by
$$ G = \left(\begin{array}{ccccccccccc}
8&3&3&2&3&4&4&4&4&4&4\\
3&8&4&4&4&-1&4&-1&4&4&4\\
3&4&8&0&0&-1&0&-1&0&4&4\\
2&4&0&8&2&2&2&1&2&4&0\\
3&4&0&2&8&3&4&-1&4&1&1\\
4&-1&-1&2&3&8&0&4&0&0&0\\
4&4&0&2&4&0&8&0&4&2&2\\
4&-1&-1&1&-1&4&0&8&0&0&0\\
4&4&0&2&4&0&4&0&8&2&2\\
4&4&4&4&1&0&2&0&2&8&4\\
4&4&4&0&1&0&2&0&2&4&8
\end{array}\right)\text. $$
The number of spheres in successive shells (equiv. theta function) are: norm 8, 308; norm 10, 320; norm 12, 680; norm 14, 1472. The packing density is $1/14\sqrt{7}=0.02699\ldots$ (number density for non-overlapping spheres of radius 1, compare to $0.03208\ldots$ for $K_{11}$ and $0.03125$ for $\Lambda_{11}$).
Does anybody know where I might be able to find if these lattices have been studied before?

REPLY [4 votes]: (Not an answer, but I could not find in the faq how one leaves a comment.) 
Your lattice is perfect.<|endoftext|>
TITLE: Normal form(s) for the elements of hyperbolic triangle groups
QUESTION [5 upvotes]: I'm seeking a reference or a sketch for any sort of normal form that would enable rapid enumeration without redundancies of the elements of hyperbolic triangle groups and/or von Dyck groups.

REPLY [3 votes]: The general reference for this topic is "Word processing in groups", Epstein et. al., MR1161694 (93i:20036). Your groups all have automatic structures, and that book gives an enumeration method as you ask for, which applies to any automatic structure on a group.<|endoftext|>
TITLE: Morse theory in TOP and PL categories?
QUESTION [17 upvotes]: Apparently there are topological and piecewise linear versions of Morse theory. I would like to know of references that treat these topics.
How is a Morse function defined for compact manifolds (with boundary) in the TOP and PL categories?
It is well known that smooth Morse functions always exits for compact smooth manifolds. Are there similar results in the TOP and PL categories?
It is possible to classify closed smooth surfaces via smooth Morse theory. Is there a classification theorem for closed TOP (respestively PL) surfaces via topological (respectively PL) Morse theory?
Thanks

REPLY [5 votes]: I think that Daniele and Sergey have, between the two pretty much answered my question. However, I would like to add the following:

Regarding TOP Morse theory, in [Mor1959] M. Morse laid the foundations of the theory of topological non-degenerate functions, and proved the TOP Morse inequalities. 
Also, in [Kui1961] the topological version of the Reeb-Milnor theorem for DIFF manifolds is proven. That is, a TOP n-dimensional closed manifold admiting a TOP Morse function having exactly two non-degenerate critical points is homeomorphic to the n-sphere.
Another source containing further results for topological manifolds via TOP Morse theory that parallel those obtained for differentiable manifolds is J. Cantwell's paper [Can1967].
Kirby and Siebenmann's "Foundational essays on topological manifolds, smoothings, and triangulations is freely available here. In section 3 of Essay III (p. 80) they define Morse functions in the DIFF and TOP categories for manifolds possibly with boundary.
Finally, simple examples of non-differentiable TOP Morse function are easily found. For example, the absolute value function on $\mathbb{R}$, $x\rightarrow |x|$. The origin is a non-degenerate critical point. Also the height function restricted to the double cone (i.e., the space formed by the cones $x^{2}+y^{2}=(z\pm1)^{2}$) has exactly two non-degenerate critical points (the tips of the cones).
Regarding PL Morse theory, J. Harer's slides contain an interesting approach using homology. In particular, a PL Morse function is defined using Betti numbers.

REFERENCES:
[Cant1967] J. Cantwel, Topological non-degenerate functions, Tohoku Math. Journ., 20 (1968), 120-125.
[Kui1961] N.H. Kuiper, A continuous function with two critical points, Bull. Amer. Math. Soc., 67(1961), 281-285.
[Mor1959] M. Morse, Topological non-degenerate functions on a compact manifold $M$, Journal d'Analyse Math., 7 (1959), 189-208.<|endoftext|>
TITLE: Triangularizing a matrix with function entries
QUESTION [6 upvotes]: Hi Everybody!
Given a matrix, with smooth functions as arguments is there any result which say about its triangularization?
I know that, the question is in affirmative for diagonalizing a matrix which has distinct eigenvalues. Infact, we can show by Implicit function theorem that eigenvalues can be chosen to be $\mathcal{C}^{1}$functions. We can also calculate projection operators for the matrix using Cauchy integral formula. 
But, I have not found any result for triangularization. An answer or any references would be great help. Thank you. 
ADDED LATER: I have found in the Micheal Taylor's book 'Pseudo-differential operators' Page.72 a claim that for a matrix $K(\xi)$ there is always a measurable unitary matrix $U(\xi)$ such that $U^{-1}(\xi)K(\xi)U(\xi)$ is an upper triangular matrix. In this case $K(\xi)$ is smooth matrix in $\xi.$ He does not say exactly the name of this result though.

REPLY [9 votes]: You cannot triangularize smoothly the parametrized matrix
$$A(z)=\begin{pmatrix} 0 & 1 \\\\ z & 0 \end{pmatrix}$$
about $z=0$. The eigenvalues, square roots of $z$ aren't smooth and, above all, cannot be distinguished one from the other: when $z$ runs over a circle $C(0;\epsilon)$ in the complex plane, and an eigenvalue is selected continuously, its sign changes after $2\pi$.<|endoftext|>
TITLE: Why are there no triple affine Hecke algebras?
QUESTION [17 upvotes]: This question arised after I recently stumbled upon the paper "Triple groups and Cherednik algebras". Doubly affine Hecke algebras are sort of a natural object to consider after finite and affine Hecke algebras. This makes one wonder, why are there no "triple affine Hecke algebras"? Or, if such a construction exists, why are they not useful? (The theory of doubly affine Hecke algebras has proved to have deep consequences and relations with many fields of mathematics, see this previous question.)

REPLY [6 votes]: This is a wonderful and highly suggestive question. In addition to the fascinating hints provided by Stephen's answer, there are also reasons from geometric representation theory (as well as from physics) to expect that a triple affine Hecke algebra theory might exist. 
A very faint hint in representation theory can be seen for example from my paper
Elliptic Springer Theory with David Nadler
(I am sure there are better references for this faint hint but I'm sadly ignorant
of the elliptic literature - just now learned of the beautiful paper of Ion-Sahi from reading this question!)
First recall that the Weyl group appears as endomorphisms of the 
Grothendieck-Springer sheaf, a remarkable perverse sheaf (or $D$-module) on the nilpotent cone of a Lie algebra. However, if you consider the ext-algebra of the Springer sheaf, you find instead the degenerate affine Hecke algebra - i.e., you "sprout" a polynomial ring (functions on a torus Lie algebra, or if you prefer, torus-equivariant cohomology of a point). Next we can consider the group version, the Hotta-Kashiwara sheaf, a $G$-equivariant sheaf on $G$ constructed from the group version of the Grothendieck-Springer resolution. It has an endomorphism algebra which is an affine Weyl group, i.e. we've added the group algebra of the lattice, but an Ext algebra which is a form of the degenerate DAHA (really we are missing all the interesting parameters here, one ought to consider a ``mixed" version to see something closer to the objects of interest). 
Finally there's an elliptic version, living on the "elliptic group" -- the moduli of semistable $G$-bundles on an elliptic curve (recovering the previous two when the curve degenerates to a nodal, then cuspidal, Weierstrass cubic). We discuss an obvious elliptic version of the Springer sheaf there -- in fact a special case of the theory of geometric Eisenstein series (a much more thorough study is in Dragos Fratila's thesis). Its endomorphisms are identified with a double affine Weyl group (ie two copies of the same lattice, not duals)... but its Exts will again sprout another polynomial ring, and should be considered a very degenerate form of a "degenerate triple affine Hecke algebra".
One really needs to make it much less degenerate to get something close to answering
this question - i.e., add a "q" (mixed version) and a "t" (twisted $D$-module/quantum group parameter) at the least. But it's something natural and triple.
[I can't say anything reasonable about the physics in finite length, or maybe at all, but I'll allow myself just a vague hint: double affine Hecke algebras (at least in type A) arise (again in very degenerate form) in studying the Seiberg-Witten integrable systems (i.e. moduli space after compactification on a circle) of $N=4$ super Yang-Mills in four dimensions - which is itself a reduction on two circles of the mysterious (2,0) theory in 6 dimensions, aka "theory $\mathfrak X$".. this realization shows that there are not just two, but three, circles in the picture - switching around two gives an $SL_2Z$ symmetry, aka Langlands/S/electro-magnetic duality, but switching around all three ought to give some
hint of $SL_3Z$ symmetry.]<|endoftext|>
TITLE: generalisation of Cauchy-Riemann equations to 3D
QUESTION [11 upvotes]: Hi, harmonicity in 2d is preserved under mappings that satisfy the Cauchy-Riemann equations.
What about 3D? What conditions should a mapping satisfy to preserve harmonicity?
is there a general characterization a la CR for 3D?
Here is an example of non-trivial such mapping
Let $u(x,y,z)=U(X,Y,Z)$ where $$X=xy+z,~~~~ Y= \frac{\sqrt3}4 (x^2-y^2) - \frac{xy}{2}+z
,~~~~ Z= -\frac{\sqrt3}4 (x^2-y^2) - \frac{xy}{2}+z$$
(I found this example by first assuming $X=xy+z$ then guessing for Y,Z from the overdetermined system that they satisfy... hope it's right...)

REPLY [2 votes]: I see the answer I gave yesterday as a guest is not posted, an email check revealed I did not properly register... 
Ok, now being a registered thing I can say:
The Cauchy-Riemann equations directly relate to all entire functions being harmonic. In $\mathbb{C}$ you can even factorize the Laplacian operator into two things and you see that only one of those factors slams the stuff to zero so the outcome is harmonic.
Basically the CR-equations say:
$$
\frac{\partial}{\partial y} = i \cdot \frac{\partial}{\partial x}
$$
so that
$$
\frac{\partial^2}{\partial y^2} = - \frac{\partial^2}{\partial x^2}
$$
Hence all analytic (entire) functions on $\mathbb{C}$ are also harmonic but it does not go the other way...
Now for $\mathbb{R^3}$ space there is indeed a complex multiplication possible but this does not give rise to harmonic functions. In this answer it will go too far to explain how 3D complex multiplication works, but as far as I know after 26 years of looking at that stuff:
You cannot make harmonic functions that way.
Here comes a small monkey out of the sleeve:
I have a separate website related to all 3D complex numbers things:
http://3dcomplexnumbers.net/
The 3D Cauchy-Riemann equations are beautiful yet they do not give harmonic stuff. Hope I answered your question a tiny bit...<|endoftext|>
TITLE: Applications of nonconstructive mathematics
QUESTION [17 upvotes]: What are good/interesting examples of theorems than can be proven classically, but not constructively, and have applications in e.g. physics?

REPLY [2 votes]: A famous Martin Gardner mathematical games column had a number of April Fool items, one of which was about a guy getting rich by duplicating gold spheres using the Banach-Tarski theorem.  I guess that doesn't count as a real application, but it's surely nonconstructive and I can't resist mentioning it.<|endoftext|>
TITLE: when a pseudo-differential operators to be compact?
QUESTION [7 upvotes]: In the theory of Pseudo-differential operators,when a symbol $a(x,\xi)\in S^{0}$,then the operator $a(x,D)$ defined by$$a(x,D)u=\int{e^{ix\xi}a(x,\xi)\widehat{u}}d \xi$$ is $L^2$ bounded.$ $
  My question is when add what condition (as least as possible) to the symbol $a(x,\xi)$ can assure the operator $a(x,D)$ to be compact on $L^2$ ?Or is there a equivalent condition of this ?I guess some decayed assumption on $a(x,\xi)$ (about $\xi$) is necessary.but I'm not sure.some references about this are also appreciated
Added:There is a equivalent condition of a pseudo-differential operators to be compact.Assume that $g \leq g^{\sigma}$,that g is $\sigma$ temperate.and that m is $\sigma$,g temperate.then the operators $a^{w}(x.D)$with $a\in S(m,g)$are compact (bounded) in $L^{2}$ if and only if $m \to 0 $ at$\infty$ (m is bound).When we let g to be the metric $|dx|^{2}+|d \xi|^{2}/(1+|\xi|^2)$,and $m=(1+|\xi|^{2})^{\frac{\mu}{2}}.$then the class $S(m,g)$ become the usual$S^{\mu}$,Is it implying that when $\mu<0$,then $a^{w}(x.D)$ is compact in $L^{2}$ without further assuming the kernel to be compact supported ?
More precisely, considering the symbol $a(x,\xi)=V(x)(1+|\xi|^2)^{-1}$(it appears in the scattering problem of the schr\ddot{o}dinger operators),now we have the differential condition (rather than the integral condition)$\partial^{\alpha}{V} \leq (1+|x|)^{-\beta-|\alpha|}$,with $|\beta|>\frac{1}{2}$. Then does the operator $a(x,D)$ compact in $L^{2}$ ? 
I try to prove it by decomposing both the $\xi-space$ and $x-space$ with a partition of unity as the almost orthogonality method. But it seems without the use of the decay of $x-space$,the compactness wouldn't be obtained(just think about the case $(1-\triangle)^{-1}$.

REPLY [7 votes]: First a simple remark: in the formulation of the question $\mu$ should be replaced by $\mu/2$ to get 
$S(m,g)=S^\mu_{1,0}$. 
Next the "if and only if" is correct but misleading since it is a condition on all the class of operators with symbols in $S(m,g)$. It is of course possible that a given operator is compact without that condition: take for instance the resolvent of the 2D operator
$$
-∆_{x,y}+x^2y^2.
$$
It is compact in $L^2(\mathbb R^2)$. That example is due to Barry Simon and was extended to a more general setting by Charles Fefferman (Bull. AMS, 1983) in his discussion of Schrödinger equation with polynomial potentials.<|endoftext|>
TITLE: Inequality of arithmetic and geometric means for the lattice polytopes?
QUESTION [6 upvotes]: Let $K$,$L$,$M$ be convex lattice polytopes (so their vertices are in $\mathbb{Z}^n$) in $\mathbb{R}^n$ satisfying $K+L\subseteq M+M$ (Minkowski sum). Do we always have
$$|K\cap\mathbb{Z}^n|\cdot|L\cap\mathbb{Z}^n|\le |M\cap\mathbb{Z}^n|^2 ?$$
I looked at books/papers on Erhart polynomial and Brunn-Minkowski inequality etc., but did not find an satisfying answer. Any comment? Thanks.

REPLY [6 votes]: This inequality does not necessarily hold, at least for $n\geq 3$. It is somehow connected with the fact that there is no Pick's formula in more than two dimensions since there exists a convex lattice polytope with a large volume but containing a small number of lattice points. 
So, for instance, for $n=3$ let $M$ be the convex hull of the points $(0,0,0)$, $(1,1,0)$, $(0,1,2k)$ and $(1,0,2k)$. Then $|M\cap {\mathbb Z}^3|=4$, but $(M+M)\cap {\mathbb Z}^3\supset \{(1,1,t):0\leq t\leq 2k\}$. So $K$ and $L$ can be chosen as vertical segments containing $k$ lattice points each.<|endoftext|>
TITLE: Reed-Muller-Codes
QUESTION [5 upvotes]: Let $F$ be the field with two elements, $V_m=F^{2^m}$.Let $R(r, m)\subset V_m$ be the 
binary Reed-Muller Code. Define $R_m:=R(1, m)$. Then the dimension of $R_m$ is 
$1+m$ and its minimal distance is $d(R_m)=2^{m-1}$. (Cf. for example the book of Luetkebohmert). Hence the information rate is $I(R_m)=\frac{1+m}{2^m}$ while the relative
minimal distance of $R_m$ is $rd(R_m)=\frac{1}{2}$.
Now $R_5$ has the same relative minimal distance as $R_4$ (and thus should have approximately the same error correction abilities), while the information rate of $R_4$ is much better ($I(R_4)\approx 0.31$ while $I(R_5)\approx 0.19$).
I furthermore read in Luetkebohmert (and elsewhere) that $R_5$ was used in practice, e.g. during a Mariner space shuttle mission.
Question: Given the values above, why would anybody use $R_5$ and not $R_4$?
(I hope this question is not too easy for MO, but I do not see the answer at the moment.)
Edit: It seems that the code $R_5$ might have been used for Mariner, because then one has really one code word per pixel of the image. (Compare the comments)
Still let me note that the $R_m$ all have the same relative minimal distance, that there
exists an efficient (poly time?) algorithm to decode them, and that 
$I(R_m)$ tends rapidly to zero when $n\to\infty$. And I find this a bit strange: For an "interesting" or even "celebrated" family of codes $(C_m)_m$, which all have the same relative minimal distance, I would naively have expected something like $(I(C_m))_m$ should be a decreasing (but bounded or even converging) sequence.

REPLY [6 votes]: This began as a comment on the comments by @Chris and @Jyrki but posted as an answer because I don't have sufficient reputation on this site.
The standard decoding algorithm for Reed-Muller codes uses majority-logic
decoding which in turn requires hard-decision demodulation. That is, the
input to the decoder is a $32$-bit vector that is obtained from a demodulator
a device that  processes $\mathbf x$, a vector of $32$ real numbers,
to produce a $32$-bit sequence.
However, for first-order 
Reed-Muller codes, it is possible to use soft-decision demodulation and decoding in
Euclidean space, that is, the decoder finds the (binary) codeword that is closest
in the Euclidean metric to $\mathbf x$.  This is because the $64$ codewords of $R_5$, regarded as vectors in  $\{+1, -1\}^n$ (instead of $\{0, 1\}^n$) are just the rows of a $32\times 32$ Hadamard matrix $H$ (or 
the negatives of the rows).  All that needs to be done is to compute 
$\mathbf y = \mathbf x H$.
The location of the entry with the largest magnitude in $\mathbf y$ together with
the sign of the entry, tells us which of the $64$
codewords is the nearest in Euclidean distance to $\mathbf x$.
If the codewords are arranged so that $H$ is in Sylvester form, then a fast
Hadamard transform algorithm (very similar to a Fast Fourier Transform algorithm)
can be used to compute $\mathbf x H$ using $160 = 32\log_2 32$ additions ans
subtractions.  In fact, this form of the computation was used in the Mariner 
program; the machine devised to carry out the calculations was referred to as 
the Green Machine.
An article describing many of the details of the coding for the Mariner
missions can be found in a conference proceedings published by Wiley
under the title Error-Correcting Codes, H. B. Mann (ed.), 1969. 
Some details, and a diagram of part of the Green Machine can be found
in Chapter 14 of MacWilliams and Sloane's The Theory of Error-Correcting
Codes, North-Holland, 1978.  I am sure that Jyrki has the latter on
his bookshelf.<|endoftext|>
TITLE: Looking for a "scientific" application of a recreational puzzle.
QUESTION [7 upvotes]: First of all the puzzle. 
A barman's got 15 glasses which are initially somehow divided into several stacks. The barman repeats the following process a thousand times. He takes the top glass from each (nonempty) stack and forms a new stack with these glasses. Which set of stacks (in terms of their heights) will he come up with?
It's a nice one, give it some thought =)
Having toyed with this problem and its obvious generalizations for an arbitrary number of glasses I came up with the (totally intuitive) hypothesis that such a process and its long-term behavior might emerge in some more-or-less advanced field of research (algebra/geometry/mathematical physics). Can anybody comment?
Update. One can also notice that in this special case of 15 glasses both the problem's statement and the answer are pleasantly simple. I'd be very interested and even somewhat surprised to hear an accordingly simple proof.

REPLY [6 votes]: I don't know about scientific applications, but you can solve the puzzle as follows (I hope you consider this solution simple):
Look at the Ferrers diagram of the partition of 15 you have, and consider the sum of the Manhattan distances from each dot to the corner of the diagram. At every step of the barman's process, this sum either stays the same or decreases, and it only stays the same if the number of stacks is at least as large as the size of the largest stack minus 1. In this case, we get the new Ferrers diagram by "rotating" the first column of the Ferrers diagram to make it into the first row, and shifting the rest of the diagram diagonally. For example:
x*****      xxxxx
x***        *****
x**     ->  ***
x           **
x

If we assume the existence of a cycle of partitions, then this sum must be constant along the cycle, so the Ferrers diagrams in the cycle can be produced by simply cyclically shifting each diagonal. If there are two adjacent partially filled diagonals, then since adjacent numbers are relatively prime, eventually in this cycle we will get a Ferrers diagram where a hole in the diagonal closer to the corner is next to a dot in the further diagonal, which is not allowed. For instance, if we continue the previous example, we eventually see the hole in the fifth diagonal line up with the dot in the sixth diagonal:
******     *****     ****     *****     *****     *****     ******     ****
****       *****     ****     ***       ****      ****      ****       *****
***    ->  ***   ->  **** ->  ***   ->  **    ->  ***   ->  ***    ->  ***
*          **        **       ***       **        *         **         **
*                    *        *         **        *                    *
                                                  *

Thus, there is at most one partially filled diagonal in any partition that cycles, and the only partition of $15$ satisfying this condition is $15 = 1+2+3+4+5$. Since the number of partitions of $15$ is less than $1000$, the barman enters this cycle before the end of his process.<|endoftext|>
TITLE: Stacks in modern number theory/arithmetic geometry
QUESTION [19 upvotes]: Stacks, of varying kinds, appear in algebraic geometry whenever we have moduli problems, most famously the stacks of (marked) curves. But these seem to be to be very geometric in motivation, so I was wondering if there are natural examples of stacks that arise in arithmetic geometry or number theory.
To me, as a non-expert, it seems like it's all Galois representations and estimates on various numeric measures (counting points or dimensions of things) based on other numeric things (conductors, heights, etc). 
I asked this question at M.SE (here : https://math.stackexchange.com/questions/143746/stacks-in-arithmetic-geometry please vote to close if you can) because I thought it a bit too 'recreational', but with no success. What I am after is not just stacks which can be seen as arithmetic using number fields or rings of integers, but which are actually used in number-theoretic problems, or have a number-theoretic origin. Maybe there aren't any, but it doesn't hurt to ask.
EDIT: I have belatedly made this question CW, as I've realised, too late, that there is clearly not one correct answer.

REPLY [13 votes]: Here are two applications of stacks to number theory.
1) Section 3 of this paper, which solves the diophantine equation $x^2 + y^3 = z^7$, explains the connection between stacks and generalized Fermat equations.
2) This post explains how stacks fit into the proof of Deuring's formula for the number of supersingular elliptic curves over a finite field.<|endoftext|>
TITLE: Acute triangulation
QUESTION [13 upvotes]: Assume that $S$ is a finite 2-dimensional simplicial complex equipped with a metric $d$
such that each triangle is isometric to a plane triangle (so $(S,d)$ is a polyhedral space).

Is it possible to subdivide the triangulation of $S$ so that it will have acute triangles only?

Comments:

It is well known that any triangle admits a triangulation with only acute triangels; one can see the proof in the following picture. But I do not see a way to fit such triangulations together. 

Such a triangulation would be useful in the proof of Zalgaller's theorem: any n-dimensional polyhedral space admits a length-preserving piecewise linear map to $\mathbb R^n$. Well, it would help only if $n=2$, for larger $n$, we should say that a simplex is acute if it contains its circumcenter.

REPLY [3 votes]: Just to add to @Gjergji's answer: there are also algorithms to compute such things. See, for example,Erten and Ungor, CCCG 2007 and references therein.<|endoftext|>
TITLE: When is a valued field second-countable?
QUESTION [8 upvotes]: Let $R$ be a valuation ring, with fraction field $K$ and residue field $k$. Denote by $\Gamma=K^{\times}/R^{\times}$ the valuation group (assumed nontrivial).
The valuation $v:K^{\times}\to\Gamma$ decomposes $K^{\times}$ as a disjoint union of nonempty open subsets, indexed by $\Gamma$. Each of these is homeomorphic to $R^{\times}$, which is in turn (using the reduction map to $k$) a  disjoint union of nonempty open subsets, indexed by $k^{\times}$.
  We conclude that any basis for the topology of $K$ must have cardinality at least $\kappa:=\max(\mathrm{Card}\,\Gamma, \mathrm{Card}\,k)$. 
Question: does there exist a basis of open subets of $K$ with cardinality $\kappa$? 
Remarks:
(1) It is true if $v$ is discrete, i.e. $\Gamma\cong\mathbb{Z}$. Proof: take a set $S\subset R$ of representatives of $k$, and a uniformizing parameter $\pi$. Let $X\subset K$ be the set of finite sums $\sum_i s_i\pi^{n_i}$ ($s_i\in S$, $n_i\in\mathbb{Z}$). Then the balls centered on $X$ form a basis.
(2) I am especially intereseted in the case $\kappa=\omega$. Explicitly: if $\Gamma$ and $k$ are countable, does it follow that $K$ is second-countable (or, equivalently, separable)?

REPLY [3 votes]: I now think that the answer is no. Let $K$ be the field of Hahn series for the group $\mathbb{Q}$ over a countable field $k$, in the variable $t$ (so $\kappa=\omega$). For any sequence $a=(a_i)$ of elements of $k$, and any increasing sequence $\gamma=(\gamma_i)$ of rational numbers between $0$ and $1$, we have an element $x_{a,\gamma}=\sum_{i\in\omega}a_it^{\gamma_i}$ in $K$. The distance between different such elements is less than $1$ (in the valuative sense), so if we take an open ball of radius $2$ around each we get a disjoint union of uncountably many balls.<|endoftext|>
TITLE: Expected values of traces of products of random matrices
QUESTION [10 upvotes]: Suppose I want to compute a quantity of the type: 
$\mathbb{E}\mathrm{tr}(AUBU^{\ast})$ 
where averaging is over Haar measure on the unitary group $\mathcal{U}(n)$ (one can of course consider higher order polynomials or other matrix ensembles etc.) and $A$, $B$ are some fixed matrices. Is there any standard technique for computing such averages? I'd guess people in random matrix theory or free probability compute such traces all the time, but I've been unable to find a reference. If it makes matters easier, I'm really interested in computing something for random projections (e.g. something of the form $\mathbb{E}\mathrm{tr}(APBP)$, where $P$ is a projection onto a random subspace), which of course reduces to computation of polynomials in $U$.

REPLY [3 votes]: For the unitary group, the first paper I am aware of to do these sorts of averages is:
http://link.aip.org/link/JMAPAQ/v21/i12/p2695/s1
An early paper of Collins' in 2003 expresses such averages in terms of Weingarten functions,  which are usually expressed as character expansions over $U(N)$, $O(N)$ or $Sp(N)$.
http://arxiv.org/abs/math-ph/0205010
Some early papers calculating these character expansions are:
http://jmp.aip.org/resource/1/jmapaq/v25/i6/p2028_s1
http://jmp.aip.org/resource/1/jmapaq/v43/i1/p604_s1<|endoftext|>
TITLE: Reference for: CM Hilbert Modular forms arise from Hecke characters
QUESTION [12 upvotes]: For classical modular forms, the correspondence between the form having CM by an imaginary quadratic field $K$ and it being induced from a Hecke character on $K$ is well-known. (Ribet's paper is a standard reference.)
I am looking for a reference for the analogous result for Hilbert modular forms over a totally real field F. In particular, if the form has CM then it arises from a Hecke character on a quadratic imaginary extension $K$ (over $F$.) I believe, for the converse, Yoshida/Hida is the  reference. Thanks

REPLY [4 votes]: This result is better understood in terms of automorphic representations. Let $F$ be an algebraic number field, and let $\pi$ be an automorphic representation of $\mathrm{GL}_2\left(\mathbb{A}_F\right)$, where $\mathbb{A}_F$ denotes the ring of adèles of $F$.
Suppose that there exists a nontrivial unitary Hecke character $\omega$ of $F^{\times} \backslash \mathbb{A}_F^{\times}$ such that $\pi \otimes (\omega \circ \det) \cong \pi$. Then $\omega$ must necessarily be quadratic, and the representation $\pi$ is said to be a monomial representation. This is what you call CM, but this labelling only really makes sense when $F = \mathbb{Q}$ and $\pi$ corresponds to a holomorphic modular form, for the reasons outlined in Ribet's paper. When $\pi$ corresponds to a Maaß form, I have seen such form called of CM-type, but this seems a little incongruous.
Let $E$ be the quadratic extension of $F$ associated to $\omega$ via class field theory. Then the following statement is Proposition 6.5 of L-indistinguishability for $\mathrm{SL}(2)$ by Labesse and Langlands:

If $\pi$ is a monomial automorphic representation, then there exists a Hecke character $\chi$ of $E^{\times} \backslash \mathbb{A}_E^{\times}$ such that $\pi \cong \pi(\chi)$.

Here $\pi(\chi)$ denotes the cuspidal automorphic representation of $\mathrm{GL}_1\left(\mathbb{A}_E\right)$ associated to $\chi$.
While I do not believed it is mentioned in this paper, it is worth noting the following:

The monomial automorphic representation $\pi$ is cuspidal if and only if $\chi$ does not factor through the norm map (that is, there does not exist some Hecke character $\widetilde{\chi}$ of $F^{\times} \backslash \mathbb{A}_F^{\times}$ for which $\chi = \widetilde{\chi} \circ N_{E/F}$).

The proof of these results involve automorphic representations instead of Galois representations, and in particular generalise to Maaß forms (when $F = \mathbb{Q}$ and $\pi_{\infty}$ is a principal series representation) and Hilbert modular forms (when $F$ is a totally real field and $\pi_v$ is a discrete series representation for each archimedean place $v$).<|endoftext|>
TITLE: Kuiper's theorem via approximation
QUESTION [11 upvotes]: Kuiper's theorem says that the unitary group $U(H)$ of a separable infinite dimensional Hilbert space $H$ is contractible, if it is equipped with the norm topology. 
Let's suppose, I do not know this theorem, but I do know that $U^{st*}(H)$ is contractible, where the latter group is $U(H)$ equipped with the strong$^*$ topology generated by the semi-norms $p_v(a) = \lVert av \rVert$ and $q_v(a) = \lVert a^*v\rVert$ for all $v \in H$. I have a continuous map $U(H) \to U^{st*}(H)$. 


Is there any way to see that this is a weak equivalence by an approximation argument, thereby proving Kuiper's theorem?


side remark:
The motivation for this question comes from a setup which looks completely different, but is from a certain point of view surprisingly similar: Let $\mathcal{O}_{\infty}$ be the Cuntz algebra on countably infinite generators. $Aut(\mathcal{O}_{\infty})$ carries two topologies: One from the norm via the inclusion into bounded maps on Banach spaces $\mathcal{O}_{\infty} \to \mathcal{O}_{\infty}$, the other is the so-called point-norm topology generated by the semi-norms $p_a(\alpha) = \lVert \alpha(a) \rVert$. It is known that $Aut(\mathcal{O}_{\infty})$ is weakly contractible in the latter topology, but I would like to know it for the first.

REPLY [4 votes]: This is not an answer but too long for a comment.
It was shown in
Popa, S. and Takesaki,M., The Topological Structure of the Unitary and Automorphism Groups of a Factor, Commun. Math. Phys. 155, 93-101 (1993)
that the unitary group $U(R)$ of the hyperfinite $II_1$-factor $R$ is contractible in the strong topology (which is equal to the strong-$*$-topology in this case). 
Years before, it was shown in
Araki, H., Smith, M.-S.B. and Smith, L., On the homotopical significance of the type of von Neumann algebra factors, Commun. Math. Phys. 22, 71-88 (1971)
that the first homotopy group of $U(R)$ in the norm toplogy is isomorphic to $(\mathbb R,+)$. Under this isomorphism, the element $\lambda \in [0,1]$ is represented by the loop 
$$[0,1] \ni t \mapsto \exp(2\pi i\cdot t p) \in U(R),$$ where $p\in R$ is some projection of trace $\lambda$.
This shows that the homotopy type depends heavily on the chosen topology and there is no reason (at least in general) that one should find an easy approximation argument.<|endoftext|>
TITLE: Derived functors of symmetric powers
QUESTION [8 upvotes]: What do the derived functors of the symmetric powers look like? I understand that this is related to the homology of the symmetric groups, but I don't know a reference for that. 
Namely, I'm interested in the homotopy groups of the free simplicial commutative ring on a simplicial set. Let $X_\bullet$ be a simplicial set; I'd like to know the homotopy groups of $\mathbb{Z}[X_\bullet]$. This is the symmetric algebra on the free simplicial abelian group $\mathbb{Z}X_\bullet$, which is weakly equivalent to a product of Eilenberg-MacLane simplicial abelian groups corresponding to the homology of $X_\bullet$ (and is cofibrant).
In particular, we have a weak equivalence of simplicial commutative rings
$$\mathbb{Z}[X_\bullet] \simeq \bigotimes \mathbb{L} \mathrm{Sym}^\bullet K( H_n(X_\bullet, n)),$$
which brings up the question of what $\mathbb{L} \mathrm{Sym}^\bullet$ looks like. 
 Tyler Lawson points out in answering this question that the answer is somewhat complicated and describes it in low degrees. 
Is a complete answer known?

REPLY [7 votes]: The homology of all of the symmetric groups together is well understood, as Tyler says.
Taking mod $p$ coefficients, that is the special case when $X = S^0$ of the calculation
of $H_*(CX)$ as a functor of $H_*(X)$, where $C$ is the monad on based spaces associated 
to any $E_{\infty}$ operad of spaces.  The calculation in this form is given as Theorem 4.1, 
page 40, of [Cohen, Lada, May.  The homology of iterated loop spaces, SLN Vol 533. 1976]
which is available on my web page.  The functor is not all that complicated, but you do 
have to understand the Dyer-Lashof operations, which are very much like Steenrod
operations and can be seen with those as special cases of a general construction of 
Steenrod operations [A general algebraic approach to Steenrod operations. In SLN Vol. 168.
1970] also on my web page.  The paper of Bisson and Joyal cited by Tyler gives a reformulation
of this functor in the case $p=2$.  If you want the integral homology, that is a mess to write
down in closed form, but the mod $p$ Bockstein spectral sequence of $CX$ is entirely determined by 
that of $X$, as explained in Theorem 4.13 op cit above, so that integral information is also
available.  It is worth emphasizing that viewing the homology of symmetric groups as a special
case of $H_*(CX)$ substantially simplifies both the calculation and understanding the answer.<|endoftext|>
TITLE: What is the probability that two numbers are relatively prime?
QUESTION [16 upvotes]: The basic question that I have is in the title, but let us make it more rigorous below.
Let $N=\{1, 2, ..., n\}$, and put the (normalized) counting measure, $\mu_n$, on $N\times N$. 
Let $\mathcal{S}_n= \{ (a, b)\in N\times N: gcd(a, b)=1\}$
and $x_n=\mu_n(\mathcal{S}_n).$
Then what is the assymptotic behavior of $x_n$ as $n\rightarrow\infty$.

REPLY [22 votes]: The probability tends to $\frac{1}{\zeta(2)}=\frac{6}{\pi^2}$ as was mentioned by Qiaochu. This actually generalizes to arbitrary number fields, and is a less commonly known fact. 
In fact in any number field, the probability that two ideals are relatively prime is given by $1/\zeta_K(2)$, where $\zeta_K$ is the Dedekind zeta function of the number field $K$. And is proven in a similar way to the classical result. Here is a reference: "The probability of relative primality of Gaussian integers". For example the analogous probability for Gaussian integers is $6/(\pi^2G)$ where $G=1-\frac{1}{3^2}+\frac{1}{5^2}+\cdots$ is the Catalan constant.

REPLY [13 votes]: This is a very standard counting problem in analytic number theory.
Here's a rigorous proof:
It is enough to derive an asymptotic formula for 
$$\sum_{a,b\leq n,  (a,b)=1} 1 $$
This is 
$$\sum_{a,b\leq n, d|a, d|b} \mu(d) $$
$$=\sum_{d\leq n} \mu(d)\sum_{k\leq n/d , l\leq n/d} 1$$
$$=\sum_{d\leq n} \mu(d) ((n/d)^2 + O(n/d) ) $$
$$=n^2\sum_{d\leq n} \mu(d)/d^2 + O(n\log n)$$
$$=n^2\sum_{d=1}^{\infty} \mu(d)/d^2 + O(n) + O(n\log n)$$.
$$=n^2 6/\pi^2 + O(n\log n).$$

REPLY [8 votes]: The probability is $\frac{6}{\pi^2} = \frac{1}{\zeta(2)}$. A sketch of a proof can be found in this blog post (actually I only show, more or less, that if the density exists it must be $\frac{6}{\pi^2}$).<|endoftext|>
TITLE: Les deux théorèmes d'existence en théorie formelle des modules
QUESTION [13 upvotes]: In Exposé 195 of the Séminaire Bourbaki, Grothendieck states the following two theorems of non-flat descent.

Theorem 1. Let $\Lambda$ be a noetherian ring and $C$ the category of $\Lambda$-algebras which are finite type artinian $\Lambda$-modules. Let $F:C\to (Set)$ be a covariant functor. Then $F$ is pro-representable if and only if the following two conditions are satisfied:
i) $F$ commutes with finite products,
  ii)  for each $A\in C$ and each homomorphism $A\to A'$ in $C$ such that the diagram
  $A\to A'\rightrightarrows A'\otimes_A A'$ is exact, the diagram
  $F(A)\to F(A')\rightrightarrows F(A'\otimes_A A')$ is exact.
Moreover, it is enough to check ii) when $A$ is local and when moreover we are in one of the following two cases:
a) $A'$ is a free $A$-module,
  b) the quotient $A'/A$ is an $A$-module of length $1$.

$\ $

Theorem 2. Let $A$ be a local artinian ring with maximal ideal $m$, $A'$ an $A$-algebra that contains $A$ such that $mA'\subset A$ and $A\to A'\rightrightarrows A'\otimes_A A'$ is exact (note that this is the case in particular when $A'/A$ is an $A$-module of length $1$). Let $\mathcal{F}$ be the fibered category of flat quasi-coherent sheaves on variable schemes. Then the morphism ${\rm Spec}(A')\to {\rm Spec}(A)$ is a morphism of strict $\mathcal{F}$-descent.

(See Sém. Bourbaki, Exp. 195 for undefined terms.)  
Concerning Theorem 1, Grothendieck writes: "The proof of this theorem is quite delicate and can not be sketched here". The proof of Theorem 2 is not given either.
I don't know any place where these theorems are proved. Does anyone know?

REPLY [5 votes]: You should look at A H M Levelt's notes:
Sur la pro-representabilite de certains foncteurs en geometrie algebrique
where, I think, proofs of both Theorem 1 and Theorem 2 are written out. Also, he has written a paper "Foncteurs exacts à gauche" (Inventiones + errata) which discusses the relation between Schlessinger's conditions and these. I tend to view these particular instances of non-flat descent as a pre-Schlessinger/Artin theory that has been obsoleted. On the other hand, I do find non-flat descent interesting.<|endoftext|>
TITLE: Comparing the Edelman-Greene bijection to David Little's bijection
QUESTION [18 upvotes]: In their 1987 paper "Balanced Tableaux", Edelman and Greene construct a bijection between standard young tableaux with staircase shape $(n-1,n-2, \dots , 1)$ and reduced decompositions of the reverse permutation $(n,n-1, \dots, 1)$. The bijection is constructed using jeu de taquin and evacuation paths in forward direction and a modified version of RSK in the reverse.
David Little constructed a more general bijection between reduced decompositions of an arbitrary permutation and standard young tableaux of specific shapes in his 2003 paper "Combinatorial aspects of the Lascoux-Schutzenberger tree". His construction is, to my eyes, completely different, using line diagrams and circle diagrams (I don't know if these are standard terms). In the paper, he conjectures that his bijection, when specialized to the reverse permutation, recovers the Edelman-Greene bijection. It is know that the sets on which the bijections act coincide.
My questions are the following:

What is the current research status of Little's conjecture? Is it known whether the bijections are the same? Little states that he will investigate it further, but I am unaware of any new results.
Are there any other bijections between standard young tableaux with staircase shape and reduced decompositions of the reverse permutation? I am aware of Felsner's generalization of the Edelman-Greene bijection, but I am restricting my question to the case of the reverse permutation.

REPLY [12 votes]: Question one now has an answer. Benjamin Young and I have proved that the two bijections are the same in this case. A reference is available on the arXiv. More generally, we have shown the Little map is the same map as the recording tableau of Edelman-Greene insertion, the RSK variant referred to above.<|endoftext|>
TITLE: Subgroups of GL_2 over a finite field
QUESTION [10 upvotes]: I've come across the phrase "by the classification of subgroups of $GL_2(F_q)$" in multiple papers, but never with a reference.  Here $F_q$ is a finite field of size $q$.  Does anyone know a good reference for this (ideally for someone who is not a group theorist)?

REPLY [4 votes]: Dickson is responsible for the classification of subgroups of $SL_2(\mathbb{F}_q)$ (and once you've got this the subgroups of $GL_2(\mathbb{F}_q)$ are easy). You can find a full proof in Suzuki's "Group Theory Part I". It's Chapter 6, Section 3 of that book. If you email me I'll even send you a copy :-)<|endoftext|>
TITLE: Asymptotics of the growth rate of a group
QUESTION [22 upvotes]: Let $\Gamma$ be a finitely generated group of exponential growth and $gr(S)=\lim_{k\rightarrow \infty} \sqrt[k]{|B_k(S)|}$ be the growth rate of $\Gamma$ with respect to the generating set $S$. I am confused with the following question: Does there always exist a generating set $S'$ such that
$$\frac{|B_k(S')|}{gr(S')^k}\rightarrow 1, \text{ when }k\rightarrow \infty$$

REPLY [16 votes]: There is never a (finite) generating set with that property.
Consider a generating set $S=\{x_1,\ldots,x_{\ell}\}$ of cardinality $\ell$.  Let $B_k := B_k(S)$, $S_k := B_k \setminus B_{k-1}$, and $g := gr(S)$.  Let $b_k := |B_k|$ and $s_k: = |S_k|$.  Assume for simplicity that $L := \lim_{k \to \infty} \frac{b_k}{g^k}$ exists (although it shouldn't be difficult to get general liminf bounds).  Also set
$$ L' := \lim_{k \to \infty} \frac{s_k}{g^k} = \lim_{k \to \infty} \frac{b_k-b_{k-1}}{g^k} = \left(1-\frac{1}{g}\right)L. $$
For the spheres, there's a trivial inequality $s_{m+n} \leq s_m s_n$.  (Any word of minimal length $m+n$ in the generators may be written at least one way as a product of two words of minimal length $m$ and $n$.)  This is already enough to give $L' \geq 1$, and thus
$$ L \geq \frac{1}{1-\frac{1}{g}} > 1. $$
This is unsatisfying and still leaves the possibility that $L'=1$.  Thus, I eliminate that possibility as well, by improving the trivial bound $s_{2k} \leq s_k^2$ to $s_{2k} \leq \left(1-\frac{1}{2\ell}\right)s_k^2$.  This improvement (unlike the trivial improvement above) does not hold for monoids, so the extra cancellation comes (unsurprisingly) from the existance of inverses.
Specifically, fix $k$, let $E_i$ be the set of all elements of $S_k$ which may be written as a word of length $k$ ending in $x_i$, and let $E_{\ell+i}$ be the set of all elements of $S_k$ which may be written as a word of length $k$ ending in $x_i^{-1}$. Set
$$ F_i = E_i \setminus \bigcup_{j < i} E_j, $$
so the $F_i$ are disjoint.
Let $F_i'$ be the image of $F_i$ under the inversion map, and let $n_i = |F_i|$.  The product of an element of $F_i$ and an element of $F_i'$ may be written with $2k-2$ letters (because the $x_i$ and $x_i^{-1}$ cancel), so those $n_i^2$ products are not in $S_{2k}$.  Thus, we have
$$ s_{2k} \leq s_k^2 - \sum_{i=1}^{2\ell} n_i^2. $$
By the Hölder inequality, we have
$$ s_k = \sum_{i=1}^{2\ell} n_i \leq \sqrt{\sum_{i=1}^{2\ell} 1^2} \sqrt{\sum_{i=1}^{2\ell} n_i^2} $$
$$ \frac{s_k^2}{2\ell} \leq \sum_{i=1}^{2\ell} n_i^2. $$
Combining the inequalities gives
$$ s_{2k} \leq \left(1-\frac{1}{2\ell}\right)s_k^2. $$
Dividing by $g^{2k}$ and sending $k \to \infty$ gives
$$ L' \geq \frac{1}{1-\frac{1}{2\ell}} > 1. $$
For balls, we obtain the bound
$$ L \geq \frac{1}{\left(1-\frac{1}{g}\right)\left(1-\frac{1}{2\ell}\right)}. $$
This bound is optimal for free generating sets of free groups.
I'd be interested in seeing what Kate's limit says (or don't say) about the underlying group.  Can the lower bound I just gave be achieved for non-free groups?  Can the limit generally be made arbitrarily close to $1$ by increasing the number of generators appropriately (as Misha speculated in the comments)?  I'm far from an expert in combinatorial/asymptotic group theory, so I don't have good intuition for what intrinsic information this value holds.<|endoftext|>
TITLE: representatives of the group of homotopy 7-spheres
QUESTION [11 upvotes]: In Milnor's paper "On manifolds homeomorphic to the 7-sphere" it is proven that there are manifolds homeomorphic but not diffeomorphic to the standard 7-sphere. His construction involves sphere bundles over $S^4$. Also, Kervaire and Milnor proved that there are exactly 28 h-cobordism (therefore diffeomorphism) classes of homotopy spheres in dimension 7. 
Does every class contain an exotic sphere arising as the total space of an $S^3$-bundle over $S^4$?, if not,
how can one determine the number of classes with such representatives?, how do they look like?

REPLY [16 votes]: This is done in the paper "An invariant for certain smooth manifolds" by James Eells and Nicolaas Kuiper. They introduce and study the so called $\mu$-invariant which is strong enough to classify homotopy $7$-spheres up to oriented diffeomorphism. A theorem on page 103 says that out of 28 oriented differomorphism types of homotopy 7-spheres precisely 16 are realized by $S^3$-bundles over $S^4$. I am not sure what is the best way to visualize the exotic spheres that aren't sphere bundles but e.g. if memory serves, all homotopy $7$-spheres are 
Brieskorn spheres.<|endoftext|>
TITLE: Z_2 versus second-order PA
QUESTION [12 upvotes]: These days, Peano Arithmetic ($PA$) refers to the first-order version of the axioms, where induction is only over formulas referring to natural number variables. Peano's original version of the induction axiom was actually second-order, allowing reference to sets-of-natural-numbers variables. This latter version corresponds to what a category theorist (like me) would call a natural number object. Call this 'second-order $PA$' or $PA_2$.
However, we also have second-order arithmetic $Z_2$ (wikipedia), which up to some redundant axioms, looks like $PA_2$ plus a comprehension axiom schema (details of the schema). We know that $PA_2$ has a unique model up to isomorphism, and so does $Z_2$. My first question is then

What, if anything, separates a model of $PA_2$ from a model of $Z_2$?

I imagine that there are sets of naturals in $Z_2$ that cannot be proved to exist in $PA2$, but where is the boundary?
From the Wikipedia page I see that $ACA_0$, a subsystem of $Z_2$ with restricted induction, is a conservative extension of $PA$ (and equiconsistent with $PA$), and I infer that $ACA$, the same subsystem plus unrestricted induction, would be a conservative extension of $PA_2$.

Is this true? Is $ACA$ equiconsistent with $PA_2$?

One reason I am interested in this is that McLarty has proved that the machinery of derived functor cohomology, in the setting appropriate for arithmetic geometry, is do-able in $Z_2$ (more precisely, derived functor cohomology over a Noetherian scheme, with coefficients in an arbitrary sheaf of modules on the Zariski site, only requires a system equiconsistent with second-order arithmetic). If one can get from working in $Z_2$ down to $ACA$, then these tools of algebraic geometry are available as soon as one accepts the existence of natural number objects (and, presumably, classical logic).

REPLY [10 votes]: Regarding the first question: The second order theory $PA_2$ is usually defined relative to an ambient universe $V$ of Zermelo-Fraenkel set theory, and as such it only has one model up to isomorphism. In other words, $PA_2$ is a categorical theory from the point of view of any model $V$ of $ZF$ since all of its models are isomorphic to $(\Bbb{N},\mathcal{P}(\omega))$, where $\Bbb{N}$ is the standard model of $PA$ and $\mathcal{P}(\omega)$ is the collection of all subsets of natural numbers (from the point of view of $V$).
On the other hand, $Z_2$ is a first-order approximation of $PA_2$, and by some standard theorems of model theory, it has many $2^\kappa$ nonisomorphic models of cardinality $\kappa$ for each infinite cardinal $\kappa$. In particular, there are continuum-many nonisomorphic countable models of $Z_2$. Each such countable model of $Z_2$ is of the form $(\Bbb{M},\mathcal{F})$, where $\Bbb{M}$ is a standard or nonstandard model of $PA$, and $\mathcal{F}$ is a countable family of subsets of the universe of discourse $M$ of $\Bbb{M}$
Regarding the second question: $ACA$ is much stronger than first order Peano Arithmetic $PA$ since it proves Con($PA$) (the formal consistency of $PA$) and much more (itertions of consistency statements) . 


However, $ACA$ is, in turn, much weaker than $Z_2$ since already the fragment known as $\Pi^1_1$-$CA$ of $Z_2$ can prove Con($ACA$). 


One way to see this is based on an old result (noticed by a number of people, including Takeuti and Feferman) that $ACA$ is equiconsistent with an extension $PA(T)$ of $PA$ with a distinguished predicate $T$ that codes up the full truth predicate for the ambient model of arithmetic. Note that $PA(T)$ includes induction in the extended language of arithmetic augmented by the predicate $T$.
P.S. The subsystem $ATR_0$ of $\Pi^1_1$-$CA$ already proves Con($ACA$).<|endoftext|>
TITLE: Are isospectral manifolds necessarily homeomorphic?
QUESTION [15 upvotes]: It's known that there are pairs of closed Riemannian manifolds which are isospectral but not isometric. 
Is it known if there are closed Riemannian manifolds which are isospectral but not homeomorphic?
(By isospectral, I mean that the Laplace-Beltrami operator on functions has the same spectrum on both manifolds.)
Thanks,
Dmitri

REPLY [20 votes]: There's an example due to Doyle and Rossetti ("Tetra and Didi, the cosmic spectral twins"; http://arxiv.org/abs/math.DG/0407422) of 3-manifolds that are isospectral but not even homeomorphic.  I don't know if this was the first such example.<|endoftext|>
TITLE: Does smooth and proper over $\mathbb Z$ imply rational?
QUESTION [22 upvotes]: Does smooth and proper over $\mathbb Z$ imply rational?
I think someone told me that this is a standard conjecture. Is it a widely held? held at all? Did someone in particular make this conjecture? write it down somewhere? give evidence for it? Would you like to provide evidence? Why rational and not, say, unirational?
Fontaine's non-digitized letter to Messing shows that if $X$ is smooth and proper over $\mathbb Z$, then the low degree nondiagonal Dolbeault groups vanish $H^q(X_{\mathbb Q}; \Omega^p)=0$ for $p\ne q$,  $p+q\leqslant 3$. An obvious conjecture is to remove the degree restriction. Is this expected? Does this conjecture imply the rationality one? (This hypothesis on a single space does not force rationality: there are fake projective planes that mimic the cohomology of the real projective plane, but they have finite covers that would again be smooth and proper, but which can't have such cohomology by the Atiyah-Bott fixed point formula. a little more here)

What I'm looking for: A citation of a survey would be ideal. Short of that, I would accept a citation of someone making this conjecture (which may well be the paper I cite above). In any event, I would be happy if people provide evidence for or against the conjecture.

Some definitions: propriety is a relative version of compactness; let's just say that the scheme is cut out of projective space by homogeneous equations with coefficients in $\mathbb Z$. Smoothness means that the fibers, that is, the schemes defined by the same equations considered over finite fields (or their algebraic closures) are smooth. This is stronger than regularity of the total space and more like a submersion, so by analogy with Ehresmann's theorem, all the fibers are supposed to be "the same." For example, $\operatorname{Spec}\mathbb Z[i]$ is regular, but the map to $\operatorname{Spec}\mathbb Z$ is not smooth, but ramifies at $(2)$. (Geometrically) rational means that that the field of rational functions with complex coefficients are the same as on $\mathbb CP^n$, namely $\mathbb C(x_1,\ldots,x_n)$. I think this question shows that strict rationality, a fraction field of $\mathbb Q(x_1,\ldots,x_n)$, is too much to ask for.
To complete the list of relevant MO questions, would (geometric) rationality imply a Hasse principle, answering this question?

REPLY [10 votes]: (This answer has been edited -- it used to say that a finite cover of $\overline M_{g,n}$ gives a counterexample, which no longer seems obvious.)
If you had written "Deligne-Mumford stack" instead of "scheme", then a counterexample would be given by the spaces $\overline M_{g,n}$, which are certainly smooth and proper but far from rational in the large $g$ limit (or, for $g > 0$, in the large $n$ limit). The original references here are, I guess, Deligne (for $\overline M_{1,11}$) and Harris-Mumford (for $\overline M_{25}$). 
Kevin Buzzard's hint with the Ramanujan $\Delta$ function is relevant here; indeed, $H^{11,0}(\overline M_{1,11})$ is nonzero, and the $\ell$-adic Galois representation corresponding to $H^{11,0}\oplus H^{0,11}$ is the representation attached to $\Delta$. 
My answer to the question Is the moduli space of curves defined over the field with one element? contains some more detailed information about these things.
The natural way to find a scheme instead of a stack would then be to consider finite or generically finite covers of $\overline M_{g,n}$ by smooth proper schemes. There are several closely related constructions of such covers in the literature by Looijenga, Boggi-Pikaart, Pikaart-de Jong, Abramovich-Corti-Vistoli, all using some kind of non-abelian level structure on curves, but as far as I can tell none of them work over the integers.<|endoftext|>
TITLE: what is the stringy Kähler moduli space?
QUESTION [14 upvotes]: I saw the stringy moduli space mentioned in a few papers but with little no explanation. I vaguely understand it is supposed to be the moduli space of complex structures on the mirror manifold.
Could you suggest a nice paper/blog post where to read some heuristics about it? Some examples? Connections to stability conditions/curve-counting?

REPLY [11 votes]: Let's see why the study of moduli space of complex structures is closely related to mirror pairs.
From the mirror map, we can heuristically identify the moduli space of symplectic structures $\mathcal M_{sym}$ and moduli space of complex structures $\mathcal M_{cpx}$ as follows $$\mathcal M_{sym}(M)\cong\mathcal M_{cpx}(M^\vee)$$ and $$\mathcal M_{sym}(M^\vee)\cong\mathcal M_{cpx}(M)$$
where $M$ and $M^\vee$ are mirror to each other. 
We have the following relation for tangent space of moduli space of symplectic structures 
$$T_\omega\mathcal M_{sym}(M)\cong H^2(M)$$
and from the BTT Theorem we have $$T_J\mathcal M_{cpx}(M)\cong H^1(M,TM)$$ for the tangent space of moduli of complex structures.
So from the mirror map, we have $$H^2(M)=H^1(M^\vee, TM^\vee)$$
More generally if we take extended moduli space of complex and symplectic structures then 
$$T_\omega\mathcal M_{sym}^{ext}(M)\cong H^*(M)$$
and 
$$T_J\mathcal M_{cpx}^{ext}(M)\cong \oplus_{p,q}H^q(M,\wedge^pTM)$$
and hence from the extended mirror map we can identify
$$H^q(M,\Omega^pM)\cong H^q(M^\vee, \wedge^pTM^\vee)$$ and in the case when the mirror $M^\vee$ is itself a Calabi-Yau variety then 
$$H^q(M,\Omega^pM)\cong H^q(M^\vee, \wedge^pTM^\vee)\cong H^q(M^\vee, \Omega^{n-p}M^\vee)$$ 
See this Kontsevich paper.<|endoftext|>
TITLE: How badly does compactness fail in $\mathcal{L}_{\omega_1\omega}$?
QUESTION [16 upvotes]: I would like to get a better idea of how badly compactness fails in $\mathcal{L}_{\omega_1\omega}$.
Let $\Gamma$  be an arbitrary set of sentences from $\mathcal{L}_{\omega_1\omega}$. Let the underlying signature $\tau$ also have arbitrary cardinality. Is there some cardinal $\kappa $ such that if every $\Delta\subseteq\Gamma$ where $|\Delta|\leq\kappa$ is satisfiable, then $\Gamma$ is satisfiable?
It is relatively easy to show that any such $\kappa$ would need to be $\geq \beth_{\omega_1}$, but I am unsure of how to proceed beyond there. If there is no such $\kappa$, I am also interested in weakening the question.

REPLY [4 votes]: This is really a comment, but (a) it's too long and (b) is ought to be attached to both of the answers. Not only does the characterization of the $\kappa$-compactness property not depend on the availability of infinite quantifier strings (as discussed in the comments on Joel's answer and as explicitly stated in Ioannis's answer), it doesn't depend on quantifiers at all. Propositional logic suffices. 
More precisely, consider propositional logic with countable conjunctions and disjunctions. Suppose $\kappa$ is a cardinal and, for any set $\Gamma$ of sentences in this logic, if every subset of size $<\kappa$ is satisfiable, then so is $\Gamma$. I claim that $\kappa$ is $\omega_1$-strongly compact, i.e., every $\kappa$-complete filter on any set $I$ can be extended to a countably complete ultrafilter.
To prove this, let $I$ and a $\kappa$-complete filter $\mathcal F$ on it be given. Consider the following set $\Gamma$ of sentences in the propositional logic described above, with a propositional variable $\bar A$ for every subset $A$ of $I$.  $\Gamma$ contains:
First, the sentences $(\bar A\land\bar B)\leftrightarrow\overline{A\cap B}$ and $\neg\bar A\leftrightarrow\overline{I-A}$, for all $A,B\subseteq I$,
Second, the sentences $\bar A$ for all $A\in\mathcal F$,
Third, the sentences $\bigwedge_{n\in\omega}\overline{A_n} \leftrightarrow \overline{\bigcap_{n\in\omega}A_n}$ for all countable sequences $(A_n)$ of subsets of $I$.
Then any subset of $\Gamma$ of cardinality $<\kappa$ is satisfiable. In fact, we can satisfy all the sentences of the first and third sorts along with any $<\kappa$ sentences of the second sort as follows. The $<\kappa$ sentences of the second sort are $\bar A$ for some $<\kappa$ elements $A$ of $\mathcal F$. As $\mathcal F$ is $\kappa$-complete, these $A$'s have a nonempty intersection (in fact, their intersection is in $\mathcal F$), so let $i$ be  a point in that intersection. Then give each propositional variable $\bar X$ the truth value "true" if $i\in X$ and "false" otherwise. It is easy to check that our subset of $\Gamma$ is satisfied by this valuation.
So, by hypothesis, there is a valuation $v$ making all of $\Gamma$ true. Define $\mathcal U\subseteq\mathcal P(I)$ to be 
$$
\mathcal U=\{A\subseteq I:v(\bar A)=\text{true}\}.
$$
Then $\mathcal U$ is an ultrafilter on $I$ because $v$ satisfies the first batch of sentences in $\Gamma$; it extends $\mathcal F$ because of the second batch; and it is countably complete becuse of the third batch.<|endoftext|>
TITLE: Hochschild Cohomology of Differential Operators in characteristic 0
QUESTION [14 upvotes]: In Mariusz Wodzicki's paper "Cyclic homology of differential operators," the following result is mentioned: for $D_M$ the algebra of differential operators on a smooth manifold $M$ we have that $HH_n(D_M) \cong H_{DR}^{2m-n}(M)$ where $m=\dim M$. I'm having trouble finding a reference for the Hochschild Cohomology of $D_M$. Does anyone know of a paper with the result?

REPLY [13 votes]: Hi,
This is an instance of a more general fact concerning deformation quantization of symplectic varieties.  The general theorem is:
`Let $X$ be an symplectic manifold, and let $A_\hbar$ be any quantization of the Poisson algebra $(C^\infty(X), \{, \})$.  Then we have $HH^*(A_\hbar)\cong H_{DR}^*(X)$.
See the paper:
Hochschild cohomology and Characteristic Classes by Weinstein and Xu
for an early proof.  As Jeremy noted, one proof of this goes through spectral sequences.  Morally the point, as I understand it, is that there is a canonical quantization of a symplectic vector space, which is the Weyl algebra, and which has trivial Hochschild cohomology.  Since every symplectic variety is locally symplectomorphic to a symplectic vector space, you can compute the Hochschild cohomology of $HH^*(A_\hbar)$ on some covering of $X$ via a Cech complex consisting of constant sheaves at each basic open set (constant sheaves because you choose the covering fine enough so that the symplectic form can be put in canonical form, and then you use the result about the Hochschild cohomology of the Weyl algebra).  This is also how you compute DeRham comology, so the answers coincide.
The reason this relates to differential operators is that for a smooth manifold $M$, its cotangent space $T^*M$ is a symplectic manifold, and we can regard $D_M$ as a quantization of $C^\infty(T^*M)$. as above.  So we find that $HH^*(D_M)\cong H_{DR}^*(T^*M)\cong H_{DR}^*(M)$, since $T^*M$ contracts onto $M$.<|endoftext|>
TITLE: Complexity of finding a 0-1 vector in a subspace or showing that there is none
QUESTION [8 upvotes]: This question, is a slightly different disguise (see below), came up in discussions of this question about equitable partitions
A $0,1$ vector in $\mathbb{Z}^n$ is any vector with all entries $0$ and $1$ (at least one of each.) This excludes (for temporary convenience) the all ones vector $\mathbf{1}.$ 
Q: Given a set of $k+1$ vectors in $\mathbb{Z}^n$, one being $\mathbf{1}$, How difficult is it to determine if there is a (rational) linear combination which is a $0,1$ vector?
It would be equivalent to say that we have a set of $k$ integer vectors such the $n$ entries of each vector sum to $0.$ We wonder if there is a  linear combination having only two distinct entries. 
In the context of the earlier question the $k$ vectors might be a basis for a certain eigenspace of (the adjacency matrix of) a given regular graph. A linear combination with only two distinct entries would indicate an intersetig two cell partition of the vertices of the graph. If the graph is connected then the largest eigenvalue is the common degree. Then the $k+1$ vector form of the question deals with a basis for the span of two eigenspaces.

REPLY [9 votes]: It is NP-complete.  Consider the case of $n$ vectors of length $n+1$, where the vectors are the rows of a matrix $[I | x]$ where $I$ is an $n\times n$ identity matrix and $x$ is a column vector of even integers.  Any linear combination of the rows that is 0-1 must have multipliers 0 or 1 (due to the identity matrix part).  So what we are asking is if there is a subset of the rows that sums to 0-1, i.e. is there a subset of the entries of $x$ that sums to 0 or 1.  However the entries of $x$ are even, so the only possible sum is 0.
The famous SUBSET SUM problem  is known to be NP-complete. It asks whether a set of integers has a subset summing to 0 and is well known to be NP-complete. Requiring the integers to be even doesn't make it any easier (multiply a general instance by 2). That cooks the goose.<|endoftext|>
TITLE: What is the "Lefschetz Principle" (examples) ?
QUESTION [6 upvotes]: Hi there,
 can anyone explain to me what the "Lefschetz Principle" is by some clear "classical"
examples (not relying explicitly on model theory, say).
 Thanks !

REPLY [6 votes]: The simplest example of an application of the Lefschetz principle is to prove the residue theorem (in characteristic zero): The sum of the residues of a differential on a smooth projective curve is zero. If you are given a curve and a differential over a field of characteristic zero, there is a finitely generated field over $\mathbb{Q}$ where the curve, the differential, the poles and the residues there, are all defined, so take this field and embed it in $\mathbb{C}$. The residue theorem over $\mathbb{C}$ is, of course, an immediate consequence of Cauchy's theorem. You can even deduce the positive characteristic case from this with a bit more effort.<|endoftext|>
TITLE: Functions holomorphic on a region minus a Cantor set
QUESTION [12 upvotes]: Let $X$ and $Y$ be simply connected open regions of $\mathbb{C}$, and let $Z \subset X$ be a Cantor set. Assume we have a homeomorphism $f$ from $X$ to $Y$, which is holomorphic on $X \setminus Z$. Is $f$ necessarily holomorphic on $X$?

REPLY [6 votes]: Removability with respect to homeomorphisms is different from the removability with respect to bounded functions mentioned in another answer.
In particular, it is not necessary to have Hausdorff dimension at most 1. Indeed, any quasicircle is removable with respect to homeomorphisms, as mentioned by Hrant.
For much more complicated sets, see Jeremy Kahn's thesis "Holomorphic Removability of Julia sets": He shows that many Julia sets of quadratic polynomials are in fact removable. 
http://arxiv.org/abs/math/9812164
(As above, we consider the set in question to be compact.)
In particular, he discusses the notion of "absolute area zero": A set $K$ has absolute area zero if there is no conformal isomorphism from the complement of $K$ to the complement of some set with positive area. Any such set is removable, and any Cantor set that is well-surrounded has absolute area zero.
On the other hand, as has been noted elsewhere, there are many examples of sets that are not holomorphically removable. The simplest example of a Cantor set would be a Cantor set of positive measure. More interesting examples are provided by Chris Bishop, as cited in Misha's answer.
EDIT: You may also wish to look at the paper "Removability theorems for Sobolev 
functions and quasiconformal maps" by Peter Jones and Stas Smirnov, which contains a number of sufficient conditions for conformal removability: http://www.unige.ch/~smirnov/papers/hr-j.pdf 
Graczyk and Smirnov use these criteria to prove removability of a large class of Julia sets.<|endoftext|>
TITLE: Integer matrices with a strange divisibility property
QUESTION [13 upvotes]: Fix an integer $n$. What can you say about a (not necessarily square) matrix $A$ with integer entries that has the property that for any $k$, every $k\times k$ minor of $A$ is divisible by $n^{k-1}$? Have you seen such matrices before? Do they have a name? Are they in a bijection with some set whose description does not require knowing about determinants?
The question I really care about is the same but with "many-variable Laurent polynomials with integer coefficients" replacing the integers in the above (except $k$), but I suspect that it doesn't really make a difference.
The reason I care is that I have but I don't understand an amusing (I think) generalized Alexander invariant of tangles and virtual tangles with excellent composition properties and with values in such matrices as above, and I would like to understand its target space. A handout and a video are at http://www.math.toronto.edu/~drorbn/Talks/GWU-1203/ but no writeup exists at present.

REPLY [14 votes]: If $n$ is prime, we have the following equivalences:

Your condition.
Your condition for $2 \times 2$ minors.
There are integer vectors $v$ and $w$ such that $A \equiv v w^T \mod n$.

The implications $1 \implies 2$ and $2 \implies 3$ are straightforward. For $3 \implies 1$, write $A = v w^T + n B$, and expand out $\det(v w^T + n B)$ as a sum of products of minors from $v w^T$ and from $n B$. All the terms which involve a minor of $v w^T$ larger than $1 \times 1$ are zero, so every term is divisible by $n^{k-1}$.
When $n$ is not prime, we still have $3 \implies 1 \implies 2$, and $2$ does not imply the others; look at $\begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}$ with $n=4$. Also, for $n$ nonprime, $1$ does not imply $3$, look at $\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$ for $n=4$.
From your described motivation, it sounds like it would be interesting to know whether your matrices obey $3$.<|endoftext|>
TITLE: Rabin's Tree Theorem
QUESTION [6 upvotes]: I've been reading Rabin's article on decidability in Barwise's text, and I came across Rabin's discussion of the decidability proof of his tree theory: the second-order theory with two successor functions.  The text mentions that the proof is hard and very technical owing to many extensions of automata theory, and I was wondering if someone might be able to sketch it out.
Thanks!

REPLY [8 votes]: [Edit, June 2, 2013: At the Nordic Spring School in Logic this year there was a course by Wolfgang Thomas on Logic, automata and games. You may be interested in the slides, particularly for part III (Rabin’s tree theorem). Slides and notes for all courses can be downloaded here; the slides for part III can be downloaded here.]
This is a nice result, but you are right that the known proofs are rather elaborate. Below, I just give the briefest of sketches, but it gives a glimpse of the ideas involved. The game theoretic approach is elegant, even if it takes a bit of time to understand the relevant notions. 
An excellent reference is the book "The classical decision problem", by Börger, Grädel, and Gurevich (Springer, 1997). You can find the proof in II.7.1. 
I lectured on it in Caltech a few centuries ago (2007). If you do not have access to the book, you may want to take a look at homework sets 5 and 6 at this link. There, you will find a sketch of the so called Forgetful Determinacy theorem of Gurevich and Harrington. This is the key technical result. Here, I'll just state it.
(The theorem comes from Gurevich-Harrington, "Trees, automatas, and games". 14th annual ACM Symposium on Theory of Computation, 60-65, 1982. The presentation in the book follows Zeitman, "Unforgettable forgetful determinacy", Journal of Logic and Computation, vol 4, 273-283 (1994).)

Theorem (Forgetful determinacy for tree automata). If $A$ is a $\Sigma$-tree automaton and $F$ is a $\Sigma$-tree, one of the players ($A$ or Pathfinder) has a winning strategy in
  $\Gamma(A,F)$ that is forgetful in the sense that whenever $p$, $q$ are positions from which the winner moves,
   $$ LAR(p)=LAR(q), $$
  and
   $$ \mbox{$Node(p)$-residue of $F=Node(q)$-residue of $F$}, $$
  then $f(p)=f(q)$.

(Here, the node of a position is simply the node in the binary tree that is being played. Given a $\Sigma$-tree $F$, the $v$-residue of $F$ is the $\Sigma$-tree $F_v$ coming from $F$ by only considering the tree from $v$ on, that is, $F_v(w)=F(vw)$. I define the other relevant notions below.)
Using this, it is relatively easy to check that the "Emptiness problem" is decidable for tree automata: There is an algorithm that, given a $\Sigma$-automaton $A$, decides whether there is a $\Sigma$-tree that $A$ accepts. 
Similarly, but this takes a bit of work and is really the whole point, Forgetful Determinacy implies that there is an algorithm that to each $\Sigma$-automaton $A$ assigns a $\Sigma$-automaton $C$ with the property that $C$ accepts a $\Sigma$-tree iff $A$ rejects it. (This is the "Complementation Theorem".)
Using this, it is a simple matter of induction in formulas to prove decidability, since one can effectively associate to each monadic formula $\phi(X_1,\dots,X_n)$ (of second-order theory with two successors) a $\Sigma$-automaton $A$ for $\Sigma=\{0,1\}^n$, with the property that, for any collections $W_1,\dots,W_n$ of binary words, the automaton $A$ accepts the tree $T$ they define iff $T^2$ satisfies $\phi(W_1,\dots,W_n)$, where $T^2$ is the structure given by the full binary tree with the two successor functions. 

Let me briefly review the relevant definitions, following the book closely. I'm just quoting the notes of my younger self, so there may be a bit of irrelevancy, for which I apologize. 
Let $MOVE$ be a finite alphabet. An arena $A$ is a colored
bipartite multi-digraph in the following sense:

The vertices of $A$ are divided into two disjoint sets, 
east vertices and west vertices. There are no edges
between east vertices or between west vertices. There may be 
several edges between an east and a west vertices, or between a
west and an east vertices (so edges have directions, which is
why we call the object a digraph, and there may be several edges
between the same vertices, which is why we call it a multi-digraph).
There is a distinguished vertex, the start vertex. Every
vertex is reachable from the start vertex (i.e., for any $v$
there is a finite sequence $v_0,\dots,v_n$ where $v_0$ is the start
vertex, $v_n=v$ and for each $i\lt n$ there is an edge going from $v_i$
to $v_{i+1}$). Any vertex has at least one outgoing edge.
The edges are labeled by elements of $MOVE$ in such a way that no
two outgoing edges from the same vertex have the same label.
There is a finite set $S$ of colors that partition the set of
vertices. We denote by $C^s$ the vertices with color $s$.

A game on $A$ is played between two players 0 and 1 who
alternate choosing an outgoing edge from the current vertex,
starting from the start vertex. So a play of the game defines an
infinite path through $A$ (we allow for the possibility of
revisiting vertices). A position $p$ is a finite directed path
through $A$ from the start vertex, so it is uniquely described by a
word in $MOVE^*$, with which we identify $p$. Given a position $p$,
the labels of the edges leading out of the last vertex of $p$ are
the possible moves at $p$. A play is an $\omega$-sequence
$P\in MOVE^\omega$ such that each initial segment is a position. The set
of plays over $A$ is $PLAY(A)$.
A graph game is a triple $\Gamma=(A,\varepsilon,W_\varepsilon)$ where $A$ is an arena,
$\varepsilon\in\{0,1\}$ (denoting the player that goes first) and
$W_\varepsilon$, the winning set for player $\varepsilon$, is
a Boolean combination of the sets ${}[C^s]$ where ${}[C^s]$ is the set
of plays that infinitely often pass through a vertex of color $s$.
Player $\varepsilon$ wins a play $P$ of $\Gamma$ iff $P\in
W_\varepsilon$. Otherwise, player $1-\varepsilon$ wins the play $P$.
Notice that if the start vertex is an east (resp., a west) vertex
then, playing $\Gamma$, player $\varepsilon$'s turns to move are
always at east (resp., west) vertices. Call the set of these
vertices $V_{\varepsilon}$ and the set of remaining vertices
$V_{1-\varepsilon}$.
A forgetful strategy $f$ for player $\delta\in\{0,1\}$ in
$\Gamma$ is a function $f:V_\delta\to{\mathcal P}(MOVE)$ that to each
$v\in V_\delta$ assigns a non-empty set of possible moves from $v$.
(The strategy is forgetful since it depends only on $v$ and not on
how $v$ was reached.)
The latest appearance record $LAR(p)$ of a position $p$ is an
ordering of the colors. We define $LAR$ inductively, with
$LAR(start)$ being an ordering whose last color is that of the start
vertex. If a position $q$ is obtained from a position $p$ by
adjoining an edge to a vertex of color $s$, then $LAR(q)$ is
obtained from $LAR(p)$ by moving $s$ to the last place. The coloring
of an arena $A$ is forgetful if any two positions at the same
vertex have the same $LAR$, in which case we can simply talk of the
$LAR$ at a vertex $v$ (rather than at a position $p$ whose last
vertex is $v$).
What one actually shows is the following:

Theorem (Forgetful determinacy).
  Let $\Gamma=(A,\varepsilon,W_\varepsilon)$ be a graph game with a
  forgetful coloring of the arena $A$. Then one of the players has a
  forgetful winning strategy in $\Gamma$.

One then uses this result to prove the version I stated earlier. For this, let $\Gamma(A,F)$ be a game on a $\Sigma$-tree $F$ between
a $\Sigma$-tree automaton $A$ and Pathfinder. Define from this a
graph game whose alphabet consists of the states of $A$ and names
for the two directions left and right. (So $A$ starts the game and
chooses a state according to its initial table, Pathfinder responds
by choosing the name of a direction, then $A$ chooses a state, etc.)
All positions $p$ where $A$ makes a move have the same default
color. If $A$ chooses a state $s$ at $p$ then the color of position
$ps$ is $s$. (One needs to check that this coloring is forgetful.)
One then has that either $A$ or Pathfinder has a forgetful
winning strategy in $\Gamma(A,F)$, by "transfering" the strategy that the forgetful determinacy theorem guarantees.<|endoftext|>
TITLE: Classifying space of the unit circle of a valued field 
QUESTION [10 upvotes]: Let $k$ be a field with absolute value $|\cdot |$ and let $S^1(k) := \lbrace x \in k \mid |x|=1\rbrace \le k^\times$. Since the absolute value definies a metric on $k$, $S^1(k)$ is a topological group. In particular, $S^1(k)$ has a classifying space $BS^1(k)$. 
Question 1: What is known about $BS^1(k)$ ? 
In case $k=\mathbb R$ (with the usual absolute value), $S^1(\mathbb R)=\lbrace \pm 1 \rbrace \cong \mathbb{Z}/2$ and $BS^1(\mathbb R) = P^\infty(\mathbb R)$. 
Similarly, if $k=\mathbb C$ (with the usual absolute value), $S^1(\mathbb C)=S^1$ and $BS^1(\mathbb C) = P^\infty(\mathbb C)$. This motivates:  
Question 2: Is it known under which conditions the relation $BS^1(k) = P^\infty(k)$ holds ?

REPLY [3 votes]: If $k=\mathbb{Q}_{p}^{\hat{}}$, I can say the following: $S^1 (k)$ is totally disconnected, since it is a subspace of the totally disconnected space $k$. Let $S^1 (k)^{\delta}$ be $S^1 (k)$, but equipped with the discrete topology. The natural map $S^1 (k)^{\delta} \to S^1 (k)$ is a weak homotopy equivalence, because a map from a standard sphere to $S^1 (k)$ is constant. It follows that the induced map $K(S^1 (k);1) = BS^1 (k)^{\delta} \to BS^1 (k) $ is a weak homotopy equivalence.
I am reluctant to say that $BS^1 (k) $ is an Eilenberg Mac-Lane space since it is not a CW complex.<|endoftext|>
TITLE: Subgroups of finitary symmetric groups
QUESTION [6 upvotes]: Question 1: Does there exist an intrinsic characterization of groups $G$ isomorphic to some subgroup of some finitary symmetric group (i.e. all the permutations of a given set that fix
all but finitely many elements)?  
Clearly every such $G$ enjoys local finiteness, but I see where (for a fixed $p$)  the multiplicative group $\{e^{2\pi i k/p^n}\}$ shows that this does not suffice (because all non-identity elements have $m$-th roots for every $m$).  
Question 2: Do there exist locally finite groups not isomorphic to any subgroup of any finitary symmetric group such that no non-identity element has $m$-th roots for every $m$?

REPLY [2 votes]: The notes at my site have moved to here: Finitary Permutation Groups<|endoftext|>
TITLE: Connecting Lemma in the Alexandrov's existence theorem.
QUESTION [12 upvotes]: At the moment I am polishing my lecture notes which in particular cover Alexandrov's existence theorem.
Denote by $\mathbf{P}_k$ the space of isometry classes of polyhedral 
metrics on the $\mathbb S^2$ with k-verteces and nonnegative curvature at each vertex.
In the proof of Alexandrov's existence theorem one has to show that
any space in $\mathbf{P}_k$ can be 
continuously deformed in $\mathbf{P}_k$
into a realizable polyhedral metric;
i.e. the one which  is isometric to the surface of convex polyhedron.
The original Alexandrov's proof is simple but tedious.
First he deforms the given polyhedral space into space 
with one degenerate vertex.
Second he shows that any polyhedral metric near the degenerate space is realizable. 
It seems that I found a simpler way to do this.
I proof that $\mathbf{P}_k$ is path connected 
by
constructing a path between given spaces in $\mathbf{P}_k$ using induction on $k$.
The existence of one realizable space in $\mathbf{P}_k$ is not a problem.

Was it done before?

REPLY [6 votes]: I am not aware of an elementary argument along these lines. A high-powered argument to show a priori connectedness uses Teichmuller theory (the space of polyhedral metrics fibers over teichmuller/moduli space (depending on how you identify things), with contractible fiber, see for example Troyanov's paper on cone metrics in Ens. Mathematique in the late eighties; this can also be gotten from Schwartz-Christoffel).  But this, while not difficult, is not elementary, so if you have a simple combinatorial proof, that's certainly a good thing.<|endoftext|>
TITLE: Question about decomposition of exterior product
QUESTION [7 upvotes]: In their paper "New lower bounds for the border rank of matrix multiplication", Landsberg and Ottaviani make use of the fact that 
$$\tag{$\dagger$} {\textstyle\bigwedge}^p(V\otimes W) \cong \bigoplus\nolimits_{\substack{\lambda\vdash p\\\\\ell(\lambda)\le n\\\\\lambda_1\le m }} \mathbb{S}_\lambda V \otimes \mathbb{S}_{\bar\lambda}W$$
where $\bar\lambda$ denotes the conjugate partition of $\lambda$. This isomorphism is basically Exercise 6.11 in Fulton & Harris, so there is no doubt about it. However, from what I gather, in Lemma 3.1 of the paper, they use the fact that the above isomorphism is given by the map
$$ (v_1\otimes w_1)\wedge\ldots\wedge(v_p\otimes w_p) \longmapsto \sum\nolimits_{\substack{\lambda\vdash p\\\\\ell(\lambda)\le n\\\\\lambda_1\le m }} c_\lambda(v_1\otimes\ldots\otimes v_p) \otimes c_{\bar\lambda}(w_1\otimes\ldots\otimes w_p), $$
where $c_\lambda$ denotes the Young symmetrizer corresponding to the partition $\lambda$. I cannot find a proof for this. Can someone explain to me why the above map defines 
   a) a morphism of $\mathfrak{S}_p$-modules and 
   b) a bijection? 
Since all vector spaces involved are of finite dimension and by $(\dagger)$, it would certainly suffice to show that it is either injective or surjective. 
Also, if I misunderstood the proof of Lemma 3.1 and the isomorphism is given by another elementary rule, please tell me what it is.

REPLY [2 votes]: This is well-known in the theory of symmetric functions and is one of two Cauchy identities.
You can find this in most books, for example, MacDonald Chapter I, Section 4. Orthogonality
equation (4.3').
Knuth's extension of the Robinson-Schensted correspondence gives bijective proofs of both Cauchy identities, for example see Stanley Enumerative Combinatorics 7.14.3 Theorem.<|endoftext|>
TITLE: Coefficients in the periodic part of continued fractions for real quadratic algebraic numbers
QUESTION [8 upvotes]: Given a strictly positive integer $A$, let $D(A)$ denote the set of all 
real quadratic algebraic numbers with a continued fraction having almost all coefficients 
$\leq A$.
Consider the field $Q_A$ generated by all elements of $D(A)$. One has $Q_1=\mathbb Q[\sqrt{5}]\subset Q_2\subset Q_3,\dots$.
The inclusion $Q_1\subset Q_2$ is strict since $Q_2$ contains for example $\sqrt{2}=[1;2,2,2,\dots]$ and $\sqrt{3}=[1;1,2,1,2,\dots]$.
Are there other strict inclusions? Are there cases of equality? (I ignore for example if 
$Q_2$ is a proper subfield of the field $\mathbb Q[\sqrt{\mathbb N}]$
generated by all real quadratic number-fields.)
A related question: Given a real quadratic algebraic number 
$\alpha$ with continued fraction expansion $\alpha=[a_0;a_1,a_2,\dots]$
consider the mean value $\mu(\alpha)=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{j=1}^na_j$.
Since $[a_0;a_1,\dots]$ is ultimately periodic, this mean value is a well-defined rational number $\geq 1$ if $\alpha$ is irrational. 
Are there examples of quadratic  number-fields $\mathbb Q[\sqrt{N}]$
such that $\inf_{\alpha\in \mathbb Q[\sqrt{N}]\setminus\mathbb Q}\mu(\alpha)>1$?

REPLY [10 votes]: In this paper of McMullen, he asks on p. 842 whether all real quadratic fields
contain infinitely many continued fractions with coefficients bounded by 2? In this case, one
would have  $Q_d=Q_2=\mathbb{Q}(\sqrt{\mathbb{N}})$ for all $d\geq 2$.
See also McMullen's slides from his talk at MSRI in February.<|endoftext|>
TITLE: What is a (partial) left adjoint of the Yoneda embedding called?
QUESTION [11 upvotes]: It is a fairly special property for the Yoneda embedding $A \hookrightarrow \mathcal{P}A$ of a category to have a left adjoint defined everywhere (this happens just when $A$ is total).  However, a partially defined such left adjoint can express the existence of any colimits in $A$.  For instance, if $F\colon I\to A$ is any functor, then colimits of $F$ are the same as values of such a partially defined left adjoint at the presheaf $\mathrm{colim}_i\; A(-,F(i))$.  A similar idea works for weighted limits in an enriched setting.  In fact, $A$ is small-cocomplete just when its Yoneda embedding has a left adjoint defined at all small presheaves (presheaves that are small colimits of representables).
My question is, given a presheaf $X\in \mathcal{P}A$, what is a name for the value of a (partially defined) left adjoint to the Yoneda embedding at $X$?
It is tempting to want to call it the "colimit" of $X$, except that $X$ itself is a functor $A^{op} \to \mathrm{Set}$, and we are certainly not talking about the colimit of that functor.  If it helps, note that the object in question is equivalently the colimit of the identity functor of $A$ weighted by $X$.

REPLY [5 votes]: The question is: Given a functor $F : A^{op} \to \mathsf{Set}$, how do we call an object $?(F)$ in $A$ satisfying the universal property
$\hom(?(F),X) \cong \hom(F,\hom(-,X))$
for all $X \in A$? Some people call it a corepresenting object of $F$. The reason is that a representing object of $F$ is some object $!(F)$ satisfying $\hom(X,!(F)) \cong \hom(\hom(-,X),F)$, since the left hand side simplifies to $F(X)$ by the Yoneda Lemma. Remark that every representing object is also a corepresenting object.
If $F$ is a moduli problem in algebraic geometry, then $?(F)$ with some additional assumptions is usually also called a coarse moduli space (whereas $!(F)$ is the fine moduli space). One of the many references is Definition 2.1. (2) in Adrian Langer's "Moduli Spaces Of Sheaves On Higher Dimensional Varieties", as well as Definition 2.2.1 in "The Geometry of Moduli Spaces of Sheaves" by Huybrecht and Lehn. Perhaps someone can add the original reference.<|endoftext|>
TITLE: why Riemann Hypothesis over curves is easy but 'normal' hypothesis for the Riemann Zeta function $ \zeta (s) $ is hard , 
QUESTION [6 upvotes]: despite both zetas $ \zeta (s,X) $ and $ \zeta (s)$ have the same functional equation, the same Euler prodcut and the same Riemann-Weil formula 
why one of them is 'easy' and can be solved but the other is so hard?? the zeta one $ \zeta (s)$

REPLY [3 votes]: An additional issue is that the adelic norm map in the function field has discrete image, so its Pontryagin dual is actually compact. In the case of a number field, it has non-discrete, non-compact image, so its Pontryagin dual is non-compact as well. Since Tate's thesis, we know that the Fourier analysis on these duals is intimately related to $L$-functions. 
After-thought:
I guess this boils down to the same issue, which GH already mentioned: we have not only one prime field for all residue fields, for number fields there are infinitely many. In fact, there is no residue field for archimedean fields (yet).<|endoftext|>
TITLE: Do ultrapowers of classical Banach spaces have unconditional bases?
QUESTION [11 upvotes]: I am trying to imagine (to some extent, of course) the geometry of ultrapowers of certain 'easy-to-handle' Banach spaces. Let me start with $X = \ell_p$, $p\in (1,\infty)$ or $X=c_0$.
Since the differences between partiuclar choices of ultrafilters are not clear to me enough, let me take any free ultrafilter $\mathcal{U}$ over an arbitrary infinite index set $I$.
So, my beginners question is, does the ultrapower $(X)_\mathcal{U}$ have an unconditional basis?

REPLY [7 votes]: For $\ell_1$ and $c_0$, the answer to your (implicit) question is easy.  If you take a free ultrafilter on the natural numbers (or most any ultrafilter used in Banach space theory) the ultrapower contains $L_1$ (or $C[0,1]$) and hence does not have an unconditional basis.
For $\ell_p$, $1<p\not= 2 < \infty$,  the situation might be trickier.  Enflo and Rosenthal
Some results concerning Lp(μ)-spaces. 
J. Functional Analysis 14 (1973), 325–348
proved that if $\mu$ is a finite measure s.t. the density character of $L_p(\mu)$ is at least $\aleph_\omega$, then $L_p(\mu)$ does not embed into a space that has an unconditional basis. Now such an $L_p(\mu)$ embeds into some ultrapower of $\ell_p$ but I don't know off the top of my head that when the density character of $L_p(\mu)$ is equal to $\aleph_\omega$ [EDIT: I meant $\aleph_1$ here] whether (under some or any set theoretic axioms about the size of $\aleph_\omega$ [EDIT: I meant $\aleph_1$ here]) $L_p(\mu)$ embeds into an ultrapower of $\ell_p$ when the ultrafilter is a free ultrafilter on the natural numbers.<|endoftext|>
TITLE: Möbius Randomness of the Rudin-Shapiro Sequence
QUESTION [18 upvotes]: The Rudin-Shapiro sequence (also known as the Golay-Rudin-Shapiro sequence) is defined as follows.
Let $a_n = \sum \epsilon_i\epsilon_{i+1}$ where $\epsilon_1,\epsilon_2,\dots$ are the digits in the binary expansion of $n$.
$WS(n)$, the $n$th term of the Rudin Shapiro sequence is defined by $$WS(n)=(-1)^{a_n}.$$
Question:
Prove that $$\sum_{i=0}^n WS(i) \mu (i) = o(n).$$
Here, $\mu(n)$ is the Möbius function.
Motivation
This question continues a one-year old question walsh-fourier-transform-of-the-mobius-function
. The two parts of the old question on "Mobius randomness" was settled by Green and by Bourgain, respectively. This question represent the simplest case, which is quite important in its own right, where some new idea/method may be needed.
Motivation (2)
Under the translation 0 --> 1, 1 --> -1, the "Walsh-Fourier" functions can be considered as (all) linear functions over Z/2Z. It tuned out that proving Mobius randomness for a few of them suffices to deduce Mobious randomness for $AC^0$ functions. This was the second part of our old question that was proved by Green. Bourgain showed Mobius randomness for all Walsh functions (namely all linear functions over Z/2Z.) What about low degree polynomials instead of linear polynomials? The Rudin-Shapiro sequence represent a very simple example of quadratic polynomial.
If we can extend the results to polynomials over Z/2Z of degree at most polylog (n) this will imply by a result of Razborov Mobius randomness for $AC^0(2)$ circuits. (This is interesting also under GRH).

REPLY [18 votes]: OK, it seems that a probabilistic swapping argument works (and is simpler than the two other suggestions I made above).
Firstly, by the Bourgain-Sarnak-Ziegler criterion (Proposition 1 in http://terrytao.wordpress.com/2011/11/21/the-bourgain-sarnak-ziegler-orthogonality-criterion/ ), it suffices to show that
$$ \sum_{n \leq x} WS(pn) WS(qn) = o(x)$$
for all fixed distinct primes $p$, $q$, and in the limit $x \to \infty$.  
Fix $p,q,x$.  We phrase things probabilistically. Let $n$ be a random integer between $1$ and $x$; the objective is to show that
$$ {\bf E} WS(pn) WS(qn) = o(1) \quad\quad (1).$$
The way we will do this is to construct a new random variable $n'$, coupled with $n$, which asymptotically almost surely (i.e. with probability $1-o(1)$) is such that
$$ WS(pn) WS(qn) = - WS(pn') WS(qn')  \quad\quad (2) $$
aas.  Also, the total variation distance between the distribution of $n$ and the distribution of $n'$ will be established to be $o(1)$, so that
$$ {\bf E} WS(pn) WS(qn) = {\bf E} WS(pn') WS(qn') + o(1).$$
Putting these facts together will give the claim (1).
It remains to construct $n'$ with the desired properties.  For simplicity let us begin with the case when $WS(p) = -WS(q)$.  Then to obtain $n'$ from $n$, what one does is scan the binary expansion of $n$ for a block of zeroes of length at least $\ell := 10 \max( \log_2 p, \log_2 q ) + 11$ (say); note from the law of large numbers that there are going to be about $2^{-\ell} \log_2 x$ such blocks aas.  We randomly select one of these blocks, and flip the middle zero in this block to a one to create $n'$.  A bit of thought (using the long multiplication algorithm) shows that $WS(pn') = WS(pn) WS(p)$ and $WS(qn') = WS(qn) WS(q)$, giving (2) since $WS(p) = -WS(q)$.  A rather tedious double counting involving the Chernoff inequality (which I will omit here) also shows that the distribution of n' is within o(1) in total variation to that of n, giving the claim.  (The point is that the indegrees and outdegrees of the edit graph connecting potential $n$s to potential $n'$s both concentrate around the same quantity, namely $2^{-\ell} \log_2 x$.)
Of course, it could happen that $WS(p)=WS(q)$ instead.  But a variant of the above argument will work as long as one can find at least one natural number $a$ for which $WS(ap) = -WS(aq)$, basically one has to insert in the binary expansion of $a$ in the middle of a sufficiently large block of zeroes of $n$ to make $n$, rather than flipping a single bit.  So the only way things can go wrong is if the primes $p,q$ are such that $WS(ap)=WS(aq)$ for all natural numbers $a$ (i.e. if there is absolutely no cancellation at all in $\sum_{n \leq x} WS(pn) WS(qn)$.   This can be eliminated as follows.  Without loss of generality we have $p > q$ and $p$ odd.  Then we can find $n,k$ such that $qn < 2^k < pn < 2^{k+1}$.  We observe that
$$ WS(p (n+2^{k+1}) ) = - WS(pn) WS(p)$$
and
$$ WS(q (n+2^{k+1}) ) = WS(qn) WS(q)$$
and so we cannot have $WS(ap)=WS(aq)$ for all $a$.<|endoftext|>
TITLE: How does pseudoconvexity restrict the topology?
QUESTION [5 upvotes]: A domain of holomorphy in $\mathbb{C}^n$ has vanishing de rham cohomology in real dimensions greater than $n$ - half of it's cohomology is missing.  Are there any other restrictions?  If I give you a finite dimensional graded ring, can you find a domain of holomorphy having that ring as its cohomology ring?

REPLY [8 votes]: A theorem of Eliashberg implies that an open subset of $\Bbb C^n$, $n \neq 2$, is isotopic to a Stein domain (hence to a domain of holomorphy) if and only if it admits a handlebody structure with all handles of index $\leq n$. For $n = 2$ the theorem still holds with few modifications regarding the framing of 2-handles. Is not properly the statement of Eliashberg that implies this, but the proof of the theorem itself.
So the answer of the last question is yes if you can build handles of index $\leq n$ (and dimension $n$) so that the resulting manifold has the prescribed cohomology ring (of course not any such handlebody smoothly embeds in $\Bbb C^n$). The problem is related to embedding CW complexes of real dimension $n$ in $\Bbb C^n$, realizing the graded ring (I mean smoothly embedding each cell, then pick a regular neighborhood).<|endoftext|>
TITLE: When do we have derived "Hartogs" for quasi-coherent sheaves?
QUESTION [5 upvotes]: If $X$ is a noetherian scheme , then for any quasi-coherent sheaf $\mathscr{F}$ on $X$ that satisfies Serre's (S2) condition and an inclusion of an open subscheme $j:U\to X$ with complement of codimension at least 2 then we have $$j_{\star}j^{\star}\mathscr{F}\cong\mathscr{F}.$$ This is like the Hartogs phenomenon for functions of several complex variables. See for example the explanation here: Why does the (S2) property of a ring correspond to the Hartogs phenomenon?
My question is, is there a derived version of this? Namely if one works in the derived category and replace the $j_\star$ the derived push forward $Rj_\star$, when do we have
$$Rj_\star j^\star\mathscr{F}\cong\mathscr{F}?$$
I want to know at least the case where $\mathscr{F}=\mathcal{O}_X$.

REPLY [7 votes]: Unfortunately almost never (although it obviously holds if $j^* \mathcal{F} = 0$).  In particular, it virtually never holds for $\mathcal{F} = \mathcal{O}_X$.  I'm going to assume that $\mathcal{F}$ is coherent.
Let me give the quick answer first, then I'll explain it in more detail.
Quick answer:  $R^i j_* j^* \mathcal{F}$ coincides with $H^{i+1}_{X \setminus U}(\mathcal{F})$ the $i+1$th cohomology with support at $Z = X \setminus U$ of $\mathcal{F}$.  By basic facts about local cohomology, you can basically never expect this to vanish for all $i > 0$.  
Detailed answer: Let's suppose that $X = \text{Spec} R$ for simplicity and that $U = X \setminus V(I)$ for some ideal $I$.  Further assume that $j^* M = M|_U \neq 0$.  Let $m$ be a minimal prime of $I$ such that $M_m \neq 0$.  We now can replace $R$ by $R_m$ and $M$ by $M_m$.  
Working locally, for any $R$-module $M$ we can identify
$$
R^i j_* j^* M = R^i j_* (M|_U) = H^i(U, M|_U) = H^{i+1}_m(M)
$$
for $i > 0$ where $H^*_m$ is local cohomology.  This comes from the long exact sequence
$$
H^i_m(M) \to H^i(X, M) = 0 \to H^i(U, M|_U) \to H^{i+1}_m(M) \to H^{i+1}(X, M) = 0.
$$
Now, $H^{i+1}_m(M) \neq 0$ for any $i$ such that $\dim \text{Supp} M = i+1$ by Theorem 3.5.7 in Cohen-Macaulay Rings by Bruns and Herzog.
In particular, if $\dim \text{Supp} M \geq 2$, then we have non-vanishing $R^i j_* j^* M$.  Now, if $\dim \text{Supp} M = 1$ you can still do something similar (the dimension can't be zero by our previous assumptions).  
In this case, it follows that $H^1_m(M) \neq 0$ and in fact is infinitely generated.  Thus $H^0(U, M|_U)$ is also infinitely generated since we have the exact sequence
$$
M \to H^0(U, M|_U) \to H^1_m(M).
$$<|endoftext|>
TITLE: Szőkefalvi-Nagy's unitarizability theorem in the Calkin algebra?
QUESTION [30 upvotes]: Here's a research problem, which I think interesting.
Suppose that $t$ is an invertible element in the Calkin algebra $\mathcal{Q} = \mathcal{B}(\ell_2)/\mathcal{K}(\ell_2)$ which satisfies $\sup_{n \in \mathbb{Z}} \|t^n\|<+\infty$. Does it follow $t$ is similar to a unitary element, i.e., is there an invertible element $s \in \mathcal{Q}$ such that $s t s^{-1}$ is unitary? 
Béla Szőkefalvi-Nagy has proved that it is the case when one replaces $\mathcal{Q}$ with $\mathcal{B}(\ell_2)$ (or any other von Neumann algebra, e.g., $\mathcal{Q}^{\ast\ast}$). I'm asking this question, because a counterexample (which I believe exists) would provide an amenable operator algebra which is not isomorphic to a $\mathrm{C}^\ast$-algebra---The existence of such an example is an open problem.
I will explain why a counterexample provides such an example. Let $\pi\colon \mathcal{B}(\ell_2)\to \mathcal{Q}$ be the quotient map,
$G$ an abelian group (which is $\mathbb{Z}$ in the above question) and $u\colon G\to \mathcal{Q}$ a uniformly bounded homomorphism.
Then, the operator algebra $\mathcal{A} := \pi^{-1}( \overline{\mathrm{span}}\ u(G) )$ is amenable. If $\mathcal{A}$ is isomorphic to a $\mathrm{C}^\ast$-algebra, then by the solution of the similarity problem for amenable $\mathrm{C}^\ast$-algebras, there is $S \in \mathcal{B}(\ell_2)$ such that $S \mathcal{A} S^{-1}$ is a $\mathrm{C}^\ast$-subalgebra. Thus $s \pi(\mathcal{A}) s^{-1}$ (where $s=\pi(S)$) is an abelian $\mathrm{C}^\ast$-subalgebra of $\mathcal{Q}$ and hence it consists of normal elements. Since $u$ is uniformly bounded, $s u(\cdot)s^{-1}$ is a unitary homomorphism. (The converse is also true: if $u$ is similar to a unitary homomorphism, then $\mathcal{A}$ is isomorphic to a $\mathrm{C}^\ast$-algebra.)
The obvious thing one should try is to see whether $H^1_b(G,\mathcal{Q}(\ell_2G))\neq0$.
For a starter, I looked at $H^1_b(G,\ell_\infty(G)/c_0(G))$, but it was zero for every countable exact group $G$. (Whether it is zero for every group $G$ is unclear.)

REPLY [8 votes]: Posting this purely so that Ozawa's interesting question does not stay marked as unanswered, and hence lead to unnecessary effort on the part of someone reading. (Note: could someone reading this please flag the answer for moderator attention, to make it CW.)

It turns out that every uniformly bounded representation of a discrete, countable amenable group inside the Calkin is unitarizable. This follows from techniques introduced in work of Farah and Hart, as expounded in e.g. arXiv: 1112.3898, which extend techniques of G. K. Pedersen for dealing with certain corona algebras. Details of the argument in this particular case are in Theorem 7 of the recent preprint arXiv 1309.2145 by Choi, Farah, and Ozawa.<|endoftext|>
TITLE: Embedding dimension=minimum dimension of a local embedding?
QUESTION [9 upvotes]: Let $X$ be a scheme locally of finite type over a field $k$, and let $p \in X$ be a $k$-point with ${\rm  dim}\; {\Omega_{X}}_{|p}={\rm dim}\; \mathfrak{m}_{p}/\mathfrak{m}^{2}_{p}=m$ (the 'embedding dimension' of the local ring at $p$). 
Then a paper that I'm reading asserts that there is a Zariski open neighbourhood $ p \in U \subseteq X$
and a closed immersion $ U \hookrightarrow M$, where $M$ is a regular (maybe even smooth) scheme over $k$ of dimension $m$. (Clearly $m$ is the minimal dimension for which such an embedding is possible.)
I would like a reference or explanation for this fact, which I have not been able to find in standard commutative algebra books (though it is probably there somewhere). Presumably this is the reason that 
${\rm dim}\;\mathfrak{m}_{p}/\mathfrak{m}^{2}_{p}$ is called the embedding dimension of a local ring.
I would also like a reference or explanation showing that one cannot always take $M=\mathbb{A}^{m}$.

REPLY [7 votes]: One algebraic version of this statement is that if $A$ is a local ring with embedding dimension $m$ and $A$ is the quotient of a regular local ring, then $A$ is the quotient of a regular local ring of dimension $m$. Write $A = R/I$ where $R$ is a regular local ring and suppose that $R$ has dimension $n$. Then we have a surjection of the Zariski cotangent spaces of the two rings $m_R/m_R^2 \rightarrow m_A/m_A^2$. We can find a basis for the kernel of this map consisting of $n-m$ vectors. Lift these to elements $f_1, \ldots, f_{n-m}$ in the kernel of $A \rightarrow R$ and let $S = R / \langle f_1, \ldots, f_{n-m}\rangle$ so that $A = S/J$ for some ideal $J$. Then I just have to show that $S$ is a regular local ring of dimension $e$. By construction, its Zariski cotangent space has dimension exactly $m$, and by the Krull principal ideal theorem, it has dimension at least $m$, so it must be a regular local ring of dimension $m$.
In the setting that you asked, every local ring of a scheme of finite type over a field is the quotient of a regular local ring since it is the quotient of a localization of affine space. If you apply the argument above then you get equations $f_1, \ldots, f_{n-m}$ which define a regular scheme at $p$ and thus in a neighborhood of $p$. This is your scheme $M$.
As for your question about whether $M$ can always be chosen to be $\mathbb A^m$, the answer is certainly not. Just take $p$ to be a smooth point on any non-rational variety. Then pretty much by definition, there is no neighborhood of $p$ which is isomorphic to an open subset of affine space. It would take some more work to think of an example where $p$ is singular, but it seems like a sufficiently singular point of a variety in a non-rational variety would work. (Although, I believe any zero-dimensional scheme of embedding dimension $m$ and finite type over a field has a closed immersion into affine space of dimension $m$.)<|endoftext|>
TITLE: Primes with more ones than zeroes in their Binary expansion
QUESTION [22 upvotes]: This question is also motivated by the developement around my old MO question about Mobius randomness. It is also motivated by Joe O'Rourke's question on finding primes in sparse sets.

Let $A$ be the set of all natural numbers with more ones than zeroes in their binary expansion. Are there infinitely many primes in $A$? 

More generally, for a function $f(n)$ defined on the natural numbers let $A[f]$ denotes the set of integers with $n$ digits and at least $n/2+f(n)$ ones, for $n=1,2,...$. Does $A[f]$ contains infinitely many primes?
Bourgain proved the Mobius randomness of $A$ and this seems closely related to this question. But I am not sure about the exact connection. (In fact Bourgain proved Mobius randomness for every $A$ described by a balanced monotone Boolean function of the binary digits.)
Showing infinitely many primes for sparse $[f]$ would be interesting. Proving this for $f(n)=\alpha n$ where $\alpha>0$ is small would be terrific. Of course, if $f(n)=n/2$ we are talking about Mersenne's prime so I would not expect an answer here. (Showing infinite primes for $A$ with smaller size than $\sqrt n$ will cross some notable barrier.)
A similar question can be asked about balanced (and unbaland) sets described by $AC^0$-formulas. This corresponds to Ben Green's $AC^0$ prime number theorem but also here I am not sure what it will take to move from Mobius randomness to infinitude of primes.
Another related question: Are There Primes of Every Hamming Weight?

REPLY [16 votes]: We can take $f(n)=\alpha n$ for any $\alpha<0.7375$.  In particular, the set of primes with more than twice as many ones that zeros in their binary expansion is infinite.
I posted a short article on the arXiv which deals with exactly this kind of problem.  Let $s_2(n)$ denote the sum of digits base $2$.  Since $x$ has approximately $\log_2(x)$ binary digits, we are looking at when $s_2(n)\geq \alpha \log_2 (n)$.  In that 4 page note we prove that
$$\left|\left\{ p\leq x,\ p\ \text{prime}\ : s_2(n)\geq \alpha\log_2(x) \right\} \right|\gg_{\epsilon}\ x^{2\left(1-\alpha\right)}e^{-c\left(\log x\right)^{1/2+\epsilon}}.$$
Moreover, such a result extends naturally to base $q$, yielding the bound
$$\left|\left\{ p\leq x,\ p\ \text{prime}\ :\ s_{q}(p)\geq\alpha(q-1)\log_{q}(x)\right\} \right|\gg_{\epsilon}\ x^{2\left(1-\alpha\right)}e^{-c\left(\log x\right)^{1/2+\epsilon}}$$ where $s_q(n)$ is the sum of digits of $n$ in base $q$.
The proof takes advantage of the fact that the multinomial distribution is sharply peaked.  The number $0.7375$ appears because $1-0.525/2=0.7375$, and $0.525$ is the exponent appearing in Baker Harman and Pintz's work on prime gaps.<|endoftext|>
TITLE: Recognising group actions on trees from the boundary
QUESTION [9 upvotes]: Let $G$ be a group acting on a locally finite tree $T$.  Then the boundary $\partial T$ is a Cantor set on which $G$ acts by homeomorphisms (indeed by quasi-isometries under a suitable metric).  However, even if $G$ is the full automorphism group of $T$, we can't get the full quasi-isometry group of $\partial T$, as the tree structure puts further restrictions on the action.
What ways are known of identifying those group actions by homeomorphisms on the Cantor set which arise as actions on the boundary of a locally finite tree?  Is there a nice algebraic description of $\mathrm{Aut}(T)$ as a subgroup of $\mathrm{Homeo}(\partial T)$?
Edit: Here is an example of a sufficient criterion to show what I mean.  Suppose that $G$ acts on the Cantor set $X$ by homeomorphisms, with finite orbits on the clopen subsets.  Then one can produce a tree $T$ such that $G$ acts on $T$ with a global fixed point and $\partial T$ is $G$-homeomorphic to $X$.  So an action with finite orbits on the clopen subsets is automatically 'arboreal' (we do not care about which tree appears, just that it exists).

REPLY [5 votes]: I'll focus on potential topological characterizations of the action. As I mentioned in the comments above, every element of $G$ will either lie in a maximal compact subgroup of $Aut(\partial{T})$, or will have north-south dynamics (a hyperbolic element). 
If the group $G$ acts discretely, cocompactly on $\mathcal{T}$, then there's Bowditch's characterization that this is equivalent to $G$ acting properly dicontinuously  and cocompactly on the triple points in $\partial{T}$. In fact, in this case $G$ will be a virtually free hyperbolic group, and a (finite) graph of finite groups. 
I believe that $Aut(\mathcal{T})$ is closed in $Aut(\partial{\mathcal{T}})$, with the induced topology. In this case, one may assume that $G\leq Aut(\mathcal{T})$ is closed, otherwise taking its closure $\overline{G}$ in $Aut(\partial{\mathcal{T}})$ will correspond to taking its closure in $Aut(\mathcal{T})$, and if $\overline{G}\leq  Aut(\mathcal{T})$, then $G\leq Aut(\mathcal{T})$. So assume now that $G$ is closed in $Aut(\partial{\mathcal{T}})$. 
If $G$ is a compact subgroup of $Aut(\partial{\mathcal{T}})$, then I think it should
be a profinite group, which acts on a tree (elliptically with a global fixed point). 
However, I'm not sure exactly how to tell if the action of $G$ on $\partial{\mathcal{T}}$
corresponds to this group action. Let's assume from now on that $G$ is not compact. 
If $G$ does not act cocompactly on $\mathcal{T}$, then consider the limit set $\Lambda(G)\subset \partial{\mathcal{T}}$, which is the set of accumulation points of $Gx\subset \partial{\mathcal{T}}$ for any $x\in\partial{\mathcal{T}}$. One may realize this as the closure of the limit points of
hyperbolic elements of $G$. Then $G$ should act cocompactly on the convex hull   of $\Lambda(G)=\mathcal{R}$ inside of $\mathcal{T}$. So one may replace $\mathcal{T}$ with $\mathcal{R}$, and  $\partial{\mathcal{T}}$ with $\partial{\mathcal{R}}=\Lambda(G)$.  
So now assume that $G$ is a closed noncompact subgroup of $Aut(\partial{\mathcal{T}})$, and $\Lambda(G)=\partial{\mathcal{T}}$. 
Now, I believe that there should be a generalization of Bowditch's theorem. $G$ should be hyperbolic as a topological group, generated by a compact subset. I think this should be equivalent to $G$ acting properly cocompactly on the triple point space of $\partial{\mathcal{T}}$, but I don't know if this is proven (or even correct). Compare to the action of $Isom(\mathbb{H}^n)$ on $(\partial\mathbb{H}^n)^3-\Delta$, which is proper and cocompact. 
There is another possible topological characterization in terms of actions on spaces of walls, but this is really just an encoding of the tree on $\partial{T}$ in terms of how each edge partitions $\partial{T}$ into pairs of clopen sets. If $G$ is closed, but noncompact and nondiscrete, then one might be able to encode the tree by the maximal compact subgroups of $G$ and their intersections, essentially Bass-Serre theory. But I haven't thought this through.<|endoftext|>
TITLE: c^2 = a^2 + b^2 + ab and its solutions
QUESTION [5 upvotes]: The equation $c^2 = a^2 + b^2 + ab$ is the law of cosines for a triangle with integer sides $a$, $b$, and $c$, and a 120 degree angle opposite side $c$.   By the substitution $x = (a-b)/2$, $y = (a+b)/2$  it can be transformed to $x^2 + 3y^2 = 4z^2$,  which is a more familiar equation, whose solutions are given in parametric form on page 353 (Corollary 6.3.15) of Cohen,  Number Theory volume I.   That parametrization doesn't seem to answer my question, though.
Define the "square residue" of $x$ to be the product of (one power each of) all the primes dividing x to an odd power.  For example, the square residue of 80 is 5 and the square residue of 40 is 10.    What I want to show is that for given  $N$,  there is some bound $K$ such that 
all solutions with square residue of $(ab) <= N$  have   $a$ and $b$ at most $K$.  Since $b$ can be bounded in terms of $a$ it's enough to have $a$ at most $K$ if that is easier.
In other words, how can we rule out HUGE solutions $a$ and $b$ with teeny-weeny 
$sqres(a,b)$?  I've investigated this numerically.   Here are the solutions with 
$sqres(a,b) <= 100$  and $a,b \le 6120$. 
\begin{verbatim}
(3, 5, 7)   sqres(ab) = 15
(5, 16, 19)   sqres(ab) = 5
(7, 8, 13)   sqres(ab) = 14
(9, 56, 61)   sqres(ab) = 14
(11, 24, 31)   sqres(ab) = 66
(16, 39, 49)   sqres(ab) = 39
(19, 80, 91)   sqres(ab) = 95
(32, 45, 67)   sqres(ab) = 10
(32, 175, 193)   sqres(ab) = 14
(33, 800, 817)   sqres(ab) = 66
(35, 288, 307)   sqres(ab) = 70
(49, 575, 601)   sqres(ab) = 23
(64, 735, 769)   sqres(ab) = 15
(65, 3136, 3169)   sqres(ab) = 65
(75, 112, 163)   sqres(ab) = 21
(225, 37856, 37969)   sqres(ab) = 14
(704, 92575, 92929)   sqres(ab) = 77
(725, 131043, 131407)   sqres(ab) = 87
(819, 1600, 2131)   sqres(ab) = 91
(847, 3200, 3697)   sqres(ab) = 14
(3179, 19200, 20971)   sqres(ab) = 33
(3825, 15488, 17713)   sqres(ab) = 34
(3887, 4800, 7537)   sqres(ab) = 69
(4312, 4563, 7687)   sqres(ab) = 66
\end{verbatim}
Do we know that if we keep going,  we're not going to get another solution with sqres = 10?
By the way, does the "square residue"  already have another name, perhaps well-known to number theorists?

REPLY [4 votes]: Continuing GH's thoughts, for $\alpha = 1$ and $\beta = 5$, the (Jacobian) elliptic curve (of) $z^2 = x^4 + 5 x^2 y^2 + 25 y^4$ has rank 1 and therefore infinitely many rational points, which correspond to coprime triples $(x,y,z)$ of integers solving the equation.
Then setting $a = x^2$, $b = 5 y^2$ and $c = z$, one obtains infinitely many (and therefore arbitrarily large) solutions with $\mbox{sqres}(ab) = 5$. Some of them are
$$(1,0,1), (0,5,5), (16,5,19), (25,80,95), (53361, 115520, 149521), \dots$$
(Not all of them are coprime, but there will be infinitely many coprime $(a,b,c)$ triples among them.)<|endoftext|>
TITLE: How far is a set of vectors from being orthogonal?
QUESTION [14 upvotes]: Given some vectors, how many dimensions do you need to add (to their span) before you can find some mutually orthogonal vectors that project down to the original ones?
Or, more formally...
Suppose $v_1,v_2,\ldots,v_k \in \mathbb C^m$.  Take $n \ge m$ as small as possible such that if we consider $\mathbb C^m$ as a subspace of $\mathbb C^n$ in the natural way, then there is a projection $\pi$ of $\mathbb C^n$ onto $\mathbb C^m$ such that for some mutually orthogonal vectors $\hat v_1,\hat v_2,\ldots,\hat v_k \in \mathbb C^n$, $\pi(\hat v_i) = v_i$ for each $i$.
Intuitively, this $n$ provides a measure of how "far from orthogonal" the original vectors are.  My  (deliberately open-ended) question is the following.  Does anyone recognize this $n$, or does the idea seem familiar to anyone from any other context?  I can't identify it with or connect it to anything else I've encountered, but I'm wondering if it might appear in some other guise in linear algebra, or elsewhere.

REPLY [2 votes]: As mohanravi points out, we can see this entirely in terms of the matrix $X^{T}X$. In particular, we are asked to find a positive semidefinite matrix of minimal rank $R$ such that $R+X^TX$ is diagonal.
We can subsume Nik Weaver's answer as follows: Let $\lambda$ be the highest eigenvalue of $X^TX$, then $R=\lambda I - X^T X$ is positive semidefinite and has rank no more than $k-1$.
This measurement has some strange properties. For instance, the matrix that is $1$ on the diagonal and $\epsilon$ off the diagonal is much harder to make diagonal by adding a positive semidefinite matrix than the matrix is that is $1$ on the diagonal and $-\epsilon$ off the diagonal, where adding a rank-$1$ matrix does the trick.<|endoftext|>
TITLE: Can one understand the Kelvin transform conceptually?
QUESTION [5 upvotes]: Let $U = \mathbf{R}^n - \{ 0 \}$, $n > 2$ and consider for a function $f \in C^2(U)$ the Kelvin transform
$$f^\star(x) = r^{2-n} f\left(\frac{x}{r^2}\right),$$
where $r = \lvert x \rvert$. One can verify by explicit computation that,
$$r^2 \Delta(f^\star(x))= (r^2\Delta f)^{\star}(x),$$
where $\Delta$ denotes the usual Laplace operator. In particular the Kelvin transform maps harmonic functions to harmonic functions. Is there a way to see this last fact without resorting to explicit computation? I suspect it has to do with how the group of conformal transformations acts.

REPLY [2 votes]: A conceptual view on the Kelvin transform is use the conformal invariance of the Yamabe operator :
If $(M,g)$ is a Riemannian manifold of dimension $n$, the Yamabe operator is the differential operator
$$L_g=\Delta+\frac{n-2}{4(n-1)} \mathrm{Scal}_g$$
It is a conformally invariant operator that is to say if 
$\widehat g=u^{\frac{4}{n-2}} g$ where $u$ is a smooth positive function then
$$ L_{ g}(u\varphi)=u^{\frac{n+2}{n-2}}L_{\widehat g}(\varphi)$$
(see the formula (2.7) in the article of Lee and Parker :
(The Yamabe problem , Bull. AMS, (17) n° 1 (1987) pp 37--91).
...Well this a consequence of  the  computation of the scalar curvature under conformal change and this is not really simpler than the computation involved in the proof of the Kelvin transform...
Now the Kelvin transform is a special case of this conformal transformation law for the Yamabe operator :
Let $I\colon \mathbb{R}^{n}\setminus\{0\}\rightarrow  \mathbb{R}^{n}\setminus\{0\}$ be the inversion :
$$I(x)=\frac{x}{\|x\|^2}.$$
It is a conformal map and the pull back of the Euclidean metric (we called it $\mathrm{eucl}$) by $I$ is:
$$I^*\mathrm{eucl}=\frac{1}{\|x\|^4} \mathrm{eucl}$$
Indeed in polar coordinate :
$$I^*\mathrm{eucl}=\left(d\frac{1}{r}\right)^2+\left(\frac{1}{r}\right)^2 (d\sigma)^2=\frac{(dr)^2+r^2 (d\sigma)^2}{r^4}.$$
Call $g=\mathrm{eucl}$ and $\widehat g=I^*\mathrm{eucl}$, 
the conformal transformation for the Yamabe operator yields:
$$L_{\widehat g}(\varphi)=u^{-\frac{n+2}{n-2}}L_{ g}(u \varphi)$$
where $u=r^{2-n}$.
But $I\colon  \left(\mathbb{R}^{n}\setminus\{0\}, \widehat g\right) \rightarrow \left(\mathbb{R}^{n}\setminus\{0\}, \mathrm{eucl}\right) $ 
is by definition an isometry hence the scalar curvature of the Riemannian metric $\widehat g$ is zero and
$$L_{\widehat g}(\varphi\circ I)=\left(\Delta\varphi\right)\circ I $$
Eventually we get :
$$\left(\Delta \varphi\right)\circ I=r^{n+2}L_{ g}(u \varphi\circ I)= r^{n+2} \Delta\left(r^{2-n}\, \varphi\circ I\right),$$
this formula is equivalent to the Kelvin transform.<|endoftext|>
TITLE: Complexity of finding a rational root of a polynomial
QUESTION [6 upvotes]: This is inspired by this  question. Let $f(x)=a_nx^n+...+a_0$ be a polynomial with rational coefficients. The sandard procedure of finding a rational root $p/q$ involves checking all $p$ that divide $a_0$ and all $q$ that divide $a_0$. This is not very complicated but involves factoring $a_0$ and $a_n$. The factoring problem is not known to be in P. If $n\le 4$, then the fact that the group $S_4$ is solvable and the well known formulas for roots of polynomials of degree $\le 4$ give easy polynomial time algorithm of finding rational roots. 
 Question Is the problem of finding a rational root of $f(x)$ in P for every $n$ (say, for $n=5$)?
 Update 1 After I posted the question, I noticed an answer by Robert Israel to the previous  question (of Joseph O'Rourke). That could give an answer to my question but I am still not sure how one can avoid factoring numbers $a_0,a_n$. 
 Update 2  Robert Israel's explanations (see his comment here ) convince me that his algorithm of checking whether a polynomial has a rational root (all roots rational) runs in polynomial time. 
I removed Question 2 so that I can accept Michael Stoll's answer. I will post Question 2 as a separate question.

REPLY [3 votes]: You do not need the full power of LLL algorithm for finding rational roots. In particular, Berlekamp-Zassenhaus algorithm exponential-time algorithm to compute the irreducible factorization can be used to get a polynomial-time algorithm for finding rational roots: The exponential part of this algorithm is the recombination of different modular factors to build true factors, but no such recombination is needed to build the linear factors. 
A complete description and analysis of such an algorithm has been given by Loos in 1983:  
Loos, R. Computing Rational Zeros of Integral Polynomials by P-Adic Expansion. SIAM Journal on Computing 12, no. 2 (May 1, 1983): 286–93. doi:10.1137/0212017.<|endoftext|>
TITLE: Characteristic polynomial of a generic n*n matric
QUESTION [6 upvotes]: Let $K$ be a field, and $F_K$ be the fraction field of the polynomial ring $R_K$ in $n^2$ indeterminates $X_{11},X_{12},...,X_{nn}$ over $K$.
Now set $A = (X_{ij})_{i,j} \in M_n (F_K)$, and let $\chi_A$ be the characteristic polynomial of $A$. 
Question : Is it always true that $\chi_A$ is irreducible over $F_K$ ?
Some thoughts :

Since $R_K$ is a UFD, it is sufficient to check irreducibility in $R_K[X]$.
If $ |K| \geq n$, then $\chi_A$ is separable (by evaluation of $A$ to a diagonal matrix with distinct diagonal entries).
If $K[X]$ contains some irreducible $f$ of degree $n$, then this is true (just evaluate $A$ to the companion matrix of $f$ and observe that the evaluation morphism cannot increase degrees). For example, this yields the result when $K = \mathbb{Q}$ or when $K$ is a finite field.
More generally, a non trivial factor of $\chi_A$ is unlikely to exist, since it would give a generic factorization of the characteristic polynomial of a $n \times n$ matrix with coefficients in $K$.

Even in the case $K=\mathbb{C}$, I cannot prove (or disprove) the irreducibility of $\chi_A$.

REPLY [9 votes]: If it were reducible, all $n\times n$ matrices over any field extension of $K$ would have reducible characteristic polynomials.  But consider the companion matrix of an irreducible polynomial over some not algebraically closed field extension of $K$.<|endoftext|>
TITLE: How many model category structures are there on Top?
QUESTION [10 upvotes]: I recently started learning a little model category theory and in particular I found this nice exercise.  I only know a little topology, but this prompted me to wonder how many model category structures there may be on Top.  I am aware of three: Serre fibrations and weak homotopy equivalences, Hurewicz fibrations and homotopy equivalences, and the usual model category of rational homotopy theory [Dwyer and Spalinski]. A secondary question could be how many homotopy theories there are since it is known that the first two I mention give the same homotopy theory.
*This question is a little out of my league right now. I hope that is ok. I'm not even sure how difficult this question is.

REPLY [7 votes]: Let's work in simplicial sets, where things are generally nicer (in particular, the standard model structure is combinatorial). I can refer to simplicial sets as a model for topological spaces because sSet is Quillen equivalent to Top and because simplicial sets can all be realized as topological spaces. In this setting, there are infinitely many different model structures, and this is proven in Tibor Beke's "Fibrations of Simplicial Sets." These model structures are all homotopically equivalent (by which I mean there is a chain of Quillen equivalences between any two). The difference between them is a difference of cofibrations and fibrations. So far we have countably many different structures. On page 2, Beke comments that taking products of these model structures demonstrates that there is no upper bound on the cardinality of possible cofibration classes. In a combinatorial setting like sSet, if you restrict attention to set-generated cofibration classes then all will be homotopically equivalent. This idea gives you the potential for arbitrarily many model structures, but we'd have to check that they all have different fibrations (this is the heart of his paper above).
If you want non-equivalent model structures then that's even easier. I wrote an answer here explaining truncated model structures, where $f$ is a weak equivalence if $\pi_k(f)$ is for $n\leq k\leq m$). Varying $n$ and $m$ give you infinitely many non-equivalent model structures. The non-equivalence can be seen by looking at how two different ones see wedges of Eilenberg-Mac Lane spaces. Again, this gives countably many, but I feel like there should be more.
As John Klein comments above, Bousfield localization helps you construct non-equivalent model structures. The idea now is that you are leaving the cofibrations fixed and increasing the weak equivalences by formally inverting a chosen set of maps. As sSet is combinatorial, every model category so created will also be combinatorial, and will therefore admit Bousfield localization again. There are certainly a class worth of choices for sets of maps to invert, but not all choices will give different model structures. Another natural idea would be to look at localization of the homotopy category and ask whether all come from Bousfield localizations of the model category. Dror Farjoun conjectured that this is the case. Casacuberta and Smith proved this under the assumption of Vopenka's principle from set theory (it's known that the standard axioms of ZFC cannot imply Vopenka). Casacuberta has also written other papers in the same vein which might interest you. He has several with Rosicky and/or Chorny (who sometimes frequents math overflow). The ones featuring the Orthogonal Subcategory Problem are very nice, and suggest again that questions about localization are often related to questions in set theory. Incidentally, if someone has an argument that there should be even uncountably many different Bousfield localizations on sSet I'd be curious to hear it. Getting a class worth would be great too.
Final remark: if you really want to work with compactly generated spaces or spaces homeomorphic to CW complexes instead of sSet then the situation is harder because you're no longer combinatorial. My answer in the middle paragraph about truncated model structures still holds. Tibor Beke addresses the answer in paragraph one at the very end of his paper. For Bousfield localization you need to use cellularity rather than combinatoriality, but you'll again have a class worth of possible model structures and will have to determine which are the same (and now you won't have the results of Casacuberta and coauthors because Top is not combinatorial).<|endoftext|>
TITLE: What nets fold to polyhedra?
QUESTION [13 upvotes]: There is a classic (and open) problem asking whether every polyhedron can be unfolded to give a non-overlapping net. The converse problem has been studied asking which polygons can be folded in some manner to give a polyhedron. For an overview of both problems and related discussion see:
Erik D. Demaine and Joseph O'Rourke Geometric Folding Algorithms: Linkages, Origami, Polyhedra
Cambridge University Press, July 2007. ISBN 978-0-521-85757-4
Question
I want to ask about a more direct converse. Given a net of polygons connected at their edges when can they fold to form a polyhedron?
Example
As an example take the two nets shown below. By identifying the edges as shown by colour both satisfy the topological constraints to be sphere.

Only one, however, will fold to give a polyhedron:

Personal Motivation
My motivation is to find visually appealing, simple, but non-symmetric, polyhedra. I have used equilateral triangles but would like to play with other shapes. General sufficient conditions would, therefore, be very interesting to me.

REPLY [12 votes]: Wonderful images, Edmund! :-)
A net $P$ can fold to a polyhedron iff there exists what I called in the book you cite an Alexandrov gluing, which is

an identification of its boundary points
  that satisfies the three conditions of Alexandrov's theorem
  [the subject of a recent MO question]:
  (1) The identifications (or "gluings'') close up the perimeter of $P$ without gaps or overlaps;
  (2) The resulting surface is homeomorphic to a sphere; and
  (3) Identifications result in $\le 2 \pi$ surface angle glued at every point.
  Under these three conditions, Alexandrov's theorem guarantees that the
  folding produces a convex polyhedron, unique once the gluing is specified.

Let me quote two results from the book, informally, the first quite disappointingly negative, the second
compensatingly positive:

Theorem 25.1.2 (p.382): The probability that a random net of $n$ vertices can fold to a convex polyhedron
goes to $0$ as $n \to \infty$.
Theorem 25.1.4 (p.383): Every convex polygon folds to an uncountably infinite variety of incongruent convex polyhedra.

In particular, for example, a square folds to an infinite number of convex polyhedra, whose space
consists of six interlocked continuua, as detailed in our book (Fig.25.43,p.416).
The exact question you pose—Which nets can fold to convex polyhedra?—remains open.
Although if you give me a specific net, we have an algorithm that will produce all the convex 
polyhedra to which it may fold.
But note my emphasis on "convex," an adjective you left out in your question.
That is Open Problem 25.1 in our book (p.384), on which topic I have written a separate
note subsequent to the book's publication: "On Folding a Polygon to a Polyhedron."
In a nutshell: every polygon folds to some (generally) nonconvex polyhedron,
by a result of Burago and Zalgaller. But their proof is complex enough that I have no understanding
what that polyhedron might look like.
Aside from the paper I cite above, you may be interested in this result:
Four of the five Platonic solids may be "unzipped" and "rezipped" to be doubly covered
parallelograms (which may conveniently be placed in your wallet!). See this MSE question for
the (difficult to find) icosahedron net [below]. The holdout here is the
dodecahedron, whose 43,380 edge unfoldings each may only fold back to the dodecahedron.<|endoftext|>
TITLE: Explaining Mukai-Fourier transforms physically
QUESTION [14 upvotes]: A core concept in mathematics, engineering, and physics is the Fourier Transform (FT) and its many variants (Generalized Fourier Series, Green's Function, Pontryagin duality).
The basic algorithm is to find dual sets of eigenvectors/eigenfunctions parametrized by a continuous (e.g., $\omega$ below) or discrete index (e.g., $n$ below), that satisfy completeness and orthogonality relations encapsulated in Dirac delta function resolutions such as that for the FT
$$\delta(x-y)= \int_{-\infty}^{\infty}\exp(i2\pi \omega x)\exp(-i2\pi \omega y)d\omega$$
giving
$$\int_{-\infty}^{\infty}f(y)\delta(x-y)dy=f(x)=\int_{-\infty}^{\infty}\exp(i2\pi \omega x)\int_{-\infty}^{\infty}f(y)\exp(-i2\pi \omega y) dy d\omega$$
$$=\int_{-\infty}^{\infty}\exp(i2\pi \omega x)\hat{f}(\omega) d\omega,$$
or that for the eigenvectors of Sturm-Liouville differential operators over finite domains
$$\delta(x-y)=\sum_{n=0}^{\infty }\Psi_n(x)\Psi_n^*(y)$$
giving
$$f(x)=\sum_{n=0}^{\infty }\Psi_n(x)\int_{a}^{b}f(y)\Psi_n^*(y) dy,$$
or Kronecker delta resolutions such as that for the associated Laguerre functions
$$\frac{(n+\alpha)!}{n!}\delta_{mn}=\int_{0}^{\infty}x^{\alpha}e^{-x}L_{n}^{\alpha}(x)L_{m}^{\alpha}(x)dx$$
giving
$$f(x)=\sum_{n=0}^{\infty }\frac{n!L_{n}^{\alpha}(x)}{(n+\alpha)!}\hat{f}_n$$
with
$$\hat{f}_n=\int_{0}^{\infty}x^{\alpha}e^{-x}L_{n}^{\alpha}(x)f(x)\,dx.$$
The basic "physical" operation (BPO) at work here can be regarded as destructive/constructive interference; the product at a point of the value of the function (to be resolved) with the corresponding value of an eigenfunction has a negative or positive value (or phase factor) that may sum constructively or destructively with products at other points (seen as a matched filtering or correlation by replacing $y$ with $x-z$ above). Alternatively, the BPO may be viewed as projection of vectors onto a set of orthonormal axes. In addition, if the function and operations are discretized and/or the domains restricted (in one space or its dual or both, as for the DFT) aliasing (which seems analogous to the introduction of equivalence classes) is introduced and periodicity imposed.
Can you explain the machinery behind the Mukai-Fourier transform in terms of these BPOs or close analogies?
(Edit 1/16/212) Further to Carnahan's answer below:
Kapustin and Witten in "Electric-magnetic duality and the geometric Langlands Program" state, ". . . it must be possible to understand the geometric Langlands program using four-dimensional electric-magnetic duality (which leads to this particular T-duality) and branes (the natural quantum field theory setting for interpreting T-duality as a Fourier-Mukai transform). This hint was the starting point for the present paper." The T-duality is analogous to the modular/automorphic function symmetry involving reciprocals of a parameter/variable. They refer to the paper "Lectures on the Langlands Program and conformal field theory" by Frenkel as a good intro.
Frenkel notes, "It has long been suspected that the Langlands duality should somehow be related to various dualities observed in quantum field theory and string theory. Indeed, both the Langlands correspondence and the dualities in physics have emerged as some sort of non-abelian Fourier transforms. Moreover, the so-called Langlands dual group introduced by R. Langlands that is
essential in the formulation of the Langlands correspondence also plays a prominent role in the study of S-dualities in physics and was in fact also introduced by the physicists P. Goddard, J. Nuyts, and D. Olive in the framework of four-dimensional gauge theory. ... The goal of these notes is two-fold: first, it is to give a motivated introduction to the Langlands Program, including its geometric reformulation, addressed primarily to physicists. I have tried to make it as self-contained as possible, requiring very little (lol) mathematical background. The second goal is to describe the connections between the Langlands Program and two-dimensional conformal field theory that have been found in the last few years. These connections give us important insights into the physical implications of the Langlands duality."
On pages 53 and 54, Frenkel presents an elaboration on the parallels, as noted by Carnahan, between the Dirac delta function-complex exponential duality central to the Fourier transform and the skyscraper sheaf-line bundle duality associated with the Fourier-Mukai transform.
Extrapolating, one would expect to find analogies to Poisson (or Dirac comb) summation, zeta function functional symmetry equations, and Euler product factorizations, as listed in Table I of "A Correspondence Principle" by Hughes and Ninham (subsumed, I suppose, by the analytic Langlands Program), and to log/exp, trace/determinant, moment/cumulant, and other dualities found in symmetric function theory, as well as 'gauge/conjugation' transformations. References to surveys sketching such associations, briefer than Frenkel's, would be appreciated, especially if linked to Fourier(Mellin)-Mukai transforms. (This is related to another MO_Q.)

REPLY [10 votes]: You can think of line bundles and skyscraper sheaves as sheaf-theoretic analogues to exponentials and delta functions, respectively.  The Fourier-Mukai transform on an elliptic curve takes one type of sheaf to the other (with a homological shift that I will ignore).  In higher dimension, you get some mixtures of these types, from complexes of sheaves with cohomology supported on subvarieties of positive dimension and positive codimension - you can think of these as an analogue of more general distributions.  Vector bundles on an abelian variety get "orthogonally decomposed" into skyscrapers on the "frequency space", i.e., the dual abelian variety.

REPLY [2 votes]: I'd suggest you to check the post HERE.
As you will see, the analogy enters by thinking the pullback of the sheaf $\mathcal{F}$ on the $X$ variety, i.e. $p_1^*\mathcal{F}$, as Fourier coefficients, while the sheaf $\mathcal{P}$ on the product variety $X\times Y$ plays the role of integral kernel.
Cheers.<|endoftext|>
TITLE: The Galois group and relations among the roots of a polynomial
QUESTION [7 upvotes]: Let $f(x) \in \mathbb{Z}[x]$ be a monic irreducible polynomial of degree $n$, and let $\alpha_1, \alpha_2, ... , \alpha_n \in \overline{\mathbb{Q}}$ be the $n$ distinct roots of $f(x)$.
Following Bewersdorff's "Galois Theory for Beginners" (and older sources?) I want to define the Galois group of $f(x)$ as follows.  Let $I \subset \mathbb{Q}[x_1, x_2, ... , x_n]$ be the ideal consisting of those polynomials $g(x_1, x_2, ... , x_n)$ in $n$ variables such that $$g(\alpha_1, \alpha_2, ... , \alpha_n) = 0$$
I propose to call $I$ the Galois ideal of $f(x)$.  If there is a more standard term for this, please let me know.
Now the Galois group of $f(x)$ may be defined as the group $G \subset S_n$ consisting of those permutations $\sigma \in S_n$ such that $$g \in I \Longrightarrow g_\sigma \in I$$
where $g_\sigma$ is the polynomial $g$ transformed by permuting the variables using the permutation $\sigma$.
Certain polynomials will be trivially members of $I$ for any polynomial $f(x)$.  Specifically, $f(x_1), f(x_2), ... , f(x_n) \in I$, and also the elementary symmetric polynomials minus the coefficients of $f$ will be members, e.g. $x_{1}x_{2} ... x_{n} - (-1)^{n} c_0$, where $c_0$ is the constant coefficient of $f$.
For some ("most"?!) polynomials $f(x)$, the Galois ideal is generated only by these trivial members, and for such polynomials the Galois group is the full symmetric group.  To the extent that there are non-trivial generators of $I$, the Galois group will be a proper subgroup of $S_n$.
For example, if $f(x) = x^4 - 2$, and $\alpha_1 = \sqrt[4]{2}$, $\alpha_2 = -\sqrt[4]{2}$, $\alpha_3 = i\sqrt[4]{2}$, $\alpha_4 = -i\sqrt[4]{2}$, then the non-trivial generators of $I$ are $$g_1(x_1,x_2,x_3,x_4) = x_1 + x_2$$ $$g_2(x_1,x_2,x_3,x_4) = x_3 + x_4$$  Accordingly, the Galois group is generated only by the permutations $$1234 \rightarrow 2134$$ $$1234 \rightarrow 1243$$ $$1234 \rightarrow 3412$$
My question is, is there an algorithm to find the non-trivial generators of the Galois ideal?  It seems that this ideal actually gives more information about the polynomial $f(x)$ than the Galois group, since it is trivial to find the Galois group given the Galois ideal, but not conversely.

REPLY [6 votes]: The Galois ideal is one of the prime factors of the ideal generated by the trivial elements. Thus, an algorithm for primary decomposition in $\mathbb Q[x_1,...x_n]$, of which there are several, will do the trick.
Proof: The Galois ideal is prime, which is obvious from its definition. It contains the trivial ideal. Since the vanishing set of the trivial ideal has dimension $0$, all prime ideals containing it are maximal, thus all prime ideals are prime factors.
(In terms of "most", if I remember correctly, almost all polynomials with coefficients of bounded height have Galois group the full symmetric group, but I am not an expert on such things.)<|endoftext|>
TITLE: Does $(x^2 - 1)(y^2 - 1) = c z^4$ have a rational point, with z non-zero, for any given rational c?
QUESTION [20 upvotes]: I need this result for something else. It seems fairly hard, but I may be missing something obvious.
Just one non-trivial solution for any given $c$ would be fine (for my application).

REPLY [39 votes]: [edited again mainly to add the Euler link, see last paragraph]
Yes, and indeed there are infinitely many rational points:
the birationally equivalent Diophantine equation
given by J.Ramsden in his partial answer to his own question,
$$
X + Y = Z + T,
\phantom{and}
XYZT = c,
$$
was already studied by Euler (in the equivalent form $xyz(x+y+z)=a$),
who in 1749 obtained the rational curve of solutions
$$
X = 6\frac{cr(2r^4+c)^2}{(r^4-4c)(r^8+10cr^4-2c^2)},
\phantom{and}
Y = -2\frac{r^8+10cr^4-2c^2}{3r^3(r^4-4c)},
$$ $$
Z = -3\frac{r^5(r^4-4c)^2}{2(2r^4+c)(r^8+10cr^4-2c^2)},
\phantom{and}
T = \frac{r^8+10cr^4-2c^2}{r^3(2r^4+c)}.
$$
Inverting J.Ramsden's birational transformation
$$
(X,Y,Z,T) = \left(
  \frac{x-1}{z},
  \frac{y+1}{z},
  \frac{x+1}{z},
  \frac{y-1}{z}
\right)
$$
then yields
$$
x = \frac{(r^8+2c^2)(r^8-44cr^4-2c^2)}{(r^8+10cr^4-2c^2)^2},
\phantom{and}
y = \frac{7r^4+8c}{9r^4},
\phantom{and}
z = \frac{-4(r^4-4c)(2r^4+c)}{3r(r^8+10cr^4-2c^2)}.
$$
Euler's solution is unpublished and somewhat mysterious;
he gave the formulas in a letter to Goldbach but didn't explain
how he found them, writing only that he obtained the solution
"endlich nach vieler angewandter Mühe" [at last, after applying much effort].
The curve is singular, and finding such a curve requires
going somewhat beyond the usual manipulation of
elliptic fibrations on K3 surfaces, which for these surfaces will not work unless
$c$ is a square (the case in which R.Kloosterman already found a nonsingular
rational curve of solutions).  I've lectured on Euler's surface and solution
a number of times over the past few years;
here's
the latest iteration.  Thanks to Franz Lemmermeyer for bringing
Euler's letter to my attention.
Added later:
The same lecture notes
show on page 43 a somewhat simpler curve that I found on Euler's surface,
which also yields a somewhat simpler curve of solutions of
$(x^2-1)(y^2-1)=cz^4$:
$$
x = \frac{u^{12}+48cu^8-48c^2u^4+128c^3}{u^4(u^4-12c)^2},
\phantom{and}
y = \frac{5u^4+4c}{3u^4-4c},
\phantom{and}
z = \frac{4(u^4+4c)}{u(u^4-12c)}.
$$
For $c=2$, we get at $u=2$ the solution $(26, 11/5, 6)$ which is almost
as simple as the solution $(5/3,17,4)$ that Mark Sapir noted.
Added later yet: Here's a

transcription of Euler's 1749 letter.  See the last page.<|endoftext|>
TITLE: Is there a cheap proof that (homotopy) endomorphisms are functorial?
QUESTION [7 upvotes]: This is, in some sense, the homotopy version of this question.)
If $C$ is a category with $iC$ the subcategory of isomorphisms, there is a functor $X \mapsto End(X)$ from $iC$ to the category of monoids.  Moreover, this extends to a result if $C$ has an enrichment in a monoidal category $\cal V$, and this changes the target to the category of $\cal V$-monoids.
My question is about the homotopical analogue of this.  For example, if $C$ is a simplicial model category, Dwyer and Kan's sequence of papers on the simplicial localization show that there is a functor which captures the homotopy type of the monoid $End(X)$ properly and which is functorial in weak equivalences.  (For a weak equivalence $X \to Y$, there are natural equivalences $End(X) \to Map(X,Y) \leftarrow End(Y)$, but these do not induce a map in the homotopy category of monoids.)
It is natural to ask this question in other circumstances.  For instance, I was always taught that one of the operating principles of the theory of operads is that that algebra structures can be made invariant under weak equivalence.  However, actually showing this seems to be surprisingly annoying, and requires one to make use of auxiliary constructions like the endomorphism operad of a map $f:X \to Y$.  If this map is an acyclic cofibration or an acyclic fibration in an appropriate model category, this map has good homotopical properties.  In general it seems to require constructing a factorization, and proving that we have a well-defined functor - from isomorphisms in the homotopy category of $C$ to the homotopy category of operads, or more generally some homotopy category of $\cal V$-monoids - seems to be ugly.
Is there a more direct proof that homotopy endomorphism monoids are functorial?
There are definitely obstructions to any direct constructions.  For example, there is a quasi-isomorphism of chain complexes
$$
(\mathbb Z \hookrightarrow \mathbb Z \oplus \mathbb Z) \to (\mathbb Z^2 \hookrightarrow \mathbb Z^2 \oplus \mathbb Z)
$$
such that there is simply no map of differential graded algebras in either direction between the associated endomorphism objects.

REPLY [7 votes]: What about simple category theory?
Let $C_{X,Y}$ denotes the full simplicial subcategory of $C$ whose objects are $X$ and $Y$. If you see $End(X)$ as a simplicial category with one object, then you have a full embedding $End(X)\to C_{X,Y}$. The latter is a Dwyer-Kan equivalence if there exists a weak equivalence $X\to Y$, in which case the other full embedding $End(Y)\to C_{X,Y}$ is a Dwyer-Kan equivalence as well. In other words, $End(X)$  and $End(Y)$ are canonically isomorphic in the homotopy category of simplicial categories. Note that cofibrant/fibrant replacement functors of the Bergner-Dwyer-Kan model structure can be chosen so that they don't affect the objects, so that, at the end, we can promote this into an isomorphism in the homotopy category of simplicial monoids.
The same proof applies to a dg category $C$ (or to any $V$-enriched category $C$, where $V$ is a symmetric monoidal model category such that the category of $V$-enriched categories admits a Dwyer-Kan style model category structure).<|endoftext|>
TITLE: Is the exponential map of a $C^{1,1}$ Riemannian metric a local homeomorphism?
QUESTION [10 upvotes]: Suppose that $g$ is a $C^{1,1}$ (i.e., continuously differentiable with locally Lipschitz first derivative) Riemannian metric on a smooth manifold $M$. It seems to be known that locally the exponential map $\exp_p$ corresponding to $g$ at some point $p$ of the manifold is still a local homeomorphism (and Lipschitz). I am looking for a proof of this result (if it is true). The standard proof for $C^2$ metrics uses the fact that the tangent map of $exp_p$ at $0\in T_pM$ is the identity, so that the inverse function theorem gives that $exp_p$ is a local diffeomorphism. However, for a $C^{1,1}$ metric, $\exp_p$ is only (locally) Lipschitz, so it is not clear that there even exists a tangent map of $exp_p$ at $0$ and the inverse function theorem does not apply. Is there some argument that can replace the use of the inverse function theorem in this situation?

REPLY [5 votes]: More generally the exponential map of any Lipschitz connection or spray is a Lipeomorphism (bi-Lipschitz homeomorphism) in a neighborhood of the origin. The proof uses Leach's      inverse function theorem which holds under strong differentiability. The non-trivial part is to show that the exponential map is strongly differentiable at the origin. Details are given here http://arxiv.org/abs/1308.6675<|endoftext|>
TITLE: Maximal ideals of Z[x,y]
QUESTION [12 upvotes]: we know that the maximal ideals of ${\mathbb Z}[x]$ are of the form $(p, f(x))$ where $p$ is a prime number and $f(x)$ is a polynomial in ${\mathbb Z}[x]$ which is irreducible modulo $p$.
Is it true that:
the maximal ideals of ${\mathbb Z}[x,y]$ are of the form $(p, f(x,y),g(x,y))$ where $p$ is a prime number and $f(x,y), g(x,y)$ are polynomials in ${\mathbb Z}[x,y]$ which are irreducible modulo $p$.

REPLY [17 votes]: Better yet, you can replace $f(x,y)$ with $f(x)$.  See the answer to this question.
Edited to add:  At Martin Brandenburg's request, I'm expanding this to add the details I thought were too obvious to mention:  
1)  A maximal ideal $M$ of ${\mathbb Z}[X,Y]$ is the kernel of a map to a field $k$.  
2)  Any field of characteristic zero contains ${\mathbb Q}$ and hence is not finitely generated as a ${\mathbb Z}$-algebra.
3)  Therefore the field $k$ has finite characteristic $p$; therefore $M$ contains $p$.
4)  Now $M/(p)$ is a maximal ideal in $({\mathbb Z}/p{\mathbb Z})[X,Y]$ and therefore (by the answer to the question linked above) has the form $(\overline{f}(X),\overline{g}(X,Y))$.
5)  We can lift $\overline{f}$ and $\overline{g}$ to polynomials $f,g\in M$.  
6)  It is easy to check that $p,f,g$ generate $M$.
7)  Because ${\overline f}(X,Y)={\overline f}(X,0)$, it follows that $f(X,Y)-f(X,0)$ maps to zero mod $p$.   
8)  By 7) and 6), $(p,f(X,0),g(X,Y))=(p,f(X,Y),g(X,Y))=M$, so that $M$ has generators of the advertised form.  
Edited to add further:  As Yves Cornulier points out in comments, step 2) above is less trivial than both I and Will Sawin made it out to be.  The key additional point is that a field $k$  finitely generated over ${\mathbb Z}$ must have finite characteristic because --- by the generalized Nullstellensatz --- the unique closed point in $Spec(k)$ must map to a closed point in $Spec({\mathbb Z})$.<|endoftext|>
TITLE: Elementary applications of Krein-Milman
QUESTION [30 upvotes]: This is a cross-post from MSE: Elementary applications of Krein-Milman. I'm starting to suspect that the question just doesn't really have a great answer, it's worth a try.
Recall that the Krein-Milman theorem asserts that a compact convex set in a LCTVS is the closed convex hull of its extreme points. This has lots of applications to areas of mathematics that use analysis: the existence of pure states in C*-algebra theory, the existence of irreducible representations of groups, the existence of ergodic measures...
I'm interested in applications of the theorem which are very easy to state but hard to achieve any other way. When I say "very easy to state" I mean the result should be expressible in the language of elementary Banach space or Hilbert space theory - no C*-algebras, representation theory, or measures. For an example of what I have in mind, the Krein-Milman theorem implies that C[0,1] is not the dual of any Banach space. If anyone knows an application of Krein-Milman to the theory of Fourier series, that would be ideal.

REPLY [5 votes]: One can prove by using repeatedly the Krein-Milman theorem that

If $T : C[0,1] \rightarrow X$ is an operator of norm at most one, with $T$ isometric on the $2$-dimensional subspace spanned by $x \mapsto \cos(\pi x) $ and $x \mapsto \sin(\pi x)$, then $T$ is an isometry. 
If $T$ is an endomorphism of $C[0,1]$ of norm at most one, which fixes the functions $x \mapsto \cos(\pi x) $ and $x \mapsto \sin(\pi x)$, then $T$ is the identity operator.<|endoftext|>
TITLE: Finite nilpotent orbits: GL(n,q)-conjugacy classes and a partial order on partitions
QUESTION [8 upvotes]: I have a question regarding a partial order $<$
on the set ${\rm Part}(n)$ of partitions of $n$.
Given $\lambda=(\lambda_1,\lambda_2,\ldots)\in{\rm Part}(n)$ with
$\sum_{i\geq1} \lambda_i=n$ and $\lambda_1\geq\lambda_2\geq\cdots\geq0$,
let $J_\lambda$ denote the $n\times n$ block diagonal matrix 
$\bigoplus_{i\geq1}J_{\lambda_i}$. For example,
$J_{(3,2,1)}=\left(\begin{smallmatrix}0&1&0&&&\\0&0&1&&&\\0&0&0&&&\\&&&0&1&\\&&&0&0&\\&&&&&0\end{smallmatrix}\right)$.
Consider the ${\rm GL}(n,F)$-conjugacy classes of the set ${\rm M}(n,F)$
of all $n\times n$
matrices over a field $F$. A nilpotent matrix $X\in{\rm M}(n,F)$ lies in a
conjugacy classes $\mathcal{O}_\lambda:=J_\lambda^{{\rm GL}(n,F)}$ for a unique
$\lambda\in{\rm Part}(n)$. (Nilpotent means $X^n=0$.)
If $F=\mathbb{F}_q$ is a finite field, then set
$n_\lambda:=|J_\lambda^{{\rm GL}(n,q)}|$. A formula for $n_\lambda$
is given in Fulman, Cycle indices for finite classical groups. It turns out
that $n_\lambda=n_\lambda(q)$ is a polynomial in $q$ with integer coefficients.
Define a partial order $<$ on ${\rm Part}(n)$ as follows:
$\lambda<\mu$ if and only if $n_\lambda(q)$ divides $n_\mu(q)$.
I call this the divisibility partial order.
When $F$ is the complex field $\mathbb{C}$, define $\lambda\triangleleft\mu$
if $\overline{\mathcal{O}_\lambda}\subset\overline{\mathcal{O}_\mu}$ where
$\overline{\mathcal{O}_\lambda}$ denotes the Zariski closure of
$\mathcal{O}_\lambda$.  It is shown in Collingwood and McGovern, Nilpotent
orbits of semisimple Lie algebras, pp 93--95, that $\triangleleft$ is the
dominance partial order on ${\rm Part}(n)$. That is,
$\lambda\triangleleft\mu$ if and only
if $\sum_{i=1}^{k-1}\lambda_i=\sum_{i=1}^{k-1}\mu_i$ and $\lambda_k<\mu_k$
for some $k\geq1$.
If $n\leq5$, then the partial orders $<$ and $\triangleleft$ are identical and are total orders.
However, when $n=6$ the partition $(3,2,1)$ of 6 has three partitions
divisibility larger, and has five partitions dominance larger.
Does anyone have any insight into divisibility partial order? or know of
its appearance in the literature? (I have not found a reference to $<$ in
Roger Carter's book Finite groups of Lie type: conjugacy classes and
complex characters, but $\triangleleft$ appears in 5.5 and 5.11.)
For specific $\lambda$, I can (theoretically) factor $n_\lambda(q)$ and so
can determined whether $\lambda<\mu$ for specific $\lambda$ and $\mu$, but
I have few global results.

REPLY [5 votes]: I think the right way is to factor $n_\lambda(q)$ in general. In particular, it is obviously a quotient of the order of $GL_n(q)$. The formula for the order of $GL_n(q)$ does not have very many prime factors: just $q$ and the first $n$ cyclotomic polynomials.
One could consider an alternate question, the order of the centralizer of $J$ in $GL(n,\mathbb F_q)$. This gives the reverse of the partial order, since the divisibility relation is reversed. The centralizer is the automorphism group of the corresponding $\mathbb F_q[x]/x^n$-module, $M$. The subgroup that fixes $M/x$ is a $q$-group, since it is unipotent. Its quotient is a product of copies of $GL(k,\mathbb F_q)$: one for each type of block, with $k$ equal to the number of that block that appears. 
The number of times that the $k$th elementary cyclotomic polynomial appears in the size of the centralizer is just the sum over all sizes $n$ of the floor of $a_n/k$, where $a_n$ is the number of blocks of size $n$.
This function behaves very erratically, so we can conclude that the partial order behaves erratically as well.<|endoftext|>
TITLE: Canonical form of symmetric integer matrix M
QUESTION [7 upvotes]: Let $M$, $N$ be a symmetric matrix over a ring $R$.
$M$ and $N$ are said to be equivalent if there exist an invertible
matrix $U$ (over the same ring $R$) such that $N=U M U^T$ ($U^T$ is the transpose of $U$).
A question is that what is the simple canonical form of $M$
under such an equivalent relation.
We know that when $R$ is the ring of real numbers, every
real symmetric matrix is equivalent to
an diagonal matrix with diagonal entries being 1, -1, or 0.
When $R$ is the ring of integers, do we have a similar result?
If there is no nice results, we may assume $M$
to satisfy additional conditions:
(a) $|\det(M)|=1$
(b) There exist a $J$ such that $J^2=1$ and $JMJ^T=-M$.
Thanks!
Edit:  I am also interested in finding the simple canonical form of integer symmetric matrices $M$, that satisfy
(a) $|\det(M)|=1$
(b) There exist a $J$ such that $J^2=-1$ and $JMJ^T=-M$.

REPLY [6 votes]: It's all in the correct reference. Cassels, Rational Quadratic Forms, chapter 9 "Integral Forms over the Rational Integers," pages 163-164, Examples 9-11. Example 11(i) says that, for "odd" matrices, we can cut down the dimension by 2 and write $y_1^2 - y_2^2 + g(z_1, \ldots , z_{n-2}).$ The determinant of $g$ is still $\pm 1,$ so the only problem is that $g$ may be "even." 
Next, if $f$ is "even" the quadratic form can, in fact, be written  $ 2y_1 y_2 + g(z_1, \ldots , z_{n-2}).$
So, all we really need is to show, as in Sylvester's Law of Inertia, that the resulting form $g$ continues to be indefinite. Presumably your condition with $J M J^T = -M$ can do this. 
Otherwise, without your $J$ condition, Example 11(vi) says that either $f$ or $-f,$ if "even," is equivalent to a sum of some $2x_j y_j$ terms along with a single $\mathbb E_8$ lattice. CASSELS 
I was uneasy about the possible need to mix 2 by 2 blocks of both types, despite Hahn's statement, but 
$$  
 \left(  \begin{array}{cccc}
 3 & 4 & 2 & 2 \\\
 2 & 3  & 1 & 2 \\\
 0 & 1 & 1  & 1  \\\ 
 2 & 3 & 2  & 1  
\end{array} 
  \right)
 \left(  \begin{array}{cccc}
 1 & 0 & 0 & 0 \\\
 0 & -1  & 0 & 0 \\\
 0 & 0 & 0  & 1  \\\ 
 0 & 0 & 1  & 0  
\end{array} 
  \right)
 \left(  \begin{array}{cccc}
 3 & 2 & 0 & 2 \\\
 4 & 3  & 1 & 3 \\\
 2 & 1 & 1  & 2  \\\ 
 2 & 2 & 1  & 1  
\end{array} 
  \right) = 
 \left(  \begin{array}{cccc}
 1 & 0 & 0 & 0 \\\
 0 & -1  & 0 & 0 \\\
 0 & 0 & 1  & 0  \\\ 
 0 & 0 & 0  & -1  
\end{array} 
  \right)
 $$

REPLY [3 votes]: For symplectic unimodular symmetric (or skew) matrices, such a result is shown in
Zarrow, Robert
A canonical form for symmetric and skew-symmetric extended symplectic modular matrices with applications to Riemann surface theory.
Trans. Amer. Math. Soc. 204 (1975), 207–227. 
You might be able to extend it to the nonsymplectic case (though I am a bit skeptical).<|endoftext|>
TITLE: Computing the Euler characteristic of the complex projective plane using differential topology
QUESTION [5 upvotes]: I am trying to compute $\chi(\mathbb{C}\mathrm{P}^2)$ using only elementary techniques from differential topology and this is proving to be trickier than I thought. I am aware of the usual proof for this result, which uses the cellular decomposition of $\mathbb{C}\mathrm{P}^2$ to get $\chi(\mathbb{C}\mathrm{P}^2) = 3$, but I would like to find a proof of this result that relies on concepts like indices of isolated zeros on a vector field. So for the purposes of this question, I would like to utilize the following definition of the Euler characteristic: For a closed orientable manifold $M$ we define $\chi(M) = \sum_i \mathrm{Ind}_{d_i} \mathrm{v}$ where $\mathrm{v}$ is a vector field on $M$ with isolated zeros.
In my first attempt at this problem I thought about finding a vector field on $\tilde{\mathrm{v}}$ on $S^5$, and then using the identification $\mathbb{C}\mathrm{P}^2 \cong S^5/\mathrm{U}(1)$, seeing if $\tilde{\mathrm{v}}$ descended to a vector field on $\mathbb{C}\mathrm{P}^2$ with isolated zeros that lent itself to computing the Euler characteristic. I had difficulty making this work out, so I am unsure if this is a good approach to tackling the problem. Any insights?

REPLY [4 votes]: Take a $3 \times 3$ complex diagonal matrix $A$ with distinct nonzero diagonal entries. The 1-parameter subgroup $exp(At)$ acts on $CP^2$; the fixed points are the lines in $C^3$ containing eigenvectors of $A$. There are $3$ of them and the derivative of the action is a vector field with $3$ zeroes. As the vector field is holomorphic, the index at each zero is $+1$.<|endoftext|>
TITLE: How to Tackle the Smooth Poincare Conjecture
QUESTION [33 upvotes]: The last remaining problem in this whole "everything is a sphere" business, is the Smooth Poincare Conjecture in dimension 4: If $X\simeq_\text{homo.eq.} S^4$ then $X\approx_\text{diffeo} S^4$.  Freedman showed that this holds if we replace "diffeomorphism" by "homeomorphism", so another viewpoint would be that $S^4$ has no exotic smooth structures.
Restated: If $X$ is a connected, closed smooth 4-manifold with $\pi_1X=1$ and $H_*X\cong H_*S^4$, then $X\approx_\text{diffeo}S^4$.
This was told to me by Michael Hutchings, who motivationally remarked a way to solve it (Edit: not an original suggestion of his):
Find a symplectic structure on $X-\lbrace pt\rbrace$ which is standard near the puncture-point.
Then we're done by "Recognition of $\mathbb{R}^4$": Let $(M,\omega)$ be a noncompact symplectic 4-manifold such that $H_\ast(M)\cong H_\ast (pt)$. Suppose there exist compact sets $K_0\subset M$ and $K_1\subset\mathbb{R}^4$ and a symplectomorphism $\phi:(M-K_0,\omega)\to(\mathbb{R}^4-K_1,\omega_\text{std})$. Then $\phi$ extends to a symplectomorphism $(M,\omega)\to(\mathbb{R}^4,\omega_\text{std})$, after removing slightly bigger compact sets.
I am ignorant to the size of this wall (the conjecture) and the ability to make an indent in it. But the above idea is pretty cool, even if not anymore helpful than the original statement. The Poincare Conjecture (in dimension 3) was solved using Hamilton's idea of Ricci flow. This leads me to ask:
Is there another idea proposed to tackle this conjecture? Or a failed attempt?

REPLY [22 votes]: It seems to me your question implicitly assumes that the 4d smooth Poincaré conjecture (S4PC) is true.  But if you were to take a poll of experts  I think you would find that most of them suspect it is false.  (See also Ryan's comment to the original question.)  There are many large families of potential counterexamples to the S4PC.  These are smooth 4-manifolds which are homotopy equivalent (and hence homeomorphic) to the 4-sphere, but no one knows how to prove that they are diffeomorphic to the 4-sphere.  
(Akbulut and Gompf have made recent progress in showing that some, but not nearly all, of these potential counterexamples are standard 4-spheres, so I think people are more open to the idea that the S4PC might be true than they were four years ago.)
There are similarly many suspected counterexamples to the closely related Andrews-Curtis conjecture.  Andrew Casson has some interesting (and unpublished) numerical evidence that the AC conjecture is false.
So if you are interested in this problem, you might want to spend as much time trying to prove that one of the potential counterexamples is an actual counterexample as you do trying to prove the conjecture.<|endoftext|>
TITLE: Characteristic classes for block bundles
QUESTION [18 upvotes]: Block bundles are PL analogs of vector bundles, see e.g. Rourke-Sanderson's 
 article
in Bulletin of AMS, 1966. There is a classifying space $B\widetilde{PL}_q$ for rank $q\ $ block bundles, which is a PL analog of $BO_q$, and there are similar classifying spaces $B\widetilde{SPL}_q$, $BSO_q$ for oriented bundles.
Question.  Is the rational homotopy type of 
$B\widetilde{SPL}_q$ recorded in the literature?
Long time ago Colin Rourke explained to me how to compute the rational homotopy type of $B\widetilde{SPL}_q$, but I cannot find the correspondence. If memory serves, the result
was as follows. 

If $q\ge 3$ is even, then $H^*(B\widetilde{SPL}_q;\mathbb Q)$ is a polynomial algebra on the Pontryagin classes and the Euler class. The Euler class occurs in degree $q$, while the Pontryagin classes occur in all degrees divisible by $4$ (which is different from $BSO_q$ where there is no Pontryagin classes in degrees $\ge 2q$).
If $q\ge 3$ is odd, then $H^*(B\widetilde{SPL}_q;\mathbb Q)$  a polynomial algebra on the Pontryagin classes, which occur in all degrees divisible by $4$, and a new class
in degree $2q-2$ which arises from works of Haefliger and Hirsch. 
If $q\le 2$, then $BSO_q\to B\widetilde{SPL}_q$ is a rational homotopy equivalence.

As usual, the Pontryagin classes are stable, i.e. they survive in $B\widetilde{SPL}\approx BPL$, while the Euler class and the Haefliger-Hirsch classes die in $B\widetilde{SPL}_{q+1}$. I recall that the proof of 1-2 was not too hard, and I can probably reconstruct it, but it would be much nicer if it were written somehwere.

REPLY [14 votes]: I don't know where the results are written down in one place (perhaps in the book of Madsen and Milgram?), but see the the end of this post  for a list of references.
In any case, here is a proof of the result for $q \ge 3$.
By a result of Haefliger, there is a homotopy cartesian square
$$
B\widetilde{PL}_q  \quad \to \quad BG_q 
$$
$$
\downarrow \qquad \qquad \quad \downarrow 
$$
$$
B\widetilde{PL}  \quad \to \quad BG
$$
where $BG_q$ classifies oriented $(q-1)$-spherical fibrations, $BG$ classifies stable oriented spherical fibrations and $B\widetilde{PL}$ classifies stable block bundles.  Rationally, $BG$ is trivial (since its homotopy groups are the shifted stable homotopy groups of spheres), and $B\widetilde{PL}\simeq BPL$ is rationally weak equivalent to $BO$.
Consequently, there is a rational equivalence
$$
B\widetilde{PL}_q  \simeq BO \times BG_q  .
$$
It suffices to identify the rational homotopy type of $BG_q$.
Note that $G_q$ is the topological monoid self-equivalences of $S^{q-1}$.
Let  $SG_q \subset G_q$ be the submonoid of degree one self maps. Then $BSG_q \to BG_q$
is a rational equivalences as well (they have the same rational homotopy groups).
It therefore suffices to identify $BSG_q$ rationally (note: the advantage of $SG_q$ over $G_q$
 is that the former is connected).
Case 1, $q$ is even:
If $q$ is even, then $S^{q-1}$ is rationally 
equivalent to an Eilenberg-Mac Lane space $K(\Bbb Q,q-1)$.
Using the fiber sequence $SF_{q-1} \to SG_q \to S^{q-1}$ (where $SF_{q-1}$ is the topological monoid of degree one pointed self maps of $S^{q-1}$) and the fact just noted,
we see that $SF_{q-1}$ is rationally trivial, so $SG_{q}$ is rationally $K(\Bbb Q,q-1)$.
Consequently, $BSG_q$ is rationally $K(\Bbb Q,q)$ when $q$ is even, so we get 
a rational  equivalence
$$
B\widetilde{PL}_q \simeq BO \times K(\Bbb Q,q)
$$
when $q \ge 3$ is even.
Case 2, $q$ is odd: In this instance $S^{q-1}$ is not rationally an Eilbenberg-Mac Lane space. But there is a rational fiber sequence
$$
S^{q-1} \to K(\Bbb Q,q-1) \to K(\Bbb Q,2q-2)  .
$$
Arguing similarly to case 1, we  see that $SG_{q-1}$ is rationally $K(\Bbb Q,2q-3)$. 
Hence $BSG_q$ is rationally $K(\Bbb Q,2q-2)$ and  we obtain
a rational equivalence
$$
B\widetilde{PL}_q \simeq BO \times K(\Bbb Q,2q-2) .
$$
Addendum
(1). Haefliger's theorem can be found in 
Haefliger, André : Differential embeddings of $S^n$ in $S^{n+q}$ for $q>2$. 
Ann. of Math. 83 (1966), 402–436.
The proof uses embedded framed surgery.
(2). In Wall's book, he says that the case $q=2$ follows from
Wall, C.T.C.: Locally flat PL submanifolds with codimension two. Proc. Cambridge Philos. Soc. 63 (1967) 5–8.
(3) The proof that $B\widetilde{PL}\simeq BPL$ is a consequence of Rourke and Sanderson's paper on block bundles:
Rourke, C. P.; Sanderson, B. J.:
Block bundles. Bull. Amer. Math. Soc. 72 (1966) 1036–1039. 
(4). The proof that $BO \to B\widetilde{PL}$ is rational requivalence is a consequence of Kervaire and Milnor's work (which amounts to the Browder-Novikov sequence for a sphere), since $\pi_n(\widetilde{PL}/O)$ is the group of exotic homotopy
$n$-spheres (at least if $n \ge 5$), and this is a finite group.<|endoftext|>
TITLE: a categorical Nakayama lemma?
QUESTION [51 upvotes]: There are the following Nakayama style lemmata:

(the classical Nakayama lemma) Let $R$ be a commutative ring with $1$ and $M$ a finitely generated $R$-module. If $m_1, \ldots, m_n$ generate $M$ modulo $I$, where $I \subset \mathrm{Jac}(R)$, then they generate $M$.
(the graded Nakayama lemma) See How to memorise (understand) Nakayama's lemma and its corollaries?.
(the filtered Nakayama lemma) See How to memorise (understand) Nakayama's lemma and its corollaries?.
(the topological Nakayama lemma, see [Neukirch, Schmidt, Wingberg], Cohomology of Number Fields, (5.2.18)): Let $\mathcal{O}$ be a commutative local ring complete in the $\mathfrak{m}$-adic topology with finite residue field of characteristic $p$.  Assume $G$ is a pro-$p$-group and $M$ is a compact $\mathcal{O}[[G]]$-module. If $M/\mathfrak{m}$ is a finitely generated $\mathcal{O}[[G]]$-module, so is $M$.
(Burnside's basis theorem, see also http://groupprops.subwiki.org/wiki/Burnside%27s_basis_theorem) Let $G$ be a group such that its Frattini subgroup $\Phi(G)$ is finitely generated. Then a subset of $G$ generates $G$ iff it generates it modulo $\Phi(G)$.

[tbc]
Now my question is: Is there a common categorical version, like there is a categorical generalisation of Baer's criterion (In a suitable abelian catgory, an object $I$ is injective iff for all subobjects $U$ of a generator $G$ and morphisms $U \to I$ there is a lift $G \to I$)?

REPLY [7 votes]: Let me build a little on Gjergji's answer, using his terminology, in order to get closer to how I have usually heard Nakayama's lemma stated. Suppose, in addition, that $\mathbb{Fr}_p(G)$ is finitely generated and let $S \subseteq G$. Then, if the image of $S$ $P$-generates $G/\mathbb{Fr}_p(G)$, then $S$ $P$-generates $G$. (Hmm, it's not clear to me that $\mathbb{Fr}_p(G)$ is always normal in $\mathbb{Fr}_p(G)$, although it is in all of the examples so far. So we might need to add that as a hypothesis.)
Proof: Let $x_1$, $x_2$, ..., $x_r$ $P$-generate $\mathbb{Fr}_p(G)$. Then $G=\langle S, x_1, x_2, \ldots, x_r \rangle_P$ (exercise!). Repeatedly using the definition of a nongenerator, $G = \langle S \rangle_P$. $\square$<|endoftext|>
TITLE: Rational homology spheres and knots
QUESTION [8 upvotes]: It is known that the boundary of a branched double cover of the four ball branched over a surface bounded by a knot in is a rational homology $3$-sphere. Two questions come to my mind simply out of curiosity:

Which rational homology $3$-spheres arise this way? By this I mean, is this set large or small (in the most vague terms, nothing formal) in the set of R.H.3S.'s? Is there an invariant capable of detecting when a RH3S arises this way?
If we now substitute usual knots for embeddings $\mathbb{S}^2 \hookrightarrow \mathbb{S}^4$ (knotted spheres), and look at the branched double covers of the $5$ ball branched over a $3$-manifold bounded by the knotted sphere, is their boundary a RH4S? 

Many thanks!

REPLY [8 votes]: For Question 1, I believe that the answer follows from:
Montesinos, José M. Surgery on links and double branched covers of $S^3$. Knots, groups, and 3-manifolds (Papers dedicated to the memory of R. H. Fox), pp. 227–259. Ann. of Math. Studies, No. 84, Princeton Univ. Press, Princeton, N.J., (1975; MR0380802).
Namely:

A $\mathbb{Q}$HS arises as a double branched cover of a knot in $S^3$ (equivalent to your construction) if and only if it is obtained by (rational) surgery on a strongly invertible link in $S^3$.

This is something you can detect, either by Casson-Walker invariants (see papers by Chbili- you will recall that these have surgery presentations and therefore take a rather special form if the surgery link is strongly invertible), or more directly, using result of Meeks-Yau and of Thurston to reduce the problem to one of identifying isometries on geometric manifolds, and then checking whether the whole JSJ structure admits a $\mathbb{Z}/2\mathbb{Z}$ action. This is a very strong condition of course, so the set of $\mathbb{Q}HS$s which arise via this construction is small.<|endoftext|>
TITLE: On the Hasse-Weil L-function of $P^n$
QUESTION [17 upvotes]: So let us start with the "simplest" scheme over $Spec(\mathbf{Z})$ namely $X_0=Spec(\mathbf{Z})$. Then the (reciprocal) Weil zeta function of $X_0$ at a prime $p$ is given by $Z_p(T)=1-T$ (a polynomial of degree $1$). So the Hasse-Weil zeta function of $X_0$ is given by
$$
L_{X_0}(s):=\prod_{p}Z_p(p^{-s})^{-1}=\zeta(s).
$$
Now if one lets $\psi(z)=\sum_{n\in\mathbf{Z}}e^{i\pi n^2z}$ then $\psi(z)$ is a modular form of weight $1/2$ (over a suitable congruence group of $SL_2(\mathbf{Z})$) in the sense that
$$
(-iz)^{-1/2}\psi(-1/z)=\psi(z). \;\;\;\;(*)
$$
A straight forward computation which uses the definition of the Gamma function implies that
$$
\tilde{L}_{X_0}(s):=\int_{0}^{\infty}(\psi(it)-1)t^{s/2}\frac{dt}{t}=2\pi^{-s/2}\Gamma(s/2)\zeta(s).
$$
Using the functional equation $(*)$ one may deduce the meromorphic continuation and the functional equation of $\zeta(s)$ (invariance of $\tilde{L}_{X_0}(s)$ under $s\mapsto 1-s$).
Now let us take the scheme $X_1=\mathbf{P}^1$ over $Spec(\mathbf{Z})$. Then the (reciprocal) Weil zeta function of $X_1$ at a prime $p$ is given by $Z_p(T)=(1-T)(1-pT)=1-\sigma_1(p)T+pT^2$ (a polynomial of degree $2$). It thus follows that the 
Hasse-Weil zeta function of $X_1$ is given by
$$
L_{X_1}(s)=\prod_{p}Z_p(p^{-s})^{-1}=\zeta(s)\zeta(s-1).
$$
Now let us look at the Eisenstein series of weight $2$ i.e.
$$
E_2(z)=(2\pi i)^{-2}\sum_{m,n}'\frac{1}{(mz+n)^2}:=\frac{-B_2}{2}+2\sum_{n\geq 1}\sigma_1(n)q_{z}^n,
$$
where $q_{z}=e^{2\pi iz}$. (Note that I don't get any convergence issue here since I take this $q$-expansion as the definition of $E_2(z)$). Note that $E_2(z)$ is "almost" a moldular form of weight $2$ (for the full congruence group $SL_2(\mathbf{Z})$) since 
$$
(-z)^{-2}E_2(-1/z)=E_2(z)-\frac{1}{2\pi iz}   \;\;\;\; (**)
$$
A straight forward computation similar to the one before implies that
$$
\tilde{L}_{X_1}(s):=\int_{0}^{\infty} (E_2(it)+B_2/2)t^{s}\frac{dt}{t}=2 (2\pi)^{-s}\Gamma(s)\zeta(s)\zeta(s-1).
$$
As before using $(**)$ one obtains the meromorphic continuation of $L_{X_1}(s)$ and its functional equation ($\tilde{L}_{X_1}(s)=-\tilde{L}_{X_1}(2-s)$, note the appearance of the sign $-1$). Note that this could already be deduced from what we know from $L_{X_0}(s)$.
Now there is no reason to stop here. So let $X_2=\mathbf{P}^2$ over $Spec(\mathbf{Z})$.
Then the Weil zeta function of $X_2$ at $p$ is $Z_p(T)=(1-T)(1-pT)(1-p^2T)$ (a polynomial of degree $3$). It thus follows that the 
Hasse-Weil zeta function of $X_2$ is given by
$$
L_{X_2}(s)=\prod_{p}Z_p(p^{-s})^{-1}=\zeta(s)\zeta(s-1)\zeta(s-2)=\sum_{n\geq 1}\frac{a_n}{n^s}
$$
Q1: Is it reasonable to expect the formal $q$-expansion $f(q_z)=\sum_{n\geq 1} a_n q_z^n$ to be related in some direct way to an automorphic form w.r.t. a suitable congruence subgroup of $GL_3(\mathbf{Z})$?
Q2: What about $X_n=\mathbf{P}^n$ in general?
added: Note that in the case of $X_0$ I'm really looking at $\tilde{\zeta}(s):=\zeta(2s)$ which is an $L$-function of weight $1/2$ in the sense that $\tilde{\zeta}(s)$ is related to $\tilde{\zeta}(1/2-s)$ which is in accordance with the fact that $\psi(z)$ has weight $1/2$.

REPLY [7 votes]: $\newcommand\GL{\mathrm{GL}}$
$\newcommand\SL{\mathrm{SL}}$
$\newcommand\R{\mathbf{R}}$
There is, of course, an autormorphic representation (Hecke character)
$\chi$ for $\GL(1)$ whose $p$-adic avatar is the cyclotomic character. From this, one thus has the isobaric sum (following Langlands and Jacquet-Shalika) 
$$\pi = 1 \boxplus \chi \boxplus \ldots \boxplus \chi^{n-1}$$
which is an automorphic representation for $\GL(n)$ with $L(\pi,s) = L(\mathbf{P}^n,s)$. 
But all of this is somewhat irrelevant to your action question, to which the answer is not really. The reason $q$-expansions arise for $\GL(2)$ has to do with the fact that
$\SL_2(\R)/\mathrm{SO}_2(\R)$ is the upper half plane, and $\SL_2(\mathbf{Z})$ contains the
element $z \mapsto z+1$ for which $q = e^{2 \pi i z}$ is invariant. 
$\SL_n(\R)/\mathrm{SO}_n(\R)$ is quite a different beast.
 Edit You don't seem satisfied with my answer, but I think you seem to be missing a basic principle:  if you write down something at random, there's no reason it should be interesting.  The theory of automorphic forms, however labyrinthian, has an incredibly precise structure. If you take an automorphic representation $\pi$ corresponding to a classical modular form, then the representation theory of $\mathrm{GL}_2(\mathbf{R})$ will
tell you that a lowest weight vector for the discrete series will be annihilated by some differential operator which one can compute to be the Cauchy-Riemann equations, and hence
the corresponding function on the upper half plane will be holomorphic: this is a very
specific reason why holomorphic functions  might be associated to automorphic forms. If, instead, one works with an automorphic form for $\mathrm{GL}(3)$, then the representation theory will (under appropriate conditions) produce an expansion in terms of Whittaker functions which will have a completely different flavour. 
It might also be worth remarking that already since Langlands time people have thought hard about the problem of  transfer, that is, for example, starting with an automorphic representation for $\GL(2)$ and producing one for $\GL(3)$. The methods used to prove these results are almost exclusively via the trace formula - in particular, they proceed via harmonic analysis and representation theory, rather than explicit manipulations with functions (holomorphic or otherwise). Thus (addressing your comment) the hope that one
might explicitly decompose the symmetric space of $\GL(3)$ in some way is a little too optimistic. 
Finally, you can (of course) recover $L(\pi,s)$ from $f(q)$, so $f(q)$ does carry (in some sense) all the information of $L(\pi,s)$. Moreover, one can recover $f(q)$ as the inverse Mellin transform of
$L(\pi,s)$ adjusted by appropriate Gamma factors. However, you will find that when $n > 2$ the relevant Gamma factors involved will not play well with respect to the functional equation, and thus there will be no reason to expect that $f(q)$ will not have any nice properties (such as a nice functional equation).
Even for $n = 1$ you have to cheat a little (replacing $q^n$ by $q^{n^2}$),
and moreover the theta function is (automorphically) more naturally thought of
as associated to
the metaplectic group rather than $\GL(2)$.<|endoftext|>
TITLE: A minimum set hitting every base of a matroid
QUESTION [6 upvotes]: We are given a matroid. Our goal is to find a set of elements of minimum size that has non-empty intersection with every base of the matroid. Is the problem studied before? Is it in P? For example, in a spanning tree matroid, the minimum hitting set should be a minimum cut. Thanks.
Crossposted at CSTheory.

REPLY [7 votes]: The problem is hard in general.  Note that a minimal set that intersects every base of a matroid $M$ is a dependent set in the dual matroid $M^{*}$.  Such sets are called cocircuits.  So, you are looking for the shortest cocircuit of a matroid.  
The shortest cocircuit problem (equivalently shortest circuit problem) is NP-complete in general (even for binary matroids).  See this Matroid Union post for more information on finding shortest circuits in binary matroids.<|endoftext|>
TITLE: Pursuit-Evasion on a Manifold
QUESTION [6 upvotes]: I know pursuit-evasion has been studied in many contexts, including
on a manifold (e.g., Melikyan,
"Geometry of Pursuit-Evasion Games on Two-Dimensional Manifolds"),
but I have not seen this version:
There is a pursuer, a point $p_t$ at time $t$, while the
evader is at $v_t$.  They start at $p_0$ and $v_0$.
For each $t \in \mathbb{N}$, the evader jumps from $v_t$ to $v_{t+1}$,
and then the pursuer jumps from $p_t$ to $p_{t+1}$.  Both jumps
are at most a unit distance, measured by shortest path on the manifold $\cal{M}$.
Both pursuer and evader know the other's location at all times.
The evader is captured if
$p_t = v_t$ at some time $t$.
Let $\cal{M}$ be a sphere.  Then the pursuer might fail to capture the
evader, who can just run away on a great circle.
But what if there are two pursuers, $p_t$ and $q_t$?
Specifically:

Q1.
  For $\cal{M}$ a sphere, can an evader always be captured by two pursuers
  initially at the north and south poles?

There is a general principle for $\cal{M}=\mathbb{R}^n$:
it suffices for the evader to be inside the convex hull of
the pursuers (e.g.,
Kopparty, Ravishankar, "A framework for pursuit-evasion games in $R^n$,"
Information Proc.  Lett., 2005).
It seems a natural extension that two antipodal pursuers on a sphere
suffice.

Q2.
  For $\cal{M}$ a surface homeomorphic to a sphere embedded in $\mathbb{R^3}$,
  with metric inherited from the
  Euclidean metric in $\mathbb{R^3}$, will placing $p_0$ and $q_0$
  at points realizing the diameter of $\cal{M}$ suffice to capture any
  evader? (By diameter here I mean that the shortest path from $p_0$
  to $q_0$ on the surface is maximum over all pairs of points on $\cal{M}$.)

This is much less clear to me.  A counterexample would be interesting.
But if the answer to Q2 is Yes,
I would be interested to learn if there are closed,
bounded surfaces of higher genus
for which just two well-placed pursuers suffice.
The literature on pursuit-evasion is vast, and I suspect the answers
to my questions are known.  Thanks for suggestions or pointers!

REPLY [2 votes]: I don't have enough reputation to comment. So I apologize that I write this as an answer. I don't understand the answer of Anton Petrunin. According to it, two pursuers are enough to capture an evader in the sphere. But as Barry Cipra said in the comments, actually the opposite is true: the evader is always able to escape (just consider the great circle through the two pursuers and the hemisphere defined by it containing the evader. The evader just have to go so close to the center of this hemisphere as possible. The pursuers won't be able to reach him.)
On the other hand, I actually like to think of two antipodal points in the sphere as a convex subset (convex meaning that any two points in the subset at distance $<\pi$ are connected by a geodesic). Thus the analogy with the cited paper won't apply (one needs three or more pursuers).<|endoftext|>
TITLE: Noncommutative localization of a ring: complete construction
QUESTION [12 upvotes]: I've been looking for the following construction in the literature, but I've only been able to find (very) partial proofs or proofs of special cases.
Let $R$ be a non-commutative ring and $S$ a multiplicative subset (i.e., $1 \in S$ and if $a, b \in S$ then $ab \in S$, the set $S$ can in particular contain zero-divisors).
It is known that the localization $RS^{-1}$ exists if :

for $a \in R$ and $s \in S$, there exist $b \in R$ and $t \in S$ such that $at = sb$,
if $sa = 0$ for $s \in S$ and $a \in R$, then there exists $t \in S$ such that $at = 0$.

Many sources give the complete construction in the simpler case where the set $S$ only contains regular elements (i.e. non-zero divisors).
The general case is presented in (amongst others) : Rings of Quotients : An Introduction to Methods of Ring Theory by Bo Stenström (Prop. 1.4, Chap. II, p.51) or in Algebra, Volume 3 by P. M. Cohn (Thm. 1.3, Chap. 9, p. 350) but in both cases large parts of the proof are omitted.
Does anyone know where I can find the complete construction? In particular, the fact that the multiplication is well-defined (i.e., does not depend on the representing objects of the classes)?

REPLY [6 votes]: From the apparent lack of complete answers in the literature, I figured I'd write the whole proof of the well-definedness of multiplication since it is only boring, not hard.  I will write the equivalence class of a fraction $as^{-1}$ as $(a,s)$; and the equivalence relation as $\sim$.  I'm assuming that you have full proofs that $\sim$ is a well-defined equivalence relation, and two classes can be brought to a common denominator. (For instance, you can find such proofs in Lam's "Lectures on Modules and Rings".)
Given classes $(a_1,s_1)$ and $(a_2,s_2)$, by the Ore condition we have $s_1 r=a_2s$ for some $r\in R$ and $s\in S$ (where $S$ is the multiplicative set we are inverting).  Then we want to define $(a_1,s_1)*(a_2,s_2)=(a_1r,s_2s)$.
Claim 1: This is independent of the choice of $r$ and $s$.
Indeed, suppose we also have $s_1 r'=a_2s'$ for some $r'\in R$ and $s'\in S$.  Since $(a_1r,s_2s)=(a_1rx,s_2sx)$ for any $x\in R$ such that $sx\in S$ (in fact, more generally for any $x\in R$ such that $s_2sx\in S$) from how we defined the equivalence classes, after replacing $s$ and $s'$ by a common denominator, we may as well assume $s'=s$.  Then $s_1r=a_2s=a_2s'=s_1r'$, hence $s_1(r-r')=0$.  By the zero-divisor condition, $(r-r')t=0$ for some $t\in S$.  Thus $(a_1r,s_2s)\sim (a_1rt,s_2st)\sim (a_1r't,s_2s't)\sim (a_1r',s_2s')$.
Claim 2: Multiplication is independent of the choice of representative for the class $(a_1,s_1)$.  
It suffices to prove this for another representative of this class of the form $(a_1x,s_1x)$ for some $x\in R$ with $s_1x\in S$.  (Why?  Because any two representatives can be brought to a common denominator.)  The Ore condition gives us $(s_1x) r'=a_2s'$ for some $r'\in R$ and $s'\in S$.
By Claim 1, we may as well assume $s=s'$.  Then $s_1xr=a_2s'=a_2s=s_1r$, hence $s_1(r-xr)=0$.  By the argument at the end of claim 1 we are done with claim 2.
Claim 3: Multiplication is independent of the choice of representative for the class $(a_2,s_2)$.
Again, it suffices to do this for a representative $(a_2x,s_2x)$ for some $x\in R$ with $s_2x\in S$.  By the Ore condition, write $sr'=xs'$ with $r'\in R$ and $s'\in S$.  (This is a little different than the other uses of the Ore condition.) Then $s_1rr'=a_2sr'=a_2xs'$.
We just need to see that $(a_1r,s_2s)\sim (a_1rr',s_2xs')$.  This is obvious, since if we multiply the first set of reps on the right by $r'$ we get the second set of reps.<|endoftext|>
TITLE: Van Den Berg-Kesten-Reimer inequality
QUESTION [8 upvotes]: Van Den Berg-Kesten-Reimer inequality
Let $n$ be a positive integer. For $i\in[n]$, let $\Omega_i$ be a finite set and $\mu_i$ a probability measure on it. Set $\Omega=\Omega_1\!\times\!\ldots\!\times\!\Omega_n$ and $\mu=\mu_1\!\times\!\ldots\!\times\!\mu_n$.
For $A\!\subset\!\Omega$ and $\sigma\!\subset\![n]$, let
$$
A_\sigma=\lbrace\omega\!\in\!A\!:\,\forall\upsilon\!\in\!\Omega,\,(\forall i\!\in\!\sigma,\,\upsilon_i\!=\!\omega_i)\!\implies\!\upsilon\!\in\!A\rbrace
$$
In other terms, the occurrence of the event $A_\sigma$ is solely controlled by $\sigma$.
Given two events $A,B\subset\Omega$, the notion of disjoint occurrence of $A$ and $B$ is defined by the following event
$$
A\!\circ\!B=\lbrace\omega\!\in\!A\!\cap\!B\!:\,\exists\,\sigma,\tau\!\subset\![n],\,\sigma\!\cap\!\tau\!=\!\emptyset \wedge\omega\!\in\!A_\sigma\!\cap\!B_\tau\rbrace
$$ 
Van Den Berg-Kesten-Reimer inequality states that 
$$
\forall A,B\!\subset\!\Omega,\,\,\mu(A\circ B)\le\mu(A)\cdot\mu(B)
$$
Question 
Are there non-trivial events that turn the inequality stated above into an equality?

REPLY [6 votes]: In the case of increasing events, the standard proof of the BK inequality works like this. Start with your set $\Omega$, and successively replace each $\Omega_i$ by a disjoint union of two copies $\Omega_i^1$ and $\Omega_i^2$. After step $i$, the event ${(A\circ{B})}(i)$ is defined by saying that $A$ and $B$ need to occur disjointly on $\Omega_{i+1}$ up to $\Omega_n$, and $A$ (resp. $B$) is allowed to use the $\Omega_k^1$ (resp. $\Omega_k^2$).
Then, $(A\circ B)(0)$ is just disjoint occurrence, and $(A\circ B)(n)$ is the occurrence of two independent events, its probability is $P(A)P(B)$. Besides, for every $i$, we relax the restrictions by allowing the events to be less and less disjoint - hence the BK inequality.
If there exists an $i$ that is pivotal for both $A$ and $B$ with positive probability, then when you split the corresponding $\Omega_i$, the inequality between the probabilities of $(A\circ B)(i-1)$ and $(A\circ B)(i)$ is strict, and so is BK. In other words, BK is an equality if and only if you can partition $[n]$ into two subsets, $A$ depending only on the first one and $B$ on the second one. 
In yet other words, in the case of increasing events, there is no non-trivial (in the sense, where disjoint occurrence is not automatic) pair of events realizing the equality.
Without monotonicity, the argument does not work directly ...<|endoftext|>
TITLE: Quick definition of the tangent space
QUESTION [7 upvotes]: I am looking for a quick definition of the tangent space $T_p M$, where $p$ is a point of a smooth manifold $M$. I mean a definition that allows easily to prove that: (1) $T_p M$ is a vector space of the same dimension of $M$; (2) if $x^1, x^2, \ldots,x^d$ are local coordinates functions then $\partial / \partial x^i|_p$ is a basis of $T_p M$; (3) The elements of $T_p M$ are the same thing of derivations of functions definited on a neighborhood of $p$ (W. Warner, Foundations of Differentiable Manifolds, called them "germs" and denoted their set by $\tilde{F}_p$).
Surely the old-style definition of $T_p M$ by equivalence classes of curves which passes through $p$ is inadequate.
The definition of $T_p M$ like the vector space of derivations of germs is, obviously, OK for the point (3) but seems to me that points (2) and (1) require to look at the cotangent space $T_p^* M$. I do not like to talk about the cotangent space before I have finished to talk about the tangent space.
I saw that someone defines the cotangent space first and then the tangent space, however the Warner definition of cotangent space like $\tilde{F}_p / \tilde{F}_p^2$, where $\tilde{F}_p^2$ is the ideal of finite linear combinations of product of two germs, I think it is rather artificial, and justifiable only in retrospect.
In conclusion I really appreciate any advice on how to define $T_p M$ so that (1), (2) and (3) are easy to prove. Thanks.

REPLY [5 votes]: Let $p\in M$ and $q\in N$ be points. Given two smooth maps $M\to N$ both taking $p$ to $q$ (each maybe only defined in a neighborhood of $p$), call them first-order equivalent if when expressed in terms of coordinates in both manifolds they have the same first derivatives at $p$. Call the equivalence classes $1$-jets $(M,p)\to (N,q)$.
Define cotangent vectors of $M$ at $p$ to be $1$-jets $(M,p)\to(\mathbb R,0)$. These clearly form a vector space. Show that it is isomorphic to $\mathfrak m/\mathfrak m^2$ where $\mathfrak m$ is the ideal of germs of functions on $M$ vanishing at $p$, and that its dimension is that of $M$.
Define tangent vectors of $M$ at $p$ to be $1$-jets $(\mathbb R,0)\to (M,p)$. These do not obviously form a vector space until you note that there is a canonical bijection between them and the linear functionals on the cotangent space (given by composing $\mathbb R\to M\to\mathbb R$).<|endoftext|>
TITLE: Textbook source for finite group properties deducible from character table?
QUESTION [9 upvotes]: Various questions have been posted on MO (some answered, some not) involving the character table of a finite group $G$ over a splitting field such as $\mathbb{C}$ of characteristic 0.    My basic question is more about references:

1) Are there textbook or other convenient sources summarizing properties of $G$ which can or can't be deduced from a knowledge of its character table?   

By itself this is a rather artificial question, since one doesn't usually know a character table without already knowing a lot about the group, but it provides good exercise material.    Recently I was going over some standard theory with a graduate student and came across old notes from a course I taught decades ago, but I can't recall which books I consulted at the time.
Two sorts of information are typically deduced from a character table using the orthogonality relations:  (a) numerical data, such as the order of $G$ or more generally all orders of centralizers and hence classes; (b) normal subgroup data, starting with the fact that normal subgroups are intersections of the kernels of characters and then deducing orders and inclusions of such subgroups.   In particular, one can pinpoint the center and derived group, as well as determine whether or not $G$ is simple.  On the other hand, it's well known that nonisomorphic groups can have the same character table (e.g., the two nonabelian groups of order 8); in particular, the character table can fail to predict the orders of the class representatives labelling columns.   
One substantive question of this type which I'm unclear about is this:

2) To what extent does the character table determine properties of $G$ ranging from solvability to nilpotency?

REPLY [14 votes]: Lots of properties related to solvability can be deduced from the character table of a group, but perhaps it is worth mentioning one property that definitely cannot be so determined: the derived length of a solvable group. Sandro Mattarei constructed such examples, including examples of $p$-groups with identical character tables but different derived lengths. I think that no examples are known where the difference in derived lengths exceeds 1, however.<|endoftext|>
TITLE: Reconstructing an Euclidean point cloud from their pairwise distances
QUESTION [7 upvotes]: I have a collection of points $P_1, ..., P_N$ in some Euclidean space $\mathbb R^m$ and the coordinates $x_1, x_2, ..., x_N$ respectively associated with them, where $x_i$ is the usual Cartesian tuple $x_i = (x_{i1}, x_{i2}, ..., x_{im})$. I can immediately calculate the pairwise distances between these points $r_{12}, r_{13}, ..., r_{N-1,N}$ under the usual Euclidean norm using Pythagoras' theorem (in $m$ dimensions), i.e. $r_{ij} = \left\Vert x_i - x_j \right\Vert$.
Suppose now I have the converse situation where I have the points $P_1, ..., P_N$ and all their associated pairwise distances $\{r_{ij}\}$, and I don't know their coordinates. What is known about the embeddability of these points in an Euclidean space, and is it possible to reconstruct the Cartesian coordinates for them? In other words, given an arbitrary collection of nonnegative numbers $\{r_{ij}\}$, how do I find all positive integers $m$ and enumerate all the possible  sets of $N$ coordinates $x_1, x_2, ..., x_N \in \mathbb R^m$ that are consistent with the interpretation $r_{ij} = \left\Vert x_i - x_j \right\Vert$?

REPLY [2 votes]: It is worth mentioning the following basic material.
Definition (Euclidean Distance Matrix or EDM). A real, symmetric, $n\times n$ matrix  with elements $-\tfrac{1}{2}d_{ij}^2$ and a zero diagonal is called an EDM, if there exist $n$ points in some Euclidean space whose interpoint distances are $d_{ij}$.
Theorem (Schoenberg, 1935). A real, symmetric, $n\times n$ matrix with a zero diagonal is an EDM, if and only if the matrix $$M := (I-\tfrac{1}{n}ee^T)D(I-\tfrac{1}{n}ee^T),$$ is positive semidefinite (here $e$ denotes the vector of all ones).
Note 1: The above result opens paths into the study of radial positive definite functions.
Note 2: Igor mentioned the actual interesting problems related to EDMS are more along the lines of EDM Completion Problems (EDMCP), or multidimensional scaling (obtaining an embedding into Euclidean space with some error, in case an exact embedding does not exist, and so on).<|endoftext|>
TITLE: Mayer-Vietoris implies Excision
QUESTION [18 upvotes]: This is going to basically be a cross-post of a MSE question: https://math.stackexchange.com/questions/147146/mayer-vietoris-implies-excision.  I suspect that the answer to this question will turn out to be fairly straightforward, but (a) I can't figure it out myself, (b) it's not in any of the obvious references, (c) it hasn't gotten any answers on MSE, and (d) the question honestly is relevant to my research.
Let $H_n$ be a sequence of functors from the category of locally compact Hausdorff spaces to the category of abelian groups which satisfy all of the usual Eilenberg-Steenrod axioms for a generalized homology theory, except the excision axiom is replaced by a Mayer-Vietoris axiom: whenever $X$ is the union of two closed subspaces $A$ and $B$ there is a (functorial) exact sequence
$$\to H_n(A \cap B) \to H_n(A) \oplus H_n(B) \to H_n(X) \to H_{n-1}(A \cap B) \to$$
According to the slogan "Mayer-Vietoris is equivalent to excision", one ought to be able to prove an excision theorem for $H_n$.  This probably takes the form of a long exact sequence
$$\to H_n(A) \to H_n(X) \to H_n(X - A) \to H_{n-1}(A) \to$$
where $A$ is a closed subset of $X$, though if this isn't the right formulation I welcome corrections.  I know how to go the other direction (excision implies Mayer-Vietoris), but this direction eludes me.  Can anyone help?
EDIT:
I guess I was a bit sloppy about my use of the term "Eilenberg-Steenrod Axioms" which require that $H_n$ be a functor of pairs.  So let me be more precise.  We'll work in the category of pointed locally compact (second countable if desired) Hausdorff spaces where the morphisms between $X$ and $Y$ are basepoint preserving continuous maps from the one point compactification of $X$ to the one point compactification of $Y$.  Define a homotopy of maps between $X$ and $Y$ to be a morphism between $X \times [0,1]$ and $Y$. 
I will declare that a sequence of covariant functors $H_n$ from this category to the category of abelian groups defines a generalized homology theory if:

Each $H_n$ is a homotopy functor
There is a functorial long exact sequence of the form above associated to any closed subset $A \subseteq X$.

Under these circumstances there is a Mayer-Vietoris sequence associated to any decomposition of $X$ into closed subspaces.  A precise statement of my question is: if we replace the second axiom with a Mayer-Vietoris axiom, can we reconstruct the long exact sequence associated to a single closed subset?
However, I am flexible.  In the setting that motivates this question, I really do have a notion of homology of pairs, so I would really be happy if somebody could deduce any version of the excision theorem from the existence of a Mayer-Vietoris sequence (perhaps with some extra conditions on the boundary map if it helps).

REPLY [9 votes]: I must admit, I don't know whether you can just replace excision by Mayer-Vietoris in the usual Eilenberg-Steenrod axioms. But you can do something slightly different:
While the Eilenberg-Steenrod axioms speak of homology of pairs of spaces, you can define a homology theory on spaces also as a sequence of homotopy invariant functors $h_n$ fulfilling a Mayer-Vietoris sequence (as Steven pointed out, you have just some functorial map $h_n(X) \to h_{n-1}(A\cap B)$, which replace as a datum the boundary maps of the long exact sequence of a pair). Then you define the homology of a pair $(X,A)$ as $h_n(X,A) := \ker(h_n(C_AX) \to h_n(pt))$. Here, $C_A X$ stands for the mapping cone of the inclusion $A\hookrightarrow X$. The Mayer-Vietoris sequence (based on the covering $CA \cup X = C_AX$) implies the long exact sequence of the pair. Excision (i.e. $h_n(X,A) \cong h_n(X-B, A-B)$ for $\overline{B} \subset A^\circ$) is a homotopy fact in the sense that the inclusion $C_{A-B}(X-B) \to C_A X$ is a homotopy equivalence. I must admit, I didn't check the last point in detail, but it should be at least true if $B\to A$ is a closed cofibration and in general you might replace the inclusion via a cylinder construction by a closed cofibration. 
Thus, you get a different axiomatization of a homology theory using the Mayer-Vietoris sequence [There is also a third axiomatization of a homology theory, using pointed spaces as input and having a sequence for the cone and a suspension isomorphism as axioms]. Of course, if you have already a functor defined on pairs of spaces, you have to check that something like $h_n(X,A) = h_n(C_AX, pt)$ holds. 
I want to comment, I know this "absolute" axiomatization primarly from bordism theory, where you can show Mayer-Vietoris quite easily.<|endoftext|>
TITLE: The first odd degree-2 Artin representation for which the Artin conjecture was proved
QUESTION [10 upvotes]: At the DeKalb conference on Hilbert's problems, John Tate gave a masterly survey of Problem 9, the General Reciprocity Law.  He ends with a discussion of the Langlands Programme, especially the case of odd Artin representations $R$ of degree $2$.  Let me quote from the final paragraph of the written version of his talk :

Another reason for the relationship's eluding Artin and Hecke may be the fact that explicit non-dihedral numerical examples are hard to find.  Indeed at the time of the DeKalb conference, none was known !  I concluded the oral presentation of the paper there by explaining that, in the hope of finding one, I had looked for non-dihedral $R$'s of low conductor, $N$, and had found an $R$ with $N=133=7\cdot19$  which I hoped might be amenable to computation.  After the talk, Atkin suggested that the labor involved might be considerably reduced by systematic use of $w_7$ and $w_{19}$.  Armed with his theory of the $w$'s, four Harvard students, D. Flath, R. Kottwitz, J. Tunnell, J. Weisinger and I succeeded in the next month in proving (by relatively easy hand computation) the existence of the corresponding new form $f_R$ of weight $1$ and level $133$ predicted by Langlands.

My first question is : what was $R$ ?  The second question is :  how would you verify today (on a computer) the existence of $f_R$, without invoking any of the theorems which have been proved in the meantime ?

REPLY [17 votes]: $R$ is a 2-dimensional conductor 133 representation of the absolute Galois group of the rationals into $GL(2,\mathbf{C})$, whose associated representation to $PGL(2,\mathbf{C})$ cuts out the $A_4$ extension of the rationals which is the splitting field of $x^4 + 3x^2 - 7x + 4$. To verify the existence of the weight 1 form of level 133 you can just fire up a magma session (magma is a computer algebra package) and ask it to compute the weight 1 level 133 forms, and the dihedral weight 1 level 133 forms, and then note that there are more weight 1 level 133 forms than dihedral ones. So it's pretty easy now. Or you can write such programs yourself! [which is what I did and which is why magma can do it ;-) ]. More shameless self-promotion available (including the algorithm) at
http://www2.imperial.ac.uk/~buzzard/maths/research/papers/wt1.pdf
(where I e.g. explain that there's even a level 124 non-dihedral form, and give a description of the $S_4$ form of smallest conductor). However the ideas all first appeared in print in Joe Buhler's thesis donkey's years ago, where he finds an $A_5$ form: Springer LNM654.<|endoftext|>
TITLE: Groups with irreducible representations of the largest possible dimension
QUESTION [8 upvotes]: I thought that the interesting question Gerry Myerson asked in the comments of this question deserved to be asked in a non-closed mathoverflow question.
What can we say about groups of order $n$ with an irreducible representation of dimension $d$ such that $(d+1)^2\geq n$?
To ask a concrete question, are there infinitely many such groups?

REPLY [15 votes]: Just to give correct references. Let $d$ be the degree of an irreducible character of a finite group $G≠1$. Then $|G|=d(d+e)$ for some $e > 0 $ (that is because $d$ divides $|G|$ and $d^2 < |G|$). Therefore the condition $(d+1)^2 > |G|$ means $d(d+e)=|G|$ with $e=1$ or $2$. If $e=1$, then $G$ is a doubly transitive Frobenius group or of order 2 by Berkovich, Yakov Groups with few characters of small degrees. Israel J. Math. 110 (1999), 325–332. If $e=2$, then $G$ is a cyclic group of order 3 or non-Abelian group of order $8$ by Snyder, Noah
Groups with a character of large degree. Proc. Amer. Math. Soc. 136 (2008), no. 6, 1893–1903. In general the order of $G$ is bounded by $((2e)!)^2$ by Snyder's paper. That estimate was greatly improved to $O(e^6)$ in Isaacs, I. M. Bounding the order of a group with a large character degree. J. Algebra 348 (2011), 264–275 (using the Classification of finite simple groups) and then to  $e^6-e^4$ (if $e\ge 2$) by Durfee, Christina, Jensen, Sara
A bound on the order of a group having a large character degree.
J. Algebra 338 (2011), 197–206 (without the Classification). On the other hand, by Snyder's remark, a finite version of the Heisenberg group gives a low bound $e^4-e^3$ and $O(e^4)$ is conjecturally also the upper bound (see Isaacs' paper where this upper bound is proved in many cases).<|endoftext|>
TITLE: Multiple zeta values at negative integers
QUESTION [10 upvotes]: For a tuple of integers $\underline{s}:=(s_1,\dots, s_d)$, the multiple zeta value (MZV) at $\underline{s}$ is defined as:
$$ \zeta(s_1,\dots, s_d):= \sum_{n_1>\dots>n_d>0}\frac 1 {n_1^{s_1}\dots n_d^{s_d}}.$$
In all the references I have seen, people restrict to tuples $\underline{s}$ such that $s_1 > 1$ and $s_i \geq 1$ for all $i\geq 2$, mentionning convergence of the series.
However, it seems to me that this series converges as soon as $s_1>1$, $s_1+s_2>2$, \dots, $s_1 + \dots s_d>d$ which allows more MZV at integers.
For example, $\zeta(5,-1)$ converges.
My question is twofold:
1) Have these MZV at "negative integers" been studied and are they "number theoretically interesting"?
2)I also understand that there should be a reason to exclude these values and that they don't fit in the general philosophy about MZV.
For example, we would have a non trivial relation such as $\zeta(s,0)=\zeta(s-1) -\zeta(s)$ between MZV of different weights.
(The weight of $\underline{s}$ is by definition $s_1+\dots s_d$. As far as I understood, the general belief is that all relations between MZV are generated by relations between MZV of the same weight but I don't think that the one above can.)
Is there an a priori good reason to restrict to positive integers?

REPLY [2 votes]: In response to 1: Yes, these have been studied, for example, in Multiple zeta values at non-positive integers
(2001) and (2014), see also The first derivative multiple zeta values at non-positive integers (2010) and Analytic continuation of multiple zeta-functions and the asymptotic behavior at non-positive integers (2013).<|endoftext|>
TITLE: Weighted limits and completeness
QUESTION [10 upvotes]: Every weighted limit can be constructed from conical limits and cotensors. However, yesterday, a friend of mine, asked a question that may be rephrased as follows.
What is the reason that in the world of $\mathbf{Set}$-enriched categories every weighted limit can be constructed from conical limits (and trivial cotensors with $1$), and in the world of $\mathbf{Cat}$-enriched categories every weighted limit can be constructed from conical limits and cotensors with $2$?
Is it directly related to the fact that every set can be built upon $1$ and every category can be built upon $2$?
Is it possible to generalise these results to arbitrary (sufficiently well-behaved) monoidal category? For example, let us say that a symmetric monoidal closed category $\mathbb{V}$ is cocomplete and there exists a set $F$ of objects from $\mathbb{V}$ such that every object in $\mathbb{V}$ is a colimit of some objects from $F$. Is it true that every $\mathbb{V}$-weighted limit can be expressed via conical limits and cotensors with objects from $F$?

REPLY [12 votes]: Yes, is directly related to that fact, as you surmise.  The cotensor $X^K$, for $K\in \mathbb{V}$, preserves (co)limits in the variable $K$, that is we have
$$ X^{\mathrm{colim}_i K_i} \cong \lim_i X^{K_i}. $$
Even better, if $\lim_i X^{K_i}$ exists, then it automatically has the universal property to be $X^{\mathrm{colim}_i K_i}$.  Therefore, if $\mathbb{C}$ is a $\mathbb{V}$-category with all small conical limits, then the class of objects in $\mathbb{V}$ for which $\mathbb{C}$ admits cotensors is closed under (conical) colimits.  For instance, $\mathbf{Set}$ is the colimit-closure of $\{1\}$ and $\mathbf{Cat}$ is the colimit-closure of $\{2\}$.  This is why those particular cotensors suffice.
A good way to check that $\mathbb{V}$ is the colimit-closure of a subcategory, by the way, is to show that that subcategory is a strong generator.  As long as $\mathbb{V}$ is complete and cocomplete and extremally well-copowered, that's sufficient, as sketched in these notes.  Note that being the colimit-closure of a subcategory is a weaker statement than every object being a colimit of objects in the subcategory; it allows iterated formation of colimits.<|endoftext|>
TITLE: On the set of divergence to infinity for sequences of positive continuous functions
QUESTION [8 upvotes]: Hi,
I have asked this question on math.stackexchange but it has not received much attention, so I ask it here.
This question is partly motivated by this one, which contains an example of a sequence $(f_n)$ of positive continuous functions on $\mathbb{R}$ such that
$$f_n(x) \rightarrow \infty$$
if and only if $x \in \mathbb{Q}$.
My question is the following :
For a given sequence of positive continuous functions $(f_n)$ on $\mathbb{R}$, denote by $S((f_n))$ the set of divergence to $\infty$ :
$$S((f_n)):=\{ x \in \mathbb{R}: f_n(x) \rightarrow \infty \}$$
Is there a necessary and sufficient condition for a given set $S$ to be $S((f_n))$ for some sequence $(f_n)$?
As noted in the comments at math.stackexchange, a necessary condition is that $S$ must be a countable interesection of $F_{\sigma}$'s... But is this sufficient?
Thank you,
Malik

REPLY [3 votes]: What follows are some additional comments about this topic.
Sierpinski's 1921 paper was written without knowledge of Hahn's 1919 paper, this being a time (end of WWI) when the flow of information and journals was intermittent and/or temporarily suspended.
p. 348 in Volume 2 of Hans Hahn's Collected Works (1996) includes these remarks about Hahn's 1919 paper:

"... Hahn sets himself in Über die Menge der Konvergenzpunkte einer Funktionenfolge the task of finding out if this property gives a complete characterization of such sets. He not only proves it, but also enters the study of Baire functions by finding a characterization of the sets of convergence of functions of any given Baire class.

Jolanta Wesolowska has published versions of Hahn/Sierpinski's result for sequences of functions belonging to various other classes of functions, for example On sets of convergence points of sequences of some real functions (MR 2001d:26003; Zbl 1035.26006) and On sets determined by sequences of quasi-continuous functions (MR 2002i:26002; Zbl 1002.26004) and On sets of discrete convergence points of sequences of real functions (MR 2005f:26010; Zbl 1070.26005). Incidentally, I was in attendance when she first presented her work (represented by the first paper above) to people outside her immediate research group [1] (this work was her 2000 Ph.D. Dissertation at Uniwersytet Gdański), and it created a minor buzz among those attending, who found it simply amazing that the kinds of questions she was working on had not been thoroughly worked over before. (I should point out that the literature of real functions and point set theory is pretty much everywhere dense with minutia on most anything you can imagine, and more.)
[1] She gave this talk on 26 May 2000 at Summer Symposium in Real Analysis XXIV, held at The University of North Texas (Denton, Texas).
(Next Day) Yesterday I forgot to post a couple of references in English to a proof of the Hahn/Sierpinski result. A proof can be found on pp. 307-308 of the 1978 3rd English edition (and presumably, on the same pages for any of the other English editions) of the 1935 3rd edition of Hausdorff’s Set Theory [1957 (MR 19,111a; Zbl 81.04601); 1962 (MR 25 #4999); 1978 (Zbl 488.04001); 1991 (Zbl 896.04001); 2005] and on pp. 185-186 of Kechris’ Classical Descriptive Set Theory [MR 96e:03057; Zbl 819.04002].<|endoftext|>
TITLE: left derived functors commute with filtered  colimits
QUESTION [7 upvotes]: Let $\mathcal{A}$ be an $\mathbf{AB5}$ category with enough projectives and let $F:\mathcal{A}\rightarrow\mathcal{B}$ be a right exact functor into abelian category that commutes with filtered colimits. What reasonable assumptions should one imposed on $\mathcal{A}$ and $F$ to obtain that letf derived functors of $F$ commute with filtered colimits?

REPLY [2 votes]: This is only a comment on some aspect of the Brandenburg answere. 
I wish elucidate the aspect about the $I$-naturality of projective resolutions:
considerind a example of a single morphism $f: A\to B$ and two projective resolution $P_\ast\to A$ and  $Q_\ast\to B$ then is a well knowed fundamental lemma of homological algebrathat exist (homotopically unique) a extentions of $f$ to the (augmented) chain complexes.
In the basic homological algebra text there is also the example of a resolution of exat sequences. 
But what if insted a single morphism (or a exat sequence) we consider a small diagram?
From  Weibel, "An introduction to homological algebra", 2.3.13 on p.43 (I see it from link text) follow that $\mathcal{A}^I$ has enought projectives, but isnt clear if for a projective $P\in \mathcal{A}^I$ each $P(i)\in \mathcal{A}$ is projective as we need in the context of the above Brandenburg proof 
(this is true if for any $i\in I$  the right Kan extention of the $i$-valutation $v(i): \mathcal{A}^I\to \mathcal{A}$ is exact, I dont know if this follow from the "$I$ is filtrant" hypothesis).
From the book  "Theory of Categories (BArry Mitchell)  cor.7.6 p. 138,   let  $T_i: \mathcal{A}^I\to \mathcal{A}$ ($i\in I$) the $i$-valuation, and $S_i$ its left adjuction (the left Kan extention), now for a projective $P\in \mathcal{A}$ the object $S_i(P)(j),\ j\in I$ is a sum of copies of $P$ (see the Weibel reference above) then is projective. THe above  corollary assert that  projectives of $\mathcal{A}^I$ are objects of the form $\bigoplus_{i\in I}S_i(P_i)$ (where $P_i\in \mathcal{A}$ is a prjective) and all its retracts. 
This is enought for ensure the existence a projective resolution of a  $(X_i)_i\in \mathcal{A}^I$ with projective arguments.<|endoftext|>
TITLE: Representations of infinite dimensional Lie algebras as vector fields on manifolds
QUESTION [5 upvotes]: Suppose I have e.g. the Witt algebra, 
$\left[l_n,l_m \right] = -(n-m)l_{n+m}$. 
I want to realize the $l_n$ as vector fields on some manifold. The classical example is when the $l_n$ span the Lie algebra of diffeomorphisms of the circle, i.e.
$l_n = -i e^{i n \phi} \partial_\phi, \ \ \ 0 \leq \phi < 2\pi.$
Now I'm interested in actions on higher dimensional manifolds, e.g. $S^1 \times$ something. 
As a practical approach I could try an ansatz $l_n = -i e^{i n \phi} \partial_\phi + e^{i n \phi} f_n (x) \partial_x$ and demand that the commutation relation is satisfied, which would lead to differential-difference equations for the $f_n$. 
I'm sure there's lots of theory about this somewhere... I'd really appreciate some pointers to the right direction...
P.S. Sorry if I seem lazy for not researching this myself, but maybe I'll find the answer faster with some help from The Community?

REPLY [7 votes]: You could try É. Cartan's papers on infinite pseudogroups (mostly appearing 1904-05).  In particular, see Paragraph 57 of Sur la structure des groupes infinis de transformation (suite). There, for example, he proves that the (pseudo-)group of diffeomorphisms of the line has three distinct (i.e., nonequivalent) $2$-dimensional homogeneous spaces and seven distinct $3$-dimensional homogeneous spaces, etc.  Obviously, these induce representations of the Lie algebra of vector fields on the line in dimensions $2$ and $3$.  A similar statement can be made for the diffeomorphisms of the circle.  
For example, if $M$ is a $1$-dimensional manifold, then Diff($M$) acts transitively on $T^\bullet M$ (the punctured tangent bundle of $M$), the space $A(M)$ (the $0$-jets of affine connections on $M$), and the space $P(M)$ (the $0$-jets of projective connections on $M$).  (Of course, these are all bundles over $M$.) 
Added information:  If you take a (possibly periodic) coordinate $x$ on $M$, the vector fields are in one-to-one correspondence with functions, say $V_f = f(x)\partial_x$.  Then one has the corresponding homomorphisms $\phi_i$ from the vector fields on $M$ to vector fields in two dimensions of the form

$\phi_1(V_f) = f(x)\partial_x - \bigl(f'(x) y\bigr)\partial_y$.  (Take $y\not=0$.)
$\phi_2(V_f) = f(x)\partial_x - \bigl(f'(x) y + f''(x)\bigr)\partial_y$.
$\phi_3(V_f) = f(x)\partial_x - 2\bigl(f'(x) y + f'''(x)\bigr)\partial_y$.

The $3$-dimensional homogeneous spaces are a little harder to describe.  Obvious examples are the spaces of $1$-jets of sections of the above bundles, but these are only three of the seven possibilities.
In high enough dimension (I think Cartan says that it starts in dimension $n=5$), it turns out that there are continuous families of inequivalent $n$-dimensional homogeneous spaces of Diff($M$).<|endoftext|>
TITLE: Graph decomposition into paths
QUESTION [6 upvotes]: Let $G$ be a $(2k+1)$-regular graph on $n$ vertices. A theorem by Lovasz implies that $G$ decomposes into $\frac{1}{2}n$ disjoint paths. 
Since the number of edges is $\frac{1}{2}n(2k +1)$, is it possible to require that each path in the decomposition be of size $(2k +1)$.
I found an argument for the case when the graph $G$ has a perfect matching and its girth $\geq 2k +1$. Is there any literature on this.

REPLY [8 votes]: The question is "no" in general, as Gjergji pointed out.  The question of when the answer is "yes" is difficult and only partial results are known. It is conjectured that having a 1-factor is sufficient. A recent paper that discusses the issue is this paper of Odile Favaron, François Genest and Mekkia Kouider. Here is the abstract:
Kotzig asked in 1979 what are necessary and sufficient conditions
for a $d$-regular simple graph to admit a decomposition into paths
of length $d$ for odd $d\gt 3$. For cubic graphs, the existence of a 1-factor is
both necessary and sufficient. Even more, each 1-factor is extendable to a
decomposition of the graph into paths of length 3 where the middle edges
of the paths coincide with the 1-factor. We conjecture that existence of a
1-factor is indeed a sufficient condition for Kotzig’s problem. For general
odd regular graphs, most 1-factors appear to be extendable and we show
that for the family of simple 5-regular graphs with no cycles of length
4, all 1-factors are extendable. However, for $d\gt 3$ we found infinite families of $d$-regular simple graphs with non-extendable 1-factors. Few authors
have studied the decompositions of general regular graphs. We present
examples and open problems; in particular, we conjecture that in planar
5-regular graphs all 1-factors are extendable.
Finally, I should mention that I'm not a specialist in this area so my assertion that the problem is still unsolved is not gospel.<|endoftext|>
TITLE: Necessary conditions for cofibrancy in global projective model structure on simplicial presheaves
QUESTION [10 upvotes]: Consider the global projective model category
of simplicial presheaves on some category
(the category of smooth manifolds is particularly interesting to me).
In Section 9.1 of Dugger's paper “Universal homotopy theories”
one can find a sufficient condition for a simplicial presheaf to be cofibrant,
together with descriptions of two different cofibrant replacement functors.
Is there a nontrivial necessary condition for a simplicial presheaf to be cofibrant in the global projective model structure that is easier to check than the lifting property?
Such a condition could be used to easily establish noncofibrancy of some simplicial presheaves.

REPLY [18 votes]: Let $\mathscr C$ be a small category. Necessary and sufficient conditions for a presheaf $F$ to be cofibrant in the global projective model structure on $[\mathscr C^\mathrm{op},  [\Delta^\mathrm{op}, \mathbf{Set}]]$ are that:
(1) Each $F(-)(n) \colon \mathscr C^\mathrm{op} \to \mathbf{Set}$ is projective (i.e., a coproduct of retracts of representables; if $\mathscr C$ is Cauchy-complete, then equivalently a coproduct of representables).
(2) $F \colon \mathscr C^\mathrm{op} \to [\Delta^\mathrm{op}, \mathbf{Set}]$ factors through the subcategory $[\Delta^\mathrm{op}, \mathbf{Set}]_{\mathrm{nd}}$ of $[\Delta^\mathrm{op}, \mathbf{Set}]$ whose objects are simplicial sets and whose maps are simplicial maps which sends non-degenerate simplices to non-degenerate simplices.
On the one hand, it's easy to show by induction that any cofibrant object satisfies (1) and (2). Conversely, condition (2) means that each $F(-)(n)$ breaks up as a coproduct $G_n(-) + H_n(-)$ of non-degenerate and degenerate simplices, and the object $G_n(-)$ of non-degenerate simplices is projective since $F(-)(n)$ is. This means that any object satisfying (1) and (2) can be built up dimension by dimension using the generating cofibrations: first construct the $0$-skeleton using pushouts of maps
$\partial \Delta_0 \cdot \mathscr C(-, W) \to \Delta_0 \cdot \mathscr C(-, W)$ (for $W \in \mathscr C$)
one for each representable summand in the projective object $F(-)(0)$; then construct the $1$-skeleton by using pushouts of maps 
$\partial \Delta_1 \cdot \mathscr C(-, W) \to \Delta_1 \cdot \mathscr C(-, W)$ (for $W \in \mathscr C$)
one for each representable summand in the projective object $G_1(-)$ of non-degenerate $1$-simplices; and so on.<|endoftext|>
TITLE: What is the precise relationship between "prodsimplicial sets" and rooted trees?
QUESTION [11 upvotes]: In Keven Walker's answer to the question, Cubical vs. simplicial singular homology  it is written:


Personally, I think it is more convenient to do singular homology with the larger collection of polyhedra which is closed under both cones and products. (The n-dimensional polyhedra in this collection are indexed by rooted trees with n edges. The simplices correspond to maximal depth trees where the valence of a vertex is at most 2, while the cubes correspond to minimal depth (star-shaped) trees where the root vertex has valence n and all the other vertices have valence 1.)


This raises a few questions. I see how simplices are the "linear trees", and I see how a map of two simplices will become a map between two such trees. I also see how to send the n-cube to the appropriate star-shaped tree. It remains mysterious to me what we send a map between two cubes. To make this a bit more precise:

Is their a small category whose objects are rooted trees, such that the morphisms give the right morphisms of prod simplices (or more aptly name prisms). The sort of prisms I have in mind are the objects in Gugenhiem's paper, "On Supercomplexes".

In the comments of said answer, it was said that the combinatorics are related to blob homology and dendroidal sets. Any references either on point or even slightly off point will be appreciated.

REPLY [5 votes]: There are a short list of operations described as generating the desired polyhedra:  

$ X : \mathrm{Prism} \vdash C X : \mathrm{Prism} $  
$ l : \mathrm{list}\ \mathrm{Prism} \vdash \Pi l : \mathrm{Prism} $  

There are a short list of operations needed to generate the family of rooted trees:

$ T : \mathrm{Tree}\ \vdash \mathrm{Stem}\ T : \mathrm{Tree} $
$ F : \mathrm{list}\ \mathrm{Tree}\ \vdash \mathrm{graft}\ F : \mathrm{Tree}$

That makes the correspondence obvious; the distinction is in the semantics of prisms vs. trees.
Now, the generators I've given would seem to distinguish between $ \Pi (\Pi (A, B), C) $ and $ \Pi (A, B, C) $, just as they seem to distinguish between ... , well, I'd have to draw pictures, and I don't really want to. I think that's quite alright, I don't mind many isomorphic polyhedra having distinct descriptions.  But if that worries you, you can argue that the strange plurality of trees is isomorphic to the strange plurality of products in a natural way.  And that should be enough.<|endoftext|>
TITLE: Eigenvalues of non-symmetric matrix and its transpose
QUESTION [9 upvotes]: What more can be said about the eigenvalues (especially the spectrum) of the $N \times N$ matrix ${\bf M} = {\bf A} + {\bf A}^T$ in terms of $\bf A$ if $\bf A$ is not symmetric and its eigenvalues are not necessarily positive ($\bf A$ is not necessarily positive semi-definite)?

The eigenvalues of $\bf M$ are real since $\bf M$ is symmetric.
$\sum_{i=1}^{N} \lambda_i({\bf M}) = {\rm trace}({\bf M}) = 2~{\rm trace}({\bf A}) = 2\sum_{i=1}^{N} \lambda_i({\bf A})$

The problem originates from principal component analysis. The principal components are the largest eigenvalues of the sample covariance matrix (SCM) $\bf C = ({\bf X}+{\bf W})^T({\bf X}+{\bf W})$ where $\bf X$ is a data matrix and $\bf W$ is a noise (white Gaussian) matrix.
Usually it is assumed that $\bf X$ and $\bf W$ are uncorrelated (when many samples are used) and the SCM can then simply be written as ${\bf C} = {\bf X}^T{\bf X} + {\bf W}^T{\bf W}$. The eigenvalue spectrum can under these conditions $quite~easily$ be determined using Weyl's inequalities since ${\bf X}^T {\bf X}$ and ${\bf W}^T{\bf W}$ are both Hermitian (or real symmetric).
However, when the number of samples is limited, the error (cross) terms ${\bf E} = {\bf X}^T{\bf W} + {\bf W}^T{\bf X}$ begin to influence the eigenvalues significantly. Since $\bf E$ is Hermitian, Weyl's inequalities can again be used to determine the eigenvalue spectrum of $\bf C$. Though, the eigenvalue spectrum of $\bf E$ is required to do so.
Setting ${\bf A} = {\bf X}^T{\bf W}$, it is clear that ${\bf A}^T = {\bf W}^T{\bf X}$, such that ${\bf M} = {\bf E}$.

REPLY [6 votes]: Perhaps the best relation one could expect between the eigenvalues of $A$ and those of $A+A^T$ is the following majorization relation
$$\Re \lambda(A)\prec \lambda((A+A^T)/2).$$
It is due to Ky Fan, see p. 360 of [F. Zhang, Matrix Theory: Basic Results and
Techniques, Springer, New York, 2nd ed., 2011.]<|endoftext|>
TITLE: Some calculations with the Adams spectral sequence and the cobar complex
QUESTION [12 upvotes]: I am trying to 'get my hands dirty', so to speak, with some of the calculations with the Adams spectral sequence in Ravenel's Complex Cobordism book, and I have a few questions (I hope it is OK to ask them in the one question). 
In great generality for a Hopf algebroid $(A,\Gamma)$ we can define the cobar complex $C_\Gamma^*(M)$ by $C_\Gamma^*(M)=\overline{\Gamma}^{\otimes s} \otimes M$ where $\overline{\Gamma}=\text{ker} \epsilon: \Gamma \to A$ with coboundary map
$$d_s(\gamma_1 \otimes \cdots \otimes \gamma_s \otimes m) = \cdots +  \sum_{i=1}^s \gamma_1 \otimes \dots \otimes \gamma_{i-1} \otimes \psi(\gamma_i) \otimes \cdots \otimes \gamma_s \otimes m + \cdots$$
(where $\psi(\gamma_i)$ is the coproduct and I have omitted the first and last term for brevity) and then $\text{Ext}_{\Gamma}(A,M)$ is the cohomology of this cobar complex. 
The first calculation (pp. 64-66) is the $E_2$ term for the calculation of $\pi_*(bo)$, which is equal to $ \text{Ext}_{\mathscr{A}(1)_*}(\mathbb{F}_2,\mathbb{F}_2) $. This is abutted to by a Cartan-Eilenberg Spectral sequence which has $E_2$ term equal to $\mathbb{F}_2[h_{10},h_{11},h_{20}]$, where $h_{i,j}$ corresponds to the class $[\overline{\xi}_i^{2^j}]$ in the cobar complex. The first claim is that $d_2(h_{20}) = h_{10}h_{11}$, and this follows from the fact that in the cobar complex $d(\xi_2) = \xi_1 \otimes \xi_1^2$, which in turn follows from the coproduct of the mod 2 Steenrod algebra. This gives $E_3$ term $\mathbb{F}_2(u,h_{10},h_{11})/(h_{10}h_{11})$ where $u$ corresponds to $h_{20}^2$. Again we can calculate $d(\overline{\xi}_2 \otimes \overline{\xi}_2) = \overline{\xi}_2 \otimes \xi_1 \otimes \xi_1^2 + \xi_1 \otimes \xi_1^2 \otimes \overline{\xi}_2 $ in the cobar complex. Ravenel then states

...the cobar complex is not commutative and when we add correcting terms to $\overline{\xi}_2 \otimes \overline{\xi}_2$ in the hope of getting a cycle we get instead $d(\overline{\xi}_2 \otimes \overline{\xi}_2 + \xi_1 \otimes \xi_1^2 \overline{\xi}_2 + \xi_1 \overline{\xi}_2\otimes \xi_1^2) = \xi_1^2 \otimes \xi_1^2 \otimes \xi_1^2$

which is used to conclude $d_3(u)=h_{11}^3$
Finally my questions:

1) Why are the correcting terms $\xi_1 \otimes \xi_1^2 \overline{\xi}_2$ and $\xi_1 \overline{\xi}_2\otimes \xi_1^2$?
2) (This may be answered by 1) Why does $\overline{\xi}_2 \otimes \overline{\xi}_2 + \xi_1 \otimes \xi_1^2 \overline{\xi}_2 + \xi_1 \overline{\xi}_2\otimes \xi_1^2$ represent $u$ in the cobar complex?
3) How can I calculate this differential? For example how do we calculate $d(\xi_1 \otimes \xi_1^2 \overline{\xi}_2$)?


That part 2 of this question concerns the May spectral sequence for calculating $\text{Ext}_\mathscr{A}(\mathbb{F}_2,\mathbb{F}_2)$. One can compute the $E_2$ term of the May spectral sequence to have generators (in the region $t-s \le 13$) $h_j = h_{1,j}$, $b_{i,j} = h_{i,j}^2$ and $x_7 = h_{20}h_{21} + h_{11}h_{30}$. There are some relations given without proof; $h_jh_{j+1} = 0, h_2b_{20} = h_0x_7$ and $h_2x_7 = h_0b_{21}$. I think that the relation $h_jh_{j+1}=0$ comes from the fact that $d_1(h_{2,j}) = h_jh_{j+1}$, but I am unsure where the other relations are coming from.

REPLY [16 votes]: One quick answer to 1 is trial and error; 2 is a typo: you don't mean $u = h_{11}^2$. 
For 3, unhelpfully, you just have to do the computation from the definitions.  But I'm
really answering the last question.  I wish that people would use my original notations, and I wish that Doug had done so, since my notations are suggestive of the relevant powers and since this old man can't remember other people's notations.  More importantly, the original notations make answers to questions like the one asked transparent. Write $R_j^i$ instead of $h_{ji}$ (note the order).  Take the polynomial algebra over $\mathbf{F}_2$ on the $R_j^i$ and give it the differential $d(R_j^i) = \sum_{0 < k < j}R^{i+k}_{j-k}R^i_k$.  I prove that the resulting homology is $E_2$ of the MSS.  Then $h_i$ is represented by $R_1^i$ and $b^i_j$ is represented by $(R^i_j)^2$ (so $b^i_1 = (h_i)^2$).  The differential on $R^i_2$ gives that $h_{i+1}h_i=0$. I don't know where the notation $x_7$ comes from; this is the guy I called $c_0$ that lives in the $8$-stem.  I also called him $h_i(1)$ as part of a systematic naming of what I believe are ALL indecomposables in $E_2$.  He is represented in $E_2$ by $R^0_2R^1_2 + R^1_1 R^0_3$. Compute $d(R^0_2R^0_3)$ and $d(R^1_2R^0_3)$. The relations $h_0c_0 = h_2b^0_2$ and $h_2c_0 = h_0 b^1_2$  in $E_2$ will appear before your eyes.  Incidentally, I prove that $d_2(b^i_j) = h_{i+1}b^{i+1}_{j-1} + h_{i+j}b^i_{j-1}$ in the MSS.  (See page II-6.3 of my thesis, which seems to be on Math Reviews now).<|endoftext|>
TITLE: Capitalization of named theorems
QUESTION [8 upvotes]: I would like to extend a previous question on the capitalization of theorem names . This time, it's not about numbered theorems like "Lemma 1.1", but about theorems that carry a name.

Question: Which of the following variants is to be preferred?


"By Zorn's lemma, we can conclude..."
"By Zorn's Lemma, we can conclude..."

Additionally, I would like to know how to deal with the corner case where the name of the person after which the theorem is named starts with a lowercase letter

Question: Which of the following variants is to be preferred?


"The proof of van der Waerden's theorem can be extended to ..."
"The proof of Van der Waerden's theorem can be extended to ..."
"The proof of van der Waerden's Theorem can be extended to ..."
"The proof of Van der Waerden's Theorem can be extended to ..."

REPLY [2 votes]: I'd say that in general, such questions about standardization in mathematics (not only in mathematics, of course) are quite important today, because they make easier the use of search engines. So, it is relevant that we all agree to call "Zorn's lemma"  a certain result, instead of e.g. "Chain's Upper Bound Theorem", much more relevant than in the past.
In particular, spelling of names is an important issue, of course, and creates a problem, for several different transliterations from other alphabets are used: e.g., Чебышёв is romanized as Chebyshev (English), Tchebychev (French), Tschebyschow (German), Chebyshov (Spanish), Tsjebysjev (Dutch), Čebyšëv (ISO 9),...&c. Here, we should at least care to write correctly the names in their original form, which is also a point of respect (yet we meet wrong forms like Lebesque or Holder or Erdös even here at MO).
On the contrary, the issue of upper vs. lower case does not seem that relevant, as search engines are usually case-insensitive. Also, Google does not care so much about the Saxon genitive, even for queries into commas.
In conclusion, as to your questions, for your thesis I would mainy care to respect an internal coherence, and of course, the correct spelling of personal names.<|endoftext|>
TITLE: Prescribing areas of parallelograms (or 2x2 principal minors)
QUESTION [17 upvotes]: Let $(a_{ij})$ be a $n\times n$ symmetric matrix such that $a_{ij}\geq 0$ for all $i,j$ and $a_{ii}=0$ for all $i$. Under which conditions on the $a_{ij}$'s can one find $n$ vectors $v_1,\ldots,v_n\in{\mathbb R}^n$ such that for all $i,j$ the area of the parallelogram spanned by $v_i$ and $v_j$ equals $a_{ij}$:
$\forall i,j:\quad\|v_i\|^2\|v_j\|^2-\langle v_i,v_j\rangle^2=a_{ij}^2$ ?
Here is the only and obvious necessary condition I know about: if $a_{ij}=0$ for some $i\neq j$, then  $a_{ik}a_{jl}=a_{il}a_{jk}$ for all $k,l$. 
What if $a_{ij}>0$ for all $i\neq j$ ?
Thank you. 
Edit. As Noah Stein suggested, a useful reformulation of the question is: can one prescribe the $2\times2$ principal minors of a symmetric positive semidefinite matrix?     
Edit 2. See also George Lowther comment. It is always possible (and easy!) to prescribe the $2\times 2$ principal minors of a symmetric $n\times n$ matrix. If the $a_{ij}$'s, $1 \leq i < j \leq n $,
 are to be those minors, we simply need to choose $n$ numbers $g_{ii}$ such that $g_{ii}g_{jj}\geq a_{ij}$ for all $i\neq j$. Then we are done with the symmetric matrix $G=(g_{ij})$ whose off diagonal entries are given by $g_{ij}=\epsilon_{ij}\sqrt{g_{ii}g_{jj}-a_{ij}}$, where $\epsilon_{ij}=\pm 1$.
So the initial question becomes : under what conditions on the $a_{ij}$'s, can one find $n$ real numbers $g_{ii}\geq 0$, and $\epsilon_{ij}=\pm 1$, so that the matrix $G$ defined above is positive semidefinite?

REPLY [7 votes]: In the case where the matrix $(a^2_{ij})\_{i,j=1,\ldots,n}$ is nonsingular, then the problem reduces to the condition that it has a single positive eigenvalue. In fact, we have the following for any $n\times n$ nonzero symmetric matrix $A$ with nonnegative components and zero diagonal (I'll remove the square from $a_{ij}$, as it doesn't seem to help).

If there exist $\nu_i\in \mathbb{R}^n$ such that
  $$
\begin{align}
A_{ij}=\lVert\nu_i\rVert^2\lVert\nu_j\rVert^2-\langle\nu_i,\nu_j\rangle^2&&{\rm(1)}
\end{align}
$$
  then $A$ has a single positive eigenvalue (counting multiplicities).
Conversely, if $A$ is nonsingular and has a single positive eigenvalue, then there exists $\nu_i\in\mathbb{R}^n$ satisfying (1).

First, suppose that (1) holds. Then, there exist nonnegative reals $\lambda_i$ and a positive semidefinite matrix $S$ such that $A_{ij}=\lambda_i\lambda_j(1-S_{ij}^2)$, simply by taking $\lambda_i=\lVert\nu_i\rVert^2$ and $S_{ij}=\langle\hat\nu_i,\hat\nu_j\rangle$, where $\hat\nu_i=1_{\lbrace\nu_i\not=0\rbrace}\nu_i/\lVert\nu_i\rVert$. Let $\lambda=(\lambda_i)\_{i=1,\ldots,n}\in\mathbb{R}^n$. Using the fact that the componentwise square of a positive semidefinite matrix is itself positive semidefinite,
$$
x^{\rm T} A x= \langle x,\lambda\rangle^2-\sum_{ij}(\lambda_ix_i)S_{ij}^2(\lambda_jx_j)\le\langle x,\lambda\rangle^2.
$$
In particular, $x^{\rm T}A x\le0$ for all vectors orthogonal to $\lambda$. So, the space generated by eigenvectors with positive eigenvalues cannot contain any nonzero members orthogonal to $\lambda$, and has dimension at most one. But, as the trace of $A$ is zero, it must have at least one positive eigenvalue. 
Conversely, suppose that $A$ has a single positive eigenvalue and is nonsingular. Diagonalization gives
$$
A=u u^{\rm T}-\sum_{\alpha=1}^{n-1} v_{\alpha}v_{\alpha}^{\rm T}
$$
for nonzero orthogonal $u,\nu_\alpha\in\mathbb{R}^n$. As the diagonal of $A$ is zero,
$$
u_i^2=\sum_\alpha\nu_{\alpha,i}^2.
$$
Using Cauchy–Schwarz,
$$
A_{ij}\le u_iu_j+\sqrt{\sum_\alpha v_{\alpha,i}^2\sum_\beta v_{\beta,j}^2}
=u_iu_j+\vert u_iu_j\vert.
$$
So, the $u_i$ are all nonzero, otherwise $A$ would have a row with no positive elements. Then, $u_iu_j > 0$.Writing $S=\lbrace i=1,2,\ldots,n\colon u_i > 0\rbrace$ we would have $A_{ij}=0$ for $i\in S$ and $j\not\in S$. Breaking $A$ down into two blocks on which the row and column indices are respectively in $S$ and not in $S$, we can break the problem down to the case where $u_i$ are all of the same sign. W.l.o.g., take $u_i > 0$. For $1 > \epsilon > 0$, define the matrix
$$
\begin{align}
S_{ij}&=\sqrt{1-\epsilon u_i^{-1}u_j^{-1} A_{ij}}\cr
&=1-\frac12\epsilon u_i^{-1}u_j^{-1} A_{ij}+O(\epsilon^2)\cr
&=1-\epsilon/2+\frac\epsilon2\sum_{\alpha}u_i^{-1}u_j^{-1}v_{\alpha,i}v_{\alpha,j}+O(\epsilon^2)
\end{align}
$$
As the vectors $u,v_{\alpha}$ are linearly independent, the same is true of the vectors $\tilde u=(1,1,\ldots,1)$ and $\tilde v_\alpha=(u_1^{-1}v_{\alpha,1},\ldots,u_n^{-1}v_{\alpha,n})$. Then,
$$
x^{\rm T}Sx=(1-\epsilon)\langle\tilde u,x\rangle^2+\frac\epsilon2\left(\langle\tilde u,x\rangle^2+\sum_\alpha\langle\tilde v_\alpha,x\rangle^2\right)+O(\epsilon^2\lVert x\rVert^2)
$$
is nonnegative for all $\epsilon$ small enough and $x\in\mathbb{R}^n$. In this case $S$ is positive definite and (by Gram-Schmidt, for example) there are $\hat\nu_i\in\mathbb{R}^n$ with $S_{ij}=\langle\hat\nu_i,\hat\nu_j\rangle$. Setting $\nu_i=\epsilon^{-1/4}u_i^{1/2}\hat\nu_i$ gives (1).

That concludes the case where $A$ is nonsingular. The singular case is, I think, considerably more complicated.<|endoftext|>
TITLE: Approximation theory reference for a bounded polynomial having bounded coefficients
QUESTION [7 upvotes]: Let $P(x)$ be a real polynomial of degree at most $d$.  Assume $|P(x)| \leq 1$ for $|x| \leq 1$.  I would like a bound saying that each coefficient of $P(x)$ is at most $C^d$ in magnitude, for some absolute constant $C$.
This is surely a well-known, basic fact in approximation theory and I'm looking for a proper reference.  I know one very recent paper which writes out a proof using the standard simple idea (Lagrange interpolation) -- Lemma 4.1 from a paper of Sherstov here: 
http://eccc.hpi-web.de/report/2012/037/download 
Sherstov obtains $C = 4e$; I don't think either of particularly cares about getting the sharpest constant.
In any case, Sherstov and I agree that this must have appeared somewhere long ago.  Could anyone provide a reference?   Thanks!

REPLY [2 votes]: I think that an interesting historical point to mention is that this problem was first posed and solved for quadratic polynomials by Medeleev (a chemist). There is a nice little article about the problem, including its origins, in the American Mathematical Monthly:

R.P. Boas, Extremal problems for polynomials,  Amer. Math. Monthly 85 (1978), No. 6, 473--475.

You can also find Markov's theorem written up (including generalizations to polynomials of several variables) in the following textbooks:

P.B. Borwein and T. Erdelyi, Polynomials and Polynomial Inequalities, GTM 161, Springer, 1995.
P. Borwein, Computational Excursions in Analysis and Number Theory, CMS Books in Mathematics, Springer, 2007.<|endoftext|>
TITLE: Cake-cutting and amenable groups
QUESTION [8 upvotes]: I recently heard Alan Taylor speak about envy-free fair division and started wondering if questions like these make sense if we replace finitely additive measures with invariant means on amenable groups. Valerio has suggested a few tweaks for this question (thanks!) so I'll post a broad version and the earlier narrow one.
Along these lines, I'll ask the following broad question:

What are some fruitful modifications of cake-cutting fair division problems which replace the cake by an amenable group and the partygoers' preferences by invariant means?

Here's an attempt at such a modification in the discrete case, which was the body of the original question:

Let $G$ be an infinite discrete amenable group with $n$ given distinct left-invariant means $\mu_{1},...,\mu_{n}$. Is it possible to partition $G$ into $n$ parts $\lbrace K_{i} \rbrace_{i=1}^{n}$ so that $\mu_{j}(K_{l})\leq\mu_{j}(K_{j})$ for all $l,j\in \lbrace 1,2,...,n \rbrace$ and $l \neq j$?

REPLY [2 votes]: Can't you just use the Lyapunov convexity theorem directly?
As usual, identify $\ell^\infty(G)$ with $C(\beta G)$, and work with $\beta G$ the Stone-Cech compactification.  As this is a compact Hausdorff space, if $\mu$ is a regular measure on $\beta G$ then an atom of $\mu$ must be a point.  So we can decompose $\mu$ as something in $\ell^1(\beta G)$ together with an atom-less measure, say a member of $M_c(\beta G)$ (continuous measures).
(Left) translation by members of $G$ give automorphisms of $\beta G$, and hence leave $\ell^1(\beta G)$ and $M_c(\beta G)$ invariant.  I claim that nothing in $\ell^1(\beta G)$ can be left invariant.  Let $\mu\in\ell^1(\beta G)$ be left invariant.  Write $\beta G$ as the disjoint union of $G$-orbits, say $\bigcup_i G u_i$.  Then $\mu$ must be supported on finite orbits (else we couldn't sum the coefficients, so we wouldn't be in $\ell^1$).  If $u\in\beta G$ with $Gu$ finite, then there is $s\not=e$ in $G$ with $su=u$.  Realise $u$ as an ultrafilter.  Let $A\subseteq G$ be maximal with $A\cap s^{-1}A=\emptyset$.  This means that if $r\not\in A$ then there is $t\in (A\cup\{r\}) \cap (s^{-1}A\cup\{s^{-1}r\})$, which implies that $t=r\in s^{-1}A\cup\{s^{-1}r\}$, that is, $sr\in A$.  So $r\not\in A\implies sr\in A \implies r\in s^{-1}A$, so $G=A\cup s^{-1}A$.
So Zorn implies there is $A\subseteq G$ with $A \cap s^{-1}A=\emptyset$ and $A\cup s^{-1}A=G$.  Then either $A\in u$ so $A\in su$ so $s^{-1}A\in u$, contradiction; or 
$s^{-1}A\in u$ so $A\in su=u$ contradiction.
So I (hope!) I've shown that actually for any $u\in\beta G$, the orbit map $G\rightarrow\beta G; s\mapsto su$ is injective.
In particular, invariant means live in $M_c(\beta G)$, and so are atom-less, and so now we can just apply Lyapunov.
Edit: As Valerio points out, this shows that $X=\{ (\mu_1(A),\cdots,\mu_n(A)) : A\subseteq\beta G \text{ is Borel}\}$ is a convex set in $[0,1]^n$.  Now, each $A\subseteq G$ induces the clopen set $O_A=\{ u\in\beta G: A\in u \}$, and these sets $O_A$ form a base for the topology.  Now each $\mu_i$ is regular, so given $\epsilon>0$ and $A\subseteq\beta G$ Borel, we can find $B,C\subseteq G$ with $O_B \subseteq A\subseteq O_C$ and with $\mu_i(C)-\mu_i(B)<\epsilon$, for all $i$ (under the obvious abuse of notation).  (This follows as any open set is a union of sets of the form $O_C$, and then approximate with a finite union.)  So $Y=\{ (\mu_1(A),\cdots,\mu_n(A)) : A\subseteq G\}$ is a subset of $X$, and is dense in $X$.  I don't see right now why $Y$ need be convex.<|endoftext|>
TITLE: determining symplecticity (if that's a word)
QUESTION [5 upvotes]: Suppose you have a matrix $M$ in $SL(n, \mathbb{Z}).$ Question: is there a necessary and sufficient condition for $M$ to be conjugate to $N \in Sp(n, \mathbb{Z}).$ It is clearly necessary that the characteristic polynomial of $M$ be palindromic, but I would assume that this is not sufficient.

REPLY [4 votes]: You forgot the condition that $n$ is even. I do not think there is a better criterion than the tautology.
The unipotent matrix
$$\left[ \begin {array}{cccc} 1&1&2&3\\\ 0&1&4&5\\\ 0&0&1&1\\\0&0&0&1\end{array}
 \right] $$ is not conjugate to a symplectic matrix in $SL(4,\mathbb{Z})$ (although it is conjugate to a symplectic matrix in $SL(4,\mathbb{C})$. 
On the other hand the matrix
$$\left[ \begin {array}{cccc} 1&0&0&1\\\ 0&1&0&0\\\ 0&0&1&0\\\0&0&0&1\end{array}
 \right] $$
is conjugate to a symplectic matrix. 
Both facts can be easily verified by using Maple.<|endoftext|>
TITLE: On connection between Knot theory and Operator algebra
QUESTION [17 upvotes]: What exactly is the connection between knots and operator algebra? I heard that Jones established such a connection while discovering the celebrated Jones Polynomial. 
Now Jones Polynomial is probably understood out of that context on its own, but what was it with Operator algebras in this space ? Can someone explain in plain English?
As is probably very evident, I am a complete newbie to this area.

REPLY [2 votes]: you may looks into David Evans monumental book on quantum symmetries on operator algebras<|endoftext|>
TITLE: Error in Polynomial Root Finding Algorithm with Synthetic Division
QUESTION [6 upvotes]: I have written a program which finds the roots of polynomial using Newton's Method. After finding the first root to within a tolerance (note that this also finds complex roots), I use synthetic division to remove that root from the original polynomial (f = f/(x-root))
My question is, how does this affect the error? I can tell I get some shift as I look at my 20th root, but exactly how would I quantify this, and how would I ensure that the max error is still less than my tolerance?

REPLY [5 votes]: It is a terrible idea to divide out roots as they are found.  There will be examples where the later roots are lost almost completely. See this wikipedia article for a famous and remarkably simple example of a polynomial whose zeros are very sensitive to the coefficients.  As soon as you divide out one zero approximately, you perturb the coefficients and the other zeros may have moved a lot.  At the very least, all the roots should be refined using the original polynomial. Textbooks usually advise trying to find a way to solve your problem that does not involve root finding in a polynomial.  (For example, getting the eigenvalues of a matrix by finding its characteristic polynomial first is nearly always a bad idea.)<|endoftext|>
TITLE: A recommended roadmap to Fermat's Last Theorem
QUESTION [19 upvotes]: I was inspired to undertake math as a career after watching a documentary on the proof of Fermat's Last Theorem. As such it's been a small goal of mine to understand Wiles et al's proof. 
In a similar vein to this question, I was hoping to get a roadmap as to the required topics, with either suggested books or papers to read, I would be required to learn undertake this task. I am, in particular, looking for expository papers on Galois representations of elliptic curves and deformations of Galois representations.
As for my background I am currently a first year graduate student with the usual algebra, analysis, and topology prerequisites. I also have a course in algebraic number theory (up to the proof of the finiteness of class numbers), modular forms, and algebraic curves (up to Riemann-Roch) under my belt. I am also currently working through Silverman's AEC.
Thank you in advance for any advice given.

REPLY [15 votes]: The book edited by Cornell, Silverman and Stevens is terrific (though you'll of course find some articles more readable than others), but a less demanding alternative is Alf van der Poorten's Notes on Fermat's Last Theorem, which is really great fun to read, or to dip into.   I see that there's a second edition due out in September, so you might or might not want to wait.  
Edited to add:  Here is Andrew Granville's review.<|endoftext|>
TITLE: nontrivial $\pi_2(\textrm{Diff}(M))$
QUESTION [21 upvotes]: Homotopy groups of Lie groups
I asked it also there, and I still don't know the answer, so I try again.
I would like to know a closed manifold (possibly of low dimension) such that $\pi_2(\textrm{Diff}(M))\neq 0$.

REPLY [5 votes]: $\newcommand{\Diff}{\mathrm{Diff}}$The previous examples have $\pi_2 \Diff(M)$ non-trivial, but finitely generated.
Here is an example of a manifold $M$ where $\pi_2 \Diff(M)$ is not finitely generated.
The manifold is $S^1 \times D^5$, and $\Diff(M)$ denotes the group of diffeomorphisms that acts trivially on the boundary.  David Gabai and I proved this in the last year, in Knotted 3-balls in $S^4$, arXiv:1912.09029
In general we can show that $\pi_{n-3} \Diff(S^1 \times D^n)$ is not finitely generated, for all $n \geq 3$.  We construct a non-trivial family $S^{n-3} \to \Diff(B)$ where $B$ is what we call a `barbell manifold'.  This manifold is a boundary connect-sum of two copies of $S^{n-1} \times D^2$.   We then embed the barbells in $S^1 \times D^n$, and extend the diffeomorphisms.
The map $\pi_{n-3} \Diff(S^1 \times D^n) \to \pi_{n-3} \mathrm{Emb}(D^n, S^1 \times D^n)$ detects our diffeomorphisms, and one can further descend via a Cerf `scanning' type construction to detect these elements in the homotopy-group $\pi_{2n-4} \mathrm{Emb}(I, S^1 \times D^n)$, which rationally looks somewhat like a two-variable Laurent polynomial ring.
If you want the manifold to be closed, I believe $\pi_{n-3} \Diff(S^1 \times S^n)$ will also not be finitely-generated.  The $n=3$ case appears in the above paper, and the $n>3$ case will be in a follow-up.<|endoftext|>
TITLE: Network flows with capacities on pairs of edges
QUESTION [10 upvotes]: Take a standard network flow problem: a directed graph with nonnegative capacities on each edge, a source $s$, a sink $t$. We all know how to find the maximum flow from $s$ to $t$.
Now add edge-pair capacities: for each pair of edges, there is some maximum total flow for those two edges.
This new problem is solvable in polynomial time since it is a linear program, but I don't want to use such a sledge-hammer.

Is there a way to convert this problem into an at-most-polynomially-larger ordinary network flow problem?
Can it be solved using an augmenting-flow type of approach?
In the original problem, if the capacities are integers then there is an optimum flow which is integer.  This is not true in the new problem, but is it true that there is an optimum flow in which the edge flows are integer or half integer? 
The polytope of all flows is interesting in the original problem; is it still interesting in the new problem?  (Choose your own definition of "interesting".)

UPDATE: Q3 was answered in the negative by Douglas Zare. Let me now weaken the hypothesis.  Say that a pair-capacity $c(e_1,e_2)$ bites if $c(e_1,e_2)\lt c(e_1)+c(e_2)$. What happens in Q3 if no two biting pair-capacities have an edge in common? [Also negative per Douglas.]

REPLY [4 votes]: Here is a simple counterexample to 3. Let the vertices be $\lbrace s, v, t\rbrace$. Let there be two edges $e_1$ and $e_2$ from $s$ to $v$, and one edge $e_3$ from $v$ to $t$. 
$$s \overset{e_1}{\underset{e_2} \rightrightarrows} v \overset{e_3} \rightarrow t$$
Let the conditions be $c(e_1,e_3) \le 1, c(e_2,e_3) \le1$. There is a flow of size $2/3$ possible by letting $1/3$ flow through $e_1$ and $e_2$ and letting $2/3$ flow through $e_3$. There is no way to have a flow of size $1$, so there is no integer or half-integer flow which is at least as large as $2/3$.
The same idea works on a larger graph without a doubled edge.

Update: A slight modification of this counterexample has no intersecting biting pairs.
$$s \overset{e_1}{\underset{e_2} \rightrightarrows} v \overset{e_3} \rightarrow w \overset{e_4} \rightarrow t$$
Then $c(e_1,e_3) \le 1, c(e_2,e_4) \le 1$ has the same effect as in the previous example, and the maximum flow is also $2/3$.<|endoftext|>
TITLE: Rationality conditions for determining Galois groups
QUESTION [8 upvotes]: Let $F$ be a field and $h \in F[x]$ be an irreducible, degree $n$ monic polynomial. Let $G$ denote the Galois group of $h$.
It is well known that $G \subset A_n $ if and only if the discriminant of $h$, which we'll denote by $D(h)$, is a square in $F$. We could think of this as being a rationality condition: we are demanding an $F-$rational solution to the equation $y^2 = D(h)$. 
My question is if this is always possible for any subgroup $H \subset Sym(n)$. That is, does there exist a polynomial $f\in F[a_0,\ldots,a_{n-1}]$ in the coefficients of $h$ and a polynomial $\phi \in F(y) $ such that $G \subset H$ only if $\phi(y) = f$ has a solution in $F$? Is it possible to make this a sufficient condition also? (I suspect that the answer is yes to the former and no to the latter).
Further, if such a $\phi$ exists, can we control its degree? Is such a condition unique and if there are many, is there a simplest?      
Thanks!

REPLY [10 votes]: More or less, yes. Fix a transitive subgroup $H \subset S_n$. Let $S_n$ act in the usual way in the field $\mathbb{Q}(x_1,\ldots,x_n)$ where the $x_i$ are algebraically independent. Then the fixed field $\mathbb{Q}(x_1,\ldots,x_n)^{S_n} = \mathbb{Q}(a_1,\ldots,a_n)$ where $\prod(X-x_i) = \sum a_iX^{n-i}, a_0=1$. Now, write $\mathbb{Q}(x_1,\ldots,x_n)^{H} = \mathbb{Q}(a_1,\ldots,a_n,z)$, by the primitive element theorem and $z$ satisfies some equation $f(a_1,\ldots,a_n,z)=0$, for some polynomial $f$ whose degree can probably be figured out from the order of $H$. Now, given $\alpha_1,\ldots,\alpha_n \in F$, where $F$ is a field of characteristic zero, the splitting field of $\sum \alpha_iX^{n-i}, \alpha_0=1$ is contained in $H$ if and only if there is $\beta \in F$ with $f(\alpha_1,\ldots,\alpha_n,\beta)=0$. When $H=A_n, f=z^2-$ discriminant.<|endoftext|>
TITLE: Uniform bounds for the order of a rational torsion point on CM elliptic curves
QUESTION [7 upvotes]: Let $K$ be an imaginary quadratic field and $E$ an elliptic curve with CM by the maximal order of $K$, such that $E$ is defined over the Hilbert class field $H$. Is it known whether there is a bound (independent of the degree of $H$) on the order of a $H$-rational torsion point on $E$?

REPLY [5 votes]: Hey, I probably should have answered this one some time ago. It was proved in 1989 by J.L. Parish that the order of an $H$-rational torsion point is 1,2,3,4 or 6, and this also can be deduced from work of either Silverberg or Prasad-Yogananda. In any case the statement you want is at the beginning of section VI in the paper below and the proof is done in section V.
http://www.sciencedirect.com/science/article/pii/0022314X89900127<|endoftext|>
TITLE: Weyl inequalities for largest eigenvalue of matrix sum
QUESTION [6 upvotes]: The $k^{\rm th}$ largest eigenvalue (arranged in decreasing order) of the sum of two $N \times N$ Hermitian (real symmetric) matrices $\bf{A}$ and $\bf{B}$ can be stated using the Weyl inequalities as
$L_k \leq \lambda_k({\bf A} + {\bf B}) \leq U_k$
with the lower and upper bounds given by
$L_k = {\rm max}\left[\lambda_i({\bf A}) + \lambda_j(\bf B)\right]$ with $i+j=k+N$
$U_k = {\rm min}\left[\lambda_i({\bf A}) + \lambda_j(\bf B)\right]$ with $i+j=k+1$.
The bounds on the largest eigenvalue $\lambda_1$ of the matrix sum ${\bf A}+{\bf B}$ are therefore
$L_1 = {\rm max}\left[\lambda_i({\bf A}) + \lambda_j(\bf B)\right]$ with $i+j=1+N$
$U_1 = {\rm min}\left[\lambda_i({\bf A}) + \lambda_j(\bf B)\right]$ with $i+j=2$.
Finding the upper bound is very simple; $U_1 = \lambda_1({\bf A})+\lambda_1({\bf B})$ since there is only one solution to the index equation $i+j=2$. However, finding the lower bound is much more difficult; the maximum of
$\left[\lambda_1({\bf A})+\lambda_N({\bf B}),
\lambda_2({\bf A})+\lambda_{N-1}({\bf B}), \ldots, \lambda_N({\bf A})+\lambda_1({\bf B})
\right]$
must be found.
Are there any simplifications or theorems that can be applied to obtain a simpler expression for the lower bound (without restricting $\bf B$ to be a slight variation or permutation to $\bf A$) ?

REPLY [8 votes]: Weyl's inequalities are not the full story. The characterization of the possible spectra of $A+B$, given the spectra of $A$ and $B$ is the object of A. Horn's conjecture. This is  now a theorem, after hard works by Fulton, Klyachko, Knutson, Terry Tao and others. The conjecture consists in linear inequalities (the simplest of these being Weyl's) which are found recursively on the dimension $n$. After Weyl's inequalities, there are Ky Fan or Wielandt's inequalities, etc ...<|endoftext|>
TITLE: Finite dimensional homogeneous spaces of $Diff(S^1)$
QUESTION [5 upvotes]: This question is a refined version of Representations of infinite dimensional Lie algebras as vector fields on manifolds
I'm interested in the finite dimensional homogeneous spaces of $Diff(S^1)$. There are several papers concerned with infinite dimensional homogeneous spaces such as $Diff(S^1)/S^1$, but I can't find anything of the finite dimensional spaces, except the paper by Cartan from 1905, as hinted by Robert Bryant in the above link... From R.B's reply in the link: 
"You could try É. Cartan's papers on infinite pseudogroups (mostly appearing 1904-05). In particular, see Paragraph 57 of Sur la structure des groupes infinis de transformation (suite). There, for example, he proves that the (pseudo-)group of diffeomorphisms of the line has three distinct (i.e., nonequivalent) 2-dimensional homogeneous spaces and seven distinct 3-dimensional homogeneous spaces, etc."
Basically I'm interested in the two dimensional spaces... it's just that I don't know French, and the exposition in Cartan's paper seems quite brief anyway. Surely there must be more modern works about the subject?
Also, it's not very clear to me how to construct these homogeneous spaces as cosets...
EDIT: Here's an example from Cartan's 1905 paper (I still don't know French but I was able to decipher that much):
Suppose we have vector fields as a Lie algebra of $Diff(\mathbb R)$ (Cartan is doing it on the line, but the circle is similar), 
$l_n = x^{n+1} \partial_x$.
Cartan obtained corresponding vector fields on three dimensional homogeneous spaces of $Diff(\mathbb R)$. On one of the seven homogeneous spaces they are
$l_n = x^{n+1}\partial _x-(n+1) x^n y\partial _y-(n+1)(n+x z)x^{n-1}\partial _z$,
but what is the homogeneous space (by which I mean, can it be expressed as a quotient)?!
Here's a link to to the relevant pages in Cartan's paper: http://goo.gl/bJXfm
EDIT: Fixed a misunderstanding regarding Cartan's notation in the formula (see RB's answer below). I expanded the vector fields in a basis by Taylor expanding the function $f(x) = \sum c_n x^{n+1}$. That way it's easy to see that they both satisfy the Witt algebra, $[l_n, l_m] = -(n-m)l_{n+m}$...

REPLY [5 votes]: I'm afraid that you are mistranslating Cartan's notation.  The first 3-dimensional example that Cartan gives is this:  Every diffeomorphism of the line, written in the form $X = f(x)$, can be 'lifted' to a diffeomorphism of 3-space:
$$
X = f(x),\qquad Y = \frac{y}{f'(x)},\qquad Z = \frac{z}{f'(x)} - \frac{f''(x)}{f'(x)^2}\ .
$$
These diffeomorphisms act transitively on the complement of the plane $P$ defined by $y=0$.  This defines a homomorphism from $\textrm{Diff}(\mathbb{R}^1)$ to $\textrm{Diff}(\mathbb{R}^3\setminus P)$ that is Cartan's first example of a 3-dimensional homogenous space of $\textrm{Diff}(\mathbb{R}^1)$.  The corresponding homomorphic lifting of vector fields is given by
$$
\Phi\left(h(x)\frac{\partial\ }{\partial x}\right)
= h(x)\frac{\partial\ }{\partial x} -  h'(x)y \frac{\partial\ }{\partial y}
- \bigl(h'(x)z + h''(x)\bigr)\frac{\partial\ }{\partial z}
$$
(and not the formula that you gave above).
This particular example is actually a fibered product of two $2$-dimensional homogeneous spaces of $\textrm{Diff}(\mathbb{R}^1)$, namely, the actions
$$
X = f(x),\qquad Y = \frac{y}{f'(x)}\ .
$$
(you should think of this $\mathbb{R}^2$ as the bundle of $1$-forms on the line) and 
$$
X = f(x),\qquad Z = \frac{z}{f'(x)} - \frac{f''(x)}{f'(x)^2}
$$
(you should think of this $\mathbb{R}^2$ as the bundle of $0$-jets of affine connections on the line), fibered over their common $1$-dimensional homogeneous space $\mathbb{R}^1$, with the original action $X = f(x)$.  
Three of the seven $3$-dimensional homogeneous spaces of $\textrm{Diff}(\mathbb{R}^1)$ listed by Cartan are fibered products of a pair drawn from the three $2$-dimensional homogeneous spaces of $\textrm{Diff}(\mathbb{R}^1)$ in this way, and three of them are the spaces of $1$-jets of sections of the $2$-dimensional homogeneous spaces, when they are regarded as bundles of rank $1$ over $\mathbb{R}^1$ (the $1$-dimensional homogeneous space).  That leaves only one $3$-dimensional homogeneous space 'unexplained' in some geometric way.  (In Cartan's list, this is the sixth example.  It is some kind of natural bundle over the bundle of $0$-jets of affine connections on the line, but I don't yet see a simple way to describe it.)  
Cartan's exposition is brief, as you say.  This is typical of Cartan; after all, he is only working out an example of the general theory.  I'm not aware of any place in the modern literature that describes this particular classification, but, on the other hand, once you know how to read Cartan, his argument fairly straightforward to follow.
If there is interest, I can sketch out a paraphrase of Cartan's argument in currently fashionable language.<|endoftext|>
TITLE: Conjugacy class splitting in double cover of alternating, symmetric group
QUESTION [5 upvotes]: Let $n \ge 4$ be a natural number.
Consider the quotient map from the double cover $2 \cdot A_n$ of the alternating group $A_n$ to $A_n$. For any conjugacy class in $A_n$ of size $r$, the inverse image in $2 \cdot A_n$ has size $2r$. This inverse image is either a single conjugacy class or "splits" as a union of two conjugacy classes each of size $r$. Which case occcurs depends on the original conjugacy class we chose.
I want a combinatorial criterion (see Note 2) on the cycle type of the conjugacy class in $A_n$ that helps determine which of these cases occur (actually, there may be one or two conjugacy classes in $A_n$ of a given cycle type, but even if there are two, they will behave the same way).
I'm also interested in the corresponding question for the double cover of $S_n$. Note that the double cover is not unique (there are two possibilities for each $S_n$). However, the behavior for fibers over conjugacy classes that I'm interested in should not depend on the choice of double cover (if my understanding is correct -- I haven't worked out a formal proof).
Note 1: I'm including the case of $A_n$ for $n = 4$ just so people know what I am talking about:
$A_4$ has conjugacy classes of sizes 1,3,4,4 (the two conjugacy classes of size 4 fuse in $S_4$ and correspond to cycle type 3 + 1, whereas the conjugacy class of size 3 corresponds to cycle type 2 + 2, with representative permutation $(1,2)(3,4)$).
Its double cover is isomorphic to $SL(2,3)$ and the quotient map can be viewed as a quotient map $SL(2,3) \to PSL(2,3)$. The fibers over the conjugacy classes of sizes 1,4,4, each split in two conjugacy classes. The fiber over the conjugacy class of size 3 remains a single conjugacy class of size 6, i.e., the fiber over this conjugacy class does not split.
Note 2: By "combinatorial criterion" I mean a criterion like the criterion we have for a conjugacy class of even permutations in $S_n$ to split in $A_n$: it splits if there is any cycle of even length or two cycles of equal odd length. 
So, I want a criterion that looks at the cycle type as an unordered integer partition, then uses the combinatorial and number-theoretic structure of that partition to determine splitting in the double cover.

REPLY [5 votes]: I believe the answer is that the class of an element of $g \in A_n$ becomes a single class when lifted to the 2-fold central covering group of $A_n$ if and only if $g$ has even order and $g$ has (at least) two cycles of the same length (including cycles of length 1).
The class of an element of $g \in S_n$ lifts to a single class in either of the two covering groups of $S_n$ if and only if either the condition for $g \in A_n$ holds, or $g$ has an even number of cycles of even length.
Let $\hat{S}$ and $\hat{A}$ be covering groups of $S_n$ and $A_n$. Let $z$ be the central element of order 2 in $\hat{S}$. To avoid too many hats, I will denote the inverse image in $\hat{S}$  of $g \in S_n$ also by $g$.
I believe that to study these covering groups, all you really need to know is that if $g$ and $h$ are disjoint transpositions in $S_n$, such as $(1,2)$ and $(3,4)$, then $[g,h] = z$ in $\hat{S}$. Also, for two commuting pairs of transpositions in $A_4$, like $g=(1,2)(3,4)$ and $h = (1,3)(2,4)$, we have $[g,h]=z$ in $\hat{A}$.
Clearly, if $g$ has odd order $k$ in $A_n$ or $S_n$, then its class lifts to two classes in $\hat{A}$ or $\hat{S}$, one of order $k$ and one of order $2k$, so we only need consider elements $g$ of even order.
If $g$ has two cycles of equal even length, such as $(1,2,3,4)(5,6,7,8)$ (+ other cycles) then we can take $h = (1,5)(2,6)(3,7)(4,8)$ and get $[g,h] = z$ in $\hat{A}$, so the class of $g$ lifts to a single class in $\hat{A}$. I had some difficulty proving that, but if it was false, then you could show that the centralizer of $h$ in $\hat{A}$ mapped on to the full centralizer of $h$ in $A_n$, which is not true, because the image of the centralizer does not contain $(1,2)(5,6)$.
On the other hand, if $g$ has even order and has two cycles of the same odd length, such as $g = (1,2)(3,4,5,6)(7,8,9)(10,11,12)$ with $n=12$, then we can take $h = (1,2)(7,10)(8,11)(9,12)$ and get $[g,h] = z$ in $\hat{A}$.
But if all cycles of $g$ have different lengths, then elements $h$ of the centralizer of $g$ in $A_n$ are made up of powers of those cycles, and since $h$ must have an even number of cycles of even length, we always get $[g,h]=1$ in $\hat{A}$.
However, in $S_n$, if $g$ has an even number of cycles of even length, like $g=(1,2)(3,4,5,6)$ with $n=6$, then we can take $h=(1,2)$ and get $[g,h]=z$ in $\hat{S}$.
Let me know if I have got something wrong, or if you would like more detail!.<|endoftext|>
TITLE: what can be reached by flat degeneration of (globally) complete intersection?
QUESTION [7 upvotes]: Let $X\subset\mathbb{P}^n$ be a (globally) complete intersection, let $(X_t)_{t\in\mathbb{C}^1}$ be a flat family, with $X_1=X$. Which types of schemes can we get as $X_0$?
Or, conversely, which (embedded, projective) schemes deform to complete intersections?
e.g. some non-ACM schemes (not arithmetically Cohen-Macaulay) deform to C.I.'s. Does every pure-dimensional subscheme of $\mathbb{P}^n$ deform to a C.I.?  Two immediate obstructions are the degree and the arithmetic genus. Any other necessary conditions?
(Just for housekeeping, a somewhat related question on deformations of C.I.:)
upd: I meant the family over the germ $(\mathbb{C}^1,0)$, so there are no complications with the geometry of the base.

REPLY [5 votes]: Anything with the same Hilbert polynomial as a globally complete intersection. This is true because two fibers of the same flat projective family have the same Hilbert polynomial, and because the Hilbert scheme is connected. EDIT: Since the base of the flat family is required to be irreducible, the projective scheme must be in the same irreducible component of the Hilbert scheme as a complete intersection. I don't know any nice characterization of when this is the case.
The Hilbert polynomial of a complete intersection is easy to calculate using the Koszul complex. The coefficients will be some set of polynomials in the degrees of the hypersurfaces you completely intersect. You can set them equal to the actual coefficients of the Hilbert polynomial and check if the system of equations has a positive integer solution.
Computationally, you can just check all possible factorizations of the degree.
For curves, degree, arithmetic genus, and dimension of the embedding are the only invariants needed. The last one is important. For instance, a genus $5$, degree $8$ curve is the complete intersection of three degree $2$ hypersurfaces in $\mathbb P^4$, but not the complete intersection of any pair of hypersurfaces in $\mathbb P^3$.<|endoftext|>
TITLE: TQFT and Mapping Class Groups
QUESTION [8 upvotes]: It is well known how could we get a representation of the mapping class group of a surface S (which I assume compact, connected and orientable), given a TQFT. My question is: is there any reference talking about the inverse step: how could we get a TQFT given a representation of the mapping class group of a surface S? How should I think this problem?

REPLY [6 votes]: The following paper answers your question precisely, I think: arxiv:1408.0668
Specifically Theorem 1.3 (which is elaborated on in Section 4) describes precisely what additional data you must specify to construct a TQFT in addition to representations of the MCG of all genera. Roughly, it is the ability to "glue" together representations of different genera in a coherent manner.<|endoftext|>
TITLE: commuting elements in a reductive group
QUESTION [7 upvotes]: Does anyone know if the following holds?
Conjecture: Any two commuting elements in a reductive algebraic group G over C of rank>1 lie in a proper parabolic subgroup of G.
To make things easier, you can assume that these elements are semi-simple.
Note that if G is simply-connected then the centralizer of any semi-simple element is connected. This implies that if both are semi-simple then they lie in a torus and hence the conjecture holds.
On the other hand, the rank assumption is important since the diagonal orthogonal matrices D(-1,1,-1) and D(1,-1,-1) commute in SO(3,C) but one can prove that they don't belong to the same parabolic subgroup.

REPLY [9 votes]: Angelo has indicated the most classical type of counterexample involving two semisimple elements of $\mathrm{PGL}_n$, but the question warrants some further discussion.   Though it is posed (and maybe motivated?) over $\mathbb{C}$, the natural setting is a connected reductive algebraic group $G$ over any algebraically closed field.   If you start with two commuting elements, one of them can be viewed as lying in the centralizer of the other: a closed but not necessarily connected subgroup.    So centralizers are involved here in an essential way.    Working with arbitrary elements may get rather complicated, since for example the study of centralizers of unipotent elements tends to require case-by-case study.
So consider just two commuting semisimple elements $s,t$ of $G$.   The question is when they both lie in a proper parabolic subgroup.    If $G$ is just a torus, there are no nontrivial proper parabolics.   Otherwise $G$ has a  nontrivial semisimple derived group.  Now we might as well assume $G$ itself is semisimple (of rank at least 1), because the center of a reductive group lies in every maximal torus and hence in every parabolic subgroup.
From the standard structure theory of $G$ (Chevalley, Borel-Tits, Springer, Steinberg), the structure of the centralizer $C := C_G(t)$ of any semisimple element $t$ is understood: $C$ is reductive of maximal rank and is connected whenever $G$ is simply connected but otherwise sometimes not.   In general any semisimple element lies in some maximal torus of a given reductive group.   As a result (see Springer-Steinberg reference in my comment above), when $G$ is simply connected both $s$ and $t$ lie in a maximal torus of $C$ (hence of $G$; this in turn lies in a Borel subgroup, which is a proper parabolic subgroup of $G$.
On the other hand, when $G$ fails to be simply connected, as in Angelo's example (for $n>1$), Springer and Steinberg observe that the group $C$ will typically not be connected for some $t$ (taking some care in prime characteristic about the separability of the universal covering map for $G$).   Thus if $s$ lies outside the identity component of $C$, one may have trouble fitting both $s$ and $t$ into a proper parabolic subgroup (though I'm not sure how to construct counterexamples other than case-by-case).    In Angelo's case, the problem is that the permutation matrix chosen represents the longest element of the Weyl group, which by the Bruhat decomposition won't lie in such a parabolic along with $t$.
ADDED: To respond to Adam's last sentence (where "proper parabolic" is intended), rank 1 is not the problem but rather the fact that his group fails to be simply connected.    His special orthogonal group (Lie type $B_1 = A_1$) is just a matrix realization of the 3-dimensional irreducible representation of $\mathrm{SL}_2(\mathbb{C})$: the adjoint representation using an orthonormal basis of the Lie algebra relative to its Killing form.   This adjoint group is isomorphic to $\mathrm{PGL}_2(\mathbb{C})$, while his commuting pair of semisimple elements is a special case of Angelo's example.   Here the centralizer of one element corresponds to the group of monomial matrices, with a Weyl group representative outside the identity component.   This disconnected group fails to lie in any Borel subgroup, here the only type of proper parabolic.<|endoftext|>
TITLE: Reference request: 2-dimensional Schonflies theorem
QUESTION [5 upvotes]: Does anyone know a reference for the 2-dimensional version of the Schoenflies theorem?  To be precise, I'd like a reference for the fact that every continuous, 1-1 map $S^1\rightarrow \mathbb{R}^2$ extends to a homeomorphism $\mathbb{R}^2 \rightarrow \mathbb{R}^2$.  The discussions of the Jordan Curve Theorem that I can remember don't prove this stronger statement.
This statement is mentioned on the Wikipedia page for the  Schoenflies problem .  I looked through several papers on the generalized Schoenflies problem (which requires extra hypotheses in higher dimensions to rule out things like the Alexander Horned Sphere), but no luck...

REPLY [2 votes]: Thomassen's paper on triangulating surfaces addresses this as well. 
See: Triangulating surfaces<|endoftext|>
TITLE: Perspective on the diamond lemma in ring theory
QUESTION [11 upvotes]: As the title indicates, I am seeking some perspective (explained below) on Bergman's diamond lemma (DOI link to Bergman's paper) in ring theory.
I know that it can be used, for instance, to prove the Poincaré-Birkhoff-Witt theorem for universal enveloping algebras.  This is shown in Bergman's original paper.  And then it can be used also for quantized enveloping algebras, quantized function algebras, etc.
I am also aware of this page at Secret Blogging Seminar, which discusses the diamond lemma in the graph-theoretic setting.  It is stated there that this version can be used to prove the Jordan-Hölder theorem on composition series for a finite group (and presumably for modules over a ring, etc).
For this to be a well-formulated question, I need to make clear what I mean by "perspective".  Here is a quote from the introduction to the paper:

The main results in this paper are trivial. But what is trivial when described in
the abstract can be far from clear in the context of a complicated situation where
it is needed.

And then, later, after describing the results, he says:

This fact has been considered obvious and used freely by some ring-theorists
(e.g., [17, Sect. !5l), but others seem unaware of it and write out tortuous
verifications.

So, what I am looking for is examples of situations where the formalism of the diamond lemma really clarified or simplified some algebraic construction, or examples of people writing out "tortuous verifications" when they could have appealed to the diamond lemma instead.
PS  If you've never looked at Bergman's paper before, you should!  It's one of my favorites.  Everything is so clear and well-motivated.  If only I could write like that...

REPLY [2 votes]: What I totally did not have in mind when writing my first answer to this in 2012 is that the confluence condition of the Diamond Lemma can be interpreted quite nicely in terms of the Maurer-Cartan equation in the resolution of the corresponding monomial algebra, see https://arxiv.org/abs/2010.14792 .<|endoftext|>
TITLE: Is the Turing equivalence relation the orbit equiv. relation of the action of a countable group?
QUESTION [5 upvotes]: The Turing equivalence relation on $\cal P(\mathbb{N})$ is defined by $A\equiv_T B$ iff $A\leq_T B$ and $B\leq_T A$.  This is a countable Borel equivalence relation on the polish space $\cal P(\mathbb{N})$.  And as stated, the answer to my question is yes since a theorem by Feldman and Moore says: for any countable Borel equivalence relation $E$ on a standard Borel space $X$, there is a countable group $G$ and a Borel action of $G$ on $X$ such that $E$ is the orbit equivalence relation of the action.
But from where I have looked, the Turing equivalence isn't the orbit e.r. of any countable group that is "found in nature".  I was wondering if anything else was known about this, as well as any other instances of equiv. relations that are provably orbit equivalences, but with no known "natural" group doing the acting.  It seems like for many important equivalences (like $E_0, E_{\infty}$, hyperfinite e.r's) we do in fact have a natural group that is acting, but perhaps that is the reason why we know so much about them.
Any references or other information would be great, thanks.

REPLY [6 votes]: Any countable Borel equivalence relation can arise from a Borel action of the free group on $\omega$ generators, which is a natural enough group. You should probably ask instead whether the action can be natural. Can the action be free, for example, as Joel discusses.
The field of countable Borel equivalence relations has gained a tremendous amount from studying equivalence relations arising from natural group actions. However, for the case of Turing equivalence, viewing this relation as arising from the Borel action of a countable group does not appear to be a helpful way of thinking.
You might be interesting in the following open question (from my thesis). The question essentially asks whether the Turing reductions are the unique way to generate Turing equivalence:
Suppose that $\{\psi_i\}_{i \in \omega}$ is any countable collection of partial Borel functions closed under composition such that $x$ and $y$ are Turing equivalent if and only if there exists an $i$ and a $j$ such that $\psi_i(x) = y$ and $\psi_j(y) = x$. Suppose also that $\{\varphi_i\}_{i \in \omega}$ is an enumeration of the Turing reductions. Then must there be a pointed perfect set $P$ and a function $u: \omega^2 \rightarrow \omega^2$ such that for every pair of Turing reductions $\varphi_i$ and $\varphi_j$, if $x,y \in P$ and $\varphi_i(x) = y$ and $\varphi_j(y) = x$, then $\psi_k(x) = y$ and $\psi_l(y) = x$, where $u(i,j) = (k,l)$?
It is probably optimistic to hope that the answer to the above question is "yes", when we have no idea how to approach this problem. However, beyond the question being nice in and of itself, a positive answer would have a large number of consequences. It would imply Martin's conjecture on Turing invariant functions is true, for example, which in turn would have a lot of consequences for the theory of countable Borel equivalence relations. 
A positive answer would also imply any universal countable Borel equivalence relation must be able to witness its universality uniformly, regardless of the way it is generated. A nice implication of this would be that any increasing union of non-universal countable Borel equivalence relations must be non-universal.<|endoftext|>
TITLE: What is the tropical Robinson-Schensted-Knuth correspondence?
QUESTION [8 upvotes]: And what are it's applications? A conceptual explanation would be great! Is there an expository note about this somewhere? 
Some references have already appeared in the answers and comments below. To make the question more specific, classical RSK has combinatorial interpretations in terms of symmetric functions, for example. If RSK gives the Cauchy identity:
$$\sum_{\lambda} s _{\lambda}(x)s _{\lambda}(y)=\prod _{i,j} \frac{1}{1-x_iy_j}$$
what is an analogous interpretation for tropical RSK? (From some buzzwords I've heard in a few talks recently, it probably has something to do with shifted or elliptic Schur functions.)

REPLY [12 votes]: I think the clearest write up is in Danilov and Koshevoy. Let me try to get you off on the right foot by explaining what it is we are proving.
RSK (for these purposes) is a bijection between $n \times n$ nonnegative integer matrices and pairs of SSYT of the same shape, filled with entries from $1$ to $n$. By the bijection between SSYT and Gelfand-Tsetlin patterns, this is the same as pairs of GT-patterns of size $n$ with the same bottom row. Remember that a GT-pattern is a triangular array of numbers so, if we glue these two GT-patterns together along their common entries, we get a square. 
Thus, we have encoded both sides of the RSK correspondence as $n \times n$ arrays of integers obeying certain inequalities. On one side, the inequality is that the entries are nonnegative, on the other side we have the GT inequalities. Then the key facts are the following: 

This map is piece-wise linear.
As a piece-wise linear function, it is (once you get all the coordinates right) the tropicalization of the LU factorization map: The birational map which takes an $n \times n$ matrix and writes it as a product of a lower triangular and an upper triangular matrix.

REPLY [3 votes]: I believe this paper is the starting point for this topic:
MR1872253 (2003j:05128)
Kirillov, Anatol N.
Introduction to tropical combinatorics.<|endoftext|>
TITLE: Has the cotangent complex been used in context other than morphism of schemes?
QUESTION [5 upvotes]: Here is what I know about the history of the cotangent complex:
Quillen did it over a point (i.e. for morphism of rings), Illusie did it in a topos (i.e. for sheaves of rings in a topos). And proved general theorems on deformation/obstructions. People have been using that to study various deformation problems of schemes. E.g. constructing obstruction theory. There are also generalization to 1-morphisms to stacks or to the logarithmic world. It seems to be a crucial object in the derived algebraic geometry setup.
My question is, has the cotangent complex used in other context with the full power of topoi? I understand that it is related to the solution of Serre's conjecture and the definition Andre-Quillen cohomology. I also understand to "glue" the affine construction is hard and you really need the power of topoi. I just want to know if this has been used in general on a morphism of rings in a topos. Or more generally, for a morphism of ringed topoi. 
(I hope this question is not too vague, a more "concrete" example might be: does infinitesimal thickening of the first order of the fppf topos of a scheme $X$ (by a module in it) over that of a scheme $Y$ contains some interesting information?)
Just to be clear, by topos I always mean a Grothendieck topos, i.e. a category equivalent to the category of sheaves of sets over a site.

REPLY [6 votes]: One does not actually need the "power of topoi" to make Illusie's definition of the cotangent complex work.  One does need sheaves, though.  Illusie's constructions would have worked equally well had he used ringed spaces instead of ringed topoi.  Some constructions related to diagrams of schemes might be easier to phrase in terms of topoi, and of course they become essential if one wants to work in, say, the étale topology.
Illusie makes use of the generality of his construction in a number of places.  By treating ringed topoi that are not the étale or Zariski topoi of schemes he is able to treat the deformation theory of diagrams, which he applies to deformations of group schemes, commutative group schemes, and modules.
I do not know if it would be useful to study deformations of the structure sheaf on the fppf topos of a scheme $S$.  Such a deformation would correspond to a consistent collection of deformations of every scheme over $S$.<|endoftext|>
TITLE: Commutativity of the fundamental group of any Lie Group 
QUESTION [10 upvotes]: How do we formally prove that the fundamental group of any Lie group is always commutative?

REPLY [17 votes]: Geometric proof: A connected Lie group $G$ is homotopy equivalent to a maximal 
compact subgroup, so we may assume $G$ is compact. Being compact, $G$ admits a bi-invariant 
Riemannian metric with respect to which it is a symmetric space, the symmetry $s$ at the 
identity being just the inversion map. Now a homotopy class in $\pi_1(G,1)$ can be represented
by a closed geodesic $\gamma$ (of minimal length in its homotopy class, by a shortening process). Since the differential of $s$ at $1$ is minus identity, $s$ sends $\gamma$ to itself
parametrized backwards. It follows that the homomorphism induced by $s$ on the $\pi_1$-level is inversion. However, the inversion map in a group is a homomorphism if and only if the 
group is Abelian.<|endoftext|>
TITLE: computational complexity of primitive recursive functions
QUESTION [6 upvotes]: If we have a rewrite system for primitive recursive functions, which simplifies each term according to how the function was defined, then what is the computational complexity of this calculation? That is, what is the coplexity of the normalization procedure? I have heard a claim that for a closed term calculating the value of the function requires transfinte induction up to $\epsilon_0$. Is this true and where can I find a proof of this? 
For example in (Schwichtenberg & Wainer 2012) there is a lemma which says that a primitive recursive function is computable in $F_{\alpha}$-bounded time, for some $\alpha<\omega$, where the set of all $F_{\alpha}$ for $\alpha<\epsilon_0$ is the Fast Growing Hierarchy. 
Is the measure of transfinite induction related to this way of bounding the complexity?

REPLY [3 votes]: Dear Andreas,
yes indeed. We have the follwing (not too difficult) fact: Let S^m(0) be the numeral for $m$.
For any PRA term $t(x_1,\ldots,x_n)$ The derivation length
of $t(S^{m_1}(0),\ldots,S^{m_n}(0))$ will be bounded by a primitive recursive
function (depending on $t$) with arguments $m_1,\ldots,m_n$. (Here I assume
standard rewrite systems for modelling primitive recursion.)
For terms from Gödel's T the derivation lengths become more complex depending
on the typelevel of the terms in question. 
My general conjecture is that 
for typical (not too small) subrecursive classes C the derivation lengths functions
for terms in C will not leave C.
The result for primitive recursion is not difficult whereas the result for Gödel's T is
somewhat involved.
For PRA there is also a tradeoff in terms of termination orderings.
Termination for rewrite systems for prim rec functions can be shown 
by the multiset path ordering. And termation proofs for the multiset
path ordering lead to prim rec derivation lengths (result by Dieter 
Hofbauer). The corresponding result for multiple recursion is one
of my earliest results in term rewriting theory. Wilfried Buchholz 
gave a nice prooftheoretic proof for these results in APAL.
There is also some nice application of term rewriting to more involved schemes of primitive
recursive functions. For example it can be used to show that prim rec functions
are closed under parameter recursion, or simple nested recursion, or even
unnested multiple recursion.
Best, 
Andreas<|endoftext|>
TITLE: Certain functional equations for the Riemann Zeta function? 
QUESTION [16 upvotes]: Referring to this question I asked on math.SE.
I am posting a more generalized question here, for answers and further inquiry.

For the Riemann zeta function, we know of the standard functional equation that relates $\zeta(s)$ and $\zeta(1-s)$. I wanted to know whether there are functional equations that relates $\zeta(s)$ and $\zeta(s+1)$, or $\zeta(s)$, $\zeta(s+1)$, and $\zeta(s+2)$ or in general $\zeta(s)$, $\zeta(s+1)$, $\zeta(s+2)$, ..., $\zeta(s+n)$ for $\Re(s) > 0$ and for $n \in \mathbb{N}$?

My main motivation behind asking this question is I have found such an equation, but I do not know whether such an equation exists in literature. Also, I do not want to appear as if I am promoting my formula here, but rather I am more interested in the works that have been done in such directions.
My formula goes as follows: For any $n \in \mathbb{N}$ and $\Re(s) > 0$ we have
$$\frac{1}{s-1} + \sum_{r=1}^n \binom{n}{r} (-1)^r \left(\frac{\zeta(s+r-1)}{s+n-1} + \int_1^\infty\frac{\rho(x)^r}{x^{s+r}}\mathrm{d}x\right) = 0$$
where $\binom{n}{r}$ is the binomial coefficient and $\rho(x)$ is the fractional part of $x$ 
For example, putting $n=1$ we get the well known identity,
$$\frac{1}{s-1} - \frac{\zeta(s)}{s} - \int_1^\infty\frac{\rho(x)}{x^{s+1}}\mathrm{d}x = 0$$
putting $n=2$,
$$\frac{1}{s-1} - 2\left(\frac{\zeta(s)}{s+1} + \int_1^\infty\frac{\rho(x)}{x^{s+1}}\mathrm{d}x\right) + \left(\frac{\zeta(s+1)}{s+1} + \int_1^\infty\frac{\rho(x)^2}{x^{s+2}}\mathrm{d}x\right) = 0$$
and so on...
EDIT: Classic answer by Juan. This question is now solved.

REPLY [15 votes]: Equations of this type are known. You may see, for example, the classical
book "Primzahlen" by Landau paragraph 67. "Continuation of the zeta function by partial integration"
There it is proved the formula
$$ (s-1)(\zeta(s)-1)-1=-\frac{(s-1)s}{2!}(\zeta(s+1)-1)-
\frac{(s-1)s(s+1)}{3!}(\zeta(s+2)-1)-\cdots$$
$$\cdots-\frac{(s-1)s\cdots(s+q)}{(q+2)!}(\zeta(s+q+1)-1)
-\frac{(s-1)s\cdots(s+q+1)}{(q+2)!}\int_1^\infty \frac{\rho(x)^{q+2}}{x^{s+q+2}} dx.$$
From this we may get the beautiful formula
$$1=\binom{s-1}{1}(\zeta(s)-1)+\binom{s}{2}(\zeta(s+1)-1)+\binom{s+1}{3}(\zeta(s+2)-1)+
\cdots$$
(See Titchmarsh 2.14, formula (2.14.1) and also (2.14.2))
From the formula in Landau you see that the integral 
$\int_1^\infty \frac{\rho(x)^r}{x^{s+r}}\,dx$ can be expressed in terms of the 
$\zeta(s+k)$.<|endoftext|>
TITLE: Touching-tetrahedra graphs
QUESTION [5 upvotes]: Have the graphs representable by touching tetrahedra been explored?
Let $\cal T$ be a collection of tetrahedra in $\mathbb{R}^3$
with pairwise disjoint interiors.
Define a graph $G_{\cal T}$ to have a node for each tetrahedron
in $\cal T$, and an arc between $T_1$ and $T_2$ if those two
tetrahedra share one or more boundary points:
$T_1 \cap T_2 \neq \varnothing$.
[Added]
I neglected to add the significant qualification (of most
interest to me) that each arc of $G_{\cal T}$
should be able to be associated with a unique point. My oversight only became clear
with Aaron's example.  Apologies!  Of course one can make many definitions (such as Igor's variation), and all are of interest.

Which graphs $G$ are equal to $G_{\cal T}$ for some $\cal T$?

For example,
$K_6$ is a touching-tetrahedra graph:

          


In contrast, responses to an earlier MO question,
"Extensions of the Koebe–Andreev–Thurston theorem to sphere packing?"
showed that $K_6$ is not a ball-touching graph.
I know triangle-touching graphs in $\mathbb{R}^2$ have been
studied, often called triangle contact representations,
e.g., the recent paper by
Gonçalves, Lévêque, and Pinlou,
"Triangle Contact Representations and Duality."
But I haven't found literature on the generalization to tetrahedra.
I would be interested in any pointers to the literature,
or classes of graphs that either are or are not
touching-tetrahedra graphs.  E.g., is $K_7$ a touching-tetrahedra graph?
Thanks!

REPLY [2 votes]: That is an impressive $K_6$ graphic but any complete graph can be realized in a less exciting manner: take $n$ thin triangles in the plane with a single common vertex. This shows that $K_n$ is a planar triangle contact graph and hence a tetrahedral contact graph. The paper you reference requires at most one shared point per pair of triangles. It goes on to study planar networks getting simultaneous triangle representations of the graph and planar dual with pleasing properties. Maybe more restrictions would lead to interesting questions. Graphs with no $3$-cycles might be a challenge.<|endoftext|>
TITLE: covering disks with smaller disks
QUESTION [11 upvotes]: How many disks with radius 1/2 are needed to cover a disk with radius 1? It certainly cannot be done with less than 5 small disks, and some non-rigorous drawings of mine suggest it can be done with 7 small disks. Can it be done with less? 
Slightly more generally, can anybody point me in the direction of any work on covering bounded bodies with disks? Am i right in thinking that covering disks with smaller disks is the most inefficient kind of covering?

REPLY [5 votes]: Erich Friedman's packing center claims that you can't cover with 6 disks, and that this was proved by Károly Bezdek in 1979.  If you want a more exact reference, ask Erich Friedman in email.<|endoftext|>
TITLE: Haar measure for profinite groups (reference needed)
QUESTION [5 upvotes]: I was wondering if anybody knows a good reference book or exposition for Haar measures over profinite groups (with some concrete examples and computations)?

REPLY [5 votes]: Bill Casselman has a very nice set of notes.<|endoftext|>
TITLE: Chromatic convergence of E(n)-localized homotopy categories
QUESTION [7 upvotes]: Given the Chromatic Convergence Theorem, can we state this globally as some convergence in the category of categories? That is, we have the subcategories of finite spectra in each stable homotopy category of $E(n)$-localized spectra, say $\mathcal{L}_n^{fin}$.  Is it correct that there should be "inclusion" functors $\mathcal{L}_n^{fin}\to\mathcal{L}_{n+1}^{fin}$? Can we say that if we take some kind of limit (maybe this should be phrased in terms of quasi-categories or something?) we get the homotopy category of finite spectra?

REPLY [8 votes]: As Fernando showed, you do have a functor from the category of finite spectra to an appropriate limit of the categories of $E(n)$-local finite spectra, and this recovers the mapping space between any pair of finite spectra:
$$
{\rm Map}(Y,X) \simeq {\rm holim}_n {\rm Map}(L_{E(n)} Y, L_{E(n)} X).
$$
However, this can't be made into an equivalence of categories because some non-finite objects show up as limits of finite objects.
For each $n$, choose a finite type $n$ complex $X_n$ whose $n$'th Morava $K$-theory is nontrivial (hence the $m$'th Morava $K$-theory is nontrivial for all $m \geq n$).  Being of type $n$ implies that $L_{E(k)} X_n \simeq *$ for $k < n$.  We therefore get a system of objects
$$
Y_n = L_{E(n)} (X_0 \vee X_1 \vee \dots \vee X_n)
$$
together with natural weak equivalences $L_{E(n-1)} Y_n \to Y_{n-1}$ induced by projecting off the factor $X_n$.  This assembles into an object in an appropriate homotopy limit category.
However, there is no finite complex $Y$ such that $L_{E(n)} Y \simeq Y_n$ for all $n$, because the number of cells necessary to build $Y_n$ grows in an unbounded fashion.  You can check this by observing that the $n$'th Morava $K$-theory of $Y_n$ has rank at least $n$ because each wedge factor contributes rank at least 1.<|endoftext|>
TITLE: classification of cyclotomic fewnomials
QUESTION [6 upvotes]: This is inspired by this question.
 Suppose I give you an integer $k.$ Is there a classification of cyclotomic polynomials with exactly $k$ nonzero terms? For example, if $k=1$ there are not any, if $k=2,$ you get just the standard $x^n-1$ polynomials, if $k=3,$ I assume you only get the squares of the standard polynomials, for $k=4$ presumably you only get the products of distinct standard cyclotomics, but after that it becomes a bit puzzling...
Edit by cyclotomic polynomial I mean one all of whose roots are roots of unity.

REPLY [2 votes]: Some of the following is mentioned in the OEIS references given by Robert:
Let $a_n$ be the number of terms in $\phi_n(t)$ , the $n$th cyclotomic polynomial. Then $a_n=a_m$ where $m$ is the product of the prime divisors of $n$ because $$\phi_n(t)=\phi_m(t^{n/m}).$$ So we only need to figure out $a_m$ for $m$ square-free.
For odd $n$ $a_{2n}=a_n.$
For odd prime $p$, $a_p=p$ and, if $q \ne p$ is prime, then $$a_{pq}=\frac{2(p-u)(uq+1)}{p}-1$$ where $1 \le u \le p-1$ and $uq \equiv -1 \mod p.$ This should allow one to determine all cases where $a_n=k$ and $n$ has at most two distinct odd prime factors. Quite a bit more can be said in this special case, the non-zero coefficients are all $\pm 1$ and one can say exactly where they appear.
I'm not sure just how much is known about $a_{pqr}$ for $p,q,r$ distinct odd primes. Perhaps lower bounds can be given. Here are a few cases: $[105, 33], [165, 57], [195, 59], [231, 57], [255, 73], [273, 99], [357, 125], [385, 177], [429, 135]$$ [455, 189], [561, 107], [595, 253], [663, 129], [715, 213], [935, 429], [1001, 321]$ So it would seem that perhaps for up to $30$ terms or so everything can be said, but that is just a guess on my part.<|endoftext|>
TITLE: Computational complexity of Knot polynomials
QUESTION [6 upvotes]: What's known about computational complexity of different types of knot invariant polynomials?
For example, Evaluating Jones Polynomial is known to be #P hard.
Is there any reference that surveys such complexity results on other knot polynomials?

REPLY [4 votes]: The old (1990) paper, "On the computational complexity of the Jones and Tutte polynomials" (Cambridge link), shows that determining the Jones polynomial of alternating links is #P-hard, as the OP notes.
Then, much later, the 2012 book Quantum Triangulations (eds.: Carfora, Marzuoli), says this (p.233):
 
See the Wikipedia entry on BPQ for a definition: essentially, solvable in polynomial time on a quantum computer, with bounded error probability. It is conjectured that BPQ $\supset$ P.
In the same book, there follows a section entitled "Efficient Quantum Processing of Colored Jones Polynomials," with several references.
(Added: All of this circa 2012 when originally posted.)<|endoftext|>
TITLE: Covering a unit ball with balls half the radius
QUESTION [32 upvotes]: This is a direct (and obvious) generalization of the recent MO question, "Covering disks with smaller disks":

How many balls of radius $\frac{1}{2}$ are needed to cover completely a ball of radius 1?

The answer may be in the paper "Covering a Ball with Smaller Equal Balls in $\mathbb{R}^n$,"
by Jean-Louis Verger-Gaugry, which I cannot immediately access (Springer link here).
If anyone knows the answer for $\mathbb{R}^3$, I'd be curious to learn of it.
And it would be especially interesting if there were a proof as satisfying as  Noam Elkies's for seven half-disks covering one in $\mathbb{R}^2$.  Thanks!
Update. [29May2012] A series of contributions by Will Jagy, Gerhard Paseman, Karl Fabian, and zy, have reduced the upper bound from $56$ to $33$ to $22$, with a lower bound of $16$.
Update. [5Aug2012] Ed Wynn settled the question with a careful analysis: $21$ balls are needed, and suffice!

REPLY [24 votes]: Here is (I'm fairly sure) an optimal solution, building on the ideas in other answers.
We know that we can generate maximum 30° caps on the sphere, by placing the half-balls at radius sqrt(3)/2.  We deduce from http://neilsloane.com/coverings/index.html that we will need 20 half-balls to cover the outer surface, plus a 21st to cover the centre.  So, 21 is a lower bound.
To get a 21-ball solution, scale the 20 coordinates of http://neilsloane.com/coverings/dim3/cover.3.20.txt to that radius, and add one at the origin.  Note that the balls at sqrt(3)/2 give caps on the central half-ball with the same angle as the outer caps, so the cover works on the inside too. 
Here are the coordinates of 21 half-balls to cover the ball: {{+0.060111,-0.479945,-0.718359},{-0.559060,+0.562759,+0.347496},{-0.164091,-0.848680,-0.053063},{-0.509173,-0.375273,-0.591535},{-0.140963,+0.208121,+0.828743},{+0.690970,+0.512659,-0.098697},{+0.377263,-0.700146,+0.342736},{-0.250648,-0.504068,+0.658097},{-0.743210,+0.007958,+0.444494},{+0.550076,+0.074555,-0.664724},{+0.521328,-0.588831,-0.362624},{+0.088527,+0.791576,+0.339956},{-0.158465,+0.221391,-0.822116},{-0.718586,-0.473046,+0.099309},{-0.339530,+0.726217,-0.327609},{+0.439846,-0.237215,+0.707294},{+0.853710,-0.140538,+0.037800},{+0.536789,+0.335659,+0.590924},{+0.243253,+0.709202,-0.433429},{-0.778146,+0.197643,-0.324694},{+0.000000,+0.000000,+0.000000}}

There is some "fat" in the surface covers (using 30° caps when 29.6..° is required), and each non-central sphere bulges convexly out of the cone that it's responsible for, so it's easy to believe that this is indeed a cover.  Also I've checked it thoroughly, though I'll be grateful if someone checks it with less makeshift methods.  At the positions stated, the radius of the balls can be as low as 0.49812.  By moving the original coordinates further inwards, to radius 0.8595, the radius of the balls can be as low as 0.49439.

Ed Wynn, 4 August 2012.

[Graphic added by J.O'Rourke:]<|endoftext|>
TITLE: Is a complete homogeneous symmetric polynomial irreducible?
QUESTION [18 upvotes]: Let $S=\mathbb{C}[x_1,x_2,\dots,x_n]$ be a polynomial ring. Let $n \geq 3$. Let $h_a$ denotes the complete homogeneous symmetric polynomial of degree $a$.
$$ h_a=\text{ sum of all monomials of degree } a.$$ 
For example: for $n=3$ and $a=2$, one has: $$h_2=x_1^2+x_2^2+x_3^2+x_1x_2+x_1x_3+x_2x_3.$$
Question: Is it true that  $h_a$ is an irreducible element in $\mathbb{C}[x_1,x_2,\dots,x_n]$. 
The $h_a$ was introduced by Sir Issac Newton in seventeenth centuary along with many other symmetric polynomials such as Power sum symmetric polynomials and elementary symmetric polynomials. 
It is known that $p_a=x_1^a+\cdots+x_n^a$ is an irreducible element in $\mathbb{C}[x_1,x_2,\dots,x_n]$ for $n \geq 3$. I am interested to know similar result for the complete homogeneous symmetric polynomial.
Thank you
Neeraj Kumar.

REPLY [17 votes]: EDIT : prompted by Will Sawin's comment, the argument now works for every $n \geq 3$. Thanks !
The polynomial $h_a(x_1,\ldots,x_n)$ is irreducible for every $a \geq 1$ and $n \geq 3$.
Recall that if $h_a = FG$ with $F$ and $G$ non constant then $F$ and $G$ have to be homogenous. By Bézout's theorem, the hypersurfaces $F=0$ and $G=0$ intersect in the projective space $\mathbf{P}^{n-1}(\mathbf{C})$ since $n \geq 3$. This gives a singular point on the hypersurface $h_a=0$. So it suffices to prove that $h_a,\frac{\partial h_a}{\partial x_1},\ldots,\frac{\partial h_a}{\partial x_n}$ have no common zero in $\mathbf{C}^n \backslash \{0\}$. This fact is true for every $a \geq 1$ and $n \geq 2$, and we prove this by induction.
For $a=1$ it is easy. For $n=2$ it amounts to the fact that the polynomial $T^a+\cdots+T+1 = (T^{a+1}-1)/(T-1)$ has distinct roots.
In general, we have
$$h_a = \sum_{a_1+\cdots+a_n=a} x_1^{a_1} \cdots x_n^{a_n}$$
so that 
$$\frac{\partial h_a}{\partial x_i} = \sum_{a_1+\cdots+a_n=a-1} (a_i+1) x_1^{a_1} \cdots x_n^{a_n}.$$
Note that $\sum_{i=1}^n \frac{\partial h_a}{\partial x_i} = (a+n-1) h_{a-1}$. Moreover $h_a=x_i h_{a-1}+R$ for some polynomial $R$ not depending on $x_i$, so that
$$\frac{\partial h_a}{\partial x_i}=h_{a-1}+x_i \frac{\partial h_{a-1}}{\partial x_i}.$$
If $x=(x_1,\ldots,x_n)$ is a common zero of $h_a$ and all its partial derivatives then $h_{a-1}(x)=0$ and $x_i \frac{\partial h_{a-1}}{\partial x_i}(x)=0$ for all $i$. By induction, we must have $x_i=0$ for some $i$. Assume for example $x_n=0$. Then $(x_1,\ldots,x_{n-1}) \in \mathbf{C}^{n-1}$ provides in fact a common zero of $h_a(x_1,\ldots,x_{n-1})$ and all its partial derivatives, so applying the induction hypothesis for $n-1$ we get $x=0$.<|endoftext|>
TITLE: Mathematician, Graciano Ricalde
QUESTION [26 upvotes]: Does anyone understand more precisely how to explain the 5th degree equation and elliptic functions accomplishments of Mathematician Graciano Ricalde?  I am his great grand-daughter and trying to accurately list his achievements on Wikipedia.  I would greatly appreciate your expert guidance.  Thank you kindly in advance for any/all responses. 
http://en.wikipedia.org/wiki/Don_Mauro_Graciano_Ricalde_Gamboa

REPLY [36 votes]: Some investigation with Wikipedia and Google books reveals the following.  
First of all, as noob explains, there is the mathematical problem that has spanned centuries of solving polynomial equations.  The problem has several interpretations, one of which is to find higher-degree versions of the quadratic formula
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
for the solutions to the quadratic equation
$$ax^2 + bx + c = 0.$$
Such solutions were found for polynomial equations of degree 3 and 4 by Cardano and Ferrari in the 16th century.  For polynomial equations of degree 5 and more, it was proven by Abel and Galois that the solution cannot be expressed in terms of radicals (i.e., expressions involving $\sqrt[n]{z}$ for several values of $n$ and $z$).  Hermite instead found a solution involving certain functions in advanced calculus that are called elliptic functions.  (They generalize trig functions like $\sin z$ and which are indeed related to elliptic curves and originally to the problem of calculating the perimeter of an ellipse.)  His solution was simplified and clarified by Kronecker and Klein.  Most of this work was before Graciano Ricalde was active as a mathematician.

Enter Ricalde.  He corresponded in a French question-and-answer journal called L'Intermédiaire des mathématiciens (which reads a lot like MathOverflow!).  In one entry that I think was published in 1898, he asked about solving the quintic equation --- he didn't solve it himself.  His question was about reducing the general quintic equation
$$ax^5 + bx^4 + cx^3 + dx^2 + ex + f = 0$$
to the special form
$$y^5 + py + q = 0,$$
which is called Bring-Jerrard form.  (In other words, Bring and Jerrard showed how to eliminate three of the terms in the quintic equation.)  Obtaining Bring-Jerrard form is a complicated part of the problem, similar to Ferrari's solution to the quartic equation, but the hardest part is to solve the Bring-Jerrard equation; that's the part that requires elliptic functions.
Much later there was a long laudatory biography of Ricalde in a regional publication called Enciclopedia Yucatanense.  This biography was cited in Spanish Wikipedia, which apparently led to the English Wikipedia entry.  It credited him with solving the quintic equation with elliptic functions, and I think some other major achievements as well.  (Although it's a bit hard for me to tell, since I only get a snippet view.)  However, in modern mathematical literature, Ricalde is barely cited for anything.  The only citation that I found in the arXiv was in a review by Lemmermeyer concerning another algebra problem called Pell's equation.  (Lemmermeyer includes Ricalde in a long list of people who solved special cases of Pell's equation, citing the 1901 issue of L'Intermédiaire des mathématiciens.)
My guess is that a local biographer of Ricalde took some topics that Ricalde studied to be Ricalde's own achievements.  Moreover, that Ricalde was a competent or aspiring research mathematician in his time, but not a standout.<|endoftext|>
TITLE: Double dual of C*-algebra
QUESTION [5 upvotes]: Suppose, that $A$ is a $C^{\ast}$-algebra. It follows, according to Sakai's book, that double dual of $A$ is also $C^{\ast}$-algebra. I'm not quite sure if I understand the proof correctly. Author proves that $A^{\ast \ast}=\pi(A)''$ where $\pi$ is universal representation showing first that 
the corresponding preduals coincide. But preduals ale only Banach spaces, so the question rises: how do we define multiplication and involution in the second dual of $A$? I've heard that can be done in natural way (not using the universal representation), using so called Goldstein theorem but I don't know what precisely this theorem states. If we follow this 'natural' procedure via Goldstein theorem and simultaneously use universal representation 
we get that double dual of $A$ and $\pi(A)''$ are isomorphic as BANACH SPACES-but is it true that they are isomorhic as $C^{\ast}$-algebras (in general it seems that this need not to be true as $\ell^{\infty}$ and $L^{\infty}[0,1]$ are isomorhic as Banach spaces but not as $C^{\ast}$-algebras). 

Thank You for Your answers-still there is one detail which I don't understand, namely-we have isomorphism of Banach spaces $T:A^{\ast \ast} \to \pi(A)''$, this isomorphism preserves multiplication when restricted to $A$ and $A$ is dense in $A^{\ast \ast}$ as well as $T(A)$ in $\pi(A)''$. Morever $T$ is $\ast$ weak continuous. But it seems to me that multiplication IS NOT continuos with respect to weak $\ast$ topology, so I'm not sure if we have enough information to conclude that $T$ preserves multiplication in general ( for example preserving $\ast$ structure can be proved via Goldstine-we define $x^{\ast}$ for $x \in A^{\ast \ast}$ as $\lim_{i}a_i^{\ast}$ where $a_i \to x$ is suitable net. As involution $\ast$ on $A$ is continuos with respect to $\ast$ weak topology, therefore it can be extended to whole $A^{\ast \ast}$ and $T$ preserves this operation-how to repeat this argument for the multiplication if we have lack of continuity?)

REPLY [5 votes]: As Yemon says, it is the Arens product on the double dual of a Banach algebra that you're thinking of. Yes, $A^{**}$ and $\pi(A)''$ are isometrically isomorphic as Banach spaces and $*$-isomorphic as C*-algebras. This is surely in Sakai's book that you're already looking at.
A related question that you didn't quite ask is whether two C*-algebras which are isometrically isomorphic as Banach spaces must be $*$-isomorphic as C*-algebras. The answer is no --- there are C*-algebras which are not $*$-isomorphic to their opposite algebras (change the order of the product and leave everything else, including norm, alone). However, isometric C*-algebras will have the same state space and this implies that they are Jordan isomorphic, and roughly speaking, switching the order of the product is the worst thing that can happen. See the book State Spaces of Operator Algebras by Alfsen and Schultz.
The "right" version of this result is that two C*-algebras that are completely isometric must be $*$-isomorphic.<|endoftext|>
TITLE: Submitting papers “out of logical order” due to different review times or rejections
QUESTION [17 upvotes]: I am asking this question anonymously and apologize for the paranoia.
Suppose you submit a paper on a topic and, while the paper is under review, continue to work on the topic and make enough progress to prepare a second paper.  I know that submitting the second paper while the first is still under review is typical, but what do you do if the first paper is rejected around the time the second is submitted?  Similarly, what if the second paper is accepted and then the first paper is rejected?  Do you resubmit the first?
Note that the second paper doesn’t simply duplicate the results of the first in a more general setting; it relies on the results of the first and uses different techniques to extend them.
I know there are cases where the right thing to do (if the timing makes it possible) is to combine the papers into one.  But suppose this would be inappropriate, because the combined paper would be too long or too cluttered with different ideas.
So, specifically:

1. If the first paper is rejected right as you are getting ready to submit the second, do you consider it unethical or otherwise a bad idea to submit the two related papers around the same time (to different journals)?  This of course assumes the rejection was not based on issues with correctness of the paper.  Is it worth waiting a few months to submit the second paper to avoid an appearance of publishing “least publishable units?”  If not, should you explain it to the two editors with the submissions (more than just sending a copy of the first along with the second to help the referee)?
2. If the first paper is rejected AFTER the second is accepted, does the first just end up never being published?  Or is it ok to continue to submit the first paper, whose results have already been used and improved upon (though they are required in the improvement)?

REPLY [12 votes]: The solution is preprints. Once you publish a preprint (through either arXiv or your own university server, although I recommend arXiv), you have a source to cite in the second paper and you can forget about the need of having it published before the other.<|endoftext|>
TITLE: Etiquette question: how to acknowledge Bugs Bunny?
QUESTION [13 upvotes]: Suppose that a mathematician such as Bugs Bunny answers one of my math questions here on MathOverflow, and then I use the idea in a research paper.  How should I acknowledge such a distinguished mathematician in my paper, submitted to a reputable journal and all that?

REPLY [21 votes]: Look right underneath where your question is posted.  Click on the "cite" link.  It will pop up a text box, from which you can copy citation data for this post in either bibtex or amsrefs format.  So you can directly cite the MO thread.<|endoftext|>
TITLE: Metric deformations from non-negative to positive curvature
QUESTION [9 upvotes]: Is it possible to deform the metric $g$ of a closed Riemannian manifold $(M,g)$ satisfying $\mathrm{Ricci}(M,g) > 0$ and $\mathrm{sec}(M,g) \geq 0$ to a metric $g_1$ satisfying $\mathrm{sec}(M,g_1) > 0$?
I understand that this question is ludicrous, at best (for instance, an affirmative would prove that $S^n\times S^m$ always carries a metric of positive sectional curvature). I'm assuming that there is a counterexample, but I can't seem to figure it out. Most examples of manifolds with positive Ricci curvature that I know can't admit metric of non-negative sectional curvature for topological reasons, for instance the Sha--Yang construction of metrics with positive Ricci curvature on connected sums of $S^n \times S^m$.
I was wondering about this in view of the following two results: The first, due to Aubin and Ehrlich, asserts that a metric $g$ with $\mathrm{Ricci}(M,g) \geq 0$ everywhere and $\mathrm{Ricci}(M,g) > 0$ somewhere can be deformed into a metric $g_1$ with $\mathrm{Ricci}(M,g_1) > 0$ everywhere.
The second, due to Gao--Yau, asserts that if $\mathrm{Ricci}(M,g) > 0$ everywhere,
then $g$ can be deformed into a metric $g_1$ with $\mathrm{Ricci}(M,g) > 0$ everywhere and $\mathrm{sec}(M,g) > 0$ somewhere.
The Gao--Yau result is a local solution to the intial question.
As a related question: Suppose the Ricci curvature of $(M,g)$ is pinched. Does this change the answer to the question if the pinching constant is particulary small or would it seem that pinching plays no role in this? Is there any theory of pinched Ricci curvature?
Aubin, T., Métriques riemanniennes et courbure, J. Differ. Geom. 4, 383-424 (1970). ZBL0212.54102.
Ehrlich, Paul, Metric deformations of curvature. I: Local convex deformations, Geom. Dedicata 5, 1-23 (1976). ZBL0345.53024.
Gao, L. Zhiyong; Yau, S. T., The existence of negatively Ricci curved metrics on three manifolds, Invent. Math. 85, 637-652 (1986). ZBL0603.53025.

REPLY [14 votes]: As Benoît Kloeckner points out this is false for non simply connected manifolds with $RP^2\times RP^2$ being a counterexample (by Synge's theorem). For simply connected manifolds this is a well known  open problem. 
BTW, Sha-Yang examples are only known not to admit nonnegative sectional curvature for connected sums of things like $S^n\times S^m$ for very large number of summands (by Gromov's betti number estimate). For, say, connected sum of 3 copies of $S^n\times S^m$ with $n,m>1$ nothing is known.
Also, there are plenty of easier examples of manifolds of positive Ricci curvature other than those of Sha and Yang. For example  homogeneous spaces and more generally biquotients of compact Lie groups with finite fundamental groups. All of them have positive Ricci curvature and nonnegative sectional curvature but almost none are known to admit positive sectional curvature.
Moreover, there are lots of such examples with quasi-positive curvature ( where sectional curvature is positive on a dense open set of points in $M$) and the question is still open for such manifolds. See for example this paper by Wilking "Manifolds with positive sectional curvature almost everywhere."
Among many other examples Wilking constructs such metrics on $S^2\times S^3$. This shows that even in the simply connected case either the Hopf conjecture (that a product of positively curved manifolds can not be positively curved) is false or the deformation conjecture for quasi-positively curvaed manifolds is false.<|endoftext|>
TITLE: Why is symplectic geometry so important in modern PDE ?
QUESTION [38 upvotes]: First, we recall that symplectic manifold is a smooth manifold, $M$, equipped with a closed nondegenerate differential 2-form, $\omega$, called the symplectic form. The study of symplectic manifolds is called symplectic geometry. In Hörmander's classic book ALPDO (The analysis of partial differential operators Ⅰ-Ⅳ) he wrote: symplectic geometry pervades a large part of the modern theory of linear partial differential operators with variable coefficients. And he had devoted the entire chapter ⅩⅩⅠ to discuss it. 
Now, with some basic background (its origins in the Hamiltonian formulation of classical mechanics), I want to know further that why it would make such a important role in modern analysis (such as fourier integral operators)?
Thanks in advance.

REPLY [4 votes]: As the cotangent space has a natural symplectic structure, we can also phrase the question as "Why is cotangent space so important in modern pde?". A nice answer for this question is given in http://math.berkeley.edu/~mjv/WhereforeCot.pdf.<|endoftext|>
TITLE: On the concept of point in category theory
QUESTION [8 upvotes]: In principle, many different abstractions of the set-theoretic notion of point / element are available in the framework of categories, but they are not equally effective and, what is more interesting to me, there doesn't seem to exist one of them subsuming all the others (hence to be preferred in every possible occurrence). Since this is a key issue in my current research on structures, I would appreciate much if 

you could provide me with references to existing literature about various interpretations of the notion of point in category theory and, if possible, a (short) historical overview of its evolution 

(I don't think it's really necessary to stress that its refers to the notion of point in category theory, and not to the theory of categories - in fact, I'm not really stressing it). Those which I'm already aware of are listed below.

Say that $\bf C$ is a category and $A$ an object of $\bf C$. Then:

If $\bf C$ has a terminal object $\top$, a point of $A$ is any arrow $a$ to $A$ such that $\mathrm{src}(a) \cong \top$ (I use ${\rm src}(\cdot)$ for the source map of $\bf C$). This notion is fine, for instance, if you work with $\bf Set$ (sets + functions), $\bf Top$ (topological spaces + continuous functions), $\bf Pre$ (preordered sets + monotonic functions), etc, but not equally effective with algebraic categories such as $\bf Grp$ (groups + group homomorphisms), $\bf Rg$ (semirings + semiring homomorphisms) and ${\bf Mod}_R$ (left modules over a ring $R$ + homomorphisms of left modules over $R$), basically due to the the fact that, for each of these, terminals are zero objects [it doesn't even work well with $\bf Sgrp$ (semigroups + semigroup homomorphisms) but for other reasons]. A point in the sense of the present definition is usually called a global element
A pont of $A$ is any morphism to $A$. In this case, one commonly speaks of generalized elements. I cannot profit from this notion, and I won't spend one more word on it.
If $\bf C$ is a closed category and $I$ is one of its units (unique up to iso), a point of $A$ is any morphism $a$ to $A$ with ${\rm src}(a) \cong I$. I have no clue how such points are usually referred to. This generalizes Definition 1 in some relevant cases (notably including closed monoidal categories), but $\bf Top$ is not closed monoidal, and let me guess that it is not even closed (I've not checked).
If $\mathcal{G}_1(\mathbf{C})$ is the class of the $1$-generators of $\bf C$, a point of $A$ is any arrow $a$ to $A$ with $\mathrm{src}(a) \in \mathcal G_1(\mathbf{C})$. I don't know how this kind of points are called in the literature.

I would be especially interested in variants, generalizations or alternatives to the previous definitions involving monomorphisms (as for the first one in the list) rather than arbitrary arrows. Thank you in advance for any help.

REPLY [8 votes]: 5.
There is a variation of the generalized elements, called members in Mac Lane's book CWM. Namely, two morphisms $x,y$ with codomain $A$ are identified when there are epis $u,v$ such that $xu=yv$. Clearly this is symmetric and reflexive; in an abelian category it is also transitive. Mac Lane uses the notation $x \in_m A$. Then some properties of members are established, which are used to prove diagram lemmas in arbitrary abelian categories. In my opinion, this is far more efficient and enlightening than proving them via Freyd-Mitchell. And for me this is a perfect translation of the "element calculus" in abstract abelian category.
For other categories, you won't have a general answer what the best "element calculus" might be. It really depends on the context. I strongly agree with  François G. Dorais' comment above. A better question would be: Given the problem xyz, how can I define elements in context abc in order to solve xyz?<|endoftext|>
TITLE: The resultant of an arbitrary polynomial and a cyclotomic polynomial
QUESTION [6 upvotes]: This is a natural generalization of this question.
Let $f$ be a monic irreducible polynomial over $\mathbb Z$. Let $S_f$ be the set of natural numbers $n$ such that one of the three equivalent conditions hold:
a. There exists some prime $p$ such that a root of $f$ in $\bar{\mathbb F}_p$ is also a primitive $n$th root of unity.
b. The ideal in $\mathbb Z[x]$ generated by $f$ and the $n$th elementary cyclotomic polynomial is not the unit ideal.
c. The resultant of $f$ and the $n$th elementary cyclotomic polynomial is not $\pm 1$.
Is $S_f$ cofinite for all $f$ with Mahler measure larger than $1$?
What I know:
It is easy to check that $S_f$ is infinite, since for each prime a root of $f$ is the $n$th root of unity for some $n$, but each prime can only show up finitely many times for a fixed $n$, since the resultant is nonzero and thus has finitely many prime factors.
If $f$ has no complex roots of norm $1$, then $S_f$ is cofinite. This is an extension of the argument in the linked mathoverflow question. The resultant of $f$ and the $n$th elementary cyclotomic polynomial can be computed by inclusion-exclusion from the resultants of $f$ and $x^{n/d}-1$, for $d$ squarefree dividing $n$. Since the resultant of $f$ and $x^n-1$ is $a^{n+o(1)}$ for $a>1$ equal to the Mahler measure of the $f$, the result of $f$ and the elementary cyclotomic polynomial is $a^{\Phi(n)+O(2^k)}$, where $k$ is the number of primes dividing $n$. Since $2^k/\Phi(n)$ is small for large $n$, the exponential is larger than $1$ for large $n$, therefore the resultant equals $1$ only finitely many times.
But this argument fails if a root is norm 1. In fact, purely analytic methods of estimating the resultant should fail in this case. There are bad points on the unit circle that get too close to roots of unity  too often. To prove $S_f$ cofinite, you would need to use some algebraic method to prove that those points are not algebraic integers.

REPLY [7 votes]: For an algebraic number $\gamma$ which is not a root of unity, Baker's theorem gives
a bound (uniform in $n$) of the form $|\gamma^n - 1| > n^{-C}$ for some constant $C$
(only depending on $\gamma$).
In particular, if $s_n:=\prod |\alpha^n_i - 1|$, then Baker's method gives the following estimate uniform in $n$ (for $\alpha_i$ not a root of unity):
$$n \log \mathcal{M}(\alpha) + A \ge \log|s_n| \ge n \log \mathcal{M}(\alpha) - A - B \log|n|,$$
where $\mathcal{M}(\alpha)$ is the Mahler measure of $\alpha$. Proof: for $|\alpha_i| > 1$,
one has $\log|\alpha^n_i - 1| \sim n \log|\alpha_i|$ up to $O(1)$, which
gives rise to the term $n \log \mathcal{M}(\alpha)$;
 the term $\log|\alpha^n_i - 1|$ for $|\alpha_i| \le 1$ is trivial to bound from above
and can be bound from below by Baker's Theorem. The logarithm of the Mahler measure is
the sum of $\log|\alpha_i|$ for the roots $|\alpha_i|> 1$. By a theorem of Kronecker this sum is positive if $\alpha$ is not a root of unity.
OTOH, if $\Phi_n(x)$ is the $n$th cyclotomic polynomial and
$t_n:=\prod |\Phi_n(\alpha_i)|$, then we deduce that
$\sum_{d|n} \log(t_n) = \log(s_n)$, and hence
$$\log(t_n) = \sum_{d|n} \log(s_{n/d}) \mu(d) \ge \varphi(n) \log \mathcal{M}(\alpha) - 
d(n)(A + B \log(n)),$$
where $d(n)$ is the number of divisors of $n$.  The bounds $\phi(n) \gg n^{1-\epsilon}$ and
$d(n) = n^{\epsilon}$ easily give the asymptotic relation
 $$\log(t_n) \sim \varphi(n)  \log \mathcal{M}(\alpha) \gg 1$$
 as $n$ goes to infinity.
FWIW, the bounds of Baker (and the other bounds used above) are effective, so for any
particular $\alpha$ one could in principle find all $n$ with $t_n = 1$.
(n.b. Gelfond's estimate would give $\epsilon \cdot n$ instead of $B \log|n|$ as an error term,
which is enough to show that $s_n \rightarrow \infty$ but not $t_n$.)<|endoftext|>
TITLE: Is the singular homology of a real algebraic set always finitely generated?
QUESTION [13 upvotes]: Here is a precise statement of my question:
Let $p\in \mathbb{R}[x_1, \ldots, x_n]$ be a polynomial, and let $Z(p)\subset \mathbb{R}^n$ be the set of zeros of $p$.  Must the singular homology $H_i (Z(p); \mathbb{Z})$ ($i\geq 0$) be finitely generated as an abelian group?  
Here I really just mean the singular homology groups of this set as a topological space with the Euclidean topology.
It's an old theorem of Whitney that $Z(p)$ has finitely many connected components, so $H_0 (Z(p); \mathbb{Z})$ is finitely generated.  Note that it is possible to triangulate $\mathbb{R}^n$ with $Z(p)$ as a subcomplex, so $H_i (Z(p)) = 0$ for $i>n$.
I'm guessing the answer is well-known (either a theorem or a counterexample), but I couldn't find an answer on Google or MathSciNet...

REPLY [10 votes]: Proposition 1.13 of these notes by Coste implies (if I've read it correctly) that any semialgebraic set $S$ is homeomorphic to a union $U$ of open simplices in some finite simplicial complex $K$.
Thus $S$ is an open subset of an ENR, hence is an ENR. Its singular cohomology therefore coincides with its Čech cohomology, which is finitely generated. Therefore its singular homology must be finitely generated.
Edit: The argument given above is incomplete, since a union of "open" simplices in a simplicial complex is not necessarily open (an easy mistake to make!). However it is easy to see that $S\subseteq\mathbb{R}^n$ is an ENR by other means (in particular, it is locally compact and locally contractible).<|endoftext|>
TITLE: What is the relation between Quasicrystals, Riemann Hypothesis, and PV numbers?
QUESTION [16 upvotes]: Could somebody explain to me, from a mathematical stand-point, what is a quasi-crystal, and how it relates to the set of Pisot numbers, and the Riemann Hypothesis?
I've heard Freeman Dyson say that the zeros of the Riemann zeta function form a quasi-crystal. But, a priori, I do not see what kind of property of the zeros, that we currently now of, would be able to confer to them more structure than to a random set of isolated numbers. 
(Notwithstanding the explicit formula in prime number theory)
To wit, my second question possibly based on a misunderstanding: why is the set of zeros of $\zeta(s)$ a quasi-crystal, while a random sequence of isolated numbers is not? Of course, I first need to fully understand what is a quasi-crystal, because Freeman's definition left me in a fog.

REPLY [19 votes]: Freeman Dyson's proposal is online, based on a talk he gave at MSRI.
Lillian Pierce's senior thesis gives a summary of Peter Sarnak's program to use properties of Gaussian Unitary Ensemble to study the zeros of the Riemann Zeta function.
N. G. Debrujin wrote about Penrose tilings and their Fourier transforms.

Crystalline structures on the line are pretty boring.   They are just evenly spaced lattices, like $\mathbb{Z}$, which might appear on different scales.
--o---o---o---o---o---o---o--
---o-----o-----o-----o-----o-
However, there are many quasi-periodic structures on the line, for example $\lfloor n\sqrt{2}\rfloor
= \{ 1, 2, 4, 5, 7, 8, 9, 11, 12, 14,\dots  \}$ which we can draw on the line.
--o--o-----o--o-----o--o--o-----o--o-----o--  
Many of these have special recursive properties.  Consider the line $y = \frac{1 + \sqrt{5}}{2} x$ which Golden ration slope.  Mark "0" if it crosses a horizontal line and "1" if for a vertical line.  You get the Fibonacci Word

Of course in 2D you get more interesting quasicrystals, which have interesting number theoretic and recursive structures.

Freeman Dyson wishes the zeros of the Riemann Hypotheses have structure like these.<|endoftext|>
TITLE: Is Mertens function negatively biased?
QUESTION [5 upvotes]: A comment in OEIS about Mertens function claims:

The graph seems to show a negative bias for the Mertens function which is eerily similar to the Chebyshev bias (described in A156749 and A156709). The purported bias seems to be empirically approximated (by looking at the graph) $- (6 / \pi^2)  (\sqrt{n} / 4)$  where $6 / \pi^2 = 1 / \zeta(2)$ is the asymptotic density of squarefree numbers (the squareful numbers having Moebius $\mu$ of 0). This would be a growth pattern akin to the Chebyshev bias. - Daniel Forgues, Jan 23 2011

Let $M(n)$ be the Mertens function.

Q1 Is there a bias (negative or positive) in Mertens function?
Q2 Is there a provable bias in some generalizations of Mertens function, e.g. in number fields or in arithmetic progression $M(an+d)$?

Plot of $M(x)$ with coloured bias:

(source)

REPLY [5 votes]: Somehow I missed this question when it was originally asked. I'm not entirely sure what you mean by a negative bias; the bits you've highlighted in the graph of $M(x)$ aren't when $M(x)$ is negative, but rather where is is, in some sense, "decreasing on average", and I don't know how to formalise that notion. If you're actually interested in the set of $x$ for which $M(x)$ is negative, then it seems clear from the graph that this happens roughly half the time. Here's how to make that notion formal.
In Nathan Ng's paper linked in Micah Milinovich's answer, Ng (very conditionally!) proves the existence of a limiting logarithmic distribution of $M(x)/\sqrt{x}$, so that there exists a measure $\nu$ satisfying
$$\lim_{X \to \infty} \frac{1}{\log X} \int_{1}^{X}{f\left(\frac{M(x)}{\sqrt{x}}\right) \ \frac{dx}{x}} = \int_{\mathbb{R}}{f(x) \ d\nu(x)}$$
for every continuous bounded $f : \mathbb{R} \to \mathbb{R}$; equivalently, for every Borel $B \subset \mathbb{R}$ whose boundary has $\nu$-measure zero,
$$\lim_{X \to \infty} \frac{1}{\log X} \int\limits_{\{x \in [1,X] : M(x)/\sqrt{x} \in B\}}{ \ \frac{dx}{x}} = \nu(B).$$
Furthermore, he calculates the Fourier transform of $\nu$ explicitly and shows, among other things, that $\widehat{\nu}$ is even about the origin. This implies that
$$\lim_{X \to \infty} \frac{1}{\log X} \int\limits_{\{x \in [1,X] : M(x) < 0\}}{ \ \frac{dx}{x}} = \nu((-\infty,0)) = \nu((0,\infty)) = \lim_{X \to \infty} \frac{1}{\log X} \int\limits_{\{x \in [1,X] : M(x) > 0\}}{ \ \frac{dx}{x}} = \frac{1}{2}.$$
To show this, we need to know that the set $\{0\}$ has $\nu$-measure zero, but this follows as Ng's explicit formula for $\widehat{\nu}$ is in $L^1(\mathbb{R})$, and hence that $\nu$ is absolutely continuous with respect to the Lebesgue measure on $\mathbb{R}$. (I don't think Ng actually includes this argument, but it's in my paper that kolik linked in his answer.)
So this shows (conditionally) that the median on $\nu$ is $0$, and hence that $M(x)$ is unbiased, in the sense that the logarithmic density of the set of points where $M(x)$ is negative is the same as that of the set of points where $M(x)$ is positive.
Interestingly, if you consider instead the weighted sum
$$M_{1/2}(x) = \sum_{n \leq x}{\frac{\mu(n)}{\sqrt{n}}},$$
then there is a negative bias; this follows from the same methods in my article that kolik linked to, together with the explicit expression
$$M_{1/2}(x) = \frac{1}{\zeta(1/2)} + \sum_{\rho}{\frac{x^{\rho - 1/2}}{(\rho - 1/2) \zeta'(\rho)}} + R(x),$$
where the sum is over the nontrivial zeroes of the Riemann zeta function, and $R(x)$ is some small error term. Other than this, I don't know much about other biases, though it should be the same in number fields, and I recently wrote a paper about this kind of thing in function fields, based on previous work of Byungchul Cha (in which he does not explicitly state the analogous result of there being no bias, though it is clear from the results).
Edit: I forgot to mention Brent and van de Lune's recent paper, where they look at a form of the Lambert series generated by $\mu(n)$ and show that it is negative as $x$ tends to $1$ from below (as juan mentioned in a comment above). But this really isn't telling you anything other than that the Riemann zeta function is negative at $s = 0$, which is a much weaker statement than, say, a pole whose residue is negative (as is the case with the summatory function of the Liouville function).<|endoftext|>
TITLE: A question about embedding hyperbolic space onto pseudosphere
QUESTION [6 upvotes]: I have a difficulty with hyperbolic geometry. 
Let $\mathbb{H}^{2}$ be a 2-dimensional hyperbolic plane.
(i.e., upper half plane in $\mathbb{R}^{2}$ with a metric $\frac{ds}{y}$)
(or, upper half plane in $\mathbb{C}$ with a metric $\frac{|dz|}{\textrm{Im}(z)}$ ) 
You may have heard about pseudosphere in $\mathbb{R}^{3}$. Let's denote the half-pseudosphere by $P$
This can be obtained by glueing both side($x=0$ and $x=2\pi$ parts) of 
$\left[(x,y)\in\mathbb{H}^{2} : 0\leq x\leq 2\pi, y>1 \right]$
Denoting this quotient space by $A$(Note that $A$ is homeomorphic to a cylinder), we can now get a "globally isometrically embedding" map $\rho:A \rightarrow \mathbb{R}^{3}$ with $\rho(A)=P$
To be specific, $\rho(x,y) = (t-\tanh(t), \textrm{sech}(t)\
\cos(x), \textrm{sech}(t)\sin(x))$ where $t=\textrm{arccosh}(y)$
Now the question I have is following 
: Is there another possible (global and isometric)embedding $\rho$ from $A$ into $\mathbb{R}^{3}$ ? 
Actually I'm interested in $\rho(A)$ and by calculating, one can easily find that if $\rho(A)$ is a "surface of revolution", then it should be $P$(up to isometry of $\mathbb{R}^3$"
Thus I'm looking for $\rho(A)$ which is different from a surface of revolution.
Any idea?

REPLY [7 votes]: You should be looking at the theory of Bäcklund transformations for surfaces of Gaussian curvature $K=-1$.  There is a large literature on this, and there are many examples of pseudospherical immersions that are not surfaces of revolution.  You should especially look at the work of Chuu-Lian Terng in this area.  She, together with her husband, Richard Palais, have developed this theory quite a bit in the past 30 years, and they have some excellent graphics for these surfaces.  Whether they can answer your specific question about isometrically embedding your particular domain $A$ into $\mathbb{R}^3$, I don't know, but that would be a good place to start.<|endoftext|>
TITLE: What is the multiplicity of a Cartier divisor at a point？
QUESTION [6 upvotes]: In the proof of nonvanishing theorem,  people used this concept, but I cant see its definiton.
The usual definition can be found in books which only deal with two dimensional case. I never see a definition on higher dimension. It seems that there are at least three descriptions concerned with multiplicity. 
Let $D$ be an effective Cartier divisor on a variety $X$, let $P\in D$. Pick a local equation $f$ of $D$.
(1) Indeicating by Hartshorne exercise V.3.4, one can define the mulitiplicity of the Noetherian local ring $\mathscr{O}_{X,P}/f$ by Hilbert-Samuel polynomial.
(2) One can define $\operatorname{mult}_P D:=\operatorname{ord}_P(f)=\max\{n\in\mathbb{N};\ f\in \mathfrak{m}^n\}$.
(3) Set $\operatorname{mult}_PD=\min\{C.D;\ C \mbox{ is a curve through P}\}$.
Does these three description coincides?  Where can I find a detailed treatment of basic notions like these? Thank you!

REPLY [2 votes]: See Fulton, intersection theory, section 4.3. more generally for a closed subscheme X of a pure dimensional scheme Y, and an irreducible component V of X, he defines the multiplicity of Y along X at V. When X and Y are varieties, this number is simply the coefficient of [X] in s(X,Y), the Segre class of the closed subscheme. When Y is an irreducible divisor and X is a codim 1 subvariety, blow up Y along X with projection p. Then $p_\ast$ (proper pushforward of cycles) of the exceptional divisor equals [X] times a number which is the multiplicity $e_XY$. When Y is not irreducible, just apply the above formula to each irreducible component and its intersection with X, then add up the numbers weighted by the geometric multiplicities of each component in Y.
More generally (e.g. when codim $n$ of X exceeds 1) you have the formula (X,Y varieties)
$$e_XY[X]=(-1)^{n-1}p_\ast(\tilde X^n)$$
and you extend to reducible Y in the same way. Here $\tilde X^n$ denotes the nth self intersection of the exceptional divisor of the blowup of Y along X.<|endoftext|>
TITLE: Is the 4-line of the E_2 term of the classical Adams spectral sequence known?
QUESTION [19 upvotes]: In other words:

What is $\mathrm{Ext}_{\mathcal{A}}^{4,t}(\mathbb{Z}/2,\mathbb{Z}/2)$?

If the 4-line is not known, how much is known about it?
Here, $\mathcal{A}$ is the 2-primary Steenrod algebra, $4$ is the homological degree corresponding to the Adams filtration, and $t$ is the internal grading degree.  Those $\mathrm{Ext}$ groups make up the fourth row of the classical Adams spectral sequence $E_2 = \mathrm{Ext}_{\mathcal{A}}^{s,t}(\mathbb{Z}/2,\mathbb{Z}/2)$ converging to the 2-adic completion of the $(t-s)^{\mathrm{th}}$ stable homotopy group of the sphere.
For context,  

the 1-line is generated by the classes $h_i$, $i \geq 0$, ($\mathrm{deg}\: h_i = (1,2^i)$),  
the 2-line is generated by the product classes $h_i h_j$, subject to the relations $h_i h_{i+1} = 0$ and $h_i h_j = h_j h_i$,
the 3-line is generated by two sets of classes, 

the product classes $h_i h_j h_k$, subject to the relations implied by $h_i h_{i+2}^2 = 0$, $h_{i+1}^3 = h_i^2 h_{i+2}$, $h_i h_{i+1} = 0$, and $h_i h_j = h_j h_i$,  
the Massey products $\langle h_{i+1},h_i,h_{i+2}^2 \rangle$.

REPLY [3 votes]: This old question caught my eye while looking at another question.   It is worth mentioning that s= 5 is known and the decomposables in s=6 are as well, though this is unpublished.
Tai-Wei Chen,
Determination of $Ext^{5,∗}(Z/2,Z/2)$,
Topology Appl. 158 (2011), no. 5, 660–A689,
DOI 10.1016/j.topol.2011.01.002.
Tai-Wei Chen,
The structure of decomposable elements in $Ext^{6,∗}(Z/2,Z/2)$, (2012) preprint.<|endoftext|>
TITLE: A question about matrices with more details
QUESTION [5 upvotes]: Let $A,B\in\mathbb{R}^{n\times n}$. Suppose that $B$ is nonsingular and that  there exists  $m$ reals pairwise distinct  $\lambda_{1},\cdots,\lambda_{r},\cdots,\lambda_{m}$  such that 
$$B^{-1}A=\begin{pmatrix}
\lambda_{1}& 1&&&&&&\cr
&\lambda_{1}&\ddots&&&&&\cr
&&\ddots&&&\LARGE{0}&&\cr
&&&\lambda_{r}&1&&&\cr
&&&&\lambda_{r}&&&\cr
&&&&&\lambda_{r+1}&&\cr
&&\LARGE{0}&&&&\ddots&\cr
&&&&&&&\lambda_{m}
\end{pmatrix}$$
(It is the canonical Jordan form)
Can we ever find reals numbers $ t_ {1}, \cdots, t_ {p} $ so that the two following assertions are true:

$A\Big(\displaystyle\prod_{i=1}^{p}(A+t_{i}B)\Big)B=B\Big(\displaystyle\prod_{i=1}^{p}(A+t_{i}B)\Big)A$
$\Big(\displaystyle\prod_{i=1}^{p}(A+t_{i}B)\Big)B\quad\mbox{is nonsingular and diagonalizable }$?

N.B :

The integer $p$ is not fixed.
This question has arisen when studying the contollability of a real discrete-time nonlinear system. This explains why the matrices are supposed to be reals.

Thanks for  help.

REPLY [3 votes]: To my taste, it seems more natural to let $A$ and $B$ play symmetric role, by asking whether there exists non-trivial factors $s_jA+t_jB$ such that 
$$A\left(\prod_{j=1}^p(s_jA+t_jB)\right)B=B\left(\prod_{j=1}^p(s_jA+t_jB)\right)A.$$
If you pose the question in an algebraically closed field $k$ (say, $k=\mathbb C$), then the answer is yes for the following reason:

There exist $2^n-1$ non-zero factors $s_jA+t_jB$ such that $\prod_{j=1}^n(s_jA+t_jB)=0$.

The proof is by induction over the rank of products $\prod_{j=1}^p(s_jA+t_jB)=0$. Suppose that exists such a product $\Pi$, with rank $r\ge1$. Let us write
$$\Pi=\sum_{j=1}^rx_ja_j^T.$$
Then
$$\Pi M\Pi=\sum_{i,j=1}^r(a_i^TMx_j)x_ia_j^T.$$
The rank of $\Pi M\Pi$ will be less than or equal to $r-1$ if $\det(a_i^TMx_j)_{1\le i,j\le r}=0$. When $M=sA+tB$, this writes $H(s,t)=0$ where $H$ is a homogeneous polynomial of degree $r$. If $r\ge1$, it does have a non-trivial zero. Then $\Pi':=\Pi(sA+tB)\Pi$ is an other product, with rank $\le r-1$. If in addition $\Pi$ has $2^{n-r}-1$ factors, then $\Pi'$ has $2^{n+1-r}-1$ factors. After $n$ steps, one obtains a product of $2^n-1$ factors whose rank is $0$.<|endoftext|>
TITLE: The simplest even Artin representations of degree 2 and the corresponding Maaß forms
QUESTION [15 upvotes]: What are the simplest numerical examples of even dihedral (resp. tetrahedral, resp. octahedral, resp. icosahedral) representations 
$\rho:\mathrm{Gal}(\bar{\mathbf{Q}}|\mathbf{Q})\to\mathrm{GL}_2(\mathbf{C})$ 
and their associated Maaß forms $f_\rho\ $ ?  The word simplest can be taken to mean that the conductor of $\rho$ is small, or a small prime.
[Serre 1977] and [Buzzard 2012] provide many simple examples of odd Artin representations $\rho$ of degree $2$ and the associated cuspidal modular forms of weight $1$.  For example, the splitting field of $T^3-T-1\ $ gives rise to an odd dihedral representation of conductor $23$ whose associated weight-$1$ modular form is 
$$
q\prod_{n>0}(1-q^n)(1-q^{23n}),
$$
as discussed by Emerton in MO11747.  For the simplest examples when the image of $\rho$ in $\mathrm{PGL}_2(\mathbf{C})$ is isomorphic to $\mathfrak{A}_4$, $\mathfrak{S}_4$ or $\mathfrak{A}_5$, see [Buzzard 2012].
I'm looking for something similar for even representations.  I'm aware of [Vignéras 1985], so my real question is whether there has been further progress in constructing even Artin representations $\rho$ of degree $2$, or the Maaß forms which are supposed to correspond bijectively to such $\rho$ ?
[Buzzard 2012]  Computing weight one modular forms over $\mathbf{C}$ and $\overline{\mathbf{F}}_p$. arXiv:1205.5077. 
[Serre 1977] Modular forms of weight one and Galois representations. Algebraic number fields: L-functions and Galois properties (Proc. Sympos., Univ. Durham, Durham, 1975), pp. 193–268. Academic Press, London. 
[Vignéras 1985] Représentations galoisiennes paires. Glasgow Math. J. 27, 223–237. 
Addendum. (2012/06/14)  Came across the short write-up Explicit Maass Forms by Kevin Buzzard which contains some nice examples.

REPLY [9 votes]: I'll leave the dihedral and octahedral case to others, but for the tetrahedral ($A_4$) and icosahedral ($A_5$) case, I can give some answer.
For the tetrahedral case, the smallest conductor is 163.  See my question:  Does anyone want a pretty Maass form?
I have some (not very well documented) code to compute Artin L-function coefficients for this even tetrahedral Galois representation and thus the (provably, by Langlands) Maass form.  I've posted this code on my webpage http://people.ucsc.edu/~weissman/
For the icosahedral case, a totally real $A_5$ extension of the rationals is given by the splitting field of $x^5 + 5 x^4 - 7 x^3 - 11 x^2 + 10 x + 3$.  In my 1999 undergraduate senior thesis, at http://people.ucsc.edu/~weissman/MWSenThesis.pdf, I found some mild numerical evidence that the associated degree-3 L-function is entire.  I can't find it written there, but I'm guessing it's the first, or among the first, $A_5$ extensions of $Q$.  I would have chosen something of minimal conductor, to minimize the compute-time.  I think I found this by looking in tables from J. Buhler's thesis.
Note that I didn't lift the projective representation $Gal \rightarrow PGL_2(C)$ to an honest 2-dimensional representation.  This is a bit subtle, and I wasn't capable of that work at the time.  Using the 3-dimensional representation (since $A_5$ has a faithful 3-dim representation) avoids this issue, capturing the adjoint square lift of the putative Maass form.  I'd guess that lifting the projective representation would be possible now, if someone wanted to do the work.<|endoftext|>
TITLE: Max of Fourier transform?
QUESTION [6 upvotes]: Let f be a real-valued function (or distribution) on $\mathbb{R}$. (You can assume it is nice in one way or another.) What would be some practical ways to bound
$\max_{\alpha \in \mathbb{R}} |\widehat{f}(\alpha)|$?
Obviously $\max_{\alpha \in \mathbb{R}} |\widehat{f}(\alpha)| \leq |f|_1$, but I am looking for something a bit better. Either a numerical method or a bag of tricks is fine, as long as the answer is rigorous.

REPLY [2 votes]: Let me reformulate your question. How can we control the $L^\infty$ norm of $u$ by some behavior of the Fourier transform? The most classical thing that could be said is 
$$
H^s(\mathbb R^n)\subset L^\infty(\mathbb R^n)\quad\text{when $s>n/2$ and then 
$\Vert u\Vert_{L^\infty}\lesssim \Vert u\Vert_{H^s}=\Vert (1+\vert \xi\vert)^s\hat u(\xi)\Vert_{L^2}$},
$$
where $H^s$ is the standard Sobolev space based on $L^2$. It is known and easy to check that 
$H^{n/2}(\mathbb R^n)\not\subset L^\infty(\mathbb R^n)$. There are some refinement of the injection above:
writing
$$
u(x)=\int e^{2i\pi x\cdot \xi}\hat u(\xi)d\xi=
\int e^{2i\pi x\cdot \xi}\omega(\xi)\hat u(\xi)\frac{1}{\omega(\xi)}d\xi,
$$
we get
$
\Vert u\Vert_{L^\infty}\le
\Vert \omega (D) u\Vert_{L^2}\left(\int\frac{d\xi}{\omega(\xi)^2}\right)^{1/2},
$
which is useful if $1/\omega$ belongs to $L^2$. In particular, 
$$
\hat u(\xi)(1+\vert\xi\vert)^{n/2}\ln(2+\vert\xi\vert)\in L^2\Longrightarrow u\in L^\infty.
$$<|endoftext|>
TITLE: Can we determine which monodromy of surface gives a fibered knot?
QUESTION [9 upvotes]: A fibered knot is a knot with a homeomorhism on compact surface with one boundary component. On the contrary, for a given homeomorohism on a compact surface with one boundary component, is there any way to determine whether it is a monodromy of a fibered knot?

REPLY [4 votes]: This answer gets around some of the pesky issues in Lee Mosher's answer, but it is "morally" the same his initial approach. However, this argument doesn't intrinsically use the fiberedness of the manifold, incorporating this condition might be a way to improve it. 
First, build a triangulation with one torus boundary component from the monodromy information.
Jaco and Sedgwick exhibit the $S^3-K$ decision problem (Algorithm (fancy) S in the reference below) for any cusped 3-manifold with real torus boundary (for ideal triangulations one can use Jaco-Rubinstein's inflation methods to reduce to this case). First they check that this manifold is irreducible and not a solid torus, then Jaco and Sedgwick enumerate the finitely many possible $S^3$-fillings and call upon $S^3$ recognition.  
William Jaco and Eric Sedgwick, MR 1958532 Decision problems in the space of Dehn fillings, Topology 42 (2003), no. 4, 845--906. arxiv version<|endoftext|>
TITLE: Are point sets of the same order type connected by continuous (order type)-preserving motion?
QUESTION [8 upvotes]: Given two general position point sets in $\mathbb{R}^2$ of the same size and order type, is it possible to continuously move the points of one set until they coincide with those of the other set in such a way that order type is preserved (equivalently, no three points become collinear) throughout the operation?

REPLY [5 votes]: The answer is indeed No.  The most economic example up to now i think is mentioned Suvorov's. Independently  examples was constructed by  P. Mani, B. Jaggi, B. Sturmfels, N. White
"Uniform oriented matroids without the isotopy property" (link to pdf)
My perhaps first by time example of simple line arrangement without isotopy property has 19 lines .
I will use language of line arrangements. 
As @joseph-orourke correctly guessed in reply to @j.c. the stronger statement holds:
Any open basic semi-algebraic set (i.e the set defined by strict polynomial inequalities with integer coefficients) can be realized up to certain trivial stabilization by realization space of a simple pseudolines arrangement. 
It was the initial target theorem due to some applications to Morse theory of vector functions. Reference is N. E. Mnev, "The universality theorems on the classification problem of configuration varieties and convex polytopes varieties" (link to pdf) Theorem A, statement 2). This old paper only briefly describes the content of unpublished phd (link to pdf) in Russian.  Unfortunately this moment was not explicitly covered in later improvements and expositions of the proof. But it is really a simple addition to the main construction modernized by P.Shor "Stretchability of pseudolines is NP-hard" (link to pdf), H. Gunzel "The Universal Partition Theorem for Oriented Matroids" (link to pdf) and U. Richter-Gebert "Mne"v's Universality Theorem Revisited" (link to pdf).  The point is that having as an input a system of strict algebraic inequalities the general machine produces “constructible” arrangement of pseudolines with realization space equivalent to the open semi-algebraic set defined by the system.  “Constructible” means “coming from a free construction using ruler”. It has an order on the pseudolines  such that every next is incident to not more then two points in general position defined as intersections of pervious pseudolines. Then such an arrangement can be perturbed  to simple with the equivalent realization space by an inductive perurbation trick applied to pseudolines in the reverse order of the construction. The trick was introduced independently by B. Sturmfels and N. White "Constructing uniform oriented matroids without the isotopy property" (link to pdf), see also Shor Lemma 4 (link to pdf) and around. 
For completeness i will add that the statement holds for uniform oriented matroids of rank >= 3 . It is not an immediate collary. Very nice reduction of any rank greater than 3 to rank 3 presented by @Arnau<|endoftext|>
TITLE: Steinmetz, Laplace and Fourier Transforms
QUESTION [9 upvotes]: I am looking for references on Steinmetz Transform and its relation with Laplace and Fourier Transforms. There is an Italian Wikipedia page about this topic but with no references.

REPLY [4 votes]: (Too long for a comment.) I think the italian Wiki page is wrong. It says the transform "was conceived by the author in 1893 [probably this text] and exposed in his treatise Theory and Calculation of Alternating Current Phenomena four years later."
However, as you can check there is not a trace in these texts (nor in the others authored by Steinmetz and available on archive.org) of what Wikipedia calls "la transformata di Steinmetz" -- i.e. the Fourier isomorphism $L^2(S^1)\to\ell^2(\mathbf{Z})$ -- which by the way, was written explicitly much before 1893.
These texts are famous for introducing complex numbers (in particular the notation $j=\sqrt{-1}$) into electrical engineering, but my impression is that naming the transform in Steinmetz's honor happened much later, perhaps in Italy, with little regard to what he himself actually did (or as the case may be, never did) with it.<|endoftext|>
TITLE: When (if ever) disclose your identity as a reviewer?
QUESTION [30 upvotes]: The peer review system is in general single blinded: The reviewers will not be known by the authors of a paper by default. One reason for this is to guarantee that the reviewer can write his opinions on the paper without expecting any drawbacks whatsoever. However, I can imagine situations in which the reviewer may think about disclosing his identity:

The reviewer is not d'accord with the editor's decision. This happened to me once - as author: the reviewer wrote some kind of open letter to the the editor and put me and my coauthor in the cc. Basically, the reviewer complained that the editor weighted some other (unqualified) review high enough to not accept the paper.
The reviewer has ideas for further research based on the paper and thinks that a collaboration would be a very good idea. He could think about contacting the author directly but this would disclose his identity (in the case that the paper is not available as a preprint - which still happens sometimes). However, waiting until the paper would be published would be a waste of time.

Since I experienced 1. myself and I am thinking about 2., I would like to hear anwers to this question:

Under what circumstances (if any) should a reviewer disclose his identity to the authors of a paper?

Edit: A small clarification for point 2.: First, the submitted paper is not publicly available and hence, writing to the author would disclose the reviewers identity.  Moreover, it is not about improving the actual paper (and wishing to become a coauthor) but about further work inspired by the paper.
In view of the current answer, I would also like to expand the current question a little bit (which is not worth a whole new question, I think):

... and what are arguments against disclosing the reviewers identity?

REPLY [2 votes]: This is actually a long comment. Concerning point 2: Here is an amusing example when the referee wanted to collaborate with the author on the paper that he/she was refereeing, and so contributed a two-page appendix to the paper without disclosing their identity:
Real reductive Cayley groups of rank 1 and 2. 
With appendices by Dolgachev and an anonymous referee. 
J. Algebra 436 (2015), 35–60.
(Excuse me for promoting my own work...)<|endoftext|>
TITLE: Hurwitz's automorphisms theorem with deformations
QUESTION [5 upvotes]: Hurwitz's automorphisms theorem bounds the order of the automorphism group of a negatively curved Riemann in terms of the genus.  
Now suppose a finite group $G$ acts faithfully on a Riemann surface $X$.  Even when the complex structure of $X$ admits deformations, it may be impossible to deform the action of $G$ along with $X$.  
But assume that the whole package, namely $X$ with it $G$ action does in fact admit an $n$-parameter deformation.  Then it seems to me that one generally ought to get a tighter bound than Hurwitz's on the order of $G$ as a function of both $n$ and the genus of $X$ as opposed to as a function of the genus alone.  My intuition is that the role of the rigid orbifold associated with $(2,3,7)$-triangle group will get replaced by orbifolds with larger Euler characteristic,  in particular those associated with hyperbolic polygons having more than three sides so that angles don't determine side lengths.
Since all these seems likely to be very classical, I'm asking if and where I can find details worked out (in particular with bound-realizing examples)?

REPLY [8 votes]: Since you asked for details worked out, here are some notes from my course, with some details, but not so many examples.  Since deformations occur by deforming the quotient curve and the branch points, as you said the rigid examples have quotient curve of genus zero, and three branch points.  These occur below as realizing 84(g-1) and 48(g-1).  These calculations also show that if there are 4 or more branch points on the genus zero quotient, hence at least a one parameter deformation, then the bound is ≤ 12(g-1).  If there are either ≥ 5 branch points or the quotient has positive genus, there are ≤ 4(g-1) automorphisms.
Now suppose there are k points p1,....,pk of X/G which are branch points of the quotient map π:X--->X/G for the action of G on X.  Then there exist k conjugacy classes of subgroups of G, containing groups of orders rj, j=1,...,k, such that over each point pj, there are |G|/rj ramified points for π, each with ramification index (rj-1).  Thus the total ramification index of the map π is ∑ (|G|/rj)(rj-1).  Hence the Hurwitz formula for the Euler characteristic reads chitop(X) = |G|chitop(X/G) - ∑ (|G|/rj)(rj-1) = 
|G|[chitop(X/G)- ∑ (1-1/rj), summed from j=1,...k.
Equivalently, 2g(X) - 2 = |G|[2g(X/G) – 2 + ∑ (1-1/rj)]. (*)
This allows one to compute bounds on the size of |G|.  Riemann surfaces of genus g(X) ≤ 1 can have arbitrarily large finite groups of automorphisms, e.g. any finite cyclic group can occur, but this is not true in higher genus.  When g≥ 2 Hurwitz gave the general bound of |G| ≤ 84(g-1) as we discuss next.
Assume g(X) ≥ 2. If g(X/G) ≥ 1 and there is no ramification, then ∑ (1-1/rj)] = 0, so 2 ≤ 2g(X) -2  = |G|[2g(X/G) – 2], so the RHS is positive, whence 2g(X/G) - 2 ≥ 2, and thus 2g(X) -2  ≥ 2 |G|, so |G| ≤ (g-1).
    If ∑ (1-1/rj)] > 0, then rj ≥ 2 implies ∑ (1-1/rj)] ≥ ½.  Then g(X/G) ≥ 1 implies RHS(*) ≥(1/2) |G|, so |G| ≤ 4g(X)-4 = 4(g(X)-1).
If g(X/G) = 0, then g(X) ≥ 2 implies again RHS(*) > 0, so ∑ (1-1/rj)] > 2.
Now if ∑ (1-1/rj) = (1 – 1/r1) + (......)+ (1 – 1/rk) > 2, then certainly k > 2, since there are k terms added, each smaller than 1.
k ≥ 5: In all cases, ∑ (1-1/rj) = k - ∑ 1/rj ≥ k/2, so if k ≥ 5, then ∑ (1-1/rj) ≥ 2 + (1/2), and 
(*) becomes 2g(X) - 2 ≥ |G|[1/2], so |G| ≤ 4(g-1).
k = 4: Then ∑ (1-1/rj) = 4 – 1/r1 – 1/r2 – 1/r3 – 1/r4.  For this to be > 2, some rj must be > 2.  Thus ∑ (1-1/rj) ≥ 4 – ½ - ½ - ½ - 1/3 = 2 + (1/6), whence |G| ≤ 12(g-1).
k = 3: If all rj ≥ 4, then ∑ (1-1/rj) ≥ 2 + (1/4), whence |G| ≤ 8(g-1).
If some rj = 3, no rj = 2, then ∑ (1-1/rj) ≥ 3 – 1/3 – 1/3 – ¼ = 2 + (1/12), so |G| ≤ 24(g-1).
If some rj = 2, no rj = 3, then ∑ (1-1/rj) ≥ 3 – ½ - ¼ - 1/5 = 2 + (1/20), so |G| ≤ 40(g-1).
r1 = 2, r2 = 3, then r3 ≥ 7, so ∑ (1-1/rj) ≥ 2 + (1/42), so |G| ≤ 84(g-1).
This is the largest possible bound when g(X) ≥ 2.  I.e. g(X) ≥ 2 implies no finite group of order > 84(g-1) can act as automorphisms of X.  If this bound does not occur, then the next best bound is r1 = 2, r2 = 3, then r3 = 8, so ∑ (1-1/rj) = 2 + (1/24),  so |G| ≤ 48(g-1).
Hence, if for example, X does not have an automorphism of order 7, then the largest finite subgroup of Aut(X) has order ≤ 48(g-1).
Exercise: The genus three “Klein” curve x^3y + y^3z + z^3x  = 0, has #(Aut(X)) = 168 = 84(g-1).  Indeed Aut(X) = PSL(2,Z/7).
Exercise:  If g(X) = 5, then no automorphism of X can have order 7, hence #(Aut(X)) = 84(g-1) cannot occur, but there is a curve X of genus 5 with 192 = 48(g-1) automorphisms.  Indeed Aut(X) = PSL(2,Z/8), where Z/8 is the ring of integers mod 8.  (See Gunning's Lectures on modular forms, p.15.)
Fact:  The next genus in which a group of automorphisms occurs achieving Hurwitz maximal bound is g = 7, where there is a curve X with 504 automorphisms and Aut(X) = PSL(2,F8), where F8 is the field with 8 elements.  This example was found by Fricke in 1899, forgotten and rediscovered by Macbeath in 1965.  Macbeath also showed that curves with the maximal number 84(g-1) of automorphisms occur for infinitely many values of g.<|endoftext|>
TITLE: Smash product of Kan complexes
QUESTION [6 upvotes]: Suppose $X$ and $Y$ are pointed Kan complexes.  Is their smash product $X\wedge Y$ also a Kan complex?
I expect the answer is probably no, but it would be nice to see a counterexample.

REPLY [9 votes]: The basic idea is the map $X \times Y \to X \wedge Y$ is a bijection on simplices except over the basepoint.  We can construct an extension problem that doesn't lift to an extension problem on the cartesian product.
Let $X$ and $Y$ both be the standard model for $E\mathbb{Z}/2$, which is the nerve of a groupoid with two isomorphic objects and trivial automorphism groups.  Call the basepoint $\ast$ and the nonbasepoint $p$; call $a$ the edge from $\ast$ to $p$ and $b$ its inverse.  In simplicial identities, $d_1(a) = \ast, d_0(a) = p$, and the reverse for $b$.
We can construct a map from the inner $2$-horn $\Lambda_1^2$ of $X \wedge Y$ with no horn filler as follows.  The $1$-simplex $b \wedge a$ satisfies
$$
d_0(b \wedge a) = \ast \wedge p = \ast = p \wedge \ast = d_1(b \wedge a),
$$
and so it defines a horn.
However, there is no horn filler.  If there was, we would have a $2$-simplex of $X \wedge Y$ filling it.  As mentioned, the map $X \times Y \to X \wedge Y$ is surjective on $2$-simplices, and injective except on those simplices living over the basepoint.  Therefore, we would have a $2$-simplex of $X \times Y$ satisfying $d_0(u) = d_2(u) = b \times a$.  However, in $X \times Y$ we have
$$
d_0(b \times a) = \ast \times p \neq p \times \ast = d_1(b \times a),
$$
and so there can be no such 2-simplex.<|endoftext|>
TITLE: Hartogs number and the three power sets
QUESTION [11 upvotes]: One of the most important constructions in ZF+$\lnot$AC is Hartogs number, defined as:
$$\aleph(X)=\min\lbrace\alpha:|\alpha|\nleq|X|\rbrace$$
We can prove that this ordinal always exists in the following way:
Consider every well-ordered subset of $X$, $\langle W,\prec\rangle$, for every $x\in W$ we can take $W_x=\lbrace y\in W: y\preceq x\rbrace$, then $W_x\subseteq W_y$ if and only if $x\preceq y$. This gives us an embedding of $(W,\prec)$ into $\mathcal{P}(X)$. We can therefore view $\langle W,\prec\rangle$ as an element of $\mathcal{P(P}(X))$. Now consider the equivalence relation of order isomorphism between the different subsets and their orders. Sending $\alpha<\aleph(X)$ to the equivalence class of all $\langle W,\prec\rangle\cong\langle \alpha,\in\rangle$ is an injective function from $\aleph(X)$ into $\mathcal{P(P(P}(X)))$.
So while $\aleph(X)$ is never smaller than $X$ it is always less or equal than third iteration of a power set.
Example 1: Suppose that $|\mathbb R|=\aleph_1$, then indeed $\aleph(\mathbb N)=2^{\aleph_0}=\aleph_1$ and we have the Hartogs is less or equal (in fact equal) to a single power set operation.
Example 2: Suppose that $D\subseteq\mathbb R$ is an infinite Dedekind-finite (e.g. Cohen's first model). We know that $\aleph(D)=\aleph_0$ for every infinite Dedekind-finite set. However since such $D$ can be mapped onto $\mathbb N$ we have that $\aleph_0<\mathcal P(D)$. We do not have equality since $\mathcal P(D)$ cannot be well-ordered so it cannot be equal to an ordinal.
Example 3: Suppose that $A$ is an amorphous set, that is an infinite set that every subset is finite or co-finite. It is immediate that $A$ is Dedekind-finite and therefore $\aleph(A)=\aleph_0$; however we also have that $\mathcal P(A)$ is Dedekind-finite, so we have to go another level and to only then we have $\aleph_0<\mathcal{P(P}(A))$.

The last example, using amorphous sets, is pretty much the "least well-orderable" set I can think of. In fact when looking for counterexamples amorphous sets are often a good place to begin with (they cannot be linearly ordered, for example).

Question: Is the bound of three iterations of taking power sets really needed?

REPLY [12 votes]: Hi Asaf,
I thought about this a while ago. Of course, the question had been asked and solved before. Digging through the FOM archives for Spring 2009, I found (April 28, 2009; I fixed a typo in what follows):

In a message dated Jan. 28, I asked whether Sierpinski's ZF result that
    $\aleph(X) < \aleph({\mathcal P}({\mathcal P}({\mathcal P}(X))))$ for all $X$, 
  could be improved by replacing the triple power set with a double power 
  set.
In a follow up dated Feb. 2, I indicated that one can, provided that
    $\aleph(X)$ is not $\aleph_\alpha$ for some infinite limit ordinal
    $\alpha < \aleph_\alpha$.
I recently found a reference that settles the other case, and wanted to 
  give an update for those curious about the question. In Theorem 11 of John 
  L. Hickman, "$\Lambda$-minimal lattices", Zeitschr. f. math. Logik und 
  Grundlagen d. Math., 26 (1980), 181-191, it is shown that for any such 
  $\alpha$, it is consistent to have an $X$ with
    $\aleph(X) = \aleph_\alpha = \aleph({\mathcal P}({\mathcal P}(X)))$.

So, yes, the triple power set is best possible in ZF. (If $\Lambda$ is an aleph (a well-ordered cardinal), a set $X$ is said to be a $\Lambda$-set iff $\aleph(X)=\Lambda$, and yet $X$ cannot be well ordered. In that case, $X$ is $\Lambda$-minimal iff for every $Y\subseteq X$, either ${}|Y|<\lambda$ or ${}|X\setminus Y|<\Lambda$.) 
Hickman's argument uses Fraenkel-Mostowski models (and the Jech-Sochor embedding theorem).
See also the appendix to these notes for the argument that two power sets suffice unless $\aleph(X)=\aleph_\alpha$ for some infinite limit $\alpha < \aleph_\alpha$.<|endoftext|>
TITLE: When is a singular point of a variety ($\mathcal{C}^\infty$-) smooth?
QUESTION [73 upvotes]: If $X$ is a nonsingular  algebraic (or analytic) variety over $\mathbb C$ or $\mathbb R$ then it is certainly $C^\infty$ over the reals.
The converse is  false for a silly reason : in the real or complex affine plane with coordinates $x,y$ the variety $x^2=0$ is  singular since it is not reduced, but set-theoretically it is the $y$-axis, a $C^\infty$ submanifold of the plane.
So let me concentrate on reduced varieties.  
Over the reals you have the disturbing phenomenon that the  plane curve $C$ defined by $y^3+2x^2y-x^4=0$ is algebraically singular at the origin, but your favourite computer graphics system  won't show you that, because the curve is actually $C^\infty$.
Worse: it is the graph of a real analytic function! (I learned this example in Milnor's Singular Points of Complex Hypersurfaces)  
I think this pathology is impossible over $\mathbb C$ but I cannot find a reference. So let me ask the official question:  
Can a reduced algebraic (or analytic) variety which has a singularity at a point be $C^\infty$ at that point? 
(Information on the more complicated real case welcome, of course)  
Edit Francesco's answer shows that indeed a singular complex variety  cannot be smooth.
This is  quite interesting, because many books will give you the Jacobian criterion for a  map $\mathbb C^n\to \mathbb C^k$ to define a submanifold but none (to my knowledge) adds that this Jacobian criterion also allows to prove nonsmoothness (modulo some technicalities), which is  paradoxically a more delicate question.    
Second Edit (June 29th,2012))
Here is a  proof  more geometric than Milnor's that the curve $X$ defined by $y^3+2x^2y-x^4=0$ is an analytic submanifold of $\mathbb R^2$.
The curve $X$ is actually rational: it is the image of $\mathbb R$ under $\phi (t)=(t(t^2+2),t^2(t^2+2))$.
The analytic morphism $\phi$ is injective, proper and immersive, hence an embedding into $\mathbb R^2$ with closed image $X$. [It  boils down to the fact that the polynomial $t^3+2t$ has positive  derivative $3t^2+2$ and is thus strictly increasing!]

REPLY [3 votes]: For the real case, here's another
example: the graph in ${\mathbb R}^2$ of
  $ y = x (1 + x^2)^{1/3} $
is real analytic and is the
${\mathbb R}$-points of a
variety defined over ${\mathbb R}$,
but the origin is 
singular; the defining equation is
 $  y^3 = x^3 + x^5 $.<|endoftext|>
TITLE: Cup products of connected sum
QUESTION [10 upvotes]: Hej,
I am interested in the cohomology ring of the connected sum $M \# N$ of two oriented manifolds $M$ and $N$ in terms of the corresponding cohomology rings of $M$ and $N$.
Mayer-Vietoris shows that in dimensions $0<k<n$ the additive structure is given by
$H^k(M \# N)=H^k(M)\oplus H^k(N)$
via the induced maps of the inclusions. And because this isomorphisms are given by induced maps the cup product translates into componentwise product whenever this is possible, i.e.:
$$([\omega_1],[\eta_1])\cup([\omega_2],[\eta_2])=([\omega_1\cup\omega_2],[\eta_1\cup\eta_2])$$
for $([\omega_1],[\eta_1])\in H^i(M)\oplus H^i(N),([\omega_2],[\eta_2])\in H^j(M)\oplus H^j(N)$ AND $i+j<n$.
But how is cup product given if $i+j=n$?
Thanks a lot for your answers!

REPLY [11 votes]: The natural description is like this.  There are augmentations $\epsilon_X:H^*(X)\to\mathbb{Z}$ (for $X\in\{M,N\}$) and orientation classes $u_X\in H^d(X)$.  Put 
$$R'=\{(a,b)\in H^*(M)\times H^*(N):\epsilon_M(a)=\epsilon_N(b)\} = H^*(M\vee N)$$
and $R=R'/(u_M,-u_N)$.  Then one can use the cofibration 
$$ S^{d-1} \to M \# N \to M\vee N $$
to identify $H^*(M\#N)$ canonically with $R$.  Thus, if $i+j=d$ with $i,j>0$ then the product of $H^i(M)$ with $H^j(N)$ in $H^d(M\# N)$ is zero, but the product of $H^i(M)$ with $H^j(M)$ is the same as in the original manifold $M$.<|endoftext|>
TITLE: Cubic graphs without a perfect matching and a vertex incident to three bridges
QUESTION [22 upvotes]: The example shown below (courtesy of David Eppstein) is a common example of a cubic graph  that admits no perfect matching:

(source: uci.edu)
Are there other examples of cubic graphs that do not admit a perfect matching and, unlike the above example, do not contain a vertex that lies at the intersection of three bridges (i.e. an edge whose removal increases the number of connected components in the graph)?

REPLY [23 votes]: Substitute your central vertex in your graph with a 3-cycle $abc$ so that the graph stays cubic. Now subdivide each edge in this 3-cycle. So we have new vertices $u$ connected to $a$ and $b$, $v$ connected to $b$ and $c$, $w$ connected to $c$ and $a$. Now add a final vertex $x$ and connect it to $u,v$ and $w$. This graph has exactly three bridges, none of which intersect the other at a vertex, and moreover has no perfect matching!
One result which relates the existence of a perfect matching in a cubic graph and its bridges is the following theorem of Petersen from "Die theorie der regularen graphen", Acta Math. 15
(1891), 163-220:

Theorem: Every cubic graph with at most two bridges contains a perfect matching. 

As well as this strengthening by Errera, "Du colorage des cartes", Mathesis 36 (1922), 56-60:

Theorem: If all the bridges of a connected cubic graph $G$ lie on a single path of $G$, then $G$ has a perfect matching.

So your instinct is true, in the sense that if the graph has no perfect matching, its bridges do not lie on a path. However the example in the beginning of this answer shows that they are not necessarily incident at the same vertex.<|endoftext|>
TITLE: Fourier decay rate of Cantor measures
QUESTION [14 upvotes]: For $0<\theta<\frac{1}{2}$, denote by $C_\theta$ the Cantor set with dissection ratio $\theta$, i.e. the Cantor set obtained from dissection parttern $(\theta, 1-2\theta,\theta)$. It is known that $C_\theta$ carries a uniform measure $\mu_\theta$ which is usually called Cantor measure. And it is not hard to show that the Fourier–Stieltjes transform of $\mu_\theta$ is (up to scaling and constant multiple)
$$\hat{\mu}_\theta(\xi)=\prod^{\infty}_{k=1} \cos(\theta^k\xi)$$ 
But unlike integrable function we do not have Riemann-Lebesgue lemma for these Cantor measures. In fact a theorem of Erdős and Salem says $\hat{\mu}_\theta (\xi)=o(1)$ as $|\xi|\rightarrow\infty$ if and only if $\theta^{-1}$ is a not PV number. On the other hand, it is known that for some $\theta^{-1}$ not a PV number, $\hat{\mu}_\theta (\xi)$ does not decay at any positive rate, even though $\hat{\mu}_\theta (\xi)=o(1)$.
My question is whether there exists $\theta$ such that $\hat{\mu}_\theta (\xi)=O(|\xi|^{-\alpha})$ for some $\alpha>0$? How much is known about the precise decay rate of $\hat{\mu}_\theta (\xi)$? 
Thanks in advance.
Edit: To make the second question precise, I was actually wondering if there exists Salem set (as pointed out by Pablo) among these Cantor sets. So the rate of decay I was expecting is (or arbitrarily close to) half of the Hausdorff dimension $\frac{1}{2}\dim_H(C_\theta)=\frac{1}{2}\frac{\log(1/2)}{\log(\theta)}$.

REPLY [3 votes]: I would start looking here:
MR1785620 (2001m:42020)
Peres, Yuval; Schlag, Wilhelm; Solomyak, Boris,
Sixty years of Bernoulli convolutions.
Fractal geometry and stochastics, II (Greifswald/Koserow, 1998), 39–65, 
Progr. Probab., 46, Birkhäuser, Basel, 2000.<|endoftext|>
TITLE: A bijective proof that the bistatistic $(\operatorname{exc},\operatorname{den})$ on permutations is Euler-Mahonian
QUESTION [6 upvotes]: In a joint project, we are currently working on an online combinatorial statistic finder in which (beside other things) want to gather information about combinatorial collections and statistic, see the website http://www.findstat.org.
In this context, I am reading through Dominique Foata and Doron Zeilberger's paper
''Denert's permutation statistic is indeed Euler-Mahonian''
from 1990. In there, they prove that the bistatistic given by the number of exceedences and the Denert index is Euler-Mahonian. Moreover, they say:
"The most natural proof of this result would be in terms of a bijection from $\mathcal{S}_n$ to itself that sends the pair (des,maj) simultaneously to the pair (exc,den). Although it is rather easy to find a bijetion that sends maj to den ..., and it is now trivial ... to find a bijection that sends exc to des, we are unable, at present, to find a bijection that does both at the same time. ... We really hope that such a bijective proof of Denert's conjecture will be found ... ."
So my question is:

Is there an explicit bijection on permutations known that sends thnumber of descents to the number of exceedences, and at the same time the major index to the Denert index?

REPLY [2 votes]: Yes. 
G.-N. Han, Distribution Euler-mahonienne : une correspondance, C. R. Acad. Sci. Paris, 310, Série I, 1990, pp. 311-314.
G.-N. Han, Une nouvelle bijection pour la statistique de Denert, C. R. Acad. Sci. Paris, 310, Série I, 1990, pp. 493-496.
G.-N. Han, Une transformation fondamentale sur les réarrangements de mots, Adv. in Math., 105(1), 1994, pp. 26-41.<|endoftext|>
TITLE: What are the truly 'global methods' in number theory?
QUESTION [9 upvotes]: I have spent some time being confused by the nature of global methods in number theory. It seems that there are in some sense (for my purposes) three levels at which algebraic number theorists operate: local (at one prime), everywhere local (at all primes simultaneously, including the infinite ones) and global (actually playing with the number field). When we talk about things like local class field theory we mean the first one, and when we talk about global class field theory I guess we mean the last one (but the extent to which the ideles seem to get involved suggests to my naive mind a strong whiff of the second one also).
When we talk about local to global principles we most definitely mean the passage from the second to the third. I guess my question can therefore by crystallised in terms of elementary number theory as follows: what global techniques do we have for proving the non-existence of solutions to diophantine equations? In other words, given a failure of the local-global principle, what are the techniques one can use to demonstrate it independently of local information?
If I am given a diophantine equation and asked to show it has no solutions, I can think of very few methods that are not in some sense `local', certainly if we count working at the infinite primes also as local (which surely we should?). 
One possibility I have been considering is that something to do with heights/descent is perhaps a global method. However, the height of a point is still measurable by concatenating local data, and descent is normally via some trick involving congruences, but perhaps the `well-orderedness' of the process is a truly global trick.
Also, returning to a pessimistic analysis, in an important classical result that I have seen called a `measure of the failure of the local-global principle': the finiteness of the class number, it seems to me that the statement doesn't involve the infinite primes in any way, so to prove it we milk the infinite primes for all the have got and get the Minkowski bound by directly studying local behaviour above infinity. Is my opinion in this regard incorrect? If so, which of the arguments are truly global?
So to conclude, are there such things as "global methods", and if there are, what are they? Apologies for posing what is probably a naive, overly-simplistic and absurdly general question, but I am hoping several people may have thought about this and have interesting things to say.
Thanks,
Tom.

REPLY [6 votes]: The following example is perhaps expressed a bit loosely but it is morally correct, and it is fundamental.
Suppose that for each prime $p$ we have chosen a decomposition group $G_p=\mathrm{Gal}(\bar{\mathbf{Q}}_p|\mathbf{Q}_p)\ $ of $G=\mathrm{Gal}(\bar{\mathbf{Q}}|\mathbf{Q})$.  If we are given a global character $\chi:G\to\mathbf{C}^\times$, then we get by restriction a family of local characters $\chi_p$ of $G_p$ almost all of which are unramified in the sense of being trivial on the inertia subgroup $I_p\subset G_p$.   Conversely, when does a family $(\chi_p)_p$ of local characters, almost all of which are unramified, come from a global character ?
The reciprocity isomorphism of local class field theory at the various primes allows us to attach  a character of finite order $\xi_p:\mathbf{Q}_p^\times\to\mathbf{C}^\times\ $ to each $\chi_p$.  Since the $\chi_p$ are almost all unramified, the $\xi_p$ are almost all trivial on $\mathbf{Z}_p^\times$, and hence give rise to a character $\xi:\mathbf{A}^\times\to\mathbf{C}^\times$ of the idèles of $\mathbf{Q}$.  Now the condition for the $\chi_p$ to come from a global $\chi:G\to\mathbf{C}^\times$ is that this $\xi$ should be trivial on the subgroup $\mathbf{Q}^\times\subset\mathbf{A}^\times$.  In other words, $\xi$ should come from a character $\mathbf{A}^\times/\mathbf{Q}^\times\to\mathbf{C}^\times$.  This is a truly global condition and cannot be expressed by any collection of local conditions.<|endoftext|>
TITLE: Does there exist a curve of degree 11 having 15 triple points?
QUESTION [19 upvotes]: Does there exist an irreducible curve of degree 11 in the projective plane which would have 15 triple points? 
For information, such a curve would be rational, if it exists, and would be smooth at all other points (compute the genus with the classical formula: $(d-1)(d-2)/2-\sum a_i(a_i-1)/2=45-3\cdot 15=0$)

REPLY [3 votes]: This seems to be study in this paper by O. Dumitrescu (which I did not read in detail, but it seems quite algorithmic).<|endoftext|>
TITLE: Fourier transform on locally compact quantum groups
QUESTION [5 upvotes]: I have read some articles on locally compact quantum groups and the Fourier Transform on them. I wonder why we define the Fourier transform as an operator valued functions from $L^1(\mathbb{G})$ to $L^\infty(\widehat{\mathbb{G}})$.
Why do not we used the intrinsic group $Gr(\mathbb{G})$ which has been defined by Mehrdad Kalantar, and define the Fourier transform from $L^1(\mathbb{G})$ to $L^\infty(\operatorname{Gr}({\mathbb{G}}))$ as a complex-valued function?
I think if we do it we can see immediately that it is an analogue of the Fourier transform in the classical case, since when we work with a locally compact Abelian group $G$ we know that $\operatorname{Gr}(L^\infty(G))=\widehat{G}= \operatorname{sp}(L^1(G))$ and $\mathcal{F}:L^1(G)\rightarrow L^\infty(\widehat{G})$.
I would really appreciate it if anybody could help me in this regard.

REPLY [5 votes]: My view is that one should treat LCQGs as a self-dual category; so there is no reason to prejudice, for a classical group $G$, the commutative case (leading to $L^1(\G)$) over the co-commutative (leading to $A(G)$).
The co-commutative is nice from the point of view of intrinsic groups-- this goes back to Takesaki and Tatsumma (and arguably Eymard, Herz etc.) where they showed that the intrinsic group of $VN(G)$ is just $G$ (with the same topology).
But in the commutative case, it's awful-- the intrinsic group of $L^1(G)$ is just the group of characters of $G$, which is rarely interesting outside of the abelian group case.  Well, "interesting" is a bit extreme, giving maximal tori etc., but it certainly wouldn't give an injective Fourier transform.
(I think here maybe I have computed things in the "dual" formalism to that of the original question).
For a quantum example, I think Mehrdad showed that for $SU_\mu(2)$, you just get the maximal torus; so again the Fourier transform fails to be injective.  That's not going to lead to an interesting theory (unless you have some specific application already in mind...)<|endoftext|>
TITLE: Is an elementary symmetric polynomial an irreducible element in the polynomial ring?
QUESTION [8 upvotes]: Let $S=\mathbb{C}[x_1,x_2,\dots,x_n]$ be a polynomial ring. Let $e_a$ denotes the elementary symmetric polynomials of degree $a$ in $S$.
For $n=2$:
$e_1=x_1+x_2$;
$e_2=x_1x_2$.
For $n=3$:
$e_1=x_1+x_2+x_3$;
$e_2=x_1x_2+x_1x_3+x_2x_3$,
$e_3=x_1x_2x_3.$
In general for any $n$ and $a$, one has
$$ e_a(x_1,x_2,\dots,x_n):=\sum_{1 \leq i_{1} < i_{2} < \cdots < i_a \leq n} x_{i_1}x_{i_2}\cdots x_{i_a}
$$
Question: Let $n \in \mathbb{N}$ with $n \geq 3$. Is it true that  $e_a$ is an irreducible element in $\mathbb{C}[x_1,x_2,\dots,x_n]$ for $a=2,3,\dots,{n-1}$.
For $n=1$, $e_1$ is an irreducible element.
For $n=2$, $e_1$ is an irreducible element.
For $n=3$, $e_1$ and $e_2$ are irreducible element. 
Fact: $e_1$ is be definition, an irreducible element. And, $e_n$ is trivially reducible. 
My Question is therefore, to know, whether $e_2,e_3,\dots,e_{n-1}$ are irreducible elements in $\mathbb{C}[x_1,x_2,\dots,x_n]$ for $n \geq 4$.  
Similar results: Power sum symmetric polynomials and complete homogeneous symmetric polynomials are irreducible elements in $\mathbb{C}[x_1,x_2,\dots,x_n]$ for $n \geq 3$.
For complete symmetric polynomial, see Is complete homogeneous symmetric polynomials, an irreducibile element in Polynomial ring?.
Therefore it is natural to ask for the elementary symmetric polynomials.
Thanks.

REPLY [11 votes]: For $\alpha\neq n$, the symmetric polynomial is of the form $f\cdot x_n + g$ where $f,g$ are non-zero elements of $A={\mathbb C}[x_1,...,x_{n-1}]$ with no common factor.
Thus $${\mathbb C}[x_1,...,x_n]/(e_\alpha)=A[x_n]/(f x_n+g)=A[g/f]\subset K$$
where $K$ is the quotient field of $A$.  It follows that ${\mathbb C}[x_1,...,x_n]/(e_\alpha)$ is a domain, so $e_\alpha$ is irreducible.<|endoftext|>
TITLE: Recognize this plane curve?
QUESTION [5 upvotes]: An aspect of my work led to a plane curve with implicit equation
$$
x^2+y^2 = 3 (y/2)^{2/3} + 1
$$
Actually, I started with the parametrization below and derived from it the
equation above:
\begin{eqnarray}
x(t) &=& t (3-2 t^2) \\
y(t) &=& 2(1-t^2)^{3/2}
\end{eqnarray}
Here is what it looks like:

          


If this falls in some classical class of curves, and perhaps even has a name, 
I would like to reference it appropriately.
Does anyone recognize this curve?
Thanks!
Answered. By Sylvain Bonnot and Francesco Polizzi: It is a type of nephroid!
Here's the Wikipedia image from the article they both cited:

REPLY [11 votes]: Your curve is a nephroid, see http://en.wikipedia.org/wiki/Nephroid.
The general equation of such a plane curve is $$(x^2+y^2-4a^2)^3=108a^4y^2.$$ Your example corresponds to the value $a=\frac{1}{2}$ of the parameter.

REPLY [7 votes]: Pretty curve...I think it is a Nephroid:
http://en.wikipedia.org/wiki/Nephroid<|endoftext|>
TITLE: Irreducibility of Schur polynomials
QUESTION [13 upvotes]: A natural question covering both this and this question would be 
Let $n>2$. Describe Young diagrams $\lambda$ with at most $n$ nonempty rows (or equivalently non-increasing sequences $\lambda=(\lambda_1\ge\lambda_2\ge\ldots\ge\lambda_n\ge0$) for which the Schur polynomial $$s_\lambda(x_1,\ldots,x_n):=\frac{\det(x_i^{\lambda_j+n-j})}{\det(x_i^{n-j})}$$ is irreducible in $\mathbb{C}[x_1,\ldots,x_n]$.
Partial results: 

Of course, for $\lambda_n>0$, this polynomial is divisible by $s_{1^n}=e_n$ , and so isn't irreducible. (This is a generalisation of the fact that $e_k=s_{1^k}$ can only be irreducible for $k< n$ (which is shown to be true in an answer to one of the questions linked above).
For $\lambda=(m)$, we have $s_\lambda=h_m$, and so one of the questions linked above shows that $s_\lambda$ is irreducible.
For $\lambda=(n-1,n-2,\ldots,0)$ we have $s_\lambda=\prod_{i< j}(x_i+x_j)$ by Vandermonde, which is very reducible. 

Of course, it is mere curiousity forcing me to ask it, but I think that this question has a potential to have a meaningful answer.

REPLY [12 votes]: Let $g=\gcd(\lambda_1+n-1,\lambda_2+n-2,\dots,\lambda_n)$. And let $V _g(X)=\prod _{i < j}\frac{x_i^g-x_j^g}{x_i-x_j}$. David already mentioned that $s_{\lambda}(X)$ is divisible by $V_g(X)$. This is in fact all that can happen.
Theorem: For every partition $\lambda$ (with $\lambda_n=0$), the polynomial $\frac{s_{\lambda}(X)}{V_g(X)}$ is either constant or irreducible.
This follows from theorem 3.1 in the paper "Newton functions generating symmetric fields and irreducibility of Schur polynomials" by Dvornicich and Zannier. 
A result which is somewhat more general is stated and proved in "On the irreducibility of irreducible characters of simple Lie algebras". They extend the irreducibility above to characters of irreducible representations of simple Lie algebras divided by a Weyl denominator type factor (analog of $V_g$ above). Schur functions are a special case corresponding to $GL_n$.<|endoftext|>
TITLE: A reducible connected scheme with pairwise disjoint irreducible components
QUESTION [10 upvotes]: Without finiteness assumptions, the irreducible and the connected components of a scheme may behave in strange ways. More precisely, let us consider a scheme $X$ and the following properties:
(1) $X$ is the sum of its irreducible components;
(2) The irreducible and the connected components of $X$ coincide;
(3) The irreducible components of $X$ are pairwise disjoint.
It is clear that (1) implies (2), that (2) implies (3), and that if the set of irreducible components of $X$ is locally finite then all three statements are equivalent (see [EGA 0.2.1.6]). However, (2) does not necessarily imply (1) in general: An affine counterexample is given by the spectrum of the product of infinitely many fields (which is non-discrete and totally disconnected). So, out of pure curiosity we may ask the following:

Is there an (affine) scheme fulfilling (3) but not (2)?

One can note that this is equivalent to the following:

Is there a nonempty, reducible, connected (affine) scheme whose irreducible components are pairwise disjoint?

(One can also note that for topological spaces that are not necessarily underlying spaces of schemes it is easy to find an example that fulfils (3) but not (2) - every connected, separated space with at least two points does so.)

REPLY [10 votes]: In the answers to Non-integral scheme having integral local rings , and in particular in t3suji's beautiful geometric construction, it is shown that there is a (non-empty) affine scheme $X=\mathrm{Spec}(A)$ whose underlying topological space is connected, such that all local rings of $X$ are integral domains and such that $X$ is however not integral. This scheme fulfills the conditions you are requiring: It is reduced and not integral, hence reducible. Assume two irreducible components $Y_1$ and $Y_2$ of $X$ met in a point $x\in X$. The point $x$ corresponds to a prime ideal $\mathfrak{q} \subset A$, and the irreducible components $Y_1$ and $Y_2$ are of the form $Y_1=V(\mathfrak{p}_1)$ and $Y_2=V(\mathfrak{p}_2)$ for  minimal  prime ideals $\mathfrak{p}_1$ and $\mathfrak{p}_2$ of $A$. As $x\in Y_1,Y_2$, we would have $\mathfrak{p}_1, \mathfrak{p}_2 \subset \mathfrak{q}$. 
Then $\mathfrak{p}_1$ and $\mathfrak{p}_2$ would correspond to distinct minimal prime ideals of $A_\mathfrak{q}=\mathcal{O}_{X,x}$, which is impossible, as $\mathcal{O}_{X,x}$ is by construction a domain, so that its only minimal prime ideal is $\{0\}$.<|endoftext|>
TITLE: power series over a polynomial ring
QUESTION [5 upvotes]: Let $k$ be a field. Consider the ring $A=k[x][[t]]$ of formal power series in a variable $t$ over the polynomial ring $k[x]$. This ring contains the ring $B=k[[t]][x]$ of polynomials in the variable §x§ over the power series ring $k[[t]]$. 
I want to understand the intersection of $A$ with the fraction field 
$F=\mathrm{Frac}(B)=k((t))(x)$ of $B$.  It is easy to see that
$A\cap F$ contains the subring $C$ which consists of fractions $f/g$, where $f, g\in B=k[[t]][x]$ with $g$ congruent to a nonzero element of $k$ modulo $t$.
Could anyone give a short argument to show the intersection $A\cap F$ equals (or not) to $C$ ?

REPLY [5 votes]: Equals. $k[[t]]$ is a unique factorization domain, so $k[[t]][x]$ is a unique factorization domain. Thus we can assume that $g$ and $f$ share no common factors. Let $p$ be a prime dividing $g$ that is not a unit mod $t$. Then since its Newton polygon is flat, and the coefficient of some power of $x$ is nonzero mod $t$, the Newton polygon must be horizontal, so the leading coefficient must be nonzero mod $t$.
We know that $p$ divides $f$ in $k[x][[t]]$, so $pa=f$ for some $a$. We will show that $a\in k[[t]][x]$. Assume not, then the degree of $x$ in $a$ at different powers of $t$ is unbounded. Let $n$ be the first power of $t$ where it gets above the degree of $f$ minus the degree of $p$. Modulo $t^{n+1}$, $p$, $a$, and $f$ are all polynomials in $\left(k[t]/t^{n+1}\right)[x]$, and $pa=f$. The degree of $p$ plus the degree of $a$ is greater than the degree of $f$, so the leading coefficients of $p$ and $a$ are zero divisors. But the leading coefficient of $p$ is a unit in $k[t]/t^{n+1}$, so this is impossible. This is a contradiction, so $a\in k[[t]][x]$, so $p|f$ in that ring, so $g$ and $f$ share a common factor, a contradiction.
Therefore, all primes dividing $g$ are units mod $t$, so $g$ is a unit mod $t$, so it is congruent to an element of $k$ modulo $t$.<|endoftext|>
TITLE: Is there any way to generalize the Laplacian to finite groups?
QUESTION [7 upvotes]: The group theoretic interpretation of harmonic analysis was born out of the observation that the discrete Fourier transform on a signal of length $n$ was precisely the Fourier transform of the finite group $\mathbb{Z}/n\mathbb{Z}.$ I was thinking about the extent to which this analogy holds, but I was having trouble with the Laplacian, which, of course, occupies a central role in harmonic analysis. In particular, there doesn't seem to be a generalization of the operator to finite groups (though certainly there is for compact connected Lie groups). 
Does anyone know of such an object?

REPLY [9 votes]: One natural generalization is the center of the group algebra (i.e., algebra of class functions - so in the abelian case you consider, just the group algebra itself). In the continuous case there are many different notions of group algebra, depending on the class of functions you consider, but if you allow a broad enough interpretation of the group algebra this will include the case of the Laplacian (and its higher analogues) in harmonic analysis. These appear from the action of the center of the universal enveloping algebra (which itself can be interpreted via invariant differential operators on the group or distributions supported at the identity). I think it's not egregious to claim that any version of Fourier transform for a group (or symmetric space) in particular simultaneously diagonalizes all of these commuting operators (in the finite abelian case that's literally all it does). However I don't think there's a canonical single operator generalizing the Laplacian (outside of the setting say of simple Lie groups where we take the quadratic Casimir, but that doesn't give something interesting in the finite context).<|endoftext|>
TITLE: Partitioning the vertices of an n-cube with random hyperplane cuts
QUESTION [14 upvotes]: An evolutionary biologist asked me a question which boils down, at least in part, to what seems to me an interesting question of combinatorial/probabilistic geometry.
It is an old chestnut of a problem to ask:  into how many pieces can an n-sphere be cut by k hyperplanes?  (Here I want the hyperplanes to be honest linear subspaces, not affine ones as in the classical "lazy caterer" problem, but the flavor is much the same.)
Now suppose that instead of the sphere, I have the 2^n vertices of the n-cube, i.e. the set {-1,1}^n.  I cut this set with k random hyperplanes.  Now I have a partition of 2^n.
What do I expect this partition to look like?  E.G. how many blocks are there?  How big is the largest blocks?  Are the biggest blocks "close to each other" in the sense that you can pass from one to the other without crossing very many of the hyperplanes?  (To formalize this, one might say that the structure one is studying isn't just a partition, but a partition in which each block is identified with an element of (Z/2Z)^k, thus providing a notion of Hamming distance between block.)
I have asked this question in a rather vague way by not specifying what range of k relative to n is in play.  This should actually be considered part of the question:  what are the threshold curves in the (n,k) plane, if any, where the partition sharply changes its expected nature?  My biologist friend is certainly interested mostly in the case k > n; I think he's most interested in the case where k is bounded between n and a constant multiple of n, but I'm not sure.  I expect he would be interested to know, for instance, how big k needs to be before all blocks of the partition are singletons almost surely.
Further remarks:  though I don't think this is relevant to MBF, one could certainly pass from a discrete to a continuous setting and ask about the statistics of the partition of the volume of the unit (n-1)-sphere by the k hyperplane cuts, which would also be interesting.  Or, instead of letting the cuts be chosen randomly from a continuous distribution, you could let them be chosen from the vertices of a cube in the dual R^n; in other words, you could choose at random from hyperplanes of the form x_1 +- x_2 +- ... +-x_n.  This last version is probably closest to what MBF is actually thinking about.
Update:  A couple of people asked about the biological context.  Here's the original paper.
http://www.ploscompbiol.org/article/info%3Adoi%2F10.1371%2Fjournal.pcbi.1000202

REPLY [4 votes]: I corrected an earlier version which assumed that only adjacent vertices need to be separated to guarantee the parts are singletons, which fedja pointed out was incorrect. Thanks.
Here is a partial answer to the question of how many cuts it takes before the nonempty pieces are singletons with high probability when the hyperplanes are chosen uniformly. Consider the roughly $4^n$ pairs of vertices of the cube. Each pair is separated by a hyperplane iff the partition separates all vertices. The expected number of pairs which intersect no hyperplane is at most $4^n$ times the probability that a particular edge is missed, since adjacent vertices are the least likely to be separated. The probability that none of the $k$ hyperplanes intersects an edge is $p^k$, where $p$ is the probability that each hyperplane misses the edge. 
What does it take for a hyperplane to separate two adjacent vertices? These points determine a great circle, and are at angle $\arccos \frac{n-2}{n} \approx \frac{2}{\sqrt n}$. The hyperplane almost surely intersects this circle in two antipodal points. If these intersect the arc of about $\frac{2}{\sqrt n}$ radians, then the hyperplane intersects the edge. So, the probability that a random hyperplane intersects an edge is about $\frac{2}{\pi\sqrt n}$. The probability the edge is missed is the complement, about $1- \frac{2}{\pi\sqrt n}$. The probability all $k$ hyperplanes miss this edge is about $(1- \frac{2}{\pi\sqrt n})^k \approx \exp(-\frac{2k}{\pi\sqrt n})$. The expected number of pairs not separated by any hyperplane is at most about $4^n \exp(-\frac{2k}{\pi\sqrt n})$.
If you choose $k\approx c n^{3/2}$ then the expected number of pairs not separated by any hyperplane is at most $1$. For much larger $k$ the expected number of pairs of vertices in the same part, hence the probability that two vertices are in the same part, becomes small.<|endoftext|>
TITLE: In search of an early picture of Max Dehn
QUESTION [12 upvotes]: I am trying to find a copy of a picture "Mathematische Gesellschaft: 
Group Portrait, Faculty, University of Göttingen (1899)."
This picture was published by Springer-Verlag as a poster in 1985,
but Springer has been unable to find a copy for me. I also sent an email
query to the Mathematics library at Göttingen, but received no response.
So, as a last resort, I would like to ask whether any MO members have
access to a copy of the picture. I want to use it in a talk about Max 
Dehn I am giving in Frankfurt in June, because Max Dehn appears in the 
picture.

REPLY [24 votes]: As a matter of fact, the poster is right on the other side of the hall in front of my office. Here is a picture of Dehn that I've made with my phone:

I could make a better picture on Monday if that would be helpful...
Edit: Ok, my next better camera produced this picture:

Probably that's already near the best you can get because you can already see the halftoning effects pretty clear...<|endoftext|>
TITLE: Existence of prime ideals and Axiom of Choice.
QUESTION [11 upvotes]: One of the must obvious equivalences of Axiom of Choice is the converse of Krull Theorem.
Bernhard Banaschewski in the Article titled by A New Proof that “Krull implies Zorn” showed a very simple proof of the following statement.
Theorem: the existence of maximal ideals in a ring with unity is equivalent to Axiom of choice.
This means that every attempt to prove the existence of maximal ideals is related to apply the  Axiom of Choice.
Another important theorem in commutative algebra is  Cohen's theorem, which tells us that if $R$ is a commutative ring with unity and  $I$ is an ideal of $R$ disjoint from a multiplicative closed subset $S\subset R$, then there exists a prime ideal $P$ so that $I \subset P$ and $P\cap S=\varnothing$. 
Cohen's theorem implies that In a commutative ring with unity there exists a prime ideal.
Notice that this prime ideal need not be a maximal ideal but we need to apply Zorn's Lemma to show the existence of it. Now Here are my Questions:

Is it true that  For Showing the existence of prime ideal in a commutative ring with unity we need the Axiom of choice or we can show the existence of it without applying this Axiom?
If the Answer of above Question is negative, what kind of Axiom weaker than Axiom of choice is needed to show the existence of prime ideals in a commutative ring?
What kind of relation is between the Axiom of countable choice and The existence of prime ideals in a commutative ring with unity?

REPLY [2 votes]: In light of Laurent Moret-Bailly's comment under Martin's answer, let us supplement that answer by showing how existence of prime ideals in finitely generated rings may be proven in ZF without a choice principle (not even the ultrafilter principle). In fact we will show that countable rings $R$ have prime ideals, using the weak König lemma. 
(Recall the weak König lemma: if $T$ is an infinite subtree of the infinite binary tree $2^\ast$ [the nodes of $2^\ast$ are finite words in letters $0, 1$; the parent of any nonempty word is obtained by omitting its last letter], then $T$ has an infinite branch. This is easily proven in ZF; see e.g. here.)  
Fix an enumeration $a_0, a_1, a_2, \ldots$ of elements of $R$. Build a tree $T$ whose nodes $w$ are labeled by ideals $P_w$ of $R$, by induction. $T_k$ will denote the set of nodes of length $k$, so $T_0$ consists of just the empty word $e$, which we label with the zero ideal. Assuming $T_{k-1}$ and the ideals $P_w$ for words $w \in T_{k-1}$ are given, define $T_k$ and the ideals there according to two cases; for this it is convenient to write $k = 2l + m$ where $m \in \{0, 1\}$ and $l$ is used to code a pair of natural numbers $(i, j)$ by the usual trick, viz. $l = \binom{i+j+1}{2} + j$. 
Case $m = 0$: for each $w \in T_{k-1}$ such that $a_i a_j$ belongs to $P_w$, put $w0, w1$ in $T_k$, and define $P_{w0} := P_w + \langle a_i \rangle$, and $P_{w1} := P_w + \langle a_j \rangle$. For any other $w \in T_{k-1}$, put $w0$ in $T_k$, and define $P_{w0} = P_w$. 
Case $m = 1$: for each $w \in T_{k-1}$ such that $1 \notin P_w$, put $w0$ in $T_k$, and define $P_{w0} := P_w$. If $1 \in P_w$, then put neither $w0$ nor $w1$ in $T_k$. 
$T$ is infinite: it is easy to prove by induction that for each $k$ there is $w \in T_k$ such that $1 \notin P_w$. If $b$ is an infinite branch through $T$, then put $P = \bigcup_{w \in b} P_w$. Then $P$ is a proper ideal by the recipe of case $m=1$, and is prime by the recipe of case $m=0$.<|endoftext|>
TITLE: Chern classes of vector bundles on a stack
QUESTION [7 upvotes]: Is there literature about chern classes of vector bundles on DM-stacks? I had a look at a lot of different papers about intersection theory on stacks and related stuff and this seems to be known, but I couldn't find a good reference on this topic. The strongest statement I could find was "there are chern classes in the Chow-cohomology satisfying all the usual properties".
A particular thing I am interested in, is the projection formula, i.e. given a proper morphism of DM-stacks $p:M\longrightarrow N$, a vectorbundle $E$ on $N$ and a cycle $\alpha\in A_*(M)_\mathbb{Q}$ is it true that $p_*(c_i(p^*E)\cap\alpha)=c_i(E)\cap p_*\alpha$? I guess this is one of the usual properties, but does it hold for any proper morphism, or only representable ones? 
What I did was trying to check the definition of the Chow-cohomology from Vitsoli's paper on Intersection theory on stacks for chern classes, but I got already stuck at the compatibility with the Gysin homomorphism, because of my poor knowledge about normal bundles and stacks... so I gave up trying to prove it myself.
Can anyone help me out with a hint to more detailed literature, or some hints how the proof of the projection formula works? I would also be happy with some precise statement under which circumstances the projection formula holds. Thank you in advance.

REPLY [8 votes]: The formula holds for arbitrary proper morphisms. If $X \to Y$ is a proper morphism of DM stacks, where $X$ has finite inertia (the hypotheses in my paper are more stringent, but the theory has been refined since then), there exists a finite map $V \to X$, where $V$ is an algebraic space, and the pushforward from the rational Chow group of $V$ to that of $X$ is surjective. This allows to reduce the non-representable case to the representable one.
About the more general question, I am reasonably certain that the construction of Chern classes in Fulton's book goes through in the formalism in my old paper without any major problem, but it's been a while since I wrote it, so it's hard to be completely positive.<|endoftext|>
TITLE: Primary ideals of the polynomial ring
QUESTION [7 upvotes]: Is it possible to classify all the primary ideals of the polynomial ring $K[X_1,\ldots ,X_n]$ where $K$ is a field.
Or, give a big class of examples of primary ideals which are not prime ideals.

REPLY [8 votes]: When you restrict to special classes like monomial or binomial ideals (those generated by polynomials with one (monomial) or two (binomial) terms) then combinatorial characterizations exist. For instance, a monomial ideal $I\subset K[X_1,\dots,X_n] =:S $ is primary if and only if in the quotient $S/I$ every image of a variable is either regular or nilpotent.  For binomial ideals the story is more complicated but things can be said.  See Eisenbud/Sturmfels "Binomial ideals", Dickenstein/Matusevich/Miller, "Combinatorics of binomial primary decomposition", and Kahle/Miller "Decompositions of commutative monoid congruences and binomial ideals".

REPLY [6 votes]: Symbolic powers give big classes of primary ideals which are not prime.  Recall that given a prime $P$, the $n$th symbolic power of $P$ (denoted by $P^{(n)}$) is the $P$-primary ideal in the minimal prime decomposition $P^n$.  It can also be described as $(P^n R_P) \cap R$.
It is perhaps surprising that $P^{(n)} \neq P^n$, but the only real large class of ideals where that always happens is for prime ideals which are cut out by regular sequences (ie, complete intersections).
In algebraic geometry, $P$-primary ideals show up frequently when $P$ is a height one ideal.  Indeed, suppose that $D$ is a prime divisor on a normal variety $X = \text{Spec} R$.  Then
$$\Gamma(X, \mathcal{O}_X(-nD)) = P^{(n)}$$
where $P$ is the prime ideal defining $D$.  For example, consider the prime $P = (x,y) \subseteq k[x,y,z]/(x^2 - yz)$, it is a good exercise to verify that $P^{(2)} \neq P^2$ in this example.  
A poor classification: Based upon the symbolic power idea, one can classify $P$-primary ideals as follows (for any prime $P$).  The set of $P$-primary ideals is equal to the set of all 
$$
Q \cap R
$$
where $Q$ runs over all ideals of $R_P$ such that $\sqrt{Q} = P R_P$.<|endoftext|>
TITLE: Asymptotics of the n-th prime using the gamma function
QUESTION [13 upvotes]: In the paper http://rgmia.org/papers/v8n2/eepnt.pdf, the author proves that proves an explicit inequality on prime numbers using the gamma function and as a corollary, he showed that.
$$
p_n = n \frac{\Gamma'(n)}{\Gamma(n)} + o(n \ln n).
$$
I obtained a stronger form of this result namely
$$
p_n = n \ln \frac{\Gamma'(n)}{\Gamma(n-1)} + O\Big(\frac{n\ln\ln n}{\ln n}\Big).
$$
The gamma function seems to beautifully approximate $p_n$. To get the same error term using the regular Cipolla's asymptotic expansion of the $p_n$  we would need three terms.
Can someone explain why the gamma function approximated the n-th prime so nicely? Is this a coincidence or is there some underlying phenomenon governing this result that can shed some new light distribution of prime numbers.

REPLY [7 votes]: Can someone explain why the gamma function approximated the n-th prime so nicely?  Is this a coincidence or is there some underlying phenomenon governing this result?

The expression which appears there is $\dfrac{\Gamma'(n)}{\Gamma(n)}=\Big[\ln\Gamma(n)\Big]'=\psi_0(n)$, see digamma function. 
This famous function has the property that $\psi_0(n)=H_{n-1}-\gamma$, where $H_n$ is the n-th harmonic  number, and $\gamma\simeq\dfrac1{\sqrt3}$ is the Euler-Mascheroni constant. But, at the same time, $H_{n-1}\simeq\ln n$, 
the small error or difference between the two tending towards the afore-mentioned constant. By 
differentiating the natural logarithm of the fundamental property of the $\Gamma$ function, $\dfrac{\Gamma(n+1)}{\Gamma(n)}=n$, 
we have that $\psi_0(n+1)=\dfrac1n+\psi_0(n)$, which is nearly identical to the recurrence relation of 
harmonic numbers, $H_{n+1}=\dfrac1{n+1}+H_n$. The only difference is the offset, $\psi_0(1)=-\gamma\neq$ 
$\neq H_0=0$, a proof of which can be found here. The conclusion then follows from the prime number theorem.<|endoftext|>
TITLE: Subgroups of amenable periodic groups 
QUESTION [12 upvotes]: Does every countable, infinite, amenable, periodic group $G$ contain an infinite locally finite subgroup? 
Remarks:

I would be happy with an infinitely generated counterexample as long as it is countable.
A counterexample cannot be elementary amenable, as
elementary amenable periodic groups are locally finite.
First Grigorchuk's group is not a counterexample: it is a 2-group, and hence it contains an infinite abelian group (by a result of Held, "On abelian subgroups of infinite 2-groups").
Amenability of $G$ is essential, due to existence of Tarski monsters whose subgroups
are finite cyclic (constructed by Olshanskii).
Any infinite locally finite group contains an infinite abelian subgroup
(see a paper of Hall-Kulatilaka here, so it is equivalent to ask whether $G$ contains an infinite abelian subgroup.
This question was asked here in 2008, so perhaps it is an open problem.

REPLY [6 votes]: @Ashot, Grigorchuk $p$-groups are branch, so one can easily construct infinite abelian subgroups as well. As generators one can just take elements from rigid stabilizers of different vertices, such that none of these vertices is a prefix of the other. So these groups don't provide counterexamples.<|endoftext|>
TITLE: Simple Lie algebras and Jordan decomposition
QUESTION [6 upvotes]: Let $F$ be an algebraically closed field and let $L$ be a simple Lie algebra of dimension $n$ over $F$. Let $ad: L\longrightarrow  End_F(L)$ denote the adjoint representation of $L$. If $F$ has characteristic zero, then it is well-known that, for every linear transformation $f\in ad(L)$, the semisimple and nilpotent parts of $f$ are also contained in $ad(L)$ (see e.g. Section 5.4 in the Humphreys's book "Introduction to Lie algebras and representation theory").
In general, the same conclusion is not true if $F$ has characteristic $p>0$. Is it possible to have the extreme case in which $ad(L)$ does not contain any nonzero semisimple linear transformation of the vector space $L$?

REPLY [15 votes]: Any finite dimensional simple Lie algebra over an algebraically closed field of characteristic $p>3$
contains a nonzero element $x$ such that $ad(x)$ is semisimple. This is a nontrivial fact, and the only proof I can think of relies on the classification. For $p>3$, the simple Lie algebras split
into three families: Lie algebras of simple algebraic groups and their quotients, filtered Lie algebras of Cartan type, and the Melikian algebras (which only occur when $p=5$). The first class is easy to sort out, of course. The third class has a very explicit presentation which is easy to find in the literature. Then one can see straight away that each Melikian algebra $\mathcal{M}(m,n)$ admits two commuting linearly independent elements $t_1,t_2$ sich that $ad(t_i)^p=ad(t_i)$ for $i=1,2$. 
The real issue with this problem (as well as with many similar problems) is the existence of filtered Lie algebras of Cartan type $H$ corresponding to Hamiltonian forms of the first kind. Such Hamiltonian forms involve parameters which are sometimes organised as sets of matrices in canonical Jordan form. Sadly, the commutator relations in the corresponding Lie algebras depend on these parameters as well, which makes computations an unpleasant experience.
However, the good news is that any finite dimensional simple Cartan type Lie algebra $L$ contains a unique maximal subalgebra of smallest codimension. It is called the  $standard\ maximal\ subalgebra$ of $L$ and often denoted $L_{(0)}$. Skryabin proved in [Comm. Algebra, 23 (1995), 1403-1453] that under our assumptions on $p$ the subalgebra $ad(L_{(0)})$ of $\mathfrak{gl}(L)$ is closed under taking $p$-th powers; see Theorem 2.1 in $loc.\ cit$. 
By maximality, $L_{(0)}$ does not consist of $ad$-nilpotent elements. Since $ad(L_{(0)})$ is closed under taking $p$-th powers, it is straightforward to see that there is a nonzero element $x\in L_{(0)}$ such that $ad(x)$ is semisimple (in fact, one can say a lot more than that).
I have no idea as to what happens when $p$ is $2$ or $3$ as we have no classification in these characteristics. This adds to a long list of open problems of which my favourite is: does there exist a finite dimensional simple Lie algebra $L$ with a finite automorphism group. This never happens when $p>3$ and there is a rather short proof of this fact in the literature.<|endoftext|>
TITLE: Regarding Cayley Graphs of Property (T) Groups
QUESTION [17 upvotes]: A well-known application of Kazhdan's Property (T) is the construction of expander graphs. Background on this is discussed, for example, in this post on Terry Tao's blog. Essentially, Cayley graphs of finite quotients of property (T) groups can give us families of expanders (see Exercise 14 of Tao's blog post). The construction seems to critically use the finite quotients to obtain the unitary representations required to employ the definition of property (T). It would be very nice to have an answer to the following:

Question: Is any of the behavior of expander graphs reflected in the (infinite) Cayley graph of a property (T) group with respect to a finite, symmetric generating set?

Please provide a reference to anything in the literature that sheds some light on this.
A somewhat broader (related) question that may be helpful:

What are some qualitative properties of the Cayley graph of a property (T) group?

For example, does the Cayley graph of a property (T) group exhibit any sort of (local) concentration of measure phenomena using the word metric w.r.t. a finite generating set? 
What are some useful intuitions for the Cayley graph of a property (T) group? (Here I'm wondering if there is anything akin to the image of "thin triangles" for hyperbolic groups.)

REPLY [5 votes]: This is not an answer but a collection of observations, some of which, are, hopefully, interesting. 
First, recall that Cheeger constant $h(X)$ for a complete open Riemannian manifold $X$ is 
$$
\inf_D \frac{|\partial D|}{|D|}
$$
where infimum is taken overall bounded domains $D\subset X$ with smooth boundary and vertical bars denote volume (of appropriate dimension). In the context of infinite connected graphs, the same definition applies. Then $h(X)=0$ iff $X$ is amenable. Cheeger constant is also defined for closed manifolds (finite graphs) where one looks at the infimum of the quantity
$$
\frac{|S|}{|M_S|}
$$
where $S$ is a cooriented  hypersurface in $M$ and $M_S$ the "positive side" of $S$ in $M$, and the infimum is taken over all cooriented hypersurfaces $S$. A sequence of closed connected manifolds $M_i$ (of "uniformly bounded geometry", i.e., constant dimension, with fixed bounds on curvature and a lower bound of the injectivity radius) is an "expander" if $h(M_i)$ is uniformly bounded below, away from zero, while diameters of $M_i$ diverge to infinity. Same definition applies to graphs where bounds on geometry are replaced by uniform bounds on valence. In other words, in the context of finite graphs or sequences of closed manifolds of bounded geometry, being an "expander" is essentially the negation of amenability. The question raised by Jon is if one can define an expander in the context of a single infinite graph or, I would say, a single open Riemannian manifold $X$ of bounded geometry. Defining an expander as before, by saying $X$ is nonamenable, is too weak. The usual link between expanders and open Riemannian manifolds/infinite graphs is via taking quotients: If $\Gamma_i$ be a infinite sequence of finite-index subgroups of a fixed discrete cocompact subgroup $\Gamma\subset Isom(X)$, then the sequence of quotients $M_i:=X/\Gamma_i$, is an expander. One then could say that $X$ itself an expander if it admit a sequence of group actions $\Gamma_i$ satisfying the above conditions. Note that the quotient manifolds $M_i$ "approximate" $X$, provided that intersection of the subgroups $\Gamma_i$ is trivial. 
The question is how to eliminate groups in this definition, since some infinite f.g. groups with property T could be simple or at least, contain no proper finite-index subgroups.  
I will now proceed in the category of Riemannian manifolds, but the discussion is equivalent to the one in the case of graphs. Let $Q_i\subset X$ be fundamental domains for the groups $\Gamma_i$. Let $S\subset M_i$ be a cooriented compact hypersurface, transversal to the projection of the boundary of $Q_i$, and let $T\subset Q=Q_i$ be preimage of $S$ in $Q$. Note that $T$ is not properly embedded in $Q$. 
For each $T\subset Q$  we have the ratio
$$
\rho_{T,i}=\frac{|T|}{|Q_T|},
$$
where $Q_T$ is "positive side" of $T$ in $Q$. Then, 
$$
\inf_{T,i} \rho_{T,i}
$$
is still bounded away from zero, provided that $\Gamma$ has property T. Here infimum is taken over all $T_i\subset Q_i$, so that $T_i$ is the preimage of a closed surface in $M_i$. (Yves exercise amounts to saying that $\Gamma$  having property T depends only on $X$ and not on the particular group acting cocompactly on $X$.) 
One then can attempt to promote this to a definition: We say that a sequence of domains $Q_i\subset X$ is "anti-Folner" if the infimum of ratios $\frac{|T|}{|Q_{i,T}|}$ defined as above (using arbitrary cooriented hypersurfaces $T\subset Q_i$) is positive and so that $diam(Q_i)\to \infty$. One then would say that $X$ is an expander itself if it admits an anti-Folner sequence of connected bounded domains $Q_i\subset X$. 
One can give a "spectral" interpretation of the anti-Folner property. Namely, for each $Q$ define
$$
\mu(Q):=\inf \frac{ \int_Q |\nabla f|^2}{\int_Q |f|}
$$
where infimum is taken over all smooth functions on $Q$. Then as in the proof of Cheeger's theorem, one gets: If $\inf \mu(Q_i)>0$ then the sequence $Q_i$ is anti-Folner.  
Note that amenability of $X$ is equivalent to $\lambda(X)=\inf \lambda(D)=0$, where $\lambda(D)$ is the lowest positive eigenvalue of a bounded domain $D\subset X$ and 
$$
\lambda(D)= \inf \frac{ \int_Q |\nabla f|^2}{\int_Q |f|},
$$
where infimum is taken over all smooth compactly supported functions on $D$. So the difference between $\lambda$ and $\mu$ is that in the definition of $\mu$ we allow functions in bounded domains in $X$ that do not have compact support. 
What's missing here is a proof that $\mu$ is bounded away from zero in some interesting examples, since, in the context of sequences of discrete groups $\Gamma_i$, I was using not all but only some hypersurfaces in $Q_i$ (the ones which came from compact hypersurfaces in $M_i$).<|endoftext|>
TITLE: K-Theory of Schemes: Monoidal vs. Exact
QUESTION [7 upvotes]: There are several ways for defining the K-Theory of a category depending on which structure it admits. The K-Theory of schemes is commonly defined as the "group completion" of the category of algebraic vector bundles with respect to its exact category structure rather than with respect to its symmetric monoidal category structure.
In general, these two are not the same and the former is a quotient of the latter. In particular, for schemes these two constructions are not equivalent since exact sequences of algebraic vector bundles do not always split (e.g. projective line $\mathbb{P}_R^1$).
My question: why is the "correct" definition to choose the exact category structure? What goes wrong when one just considers the monoidal category structure which seems more natural since that is how for instance K-Theory of rings is defined.

REPLY [10 votes]: There is no "universally correct" definition, but $K^0(X)$ usually denotes the K-theory of the exact category of vector bundles over $X$. As you say, there are various approaches for (higher) algebraic K-theory (summarized in Weibel's K-book, for instance), and which one one uses depends on the applications, but also on the realizability. Namely, there are lots of exact sequences of vector bunbles on projective varieties which give you relations in the K-theory, but these exact sequences don't split. Although this makes $K^0(X)$ relatively small compared to $K^0_{\oplus}(X)$ (the K-theory of the monoidal category of vector bundles w.r.t. $\oplus$) and perhaps doesn't contain enough information for what you want, it is at least computable! For example, $K^0(\mathbb{P}^r) \cong \mathbb{Z}[x]/(x^{r+1})$ is well-known (SGA 6), but little is known about $K^0_{\oplus}(\mathbb{P}^r)$ (cf. MO/20444): It is the free abelian group on indecomposable vector bundles on $\mathbb{P}^r$ (Atiyah), but these have not been classified yet.
Grothendieck's celebrated and important resolution theorem won't tell you anything about $K^0_{\oplus}(X)$ because of the appearance of exact seqeunces and alike, so there won't be a nice comparision between K-theory and G-theory. Many other theorems for $K^0$, such as Grothendieck's homotopy-invariance $K^0(X) \cong K^0(X \times \mathbb{A}^1)$ will probably fail for $K^0_{\oplus}$. The upshot is: Although $K^0_{\oplus}$ contains more information, it is not flexible enough for computations.
If $X$ is an affine scheme, the canonical map $K^0_{\oplus}(X) \to K^0(X)$ is an isomorphism. Because of that, many authors define $K_0(R)$ of a ring $R$ to be the free abelian group on iso-classes of f.g. proj. $R$-modules modulo the relation $[P \oplus Q] = [P]+[Q]$, but this should not be seen as the correct definition for the general case of a scheme. In particular I don't agree with your sentence "... the monoidal category structure which seems more natural since that is how for instance K-Theory of rings is defined."

REPLY [7 votes]: You don't need to go to non-affine schemes to run into trouble with your proposed definition.
In the simplest case, if $R$ is a regular ring and $f$ a non-zero element, you want a long exact localization sequence relating the $K$-theory of $R$ to the $K$-theory of $R[f^{-1}]$.  The missing term turns out to be the $K$-theory of the category of finitely generated $R/fR$-modules.  In most cases $R/fR$ won't be regular (so you can't just replace this with the category of projective modules), and this category won't have the property that all short exact sequences split.   Therefore it matters which $K$-theory you use, and it's the $K$-theory of exact categories that makes the localization theorem true.<|endoftext|>
TITLE: Affine "real algebraic geometry" of hyperbolic space?
QUESTION [15 upvotes]: Real algebraic geometry, at least to start with, traditionally studies the zero-sets of real polynomials in a given set of variables.  But treating, say, the Euclidean plane as an uncoordinatized metric space, one may still consider the ring of functions generated by all functions $D_p(\cdot)=d(\cdot,p)^2$, and also constant functions.  Then one has really lost nothing, because (bring back coordinates $x,y$ just for the moment),
$x=(D_{(0,0)}-D_{(1,0)} +1)/2$ and $y=(D_{(0,0)}-D_{(0,1)} +1)/2$.
Since the ring definition here makes sense over any metric space, the possibility of a generalized real algebraic geometry arises. 
Questions: Does this line of thought occur in the literature?  Independently of the geometric optic, does this sort of ring arise anywhere in the literature (other than in the motivating example)?  Do complexification and projectivization have well-behaved analogues in this context?
I'm particularly interested in the properties of "real algebraic curves" in the hyperbolic plane. I have narrower questions, but I'll save them until I see  response to this.

REPLY [5 votes]: Upon Andre's request, I am rewriting my comments as an answer, even though I am on a somewhat shaky ground since my experience with infinite-dimensional algebraic geometry is very limited. 
It seems that for "most" metric spaces (and hyperbolic space is one of the "most") distance functions (or their squares) will span an infinite-dimensional real vector space. I checked this in the case of distance functions on the hyperbolic space (of dimension $\ge 2$, of course) and the same should hold for the distances squared. The point is that for every metric space $X$, the map $i=i_X: x\mapsto d_x$, where $d_x: X\to {\mathbb R}$ is the function $d_x(y)=d(x,y)$, determines an isometric embedding from $X$ to a linear subspace $V$ in $C(X)$ (space of continuous functions on $X$) with the sup-norm on $V$, where $V$ is the linear span of the image of $i$. On the other hand, hyperbolic space (of dimension $\ge 2$) does not embed isometrically in any finite-dimensional Banach space. Thus, doing algebraic geometry on the hyperbolic space becomes (in my mind) a somewhat daunting task since you have to consider the ring of polynomial functions ${\mathbb R}[V]$ on the infinite-dimensional vector space $V$. In particular, you loose the Noetherian property which makes life difficult. The only context where I have seen infinite-dimensional algebraic geometry is the ind-schemes. Ind-schemes appear naturally when one works with, say, affine Grassmannians and which I had to do exactly once in my life (I mean, thinking of affine buildings in algebro-geometric terms). As far as I can tell, dealing with the ind-scheme based on the ${\mathbb R}[V]$ for the hyperbolic space, would amount to considering geometry of finite configurations of points in ${\mathbb H}^n$ (and, occasionally, lines). I could be mistaken, but for the union $Y$ of two distinct geodesic segments in the hyperbolic space, linear span of the image of $i_Y$ (where we use the restriction of the distance function from ${\mathbb H}^n$) will be infinite-dimensional, so you are not allowed to use more than one geodesic. While geometry of finite subsets of hyperbolic spaces has some uses (see the discussion of the sets $K_m$ below), it strikes me (I am a hyperbolic geometer) as mostly boring. (Maybe logicians can add something interesting here since considering finite subsets in ${\mathbb H}^n$ we are dealing with the elementary theory of the hyperbolic space.)  
Gromov (see e.g. his book "Metric Structures for Riemannian and Non-Riemannian Spaces") introduces, for every metric space $X$, the collection of subsets $K_m(X)\subset {\mathbb R}^N$, $N=\frac{m(m-1)}{2}$. The set $K_m(X)$ consists of $N$-tuples of pairwise distances between various $m$-tuples of points in $X$. Then $K_3(X)$ (under some mild assumptions on $X$, e.g., $X$ is an unbounded geodesic metric space) is defined by triangle inequalities and nothing else. The set $K_4(X)$ is quite interesting, since all the "curvature" conditions on metric spaces are defined via quadruples of points. However, Gromov could not come up with any interesting uses for $K_m(X), m\ge 5$, and I do not know what to make of these sets either. Thus, the ind-scheme approach to the algebraic geometry of the hyperbolic space might yield something geometrically interesting for quadruples of points, beyond which algebraic geometry is likely to get disconnected from (geo)metric geometry. 
Lastly, MO discussion at Is there an algebraic approach to metric spaces? is related to David's question.<|endoftext|>
TITLE: extension of surface homeomorphism
QUESTION [15 upvotes]: Can anyone give me a reference (or proof sketch) for the fact that there are psuedo-Anosov diffeomorphisms of closed hyperbolic surfaces which do not extend over any handlebody? Thanks.

REPLY [2 votes]: I accidentally found this Rice University thesis (never published, it seems) which studies the question in some depth:
Author Jamie Bradley Jorgensen
Title Surface homeomorphisms that do not extend to any handlebody and the Johnson Filtration.<|endoftext|>
TITLE: Using Exterior Algebras in combinatorics
QUESTION [10 upvotes]: As addressed in this past question, there are many applications of linear algebra to combinatorics. What about examples of applications of exterior algebras? Part 4 here is one such example. What are the heuristics for when to apply them and the intuition for how to do so?

REPLY [2 votes]: Probably the main thing about the exterior algebra is that it is intimately related to the determinats of matrices, and it can be effeciently used to obtain certain statements which includes matrix determinants like det(AB)=det(A)det(B), Cramer's formula for A^{-1},  Plucker relations, Jacobi ratio theorem, etc. (Example on MO).
On the other hand there are cerain amount of papers from combintatorial community 
(names include G.-C. Rota, D. Zeilberger, D. Foata) to prove similar relations by combinatorial means.
Hope this might be considered as an application or at least as a relation.
Before giving some concrete examples let me mention that both techniques (combinatorial and exterior algebra) allows to handle matrices with non-commutative elements, which is of certain interest in representation theory and quantum integrable systems.
Example 1. Capelli identity det(A)det(B) = det(AB+ correction)
There is a paper 
Combinatorial Proofs of Capelli's and Turnbull's Identities from Classical Invariant Theory
Dominique Foata, Doron Zeilberger
And some powerful generalization:
Noncommutative determinants, Cauchy-Binet formulae, and Capelli-type identities
Sergio Caracciolo, Andrea Sportiello, Alan D. Sokal
Both combinatorial in nature.
Grassman algebra proof and genealization were given in Algebraic properties of Manin matrices . 
Let me mention that G.-C. Rota was also interested in Capelli identies (G.C. Rota, Combinatirial theory and invariant theory, NSF advanced theory in combinatorial theory, Bowdain College, Maine, Notes by L. Guibas, 1971).
Which influenced S. G. Williamson, Symmetry operators, polarizations, and a generalized Capelli identity, 
Linear Multilinear Algebra , vol. 10, no. 2, pp. 93-102, 1981 DOI: 10.1080/03081088108817399 
Remarkable paper which is undesrvedly forgotten, results partly rediscovered by A. Okounkov and M. Nazarov decades later.  
Example 2. Matjaž Konvalinka (who was a student of Igor Pak in MIT at that time) obtained combinatorially the proofs the so-called Jacobi, Sylvester theorems
for some matrices with non-necessary commutative entries (Manin matrices). 
(These papers are motivated by other Zeilberger's&K paper - I already discussed this in another answer to this question).
Exterior algebra proofs can be found again here: 
 Algebraic properties of Manin matrices<|endoftext|>
TITLE: Geometric interpretation of the exact sequence for the cotangent bundle of the projective space
QUESTION [6 upvotes]: Edit: As Dan Petersen pointed out, this question is a duplicate of a  previous one. I would leave it for the moderators to decide if this should be closed. On the other hand, may be this should be left open on the merit of the excellent answers and comments (@Emerton: Thanks!).
I was trying to understand the following exact sequence (for $X := \mathbb{P}^n_k$, where $k$ is an algebraically closed field):
$$0 \to \Omega_X \to \mathcal{O}_X(-1)^{n+1} \to \mathcal{O}_X \to 0$$
The explanation (as in the proof of Theorem II.8.13 of Hartshorne) is given by some algebraic formulae, which I am having trouble to digest. I was trying to see in more geometric terms what is going on, and was somewhat successful in the case of the surjection $\mathcal{O}_X(-1)^{n+1} \to \mathcal{O}_X$, namely: we can regard $\mathcal{O}_X(1)$ (respectively $\mathcal{O}_X(-1)$) as the normal bundle $\mathcal{N}$ of (respectively conormal bundle) of $X$ in $Z := \mathbb{P}^{n+1}_k$. Any global section of $\mathcal{O}_X(1)$ therefore induces a map (via evaluation) from $\mathcal{O}_X(-1)$ to $\mathcal{O}_X$. The above surjection comes from taking $n+1$-linearly independent global sections of $\mathcal{O}_X(1)$. 
But I do not understand how to interpret the injection $\Omega_X \to \mathcal{O}_X(-1)^{n+1}$. How would someone 'naturally' come up with the algebraic formula?

REPLY [6 votes]: By dualizing and twisting we obtain the equivalent exact sequence of vector bundles
$$0\to \tau\to \mathbb P^n_k\times k^{n+1} \to T_{\mathbb P^n}(-1)\to 0              \quad (*)       $$
The first morphism is just the inclusion of the tautological vector bundle $\tau$ into the trivial bundle and is geometrically transparent.
To understand the second morphism geometrically, fix a point $p\in \mathbb P^n_k$ and the corresponding line $l\subset \mathbb P^n_k$ (I forgot to say I'm using the pre-Grothendieck definition of projective space as a set of lines) .
At $p$ the exact sequence  $(*)$ becomes the exact sequence of vector spaces$$0\to l\to  k^{n+1} \to T_{\mathbb P^n}[p]\otimes l\to 0$$
Exactness then translates into the canonical isomorphism  $$T_{\mathbb P^n}[p] = \mathcal L(l,k^{n+1}/l) \quad (**)$$  
So the whole problem boils down to understanding $(**)$, i.e.understanding in a canonical way  the fiber of the tangent bundle to $\mathbb P^n$ at a point $p=(a_0....:a_n)\in \mathbb P^n$.
Here is the idea  inspired by differential geometry.   
The  "curve"  $\epsilon \mapsto (a_0+\epsilon t_0,....,a_n+\epsilon t_n)\; (\epsilon^2=0)$ [algebraic geometers consider  very short curves!] gives rise to a tangent vector $t\in T_{\mathbb P^n}[p]$.
The canonically associated linear map $\lambda _t:l\to k^{n+1}/l$ is then characterized by the condition $$\lambda _t(a_0,...,a_n)=\overline {(t_0,...,t_n)}  $$
[Be careful that if you change the vector $(a_0,...,a_n)$ representing $p$ to a colinear vector $(a_0',...,a_n')$, you also have to change $(t_0,...,t_n)$ to another $(t_0',...,t_n')$]
The details are in Dolgachev's online notes , Example 13.2<|endoftext|>
TITLE: The number of group elements whose squares lie in a given subgroup
QUESTION [16 upvotes]: This number is divisible by the order of the subgroup  http://arxiv.org/abs/1205.2824.
The proof is short but non-trivial. Is this fact new or is it known for a long time?

REPLY [6 votes]: In 2017, we learned that this fact was proven by Shiro Iwasaki in 
1982.<|endoftext|>
TITLE: $L^\infty$ properties of an infinite-dimensional Gaussian semigroup
QUESTION [5 upvotes]: Let $W$ be a separable Banach space and $\mu$ a Gaussian Borel measure on $W$ which is centered and non-degenerate.  For $F : W \to \mathbb{R}$ bounded Borel and $t \ge 0$, let
$$P_t F(x) = \int_W F(x+\sqrt{t}y)\,\mu(dy)$$
be the transition semigroup of Brownian motion on $W$.

Is $P_t$ well defined as an operator from $L^\infty(W,\mu)$ to itself?  That is, if $F=0$ $\mu$-almost everywhere, must the same be true of $P_t F$?

Of course, if $W$ is finite dimensional the answer is yes, because the measures $\mu(x +  \cdot)$ are mutually absolutely continuous as $x$ ranges over $W$.  This suggests that the answer may be no in infinite dimensions, but I can't seem to find a counterexample.

REPLY [5 votes]: George Lowther's answer (thanks George!) set me on the right track.  Here's a note that the same argument can be used for arbitrary infinite-dimensional $W$.
Let $\mu_t(A) = \mu(\frac{1}{\sqrt{t}} A)$, so that $P_t F(x) = \int_W F(x+y)  \mu_t(dy)$.  Since $\mu_t$ is a convolution semigroup, we have $\int_W P_t F(x)\mu(dx) = \int_W F(x) \mu_{1+t}(dx)$.  (Or written another way, it's $P_{1+t}F(0)$.)
We can choose a sequence $f_1, f_2, \dots \in W^*$ which, viewed as random variables on the probability space $(W,\mu)$, are iid $N(0,1)$.  (Equip $W^*$ with the $L^2(\mu)$ inner product $\langle f ,g \rangle = \int_W f(x) g(x) \mu(dx)$ and use Gram-Schmidt.)  Then on $(W,\mu_t)$ they are iid $N(0,t)$.   Let
$$A_t = \left\{x \in W : \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n |f_k(x)|^2 = t\right\}.$$
Clearly the $A_t$ are Borel, pairwise disjoint, and by the strong law of large numbers $\mu_t(A_t)=1$.  If we take $F$ to be the indicator of $A_1^C$, then $F = 0$ $\mu$-a.e., but for every $t > 0$, $$\int_W P_t F(x) \mu(dx) = \mu_{t+1}(A^C) = 1.$$  Indeed, since $P_t F \le 1$, we have $P_t F = 1$ $\mu$-a.e.
This also works if $W$ is replaced by any reasonable infinite-dimensional TVS (say, Hausdorff and locally convex).<|endoftext|>
TITLE: Energy of a graph
QUESTION [12 upvotes]: Let $G=(V,E)$ be a finite graph with weights $\phi: V\to \{1,...,|V|\}$ assigned to vertices. We can view $V$ as the set of numbers $1,...,|V|$, and $\phi$ as a permutation of $V$. For every edge $e$ we assign its energy $\phi(e_-)*\phi(e_+)$ (the product of weights of its end vertices). The energy of $\phi$ is the sum of energies of all edges from $E$. We are interested in $\phi$ that maximizes the energy of $G$. Is anything known about this problem? 
For example, if $G$ is an $n$-cycle, $1-2-...-n-1$, then for $n=3$ all $\phi$'s have the same energy. If $n=4$, then the maximal energy is given by $\phi=(1,2,4,3)$, for $n=5$, we get $\phi=(1,2,4,5,3)$, for $n=6$, $\phi=(1,2,4,6,5,3)$, etc. (the new number gets inserted between two biggest numbers in the previous permutation. The interesting thing is that the sequence $1,2,3, 1,2,4,5,3, 1,2,3,4,6,5,3,...$ seems to coincide with A194983  which is defined in OEIS in a completely different way. I can prove it for $n\le 10$. Is it possible to prove the coincidence for all $n$? 
 Update.   Gjergji  answered the question about the cycle. See comments below about other interesting graphs, other collections of weights and the problem of minimalizing the energy (maximizing the cost).

REPLY [8 votes]: In the case of the $n$-cycle there will be two ways to write the optimal permutations. In the case when we write them like $$(1),(1,2),(1,2,3),(1,2,4,3),(1,2,4,5,3),(1,2,4,6,5,3),\dots$$ 
this will be the fractalization of the sequence $$1,2,3,3,4,4,5,5,6,6,\dots$$
Notice that the coincidence with the fractalization of $1+\lfloor n/\sqrt{5} \rfloor$ stops after $n=10$. For example the permutation $(1, 2, 4, 6, 8, 11, 10, 9, 7, 5, 3)$ which appears in A194983 is not optimal.
I would prefer to write the permutations with the opposite orientation, so that one gets
$$(1),(1,2),(1,3,2),(1,3,4,2),(1,3,5,4,2),\dots$$ 
which is the fractalization of $1+\lfloor n/2\rfloor$. 
This is of course just a fancy way of saying that the optimal permutations contain the two largest entries in consecutive positions and can be generated recursively by inserting $n+1$ between $n$ and $n-1$. This can be proven easily by induction.
Proof: Let C(n) be the value of $\sum \pi(i)\pi(i+1)$ (cyclic sum) for these permutations. We have $C(n)=C(n-1)+n^2-2$. Suppose the statement is true for $n-1$. Now let $\sigma\in S_n$ be a random permutation. We will prove that $\sum \sigma(i)\sigma(i+1)\le C(n)$. But $\sum \sigma(i)\sigma(i+1)=n(a+b)-ab+\sum \sigma'(i)\sigma'(i+1)\le C(n-1)+n(a+b)-ab$
where $\sigma'\in S_{n-1}$ is obtained from sigma by deleting $n$, so we just need to show 
$$C(n-1)+n(a+b)-ab\le C(n)$$ which can be written as $(n-a)(n-b)\geq 2$, so we are done. 
For the general problem, this is close to the problem of minimizing the $\lambda$-cost over all labelings $\pi$. Where the cost of an edge is $|\pi(i)-\pi(j)|^\lambda$. In fact for regular graphs your problem is equivalent to minimizing the cost for $\lambda=2$. Unfortunately, I don't think much is known about this beyond $\lambda=1$.<|endoftext|>
TITLE: Can one represent a generalized hypergeometric function 1F2 as a product of two confluent hypergeometric functions?
QUESTION [6 upvotes]: I am trying to somewhat simplify a series, whose coefficients feature generalised hypergeometric functions ${}_1F_2(1;a,a+1;z)$. I was unable to find useful functional relations for this specific combination of parameters (I tried the new NIST Handbook, third volume of A.P. Prudnikov, Brychkov & Marychev's "Integrals and Series" and all online sources I could get my hands on).
Interestingly, L.J. Slater mentions on p.47 of her book "Generalized Hypergeometric Functions" that
${}_1F_2(a;b,c;z)$ is the product of two confluent hypergeometric functions
and gives a reference to the paper of F.J.W. Whipple (1927) in J. Lond. Math. Soc., 2, p. 85, which is focussed on relationships between functions ${}_3F_2$ and ${}_4F_3$. I must be missing something here, but I cannot figure out how Whipple's paper supports the Slater's statement. Therefore, my question is

Is it really possible to represent generalized hypergeometric functions ${}_1F_2(a;b,c;z)$ with arbitrary (within reason) parameters $a$, $b$ and $c$ as a product of two confluent hypergeometric functions?
If yes, could you please point me in a direction of the relevant book/paper/formula/derivation?

REPLY [3 votes]: It seems the relevant paper is in fact C. T. Preece (1923) in Proceedings of The London Mathematical Society, Volume s2-22, Issue 1, p. 370. Identities such as
$
\left\{ {}_1\mathcal{F}_1\left(a;2a;x\right)\right\}^2 = e^x {}_1\mathcal{F}_2(a;2a,a+1/2;x^2/4)
$
are provided. There don't seem to be any with more than one free parameter, however.<|endoftext|>
TITLE: Representing a real number as the value of a countably infinite game
QUESTION [9 upvotes]: Is it true that for any real number $p$ between 0 and 1, there exist finite or infinite sequences $x_m$ and $y_n$ of positive real numbers, and a finite or infinite matrix of numbers $\varphi_{mn}$ each of which is either 0 or 1, such that 
1) $\sum x_m=1$
2) $\sum y_n=1$
3) $\forall n\;\sum x_m\varphi_{mn}=p$
4) $\forall m\;\sum y_n\varphi_{mn}=p$

REPLY [5 votes]: Generalization of my previous answer.  Is this right?  $p$ irrational...
For $0\lt p \lt 1/2$ let $x_1(p)=y_1(p)=p$, $\varphi_{1,1}(p) = 1$,
$\varphi_{1,j}(p)=\varphi_{j,1}(p)=0$ for $j>1$,
$x_n(p)=(1-p)x_{n-1}(p/(1-p))$ for $n>1$. And
$\varphi_{n,m}(p) = \varphi_{n-1,m-1}(p/(1-p))$.
For $1/2 \lt p \lt 1$ let $x_n(p)=x_n(1-p)$ and $\varphi_{n,m}(p) = 1-\varphi_{n,m}(1-p)$.<|endoftext|>
TITLE: Probability of zero in a random matrix
QUESTION [31 upvotes]: Let $M(n,k)$ be the set of $n\times n$ matrices of nonnegative integers such that every row and every column sums to $k$.  Let $P(n,k)$ be the fraction of such matrices which have no zero entries, equivalently the probability that a random matrix from the uniform distribution on $M(n,k)$ has no zero entries.
One thing to note is that $$P(n,k)=\frac{|M(n,k-n)|}{|M(n,k)|}$$
(think about subtracting 1 from every entry).  Also note that $P(n,k)$ is the fraction of integer points in the $k$-dilated Birkhoff polytope that lie in the interior.
It seems "obvious" that $P(n,k)$ is a non-decreasing function of $k$. For large enough $k$ it is strictly increasing by Ehrhart theory, but I'd like to see a proof for all $k$.  So the problem is: 
Prove that $P(n,k)$ is a non-decreasing function of $k$ for fixed $n$.
COMMENT: By a result of Stanley (see David Speyer's answer for refs) there are non-negative integers $\{h_i\}$ such that
$$|M(n,k)| = \sum_{i=0}^d h_i \binom{k+i}{d}$$
where $d=(n-1)^2$. I'm wondering if this is enough. Does every polynomial of that form have the desired properties? [Gjergji showed not.]
COMMENT2: The reciprocity theorem for Ehrhart series provides a formula for the number of points in the interior in terms of the number in the whole (closed) polytope. Making use of the above, we find that if the $H_n(k)$ is the polynomial equal to $|M(n,k)|$ for positive integers $k$, then the number of interior points (already identified as $H_n(k-n)$) equals $(-1)^{n+1}H_n(-k)$.  So what we have to prove is that
$$(-1)^{n+1}\frac{H_n(-k)}{H_n(k)}$$
is non-decreasing for integer $k\ge n$. Experimentally, it is not increasing for real $k$ until $k$ is larger.

REPLY [6 votes]: Here are a few more ways to rephrase the problem. Let $A_k$ denote a $kn\times kn$ matrix, composed of $k\times k$ blocks, where the $ij$ block is filled with copies of a standard exponential random variable $\gamma_{ij}$. The $\gamma_{ij}$'s are taken to be independent. A. Barvinok showed that $$M(n,k)=\frac{\mathbb{E}(\operatorname{per}(A_k))}{(k!)^{2n}}$$
and used this to prove that $M(n,k)$ is almost log-concave in $k$. More specifically, he showed in "Brunn-Minkowski inequalities for contingency tables and integer flows",
Adv. in Math., 211 (2007), 105-122, that the following inequality holds
$$\alpha(k)M(n,k)^2\geq M(n,k-1)M(n,k+1)$$
with $\alpha(k)=O(k^n)$. He conjectures that this inequality holds for $\alpha=1$, but as far as I know, this remains open. I'm not sure what's the best upper bound for $$\frac{M(n,k)M(n,n+k+1)}{M(n,k+1)M(n,k+n)}$$ 
that one gets using this analytic approach.

Another way of stating the problem is through RSK, and turn it into an inequality in terms of Kostka numbers. If $\tau(k)$ stands for the partition $(k,k,\dots,k)$, with $k$ parts. Then log-concavity is equivalent to
$$\left(\sum_{\lambda} K_{\lambda \tau(k)}\right)^2\geq \left(\sum_{\lambda} K_{\lambda \tau(k-1)}\right)\left(\sum_{\lambda} K_{\tau(k+1)}\right)$$
and your inequality can be written similarly. One might hope that there is a proof making use of known inequalities between Kostka numbers or known Schur positivity results.
Yet another equivalent formulation comes from Pietro's answer. We need to prove that the coefficient of $(x_1y_1\cdots x_ny_n)^k(z_1t_1\cdots z_nt_n)^{k+1}$ is non-negative in 
$$\frac{\left(z_1t_1\cdots z_nt_n-x_1y_1\cdots x_ny_n\right)}{\prod_{i,j=1}^n(1-x_iy_j)(1-t_iz_j)}.$$

In a different direction, we have $$M(n,k)=\sum_{i=0}^d h_i\binom{k+i}{d}$$ 
where $d=(n-1)^2$. In "Ehrhart polynomials, simplicial polytopes, magic squares and a conjecture of Stanley", C.A. Athanasiadis proved that $(h_0,h_1-h_0,\dots,h_{\lfloor d/2\rfloor}-h_{\lfloor d/2\rfloor-1})$ is a g-vector, so it satisfies the inequalities of Mcmullen's g-theorem. In particular $h$ is a symmetric unimodal sequence. I doubt that this is enough to conclude log-concavity of $M(n,k)$, or your property for that matter, but perhaps it gives some insight on how hard the problem is.<|endoftext|>
TITLE: Is there a topos theoretic interpretation/proof of Quillen's Theorem A?
QUESTION [17 upvotes]: I think the title says it all.  Quillen's Theorem A says that a functor $F\colon C\to D$ induces a homotopy equivalence of classifying spaces if each fiber category $F/d$ with $d$ an object of $D$ is contractible. Now Moerdijk showed that in some sense the classifying topos of a category is weakly equivalent to the classifying space of the category, so one would guess there is a topos theoretic interpretation/proof of the theorem.


Question: Is there a topos theoretic interpretation/proof of Quillen's Theorem A?

REPLY [4 votes]: I have no compelling answer to this question myself, but you may find relevant results and ideas in the work of Grothendieck, Maltsiniotis and Cisinski in homotopical algebra. Have you looked at Pursuing Stacks? In Maltsiniotis's Astérisque, there is a hint as to what a cohomological proof of Quillen's result would be. See http://www.math.jussieu.fr/~maltsin/ps/prstnew.pdf, page numbered 11 in the document. The text of Cisinski's Astérisque is available at http://www.math.univ-toulouse.fr/~dcisinsk/publications.html. It is called Les préfaisceaux comme modèles des types d'homotopie. He also has an old and never-published preprint —which may contain some typos— called Faisceaux localement asphériques, available at the same webpage. I hope this helps and is not off the point.<|endoftext|>
TITLE: Number of spanning subgraphs of the complete bipartite graph $K(m,n)$
QUESTION [6 upvotes]: Let $K(m,n)$ denotes the complete bipartite graph on parts of cardinality $m$ and $n$.
My question is; How many nonisomorphic spanning subgraphs are there of of $K(m,n)$? This is such an obvious question, it has probably been answered. I just don't know where to look. There is an obvious, but complex to use, recursion for the constructions. Given the set of nonisomorphic subgraphs of $K(m-1,n)$ -- or of $K(m,n-1)$ -- appending the $n-1$ edges from the missing vertex in the first case or $m-1$ in the second edges in all inequivalent ways will generate the set for $K(m,n)$. But this is not a numerical problem so no simple recursion seems possible -- yet it may have well been solved using Polya's counting theorem. Do any of you know the answer, or where it can be found?

REPLY [5 votes]: This is only known explicitly for $m=4$. A decent survey is here:
http://www.math.ru.nl/~bosma/Students/JannekevandenBoomen/JannekevdBoomenMScthesis.pdf<|endoftext|>
TITLE: Topos Without point, from the point of view of logic
QUESTION [14 upvotes]: I am a little troubled by the following "paradox" :
Let $X$ be a non trivial (Grothendieck) topos without Set points. 
We want to look at this situation from the point of view of logic: $X$ classifies some geometric theory $T$. The assumption on $X$ means that $T$ is consistent but has no model in Set. This is not in contradiction with Gödel's theorem because the theory $T$ might not be a "finitary first order theory".
We have been able to prove that $X$ doesn't have points, hence we have a proof (using boolean logic and possibly the axiom of choice) that the theory $T$ doesn't have any model.
Let now $Y$ be a Boolean topos with internal axiom of choice. It should be possible to apply the previous proof in the internal logic of $Y$, and then prevent $T$ to have any model in $Y$... But this is not the case: Barr's covering theorem implies there is a topos $Y$ that cover $X$ and hence which has a $T$-model.
Can someone explain me why this is not working ? Or give an example where $T$ and $Y$ are explicit?

REPLY [23 votes]: Here is a concrete example of a consistent geometric theory that has no model in $\mathbf{Set}$ but does have a model in a Boolean Grothendieck topos.
Our base theory $T$ is an expansion of the theory of linear orders with a constant $c_q$ for each rational number $q$. In geometric form, the base axioms are:
$$x = y \lor x \lt y \lor y \lt x$$
$$x \lt y \land y \lt z \Rightarrow x \lt z$$
$$x \lt x \Rightarrow \bot$$
together with the axioms $c_q \lt c_r$ for all pairs of rationals $q \lt r$.
In addition to the above, we add an axiom which says that the $c_q$'s form a dense subset
$$x \lt y \Rightarrow \bigvee_q (x \lt c_q \land c_q \lt y).$$
A model of $T$ is a linear order with a countable dense subset isomorphic to $\mathbb{Q}$. In other words, models of this theory in $\mathbf{Set}$ are, up to isomorphism, subsets of $\mathbb{R}\cup\lbrace\pm\infty\rbrace$ that contain $\mathbb{Q}$. In particular, models of this theory all have size at most $2^{\aleph_0}$.
Now expand the theory further to a theory $T_I$ by adding new constants $d_i$ for each $i$ in the set $I$, together with the axioms $d_i \lt d_j \lor d_j \lt d_i$ when $i, j$ are distinct elements of $I$. If $|I| \gt 2^{\aleph_0}$ then our new theory $T_I$ has no model in $\mathbf{Set}$. However, in the Grothendieck topos for adding $I$ many Cohen reals side by side (see below) this theory does have a model since $2^{\aleph_0} \geq |I|$ in that topos. So this is a consistent geometric theory that has a model in some Boolean Grothendieck topos but no model in $\mathbf{Set}$.
(The Grothendieck topos for adding $I$ many Cohen reals side-by-side is obtained by imposing the double-negation topology on the poset category of finite partial functions $I\times\omega\to2$. Note that the double-negation topology is always Boolean. This is not the Barr cover for the classifying topos of $T_I$ but it's perhaps easier to understand what it looks like.)<|endoftext|>
TITLE: Minimal prime ideals and Axiom of Choice(revised version)
QUESTION [9 upvotes]: From the page: 
Existence of prime ideals and Axiom of Choice.,
I have found that The existence of prime ideals in commutative rings is equivalent to the Boolean Prime Ideal theorem. But $BPI$ is weaker than Axiom of choice. this means that The existence of prime ideal in commutative rings with unity is weaker than $AC$. Know Another Question came in my mind that I think It is a bit different from that one. Let me recall the following theorem:
Theorem:For any commutative unitary ring $R$ there exists a minimal prime ideal.
To proving this result One can pickup a prime ideal, and throw it in a maximal chain of prime ideals(Zorn's lemma) and then the intersection of this chain gives a minimal prime ideal at hand.
You Know that the existence of minimal prime ideal needs to apply one of the equivalences of $AC$ (i.e.Zorn's Lemma) But I didn't see anything about the converse of Above theorem.
STATEMENT:Is it true that The existence of minimal prime ideals in commutative unitary rings is equivalent to $AC$.
I am interested in To Know if the situation changes When we give minimality Condition on Prime ideals.

I think its better to recall the difference of two following situations in topology:
The statement "product of compact Hausdorff spaces is compact", does not implies $AC$
But
The statement "product of compact spaces is compact" is equivalent to $AC$

REPLY [19 votes]: Suppose I have a set of disjoint, nonempty sets, and I want to choose one element from each. Consider the free polynomial ring generated by all the elements of all the sets, then take the quotient by the ideal generated by $xy$ for each pair $x$ and $y$ different elements in the same set. Any prime ideal must contain all but one element from each set.
We need to show that a minimal prime ideal does not contain all the elements from any set. Then a minimal prime ideal will give us a choice function. We can take the minimal prime to be generated by the elements, since every prime ideal contains a prime ideal generated by elements. Then remove one element from a set entirely contained in the prime ideal. The ideal will still be prime, and smaller, this is a contradiction.
So minimal primes give a choice function.<|endoftext|>
TITLE: Lurie's "Virtual fundamental classes" and "Geometric derived stacks"
QUESTION [9 upvotes]: In his thesis, Jacob Lurie mentioned two work in preparation (by him), namely "Virtual fundamental classes and the motivic sphere" and "Geometric derived stacks". Now that much is written in the DAG series and the two books (Higher Algebra and Higher Topoi), I'm wondering if anybody here know if these two papers are contained in the DAG series already or still yet to be written?

REPLY [8 votes]: I'm not sure I understand the question (unless it's intended as general curiosity about the progress of Lurie's work). I can see two legitimate options: 

you are interested in a particular result and want to learn about it or apply it, in which case the answer is there are no documents circulating (AFAIK) beyond the vast amount on Jacob's homepage. Are you then looking for help finding a result there, in which case a specific question on MO is likely to provide an answer, or you know what you're looking for isn't there? - in the latter case, sadly I think the answer is you have to wait patiently for future publications like the rest of us.. (This is basically repeating Noah's comment.)
you have your own ideas about a result that was announced as to appear in these manuscripts, and would like to know their status before working hard to complete or publish a result which may become obsolete. In this case I'd say realistically you should either change field or (much better option) not worry about whether your result will be subsumed by his work. Jacob's writing tends to subsume results of many people, it's an obvious danger of working in this (exciting and rapidly developing) field, but the community (judging from my experience) will still very much appreciate other takes, and given the great scope of things that Jacob is working on it's likely you can complete something before it appears. 

Of course if you're just starting to think about a problem it might be wiser to find your own questions rather than trying to prove something you saw announced in Jacob's thesis..

REPLY [4 votes]: It's worth noting that Lurie's thesis is not on his webpage, and is not regularly updated like the papers which are on is webpage.  I would take this to mean that you're supposed to be reading the current versions of this material in the DAG series, and not reading the 8 year old outline.<|endoftext|>
TITLE: Perturbation theory for the generalized eigenvalue problem
QUESTION [9 upvotes]: Is there a standard reference for the perturbation theory of the generalized eigenvalue problem?
More specifically, I would like to get a systematic expansion for the problem

$(A_0 + \epsilon A_1)v = \lambda B v$

in terms of the solutions to 

$A_0 v = \lambda B v$

where $A_0$, $A_1$ and $B$ are known, $n\times n$ Hermitian matrices,  $\epsilon >0 $ is a "small" parameter, and $\lambda\in \mathbb{C}$ and $v\in\mathbb{C}^n$ are the unknowns. When $B$ is positive-definite, the usual (Rayleigh-Schroedinger) perturbation series can easily be generalized by using $B$ to define an alternative inner product. However, in my problem, $B$ is not positive-definite (although it is still nonsingular).

REPLY [6 votes]: In addition to those already suggested by Federico, a good reference is Chapter 2 in Kato's book Perturbation Theory for Linear Operators, especially if you want to tackle the question from the higher perspective of functional analysis. Another reference, addressing the same problem from the alternative point of view of the calculus of variations (with an eye on computations), is I. Babuška and J. Osborn, Eigenvalue Problems, in Handbook of Numerical Analysis (Part 1), Vol. II, ed. by P.G. Ciarlet and J.L. Lions, pp. 641-787 (1991).<|endoftext|>
TITLE: Can the Alexander horned sphere arise as a cell boundary in a finite CW-sphere?
QUESTION [20 upvotes]: Recently, I've been wondering to what extent certain types of pathologies can arise in finite CW complexes -- notice that I do not want to assume that I'm in the PL category or that the CW complexes are regular, but do want to insist on there only being finitely many cells.
Specific question: is there a finite CW complex homeomorphic to a sphere such that one of its maximal cells has as its closure a ball whose boundary is embedded in the CW-sphere as an Alexander horned sphere?
Follow-up question: if the answer is no, is there a finite CW complex homeomorphic to a sphere such that the closure of one of the maximal cells is a ball, but the closure of its complement is not a ball?
I don't have a specific reason I need to know this, but recently have been working a lot with finite CW complexes that are not in the PL category, or at least not obviously so, and would like to understand better the interaction between the finiteness of the number of open cells and the potential weirdness of the attaching maps.  Information about other pathologies that can or cannot occur in this setting would also be much appreciated.  
Thanks in advance for your help with this!

REPLY [19 votes]: Specific question: is there a finite CW complex homeomorphic to a sphere such that one of its maximal cells has as its closure a ball whose boundary is embedded in the CW-sphere as an Alexander horned sphere?

1) No. If $K$ is the $2$-skeleton of the CW complex, then $K$ has a mapping cylinder neighborhood in $S^3$, and hence by Nicholson's theorem it is tame, that is, equivalent to a subpolyhedron of $S^3$ by a homeomorphism $h$ of $S^3$. Since $K$ is $2$-dimensional, it is not hard to show that $h$ also takes any $2$-sphere in $K$ onto a subpolyhedron of $S^3$. So $K$ cannot contain the horned sphere.
2) Yes, if you allow an unspecified wild codimension one sphere in place of the Alexander horned sphere. By Example 7.11.2 on p.419 in the Daverman-Venema book (this seems to be among a few original results in the book, perhaps the most important one), $S^n$ for $n\ge 6$ contains a wildly embedded sphere $\Sigma$ with a mapping cylinder neighborhood $N$. By construction, the complement to the interior of $N$ consists of two closed $n$-balls. It follows that the closures of the complementary domains of $\Sigma$ are the mapping cones of some self-maps of $S^{n-1}$. So we get a CW-complex with one $0$-cell, one $(n-1)$-cell and two $n$-cells, which is homeomorphic to $S^n$, and has a wild $(n-1)$-skeleton.

Follow-up question: if the answer is no, is there a finite CW complex homeomorphic to a sphere such that the closure of one of the maximal cells is a ball, but the closure of its complement is not a ball?

1) Yes. You can glue one of the complementary domains of $\Sigma$ and an $n$-ball along their boundary sphere. The result is again $S^n$, according to Proposition 7.10.1 in Daverman-Venema.
2) I should also mention a simpler but somewhat similar example. Using the Edwards-Cannon theorem, it is not hard to construct a finite regular CW-complex $K$ that is homeomorphic to $S^5$, even though the boundary of some $2$-cell (in fact, of each $2$-cell) of $K$ is wild, viewed as a copy of $S^1$ in $S^5$. In more detail, if $H$ is a traingulation of a non-simply-connected homology $3$-sphere, then the double suspension $S^0*S^0*H$ is a simplicial complex homeomorphic to $S^5$; the desired CW-complex is the 'prejoin' $(S^0*S^0)+H$, which is PL homeomorphic to $S^0*S^0*H$ and has all its $2$-cells attached to the suspension circle $S^0*S^0$. On identifying regular CW-complexes with their posets of nonempty faces, the prejoin $P+Q$ of two posets is defined by placing all the elements of $P$ below all the elements of $Q$ in the Hasse diagram, and keeping the original order within $P$ and within $Q$. The order complex of $P+Q$ is easily seen to be isomorphic to the join of the order complexes of $P$ and of $Q$. As an example, $S^0*S^0*pt$ is a simplicial complex with $4$ of $2$-simplices, whereas $(S^0*S^0)+pt$ is a cell complex with only one (quadrilateral) $2$-cell. 
Beware that $K$ itself is a PL CW-complex, with PL attaching maps; the only trouble is with the homeomorphism between $K$ and $S^5$.<|endoftext|>
TITLE: Schur functors generalization to "Jack", "Hall-Littlewood", "Macdonald" functors ? 
QUESTION [12 upvotes]: Schur functors are functors from the category of vector spaces to itself.
If we take an operator $M: V->V$ and apply a Schur functor to it and then calculate trace $Tr(M^{\Lambda})$ we will get Schur polynomial in the eigenvalues of $M$.
Question Can one generalize (deform) Schur  functors, such that 
$Tr(M^{\Lambda})$ will give polynomials which generalize (deform) Schur polynomials e.g. 
Hall-Littlewood  polynomial, or Jack  polynomial and most generally 
Macdonald polynomials ?

REPLY [9 votes]: It seems to me that this is answered, perhaps in a boring way, by Haiman's work on the $n!$-conjecture (now a theorem due to Haiman). For any partition $\lambda$, Haiman constructs a finite dimensional graded module $C_{\lambda}$ for $\mathbb{C}[x,y][S_n]$. (The group elements commute with $x$ and $y$, and $x$ and $y$ commute with each other.) The doubly graded Frobenius character of $C_{\lambda}$ is the $\lambda$-Macdonald polynomial.
Now just use Schur-Weyl duality: Define the functor $F_{\lambda}$ from vector spaces to vector spaces by
$$V \mapsto V^{\otimes |\lambda|} \otimes_{\mathbb{C}[S_n]} C_{\lambda}.$$
The result is a doubly graded $\mathbb{C}[x,y]$ module which is the sum of Schur functors corresponding to the Macdonald polynomial.<|endoftext|>
TITLE: Godel's incompleteness: can the diagonalization be done in computation rather than logic?
QUESTION [13 upvotes]: I have always had trouble understanding Godel's proof of his first incompleteness theorem, because the diagonalization part is done on the logical side, which is unfamiliar to me, rather than on the computational side, which I find more familiar.  I decided to replicate the proof but do as much of the work as possible in terms of computation, rather than logic.
Consider the Turing machine $H$ which when run on input $x$:

(firstly a triviality) checks whether $x = \lceil \phi \rceil$  for some two-place predicate $\phi$ and if not then loops forever (or indeed it could do anything –  I don't believe this part of the argument is important!)
searches for proofs in PA of

$\phi(\lceil\phi\rceil, 1)$ and
$\lnot\phi(\lceil\phi\rceil, 1)$ [A]

if it finds the former first, it halts writing $0$ on its tape
if it finds the latter first, it halts writing $1$ on its tape
if there is a proof of neither, it loops forever

Since computable functions are expressible in PA, there is a two-place predicate $h$ such that $H(x) = y$ implies $\vdash h(\underline{x}, \underline{y})$ and $H(x) \not= y$ implies $\vdash \lnot h(\underline{x}, \underline{y})$ [B].
Then what is $H(\lceil h \rceil)$?

if $H(\lceil h \rceil) = 0$ then $\vdash h(\lceil h \rceil, 1)$ (by definition of $H$) and $\vdash \lnot  h(\lceil h \rceil, 1)$ by definition of $h$
if $H(\lceil h \rceil) = 1$ then $\vdash \lnot h(\lceil h \rceil, 1)$ (by definition of $H$) and $\vdash h(\lceil h \rceil, 1)$ by definition of $h$
if $H(\lceil h \rceil)$ is anything else, including a non-terminating computation, then neither $\vdash h(\lceil h \rceil, 1)$ nor $\vdash \lnot h(\lceil h \rceil, 1)$

My conclusion is that either PA is inconsistent (first two possibities) or incomplete (third possibility).
My first question is: is this a correct proof that PA is either inconsistent or incomplete?  I don't want to be misled by a simple misunderstanding!
My second question is: if indeed it is a correct proof, why are the common expositions not done this way?  It seems much more easy to do the diagonalization part of the argument in the world of Turing machines and computation than in the world of wffs and logic.  Moreover no weakening to $\omega$-consistency is necesary.
Footnotes:
[A] In wffs, $1$ is my abbreviation for $s(0)$ of course.
[B] I use $\vdash \psi$ to mean there is a proof in PA of $\psi$.  Perhaps this is more usually denoted $\vdash_{PA} \psi$, but I wanted to conserve space and typing!

REPLY [25 votes]: There are indeed many proofs of the incompleteness theorem,
and when I teach my introductory graduate logic course, we
usually give at least four or five different proofs as they
arise naturally with the different topics. Here are a few
sketches of different proof methods that commonly arise:
Traditional proof via the fixed point lemma. This is
the traditional proof via the fixed-point lemma, by which
we know that for each formula $\varphi(x)$ there is a
sentence $\psi$ such that
$PA\vdash\psi\leftrightarrow\varphi(\psi)$.
With this lemma, one makes the Goedel sentence $\psi$ that
asserts that $\psi$ is not provable in PA. It follows that
indeed it is not provable in PA, and hence it is true and
unprovable.
Via the halting problem. This is the proof via the
halting problem, and uses some similar ideas to your proposed proof. 
First, one proves that the halting problem is not decidable
by any computable procedure. Next, one argues that there
must be a Turing machine $p$ which does not halt, but we
cannot prove it in PA, since otherwise we can solve the
halting problem as follows: given a Turing machine program,
we simulate it during the day, waiting to see if it halts,
and at night, we search for a proof that it doesn't halt.
If all instances of non-halting were provable, we would
thereby solve the halting problem.
Via the existence of undecidable problems. Indeed, if there is any computably undecidable problem that is expressible in arithmetic, then there must be instances that cannot be settled in PA (or any othere computably axiomatizable theory), since otherwise by searching for proofs we would be able to decide the problem. This observation is at the core of the connection between two common uses of the word "undecidable". Namely, any c.e. set $A$ that is not (computably) decidable must admit infinitely many natural numbers $n$ such that $n\notin A$ but this is not provable in your favorite formal system, since otherwise $A$ would be decidable. So the computably undecidable sets are saturated with non-instances that are undecidable in your formal system. 
Via Tarski's theorem. Tarski's theorem is that there
is no arithmetic predicate true of exactly the Goedel codes
of true arithmetic assertions. That is, arithmetic truth is
not arithmetically definable. This is a generalization of
the incompleteness theorem, since provability is
arithmetically definable, and so it cannot coincide with
truth. One can prove Tarski's theorem in a variety of ways,
including the fixed point lemma, or the non-collapse of the
arithmetic hierarchy, among others.
Via the non-collapse of the arithmetic hierarchy. It
is not difficult to prove the existence of a universal
$\Sigma_1^0$-set of natural numbers. This is analogous to
the arithmetization of syntax in the traditional syntactic
proof or to the existence of universal machines used in the
halting problem proof. Once one knows that there are
universal $\Sigma_1^0$ sets, it follows by an easy
diagonalization that $\Sigma^0_1\neq\Pi^0_1$ and the
arithmetic hierarchy does not collapse. This immediately
implies Tarski's theorem, which as we've said implies that
provability (which is arithmetic) does not coincide with
truth.
By the end of my course, it often becomes a running joke that every new theorem also implies the incompleteness theorem, and I usually try to make those conclusions explicit. So there are indeed many other proof methods besides the main ones I
mention above, and perhaps other users will post them. It
is generally much easier to find proofs of the first
incompleteness theorem than the second, but by now there
are also interesting proofs of the second incompleteness
theorem from various directions.<|endoftext|>
TITLE: Ultimate Maximality Principle
QUESTION [9 upvotes]: I wonder if it's possible to formulate an "ultimate" maximality principle (UMP) and prove its consistency. I envision UMP to express the idea that no matter how we enlarge the universe of set theory V (by any means e.g. set forcing, class forcing, infinite model theory), we would gain n o t h i n g. Let W be the ultimate enlargement of V. Then UMP would say that a statement is true in W iff it's true in V. So any statement that is true in W is already true in V.
Questions:
1) Are there available reference in literature concerning UMP?
2) If not, what is the prospect of UMP in foundational research?

REPLY [13 votes]: There are several maximality principles that already have
some of this flavor, with a growing literature surrounding them.
For example, the maximality principle MP as introduced in
my paper A simple maximality
principle,
and also in a paper of Stavi and Vanaanan, is the scheme
asserting that any statement $\varphi$ that is forceable by
set forcing in such a way that it remains true in all
further set forcing extensions, is already true in $V$.
This axiom MP is actually equiconsistent with ZFC. Stronger
versions of the axiom allow countable parameters (the axiom
is inconsistent with uncountable parameters, since they can
become countable by forcing). The strongest version of the
axiom is the Necessary Maximality Principle NMP, which
asserts that $\text{MP}(\mathbb{R})$ holds in all set forcing
extensions, and this has determinacy consequences, but has
strength below $\text{AD}_{\mathbb{R}}+\Theta$ is regular.
The natural analogue of MP for class forcing is either
inconsistent or not expressible in first order set theory.
Another tack on the issue is the Inner Model
Hypothesis of Sy Friedman,
which aspires more in the universal direction of your
question. Namely, the IMH asserts that if there is any
outer model of the universe having an inner model
satisfying a certain assertion, then there is already such
an inner model with that feature. This axiom has the flavor
of what you have wanted; it has a strong consistency
strength, but it itself is inconsistent with the actual
existence of large cardinals, as opposed to their existence
in inner models. The penalty for the greater universality
of the IMH is that it is not expressible in first order set
theory as an axiom about $V$. One can, however, understand
it as an external assertion about a countable model,
treating that countable model as a kind of universe
substitute.
Both the MP and the IMH are naturally expressible in modal
terms by the S5 axiom
$\Diamond\square\varphi\to\square\varphi$, which expresses
the idea that anything that could become necessarily true
is already necessarily true. Benedikt Loewe and I explored
the nature of the forcing modality in our paper The modal
logic of
forcing.
Your proposed Ultimate Maximality Principle would seem to
need a more detailed fleshing out: in the axiom you refer
to an "ultimate" enlargement $W$ of $V$, but what is this
$W$? After all, for any enlargement $W$ of $V$ we may
continue to form the forcing extensions of $W$, so strictly
speaking, there is no largest one. For example, $W$ itself
would have forcing extensions, some with CH and others with
$\neg\text{CH}$. Similarly, any set in $W$ can be made
countable by forcing, and so if you are entertaining the
idea of a single largest one, then every set in $V$ would
have to be countable in it. So the idea that one can
achieve literal maximality as you describe becomes
problematic, and this is the reason why the MP and the IMH
make use of the S5 style maximality, which asserts that
anything that could become true forever afterwards is
already true, an assertion that takes the place of an
actual maximal extension.
Meanwhile, there is current work to investigate the extent
to which we may have maximality-type principles for class
forcing and for arbitrary extensions. For example, it
appears that one may get it for extensions with the
approximation and cover properties without much
modification from the original work.<|endoftext|>
TITLE: Moore-Penrose Inverse as an adjoint
QUESTION [15 upvotes]: A Moore-Penrose pseudoinverse of a morphism $f: V \rightarrow W$ between Euclidean vector spaces is a map $g: W \rightarrow V$ in the other direction satisfying the identities
$fgf = f$
$gfg = g$
$(fg)^\ast = fg$
$(gf)^\ast = gf$
where $\phi^\ast$ denotes the adjoint of a linear map $\phi$.
Now the first two identities obviously resemble the triangle equalities of an adjunction. My question is: Can one actually understand the Moore-Penrose inverse as an adjoint?
One possibility would be to find a "nice" (compatible with composition) partial order on Hom-Sets $\text{Hom}(V,W)$ making the category of Euclidean vector spaces into a 2-category, where the notion of adjunction is defined and the triangle equality in fact would imply $fgf = f$. So a more precise question would be: Does there exist such an order?

REPLY [6 votes]: I do not think the concept of Moore-Penrose Inverse and the concept of categorical adjunction have much in common (except they both try to generalise the concept of inverse):

Equations $g = gfg$ and $fgf$ do not reaseble triangle equalities. Let me focus on the first equation. The "corresponding" triangle equality says that the composition $g \to^{\eta_g} gfg \to^{g\epsilon}g$ is the identity on $g$. Obviously $\eta_g$ is an inclusion, but generally there are no reasons for $\eta_g$ to be an isomorphism. However, in the world of 2-posets (i.e. categories enriched over posets) there is a reason. If we have a pair of 2-morphisms $\mathit{id} \le gf$ and $fg \le \mathit{id}$, then we may compose the first one on the right and the second on the left with $g$, obtaining $g \le gfg$ and $gfg \le g$, therefore $g = gfg$. Notice that I have not used any triangle equality here (just the existence of an appropriate pair of 2-morphisms).
The concept of an adjunction is inherently asymmetric --- a left adjoint is the best approximation of the identity from the left, and the right adjoint is the best approximation of the identity from the right; whilst the concept of Moore-Penrose pseudoinverse is perfectly symmetric. This means that if we had a 2-category, buit upon the category of vector spaces, where every pseudoinverse was a part of an adjunction (satisfying, perhaps, some other conditions), then every morphism would have both left and right adjoint, furthermore these adjoint functors would be isomorphic to the pseudoinverse (so isomorphic to each other). For a morphism $f$ in a 2-poset being both a left and right adjoint to $g$ is just being the inverse of $g$ --- the unit and counit from one adjunction give inequalities $\mathit{id} \le gf$ and $fg \le \mathit{id}$, whereas the unit and counit of the other adjunction give inequalities $\mathit{id} \le fg$ and $gf \le \mathit{id}$, hence $\mathit{id} = gf$ and $\mathit{id} = fg$. This fully answers your precise question, since in the category of vector spaces not every pesudoinverse is an inverse.
In any 2-category adjunctions compose --- if $f \colon A \to B \dashv f^+ \colon B \to A$ and $g \colon B \to C \dashv g^+ \colon C \to B$ then $g f \colon A \to C \dashv f^+ g^+ \colon C \to A$, but Moore-Penrose pseudoinverses --- generally --- do not. This almost answers your main question, because if $f \colon A \to B$, $g \colon B \to C$ are any maps between vector spaces then from the above property: $(gf)^+ \approx f^+ g^+$, so from this point of view pseudoinverses are not stable under isomorphisms, thus are not categorical.<|endoftext|>
TITLE: Normal generation of Torelli
QUESTION [5 upvotes]: The only normal generators of the Torelli group of a closed surface I can find is Powell's 1977 paper (where the presentation is a bit complicated and given essentially without proof). Is there any more enlightening or smaller presentation? I would also be interested in such a result for $\mbox{Aut}(\pi_1(F_g))$ (as opposed to $\mbox{Out}(\pi_1(F_g))$.
EDIT Actually, it seems that in the 1980 paper of Dennis Johnson he seems to show that Torelli is normally generated by a single element, which improves the Powell result.

REPLY [5 votes]: For the Torelli group of a surface, I gave a much easier and more geometric proof in my paper
MR2302503 (2008c:57049) 
Putman, Andrew(1-CHI)
Cutting and pasting in the Torelli group. (English summary) 
Geom. Topol. 11 (2007), 829–865. 
This proof was simplified quite a bit by Hatcher and Margalit in their paper "Generating the Torelli group" (to appear in L'Enseignement Mathématique).
For $\text{Aut}(F_n)$, modern proofs can be found in the paper 
MR2336078 (2008k:57029) 
Bestvina, Mladen(1-UT); Bux, Kai-Uwe(1-VA); Margalit, Dan(1-UT)
Dimension of the Torelli group for Out(Fn). (English summary) 
Invent. Math. 170 (2007), no. 1, 1–32. 
and in my paper "The complex of partial bases for $F_n$ and finite generation of the Torelli subgroup of $Aut(F_n)$" with Matt Day.<|endoftext|>
TITLE: non-continuous inverse Galois problem
QUESTION [5 upvotes]: Let $G=Gal(\bar{\mathbf{Q}}/\mathbf{Q})$ be the absolute Galois group over $\mathbf{Q}$. 
Q1: Is it possible to find a (necessarily non-closed) normal subgroup $K\leq G$ such that
$G/K$ is free of infinite rank ?
Q2: Let $H$ be a finite group. Is it always possible to find a non continuous surjective homomorphism $\rho:G\rightarrow H$?  
If you think that removing the continuity assumption does make the inverse Galois problem any simpler then please provide some explanations.

REPLY [5 votes]: It's an old result by R. Alperin that a compact group has no abstract homomorphism onto $\mathbf{Z}$. [Compact groups acting on trees. 
Houston J. Math. 6 (1980), no. 4, 439--441.] So Q1 has a negative answer.
Here $\mathbf{Z}$ cannot be replaced by any countable abelian group. For instance, the direct product $G$ of all finite perfect groups (up to isomorphism) has an infinite abstract abelianization. Indeed since $G$ is isomorphic to its countable power, if $G$ were perfect it would be uniformly perfect, and thus all finite groups would be together perfect with a uniform commutator width, which is false. So $G$ has nontrivial abstract abelianization, and again since $G$ is isomorphic to its countable power, its abstract abelianization is uncountable (and thus admits a countable infinite quotient). 
About Q2: I don't know any example of a profinite group $G$ and finite simple group $S$ such that $G$ admits $S$ as a quotient abstractly, but not as a quotient by any open subgroup (in all the examples I know with $G$ admitting $S$ as a quotient by a non-open subgroup, $G$ actually has infinitely many open subgroups $H$ with $G/H\simeq S$) . Edit: there's indeed an example: Nikolov and Segal in "Remarks on profinite groups...", preprint 2013 (arxiv link) have an example of a profinite group with $\mathbf{Z}/p\mathbf{Z}$ as a quotient as an abstract group, but not as a topological quotient. The idea is to find large perfect groups $G_n$ for which the proportion $p_n$ of central elements of order $p$ is big, in the sense that $p_n^{-k}\ll |G_n|$ for all $k$; (e.g., consider central extensions of $\text{SL}_2(\mathbf{F}_p)\ltimes(\mathbf{F}_p^2)^n$ by $\mathbf{F}_p^{n(n+1)/2}$, so $|G_n|\sim p^{3+2n+n(n+1)/2}$ while $p_n^{-1}\simeq p^{3+2n}$), then in the product $G=\prod G_n$, denoting $S=\{x^p[y,z]:x,y,z\in G\}$, the set of products of $k$ elements of $S$ is a closed subset with empty interior and therefore (using Baire) $S$ does not generate $G$, so if $N=\langle S\rangle$ then $G/N$ is an uncountable $p$-elementary abelian group and thus $G$ admits $\mathbf{Z}/p\mathbf{Z}$ as an abstract quotient although $G$ is topologically perfect (replace $\mathbf{F}_p$ by $\mathbf{F}_{p^2}$ if $p=2,3$). Still, I don't know any example with $S$ nonabelian.<|endoftext|>
TITLE: Can formal logic give a precise notion of "canonical"?
QUESTION [8 upvotes]: Coming off of this discussion, I'm wondering what the term "canonical" really means. In that thread, many suggested category theory as a way to formalize the concept of what "canonical" means, using the precise term "natural" (and, many suggested that the two were not the same thing). Beyond its formal equivalent in category theory, the word natural seems to have nothing other than an intuitive or even "theological" meaning.
However, I was wondering if there is some way to define the notion of canonical by using formal logic.
Here's my own idea:
After all, when we choose something we have to use some sort of logical procedure. The notion of canonicity then might mean that there actually exists some logical way to pick out a particular element, morphism, etc. When there's no canonical choice, it might mean that there is no logical way to pick out one choice over another.

REPLY [2 votes]: It might be of interest to recall how Bourbaki uses the word "canonical" -- though admittedly this is far from formal logic. When a notion is first defined, for instance the homomorphism from a group to a quotient group, Bourbaki says "this homomorphism is called canonical".
This allows him to talk about "the canonical homomorphism from $G$ to $G/H$" without any ambiguity  whenever this situation occurs.
So he does not define the term canonical, but only certain canonical maps, sets, objects ...<|endoftext|>
TITLE: How often do people read the work that they cite?
QUESTION [55 upvotes]: I have the following question:

How likely it is that an author carefully read through a paper cited by him?

Not everyone reads through everything that they have cited. Sometimes, if one wants to use a theorem that is not in a standard textbook, one typically finds another paper which cites the desired result and copies that citation thereby passing the responsibility of ensuring correctness to someone else. This saves a lot of time, but seems to propagate inaccurate citations and poor understanding of the work being cited.
The question is thus about what should authors' citing policy be, and to what extent authors should verify results they are citing rather than using them as black boxes.

REPLY [12 votes]: [offtopic] Since I cannot comment, let me just throw in an old story I heard from my professor. Some time back, a paper by Einstein and Preuss was being cited all around. Now, we all know some names that collaborated with Einsten, but this Preuss is kinda unknown. Turns out that the journal name of the reference
Einstein, A. (1931). Sitzungsber. Preuss. Akad. Wiss. ...
after some citations, got to be promoted to coauthor. NICE!
Here's some (german) reference: http://de.wikipedia.org/wiki/S._B._Preuss
Now just to account to the statistics, I know some people that skip the reading of some papers to present seminars and talks, but I think they do check the stuff before writing something up. As to me, I try to read some stuff and then check the references and references of references until I give up. But that doesn't matter, since I'm far from publish anything at all, as it seems. Cheers.<|endoftext|>
TITLE: Fractal Tiling of Rhombic Dodecahedra
QUESTION [9 upvotes]: Hello, this is my first question on Math Overflow...
Rhombic dodecahedra can be tiled in 3-space, leaving no gaps. This tiling corresponds to the close-packing of spheres. 
Consider a "nucleus" rhombic dodecahedron ("rd") surrounded by 12 rd's, such that each of the faces of the nucleus rd is shared with a face of one of the 12 outer rd's. This becomes a composite polyhedron (call it "rd1") which can be described as the union of 13 rd's.
My question is this: can rd1 then be used as a tile for a similar tiling of 13 rd1's? (call it "rd2") - leaving no gaps and no overlapping? (You can think of this as the 3D version of tiling 7 regular hexagons, and then tiling those 7 to form a shape comprised of 49 hexagons.
The next question is: can this process be applied forever? (rdn). If rdn is scaled during each iteration such that it has the same volume as the original rd, then at the limit the resulting shape could be described as a "fractal rhombic dodecahedron" - having a fractal surface. But I don't know if such a thing can exist. 
I am wondering if there is a way to arrive at a proof or a strong conjecture that such a fractal tiling ("pertiling") can exist. What kinds of geometry tools should I use to arrive at an answer? 
Thanks!
-Jeffrey

REPLY [4 votes]: Yes, although the process can't be quite as symmetric as it is in 2D.
Let's first try a simpler version of the question to warm up:
Cubes tile 3-space, leaving no gaps.
Consider a "nucleus" cube surrounded by 6 cubes, such that each of the faces of the nucleus cube is shared with a face of one of the 6 outer cubes. This becomes a composite polyhedron (call it "cb1") which can be described as the union of 7 cubes, or as a 3D plus sign.
We can see that this cb1 object is roughly octahedral in some sense.  (The 2D cases of squares or hexagons was simpler in that regular n-gons are their own duals.)  Unfortunately, unlike the approach used in the 2D cases of 5 squares or 7 hexagons, in 3D we cannot align a slightly twisted octahedron so that its vertices move onto vertices of the cb1.  Even if we could, it wouldn't help, since octahedra don't tile space.
Nevertheless the cb1 can tile space in a regular lattice with 6 "nearest" neighbors offset by the cyclic permutations of $\pm(0,2,-1)$.  (Actually, lattice member $(1,1,1)$ is closer, but I want to emphasize that the lattice of cb1 centers is a cubic lattice again, just squished in the (1,1,1) direction by a factor of $\sqrt{7}$.)  Each layer looks like this: (+/- indicates center on next/previous layer)
+ - . * * * . + - . * * * .
. * + - * . . . * + - * . .
* * * . + - . * * * . + - .
- * . . . * + - * . . . * +
. + - . * * * . + - . * * *
. . * + - * . . . * + - * .
. * * * . + - . * * * . + -
+ - * . . . * + - * . . . *
* . + - . * * * . + - . * *
. . . * + - * . . . * + - *
- . * * * . + - . * * * . +
* + - * . . . * + - * . . .
* * . + - . * * * . + - . *
* . . . * + - * . . . * + -

Now we want to combine these into some kind of "cb2" clusters.  If we just consider the cb1 centers, we see (as mentioned above) that we have a cubic lattice, squished in the (1,1,1) direction.  So we can make clusters in this lattice just like we did in the initial lattice, attaching 6 neighbor cb1s to a central 7th cb1.  The process can be repeated ad infinitum, yielding the desired type of structure, except that the (1,1,1) direction gets progressively more squished at each stage.  To avoid this, you could pick a different direction to squish at each level, so that the eccentricity of the object never gets too bad.
So with cubes it works, with the caveat that the eccentricity changes at each level.

So how about with rhombic dodecahedra?
In this case, the rd1 is something like the dual of a rhombic dodecahedron, which doesn't tile space, and again, since we are in 3D, we won't be able to "twist" a tiling-capable polyhedron so its vertices fall onto vertices of the rd1, one per attached rd0.  So again, we can't do an exact analog of the 2D case.
But as we saw with the cubes, if we can find some tiling of space by rd1 objects, then we can hopefully look at the lattice of centers of those objects as a cubic or rhombododecahedral lattice (the difference between the two being only a factor of 2 in the (1,1,1) direction), and use that to repeat the original tiling at higher and higher levels.
Here is one simple way to do it: $\ \ \ $ (+/T indicate center on next layer, -/= indicate center on previous)
+   +   T   *   *   .   *   *   -   =   =   -   +   T   T   +   .   .   *   .  
  +   T   T   +   .   .   *   .   .   =   -   -   =   T   +   +   T   *   *   .
-   =   T   +   +   T   *   *   .   *   *   -   =   =   -   +   T   T   +   .  
  =   =   -   +   T   T   +   .   .   *   .   .   =   -   -   =   T   +   +   T
.   =   -   -   =   T   +   +   T   *   *   .   *   *   -   =   =   -   +   T  
  *   *   -   =   =   -   +   T   T   +   .   .   *   .   .   =   -   -   =   T
.   *   .   .   =   -   -   =   T   +   +   T   *   *   .   *   *   -   =   =  
  *   *   .   *   *   -   =   =   -   +   T   T   +   .   .   *   .   .   =   -
T   +   .   .   *   .   .   =   -   -   =   T   +   +   T   *   *   .   *   *  
  +   +   T   *   *   .   *   *   -   =   =   -   +   T   T   +   .   .   *   .
-   +   T   T   +   .   .   *   .   .   =   -   -   =   T   +   +   T   *   *  
  -   =   T   +   +   T   *   *   .   *   *   -   =   =   -   +   T   T   +   .
-   =   =   -   +   T   T   +   .   .   *   .   .   =   -   -   =   T   +   +  
  .   =   -   -   =   T   +   +   T   *   *   .   *   *   -   =   =   -   +   T
.   *   *   -   =   =   -   +   T   T   +   .   .   *   .   .   =   -   -   =  
  .   *   .   .   =   -   -   =   T   +   +   T   *   *   .   *   *   -   =   =
T   *   *   .   *   *   -   =   =   -   +   T   T   +   .   .   *   .   .   =  
  T   +   .   .   *   .   .   =   -   -   =   T   +   +   T   *   *   .   *   *
T   +   +   T   *   *   .   *   *   -   =   =   -   +   T   T   +   .   .   *  
  -   +   T   T   +   .   .   *   .   .   =   -   -   =   T   +   +   T   *   *
-   -   =   T   +   +   T   *   *   .   *   *   -   =   =   -   +   T   T   +  
  -   =   =   -   +   T   T   +   .   .   *   .   .   =   -   -   =   T   +   +
.   .   =   -   -   =   T   +   +   T   *   *   .   *   *   -   =   =   -   +  
  .   *   *   -   =   =   -   +   T   T   +   .   .   *   .   .   =   -   -   =
.   .   *   .   .   =   -   -   =   T   +   +   T   *   *   .   *   *   -   =  

In this pattern each rd1 cluster touches 14 others, but 2 of them just barely.  Leaving those two out, the 12 neighbors are at $\pm(3,-1,0), \pm(0,-3,1), \pm(3,2,-1), \pm(2,2,2), \pm(-1,3,2), \pm(-1,0,3)$.  The lattice of rd1 centers is indeed the sort of lattice we were looking for, with basis vectors $2\vec{x}=(2,5,1), 2\vec{y}=(2,-1,3), 2\vec{z}=(-4,1,3)$.  Using these basis vectors, we can draw the lattice of cluster centers as a cubic lattice, and make "rd2" clusters of clusters using the same pattern as above, and so on up the hierarchical levels.
The resulting objects do have the flavor of a Gosper Island, although in 3D I would call it an asteroid.<|endoftext|>
TITLE: Examples of acylindrical 3-manifolds
QUESTION [10 upvotes]: Let $C$ be the compact cylinder $S^1\times [0,1]$.  A 3-manifold $M$ with incompressible boundary is called acylindrical if every map $(C,\partial C)\to (M,\partial M)$ that sends the components of $\partial C$ to essential curves in $\partial M$ is homotopic rel $\partial C$ into $\partial M$.

I'm looking, for each $g\geq 2$, for examples of compact, orientable, acylindrical, hyperbolic 3-manifolds $M_g$ with non-empty, incompressible boundary such that each component of $\partial M_g$ is homeomorphic to the surface of genus $g$.

I'm sure such things should be well known to the experts.
Here's a little motivation.  Such examples would be useful because, given an arbitrary hyperbolic 3-manifold $N$ with incompressible boundary, you can glue copies of the $M_g$ to the non-toroidal boundary components of $N$ and the result, by Geometrization (for Haken 3-manifolds, so you only need Thurston, not Perelman), is a hyperbolic 3-manifold of finite volume.

REPLY [11 votes]: (source)
The exterior of Suzuki's Brunnian graph on $n$-edges, here pictured with $n=7$, is irreducible, atoroidal, boundary incompressible, and acylindrical.  See 
Luisa Paoluzzi and Bruno Zimmermann. On a class of hyperbolic 3-manifolds and groups with one defining relation. Geom. Dedicata, 60(2):113–123, 1996 
or  
Akira Ushijima. The canonical decompositions of some family of compact orientable hyperbolic 3-manifolds with totally geodesic boundary. Geom. Dedicata, 78(1):21–47, 1999.
(I think these manifolds may be contained in Bruno's list also.)<|endoftext|>
TITLE: Question about an estimate in Hörmander's proof of Cartan's Theorem B
QUESTION [9 upvotes]: I have been working through the proof of Cartan's Theorem B that Hörmander gives in his book 'Introduction to Complex Analysis in Several Variables'. When I began, I skipped over some of the initial steps so that I could get to a specific calculation that my supervisor wanted me to look at. Now, looking back at the setup, I realise that I don't understand all of the details.
In Chapter 4, Hörmander introduces linear, closed, densely defined operators $T : L^2_{(p,q)}(\Omega, \varphi_1) \to L^2_{(p,q+1)}(\Omega, \varphi_2)$ and $S : L^2_{(p,q+1)}(\Omega, \varphi_2) \to L^2_{(p,q+2)}(\Omega, \varphi_3)$ which are defined by $\overline{\partial}$.
Some functional analysis shows that it is enough to prove that there is a positive constant $C$ such that $\|f\|^2 \leq C(\|T^*f\|^2 + \|Sf\|^2)$ for all $f \in D_{T^*}\cap D_S$. An argument is then given to show that $D_{(p,q+1)}(\Omega)$ is dense in $D_{T^*}\cap D_S$ with respect to the graph norm $f \mapsto \|f\| + \|T^*f\| + \|Sf\|$, where $D_{(p,q+1)}(\Omega)$ denotes the smooth compactly supported $(p,q+1)$ forms. The proof of this fact is where my troubles begin.
Hörmander shows that for suitable weights, and a sequence of compactly supported functions $(\eta_{\nu})_{\nu \in \mathbb{N}}$ with $0 \leq \eta_{\nu} \leq 1$ and $\eta_{\nu} = 1$ on any compact subset of $\Omega$ when $\nu$ is large (which satisfy an appropriate bound on $|\bar{\partial}\eta_{\nu}|$), we have $\|\eta_{\nu}f - f\|_{\varphi_{2}} \to 0$, $\|S(\eta_{\nu}f) - \eta_{\nu}Sf\|_{\varphi_{3}} \to 0$, and $\|T^*(\eta_{\nu}f) - \eta_{\nu}T^*f\|_{\varphi_{1}} \to 0$. I can understand why the first two are true, but not the third.
Hörmander shows that $\eta_{\nu}f \in D_{T^*}$. From there I can see how he gets, for $u \in D_T$, $|(T^*(\eta_{\nu}f) - \eta_{\nu}T^*f, u)_{\varphi_{1}}| \leq \int|f|e^{-\varphi_{2}/2}|u|e^{-\varphi_{1}/2}d\lambda$, but after this inequality, he states 

$\dots$ which implies the bound $|T^*(\eta_{\nu}f) - \eta_{\nu}T^*f|^2e^{-\varphi_{1}} \leq |f|^2e^{-\varphi_{2}}$.

I don't see how this follows. How does Hörmander obtain this (pointwise) estimate? At the moment, the best I’ve got is a messy measure theoretic argument that I’m not even sure is correct. Any help would be much appreciated.

REPLY [4 votes]: I also do not see how the desired pointwise bound $|T^*(\eta_{\nu}f) - \eta_{\nu}T^*f|^2e^{-\varphi_1} \leq |f|^2e^{-\varphi_2}$ follows from the $L^2$ estimate. The latter certainly implies by Cauchy-Schwarz that the operator norms of the commutators $[T^*,\eta_{\nu}]$ are uniformly bounded by unity, but this is not what is needed. Instead, I propose to argue as follows. The commutator  $[T^*,\eta_{\nu}]$ is the adjoint of the multiplication operator $[\overline{\eta_{\nu}},T]=-\overline{\partial}\overline{\eta_{\nu}}\wedge$. Using only pointwise estimates the desired bound follows.
Let me add two observations: (1) Two pages later Hörmander computes $T^*$ and states that this gives another proof of his formula (4.1.8) which is what needs to be proved. Actually the argument I gave above is this alternative proof without computing the commutator very explicitly. (2) In his 1965 Acta. Math. paper on the $\overline{\partial}$ operator, Hörmander does prove the analog of (4.1.8) via an operator norm estimate of the commutator. However, in the proof of Proposition 2.1.1 of that paper he has a setup which gives that the commutators converge to zero in norm, in contrast to only strong convergence as in Lemma 4.1.3 of the book.<|endoftext|>
TITLE: Strength of $\Delta_1^0$ subset of $2^\mathbb{N}$ as finite union of specific basic open sets.
QUESTION [8 upvotes]: This question is to find the Reverse Mathematical strength of writing $\Delta_1^0$ (clopen) subset of $2^\mathbb{N}$ as a finite union $\bigcup_{\sigma \in F} [|\sigma|]$ where $F \subset 2^{<\mathbb{N}}$ is finite and $[|\sigma|] = \{f \in 2^{\mathbb{N}} : \sigma \prec f\}$. 
More formally, if $\varphi(f)$ is a $\Delta_1^0$ formula in second order arithmetic, does there exists a finite set $F \subset 2^{<\mathbb{N}}$ such that for any $f \in 2^\mathbb{N}$, $\varphi(f)$ holds if and only if there exists a $\sigma \in F$ such that $\sigma \prec f$.
I am quite sure that $WKL_0$ can prove this by formalizing the usual compactness argument in Cantor space. Is this property equivalent to $WKL_0$ over $RCA_0$? Can anyone see a proof of this result from weaker systems like $WWKL_0$, $RCA_0$?
Thanks for any help you can provide. I have proved something using the clopen principle above and an idea of the strength of this result would help me pinpoint the proof theoretic strength of what I am really interested in. Thanks very much.

REPLY [8 votes]: This is equivalent to $\mathsf{WKL}_0$, with little a caveat... Note that there is no such thing as a "$\Delta^0_1$ formula." Below, I will use the most permissive meaning for $\Delta^0_1$, which is the usual one in this context. If one uses a more restrictive meaning (e.g. bounded formula, provably $\Delta^0_1$ formula) then we can possibly prove the existence of such a finite set $F$ in plain $\mathsf{RCA}_0$.
I will show that the statement implies $\Sigma^0_1$-separation, which is a well known equivalent of the Weak König Lemma. (The reverse implication is a standard compactness argument as you described.) Suppose, for the sake of contradiction, that $e:\mathbb{N}\to\mathbb{N}$ is an injection such that $\lbrace e(2s) : s \in \mathbb{N}\rbrace$ and $\lbrace e(2s+1) : s \in \mathbb{N}\rbrace$ form an inseparable pair: there is no $f:\mathbb{N}\to2$ such that $f(e(s)) \equiv s \bmod{2}$ for all $s$.
Now consider the statements $$\phi_0(f) \equiv (\exists s)(f(e(2s))=1 \land (\forall t \lt 2s)(f(e(t)) \equiv t \bmod{2}))$$ and $$\phi_1(f) \equiv (\exists s)(f(e(2s+1)) = 0 \land (\forall t \lt 2s+1)(f(e(t)) \equiv t \bmod{2})).$$
These are both $\Sigma^0_1$-formulas. Clearly, these represent disjoint subsets of $2^{\mathbb{N}}$. In fact, by our inseparability assumption, these represent complementary subsets of $2^{\mathbb{N}}$. Therefore, $\phi_0(f)$ and $\phi_1(f)$ are $\Delta^0_1$.
Now suppose $F_0$ and $F_1$ are finite sets of binary sequences such that 
$$\phi_0(f) \leftrightarrow f \in \bigcup_{\sigma \in F_0} [\sigma], \quad \phi_1(f) \leftrightarrow f \in \bigcup_{\sigma \in F_1} [\sigma].$$
Note that by definition of $\phi_0(f)$, every $\sigma \in F_0$ must be such that for some $s$, we have $\lbrace{e(2s),e(1),e(3),\dots,e(2s-1)\rbrace} \subseteq \operatorname{dom}(\sigma)$, $\sigma(e(2s)) = 1$ and $\sigma(e(2t+1)) = 1$ for all $t \lt s$. Similarly for $F_1$. Let $m$ be the maximum length of a sequence in $F_0 \cup F_1$ and let $n$ be such that $e(t) \geq m$ for all $t \geq n$ (such an $n$ must exist since $e$ is injective). It must then be the case that for every $f:\mathbb{N}\to2$, whether $\phi_0(f)$ or $\phi_1(f)$ is already determined by looking at the restriction $f{\upharpoonright}\lbrace0,\dots,m-1\rbrace$. But this is impossible since the characteristic function of the finite set $\lbrace e(2s+1) : s \lt n\rbrace$ does not have this property, for example.<|endoftext|>
TITLE: How to make an approximation of path with polynom P(x,y)=0?
QUESTION [5 upvotes]: Hi. Imagine that a user draws on the canvas any path. I want to approximate this path with a path $P(x,y)=0$ where $P(x,y)$ - is unknown polynom. May be somebody can suggest an appropriate algorithm?
1) I tried to use the method of least squares to find this polynom. Just choosed on the path a lot of points $(x_i,y_i)$ . And minimized unknown $\sum\limits_i P(x_i,y_i)^2$ among all polinoms of degree n. This problem of minimization is a problem of finding an eigen vector of huge matrix that grows as $n^2\times n^2$. Already for $n=7$ this matrix has a size $36\times36$ it's hard work for PC   to find the solution. And for $n=7$ it doesn't give appropriate result.
2) Spline doesn't work for me. Because spline - is a union of curves $P_i(x,y)=0$. To each spline  of course we can correspond $P(x,y)=\prod\limits_i P_i(x,y) $. But this union  $P(x,y)=0$ will have a lot of bifurcation points on the curve. And for my project it is very bad

REPLY [2 votes]: Your reference to eigenvalues suggests that you are minimizing $\sum_i P(x_i, y_i)^2$ subject to a constraint like $\sum_{a,b} P_{ab}^2=1$, where $P(x,y) = \sum_{a,b} P_{ab} x^a y^b$. Why not instead minimize $\sum_i P(x_i, y_i)^2$ subject to a linear constraint like $P_{00}=1$? 
This is a standard approach when fitting a conic through a cloud of points (see, for example, Paul L. Rosin, A note on the least squares fitting of ellipses, Pattern Recognition Letters, Volume 14, Issue 10, October 1993, Pages 799–808) and I don't see why it wouldn't be a good idea for higher degree.
This puts your problem in the form "minimize $\vec{x}^T A \vec{x}$, subject to $\vec{b} \cdot \vec{x}=1$", to which the solution is $\vec{x} = \frac{A^{-1} \vec{b}}{\vec{b}^T A^{-1} \vec{b}}$; no eigenvalues needed. 
I suspect Misha's comment will be more useful than this answer, though.
I also second everyone who said that $36 \times 36$ is not large. It looks like this thread might be able to help you find a better linear algebra library.<|endoftext|>
TITLE: Connectedness of the linear algebraic group SO_n
QUESTION [15 upvotes]: I apologize in advance if my question is too elementary for MO.
It is a well known fact that the linear algebraic group $G = \mathsf{SO}_n$ is connected, and there exist a few different proofs of this fact. One proof goes by showing that $G$ is generated by unipotent elements, and invoking the theorem that every linear algebraic group with this property is connected.
My question is about a different, more direct proof, involving the Cayley transform
$$ A \mapsto (I_n + A)^{-1} (I_n - A) , $$
which maps every matrix $A \in G(k)$ for which $I_n + A$ is invertible (let's write $W$ for the set of such matrices), to a skew-symmetric matrix, and in fact, this map defines an isomorphism of varieties between the non-empty open subset $W$ of $G$, and an open subset of the irreducible variety $V$ of skew-symmetric matrices.
But how does one now conclude that $\mathsf{SO}_n$ is connected? In particular, why is the closure of $W$ equal to $G$? Of course the complement of $W$ is given by the polynomial equation $\det(I_n + A) = 0$, but how is this situation different from, for instance, the fact that $\mathsf{SO}_n$ is the subvariety of $\mathsf{O}_n$ defined by the polynomial equation $\det(A) - 1 = 0$ (or $\det(A) + 1 \neq 0$) whereas $\mathsf{O}_n$ is not connected?
What subtlety am I missing?
added:

I'm assuming $\operatorname{char}(k) \neq 2$.
The underlying topology is the Zariski toplogy.
The proof I'm mentioning in the second paragraph is only valid in characteristic $0$.
It seems that the question is not too elementary...

added: I'll add a few more sentences to make clear what I mean by connected in the Zariski topology.
I'm considering $G$ as a group functor (i.e. a functor from the category of commutative $k$-algebras to the category of groups). Then $G$ is connected if and only if its coordinate algebra $k[G] = \mathcal{O}(G)$ has no non-trivial idempotents. In particular, in order to show that $G$ is connected, we may assume w.l.o.g. that $k$ is algebraically closed. [In fact, since $G$ is smooth (since $\operatorname{char}(k) \neq 2$), we may instead consider the group of $\bar{k}$-rational points $G(\bar{k})$, which brings us back to a more classical situation.]

REPLY [5 votes]: This is a more elementary solution, which would make sense in the context of Springer's book.  We consider
$SO(V,\langle,\rangle)$, or for short $SO(V)$, where $V$ is $n$-dimensional and $\langle,\rangle$ a nondegenerate
symmetric bilinear form on $V$. (Recall that any two such are equivalent.)
We'll show by induction on $n$ that $SO(V)$ is connected.
Springer's exercise shows that any $g \in SO(V)$ that doesn't have eigenvalue $-1$ is in the identity component of $SO(V)$.
When $n = 1$, $SO(V) = 1$ and when $n = 2$ we see directly that $SO(V)$ is connected. (In fact,
$SO_2 = \{\left(\begin{matrix} a & b\\ -b & a\end{matrix}\right) : a^2 + b^2 = 1\}$ and
$SO_2 \cong \mathbb G_m$ as algebraic groups via $\left(\begin{matrix} a & b\\ -b & a\end{matrix}\right) \mapsto a+bi$
with $i^2 = -1$.)
In general, fix $g \in SO(V)$ and we'll show that $g$ is contained in the identity component.
Note that $V = \oplus V_\lambda$ is a direct sum of generalised eigenspaces and that $V_\lambda \perp V_\mu$ if
$\lambda \ne \mu^{-1}$. Thus we get an orthogonal decomposition $V = V_{-1} \perp V_1 \perp \perp_{\{\lambda \ne \lambda^{-1}\}}
(V_\lambda \oplus V_{\lambda^{-1}})$. In particular, $V_{-1}$ is a nondegenerate subspace and we have the
orthogonal decomposition of $g$-invariant subspaces $V = V_{-1} \perp V_{-1}^\perp$. Also, we get that $\dim V_\lambda = \dim V_{\lambda^{-1}}$
if $\lambda \ne \lambda^{-1}$, so by considering the determinant we see that $\dim V_{-1}$ even (as remarked by Will), say it equals $2d$.
It follows that $g$ is contained in the subgroup $SO(V_{-1}) \times SO(V_{-1}^\perp)$ of $SO(V)$. By induction
we are thus reduced to the two extreme cases: $V = V_{-1}$ or $V = V_{-1}^\perp$. The second case is treated by the
Springer exercise, so we may assume that all eigenvalues of $g$ are $-1$ and $n = 2d$.
Now $-g$ is unipotent, hence by the Springer exercise we know it is in the identity component of $SO(V)$.
Multiplying by $-1 \in SO(V)$ we deduce that $g$ is in the same component as $-1$. Finally, after choosing an
appropriate basis $-1$ lies in the subgroup $SO_2 \times \dots \times SO_2$ of $SO_{2d}$, so we are done by the $n = 2$ case.<|endoftext|>
TITLE: Finite local rings
QUESTION [9 upvotes]: There is some classification of finite commutative local rings. For example how many not isomorphic finite local rings with the same order $p^k$ and the same residue field $\mathbb F_p$ exist?

REPLY [5 votes]: Corbas and Williams answered this question for local rings with p^k elements, where k is at most 5.  See "Rings of Order p^5 Part I. Nonlocal Rings" and "Rings of Order p^5 Part II. Local Rings". Both papers are in Journal of Algebra, vol. 231; the first on pages 677-690 and the second on pages 691-704.<|endoftext|>
TITLE: Hyperbolic structures on $S\times\mathbb{R}$
QUESTION [8 upvotes]: Let $S$ be a closed incompressible surface in a finite volume hyperbolic 3-manifold $M$ without cusps.  Let $N$ be the cover of $M$ associated to $\pi_1(S) \subset \pi_1(M)$.  The cover $N$ is homeomorphic to $S\times\mathbb{R}$ and has two ends.  An end is convex cocompact if it has a neighborhood whose intersection with the convex core of $N$ is bounded and degenerate otherwise.
If $S$ is a fiber, then both ends of $N$ are degenerate. If the group of isometries corresponding to $\pi_1(S)$ is quasi-Fuchsian, then both ends of $N$ are convex cocompact.
My questions:  Is it possible for one end of $N$ to be convex cocompact and the other to be degenerate?
What if we allow $M$ to have infinite volume?
Can you point me to a description of an example or an explanation of why there are no examples?

REPLY [13 votes]: It follows from Thurston's Covering Theorem that there are no such examples.
The covering theorem says that if a degenerate end is infinite-to-one under a covering map, then you are (virtually) in the fibered case. See R.D. Canary, A covering theorem for hyperbolic 3-manifolds and its applications, Topology 35:3 (1996), 751–778.

REPLY [7 votes]: To answer your question about allowing $M$ to have infinite volume, there exist such examples on $M=S \times \mathbb{R}$ itself. These were originally constructed by Bers, his "singly degenerate" groups.<|endoftext|>
TITLE: Are there Maass forms where the expected Galois representation is $\ell$-adic?
QUESTION [24 upvotes]: Recall that by theorems of Deligne and Deligne--Serre, there is the following dichotomy:

Modular forms on the upper half plane of level $N$ and weight $k\geq 2$ correspond to representations $\rho:\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q)\to\operatorname{GL}(2,F_\lambda)$ for some number field $F$ and prime $\lambda$ over $\ell$.
Modular forms on the upper half plane of level $N$ and weight $k=1$ correspond to representations $\rho:\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q)\to\operatorname{GL}(2,\mathbb C)$ satisfying $\det\rho(\sigma)=-1$ ($\sigma$ is complex conjugation).

Now I hear that Maass forms of eigenvalue $\frac 14$ are conjectured to correspond with representations $\rho:\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q)\to\operatorname{GL}(2,\mathbb C)$ satisfying $\det\rho(\sigma)=1$ ($\sigma$ is complex conjugation).  Is this still true (that is, conjectured) for Maass forms of higher weight?  Or do they "turn $\ell$-adic" in higher weight?

REPLY [45 votes]: Here's some piece of the bigger picture. Maass forms and holomorphic modular forms are both automorphic representations for $GL(2)$ over the rationals. An automorphic representation is a typically huge representation $\pi$ of an adele group (in this case $GL(2,\mathbf{A})$, with $\mathbf{A}$ the adeles of $\mathbf{Q}$). Because the adeles is the product of the finite adeles and the infinite adeles, this representation $\pi$ is a product of a finite part $\pi_f$ and an infinite part $\pi_\infty$. The infinite part is a representation of $GL(2,\mathbf{R})$ (loosely speaking -- there are technicalities but they would only cloud the water here). 
The representation theory of $GL(2,\mathbf{R})$, in this context, is completely understood. The representations basically fall into four categories, which I'll name (up to twist):
1) finite-dimensional representations (these never show up in the representations attached to cusp forms).
2) Discrete series representations $D_k$, $k\geq2$ (these are the modular forms of weight 2 or more).
3) The limit of discrete series representation $D_1$ (these are the weight 1 forms).
4) The principal series representations (these are the Maass forms).
Now what does Langlands conjecture? He makes a conjecture which does not care which case you're in! He conjectures the existence of a "Galois representation" attached to $\pi$, and this is a "Galois representation" in a very loose sense: it is a continuous 2-dimensional complex representation of the conjectural "Langlands group", attached to $\pi$. Note that there should be a map from the Langlands group to the Galois group, and in the case of Maass forms and weight 1 forms Langlands' representation should factor through the Galois group. For modular forms of weight 2 or more Langlands' conjecture has not been proved and in some sense it is almost not meaningful to try to prove it because no-one can define the group. In particular Deligne did not prove Langlands' conjecture, he proved something else.
So Clozel came along in 1990 and tried to put Deligne's work into some context and he came up with the following: he formulated the notion of what it meant for $\pi_\infty$ to be algebraic (in fact there are two notions of algebraic, which differ by a twist in this context, so let me write "$L$-algebraic" to make it clear which one I'm talking about) and conjectured that if $\pi$ were $L$-algebraic then there should be an $\ell$-adic Galois representation $\rho_\pi$ attached to $\pi$. Maass forms with eigenvalue $1/4$, and holomorphic eigenforms, are $L$-algebraic, and the $\ell$-adic Galois representation attached to the Maass forms/weight 1 forms is just the one you obtain by fixing an isomorphism $\mathbf{C}=\overline{\mathbf{Q}}_\ell$. I should say that Clozel worked with $GL(n)$ not $GL(2)$ and also worked over an arbitrary number field base.
Whether or not the image of $\rho_\pi$ is finite is something which is conjecturally determined by $\pi_\infty$: you can read it off from the infinitesimal character of $\pi_\infty$ and also from the local Weil group representation attached to $\pi_\infty$ by the local Langlands conjectures, which are all theorems (of Langlands) for real reductive groups.
Put within this context your question becomes purely local: one has to figure out what Clozel's recipe gives in each case to get a handle on what your question is asking. You're asking about principal series representations. If you work out Clozel's recipe in these cases you find that if $\lambda\not=1/4$ then $\pi_\infty$ is not $L$-algebraic (and so we don't even expect a representation of the Galois group, we just expect a representation of the conjectural Langlands group), and if $\lambda=1/4$ then, up to twist, we expect the image to be always finite, because, well, that's what the calculation gives us.
I learnt this by just doing all these calculations myself. I wrote them up in brief notes at http://www2.imperial.ac.uk/~buzzard/maths/research/notes/automorphic_forms_for_gl2_over_Q.pdf and http://www2.imperial.ac.uk/~buzzard/maths/research/notes/local_langlands_for_gl2R.pdf (both available from http://www2.imperial.ac.uk/~buzzard/maths/research/notes/index.html ).
So why is there this asymmetry? Well actually this asymmetry is not surprising because it is predicted on the Galois side as well. If you look at an irreducible mod $p$ ordinary Galois representation which is odd then its universal ordinary deformation is often known to be isomorphic to a Hecke algebra of the type defined by Hida (so in particular we get lots of interesting $\ell$-adic Galois representations with infinite image). In particular its Krull dimension should be 2 (and this was already known to Mazur in the 80s). But the calculations for these Krull dimensions involve local terms, and the local term at infinity depends on whether the representation is odd or even. If you consider deformations of an even Galois representation then the calculations come out differently and the Krull dimension comes out one smaller. In particular one only expects to see finite image lifts, plus twists of such lifts by powers of the cyclotomic character. 
So in summary you see differences on both sides -- the automorphic side and the Galois side -- and they match up perfectly! You don't expect $\ell$-adic representations to show up in the Maass form story and yet things are completely consistent anyway.
Toby Gee and I recently tried to figure out the complete conjectural picture about how automorphic representations and Galois representations were related. Our conclusions are at http://www2.imperial.ac.uk/~buzzard/maths/research/papers/bgagsvn.pdf . But for $GL(n)$ this was all due to Clozel over 20 years ago (who would have known all those calculations that I linked to earler; these are all standard).<|endoftext|>
TITLE: Building a genus-$n$ torus from cubes
QUESTION [24 upvotes]: I wonder if this has been studied:

What is the fewest number of unit cubes
  from which one can build an $n$-toroid?

The cubes must be glued face-to-face,
and the boundary of the resulting object should 
be topologically equivalent to an $n$-torus, by which I mean
a genus-$n$ handlebody in $\mathbb{R}^3$ (as per Kevin Walker's terminological correction).
For example, 8 cubes are needed to form a 1-toroid:

          


And it seems that 13 cubes are needed for a 2-toroid:

          


I know how intricate is the analogous question for minimizing the
number of triangles from which one can build a torus
(cf. Császár's Torus), but I am hoping that my much easier question
has an answer for arbitrary $n$.
Thanks for ideas and/or pointers!
Addendum. Here is Steve Huntsman's 20-cube candidate for genus-5:

REPLY [23 votes]: Theorem. Let $c(g)$ be the minimum number of cubes such that the boundary of some configuration of $c(g)$ cubes is a genus $g$ surface.  Then $c(g)/g \to 2$ as $g \to \infty$.  
Proof. We write $\chi(X)$ for the compactly supported Euler characteristic
of $X$, i.e., 
$\chi(X) = \sum (-1)^i \dim H^i_c(X, \mathbb{Q})$. 
Note this is not a homotopy invariant: the compactly supported Euler characteristic of $\mathbb{R}^n$ is $(-1)^n$.  It does however have the property that for reasonable finite disjoint union decompositions, $X = \coprod X_i$, we have $\chi(X) = \sum \chi(X_i)$. 
Let $H_g$ be a closed genus $g$ handle-body.  Then $$\chi(H_g) = 1-g.$$ 
On the other hand, let $K$ be a configuration of cubes.  We write $K^0, K^1, K^2, K^3$ for the sets of vertices, edges, squares, and cubes, respectively, 
and $k^0, k^1, k^2, k^3$ for their cardinalities. Then 
$$\chi(K) = k^0 - k^1 + k^2 - k^3$$
We will count $k^0, k^1, k^2, k^3$ by looking at the interior of the $2 \times 2$ cube around each vertex.  That is, abutting some vertex $v$, there is $1$ vertex, $6$ edges, $12$ faces, and $8$ cubes.  More to the point, each $i$-dimensional face abuts $2^i$ vertices.  We write $K_v$ to mean the configuration localized at $v$; we write $K^i_v$ for the set of $i$-dimensional faces which abut the vertex $v$, and $k^i_v$ for its cardinality.  Thus: 
$$\chi(K) = \sum_{v \in \mathbb{Z}^3} \sum_{i=0}^3 (-2)^{-i} \cdot 
k^i_v $$ 
so we estimate
$$\frac{\chi(K)}{k^3} = \frac{\sum_{v \in \mathbb{Z}^3} \sum_{i=0}^3 (-2)^{-i} \cdot k^i_v }{\sum_{v \in \mathbb{Z}^3} 2^{-3} \cdot  k^3_v } \ge 
\min_{v \in \mathbb{Z}^3} \sum_i \frac{(-2)^{-i} \cdot k^i_v }{2^{-3} \cdot k^3_v }$$
The above inequality comes from the following fact: a (weighted) average is greater than the minimum term being averaged.  Thus for any $a_i, b_i$ with $b_i > 0$, we have 
$$\frac{\sum_i a_i}{\sum_i b_i} = \sum_i \frac{a_i}{b_i} \cdot \frac{b_i}{\sum b_i}
\ge \min_i \frac{a_i}{b_i}$$
Now let us analyze the possibilities for the right hand quantity
$$\tau(K_v) := \frac{8}{k_v^3} \cdot \left(k_v^0 - \frac{k_v^1}{2} + \frac{k_v^2}{4} - 
\frac{k_v^3}{8} \right)  $$
In fact, in the $2\times 2$ cube around a vertex, there are, up to symmetry, only 9 possible configurations of cubes whose boundary is (locally at that vertex) topologically a manifold: one each for every number of boxes other than 4, and for 4 boxes, the square, and the tripod configuration where, if say $v = (0,0,0)$, the cubes are the ones with most negative coordinates $(-1,-1,-1)$, $(-1, -1, 0)$, $(-1, 0, -1)$, and $(0, -1, -1)$.  Note that the neighborhood of every point in the interior of a very porous solid is a tripod configuration.
It remains to compute in each of these cases the above quantity.  For example, in the configuration $C_1$ when there is one cube, there is one vertex abutting the central one, three edges, three faces, and one cube.  Thus this contributes
$$ \tau(C_1) = \frac{8}{1} \cdot \left(1 - \frac{3}{2} + \frac{3}{4} - \frac{1}{8}  \right) = 1$$
We tabulate the remaining cases: 
$$ \tau(C_2) = \frac{8}{2}  \cdot \left(1 - \frac{4}{2} + \frac{5}{4} - \frac{2}{8}  \right) = 0$$
$$ \tau(C_3) = \frac{8}{3} \cdot \left(1 - \frac{5}{2} + \frac{7}{4} - \frac{3}{8}  \right) = -\frac{1}{3}$$
$$ \tau(C_4) = \frac{8}{4} \cdot \left(1 - \frac{5}{2} + \frac{8}{4} - \frac{4}{8}  \right) = 0$$
$$ \tau(C_4') = \frac{8}{4} \cdot \left(1 - \frac{6}{2} + \frac{9}{4} - \frac{4}{8}  \right) = -\frac{1}{2}$$
$$ \tau(C_5) = \frac{8}{5} \cdot \left(1 - \frac{6}{2} + \frac{10}{4} - \frac{5}{8}  \right) = -\frac{1}{5}$$
$$ \tau(C_6) = \frac{8}{6} \cdot \left(1 - \frac{6}{2} + \frac{11}{4} - \frac{6}{8}  \right) = 0$$
$$ \tau(C_7) = \frac{8}{7} \cdot \left(1 - \frac{6}{2} + \frac{12}{4} - \frac{7}{8}  \right) = \frac{1}{7}$$
$$ \tau(C_8) = \frac{8}{8} \cdot \left(1 - \frac{6}{2} + \frac{12}{4} - \frac{8}{8}  \right) = 0$$
Here, $C_4'$ is the tripod configuration.  We conclude that 
$$\frac{1-g(K)}{\# K^3} = \frac{\chi(K)}{\# K^3} \ge -\frac{1}{2}$$
hence $\# K^3 \ge 2g(K) - 2$. 
On the other hand, for any family of configurations in which the fraction of $v$ with $K_v \sim C_4'$ -- i.e., the probability that (the neighborhood of a given cube is a very porous solid) -- tends to 1, the estimates above are sharp and $\# K^3 /g(K) \to 2$. 
(An explicit calculation for the very porous cube appears in the comments above.) $\blacksquare$<|endoftext|>
TITLE: Existance of certain almost invariant functions related to amenability and piece-wise transformations 
QUESTION [18 upvotes]: We would like very much to know the answer to the following question:

Let $\|\cdot\|$ be any norm on $\mathbb{Z}^d$ and let $W(\mathbb{Z}^d)$ be the group of all bijections of $\mathbb{Z}^d$ such that
$$\|g(j)-j\|\leq C_g,$$
for some constant $C_g$ which depends only on the element $g\in W(\mathbb{Z}^d)$. 
Consider the Hilbert space $L^2(\{0,1\}^{\mathbb{Z}^d},\mu)$, where $\{0,1\}^{\mathbb{Z}^d}$ comes with  the standard  Bernoulli measure $\mu$.

We are looking for a sequence of
  functions $f_n\in
> L^2(\{0,1\}^{\mathbb{Z}^d},\mu)$ with
  $\|f_n\|_2=1$ and
  $$\|g.f_n-f_n\|_2\rightarrow 0, \text{
> for every } g\in W(\mathbb{Z}^d),$$
  $$\|f_n\cdot\chi_{\{\omega_j\in
> \{0,1\}^{\mathbb{Z}^d}:\omega_0=0\}}\|_2\rightarrow
> 1.$$
  where $\chi_{\{(\omega_j)_{j\in \mathbb{Z}^d}\in \{0,1\}^{\mathbb{Z}^d}:\omega_0=0\}}$ is the characteristic function of the cylinder set $\{(\omega_j)_{j\in \mathbb{Z}^d}\in \{0,1\}^{\mathbb{Z}^d}:\omega_0=0\}$


Motivation:

The existence of such a sequence for all $d$ would disprove the conjecture of Katok, that the interval exchange transformation group contains a free subgroup.

What we know about the question above:

In the joint work with Nicolas Monod, here, we showed that for $n=1$ the following function satisfy the properties above:
$$f_n(\omega)=e^{-n \sum\limits_{j\in \mathbb{Z}} \omega_j e^{-\frac{|j|}{n}}}=\prod_{j\in Z} a_j^{\omega_j},$$
where $a_j=e^{-n e^{-\frac{|j|}{n}}}$.
We are interested in extending this result to higher dimensions. Note that the function above is the product of functions of independent i.d. random variables.
Let $G< W(\mathbb{Z}^d)$ be a finitely generated subgroup of $W(\mathbb{Z}^d)$.
In addition to above (in collaboration with Nicolas Monod and Mikael de la Salle) we know that the existence of the functions with the property above in the class of functions which can be written as the product of functions of independent i.d. random variables is equivalent to a certain property of the Schreier graph of the action of $G$ on $\mathbb{Z}^d$. Let me give more details on this.
The Schreier graph of the action of $G$ on $X$
with respect to $S$ is the graph with vertices $X$ and with an edge
between $x$ and $y$ for each $g \in S$ with $g x=y$.
We say that an infinite graph $G=(V,E)$ satisfies a Sobolev inequality
rooted at $x_0 \in X$ if the value at $x_0$ of any $c_0$-function on
$V$ is bounded by the $\ell^2$-norm of its gradient, i.e.,
there is a constant $C>0$ such that
$$\|f\|_{c_0(V)} \leq C \sum_{x \sim x' \in V} |f(x') - f(x)|^2.$$
We can show that
The functions $f_n$ with the property above can be found
in the class of products if and only if the Schreier graph of the action of
$G$ on $X$ with respect to $S$ does not satisfy a Sobolev
inequality.
Moreover, for $d=1,2$ and $G$ be a finitely generated subgroup of $W(n)$ with symmetric generating set $S$. Then the Schreier graph of the action of $G$ on $\mathbb{Z}^d$ does not satisfy a rooted Sobolev inequality. However there are subgroups in $W(\mathbb{Z}^3)$ such that their Schreier graph satisfies Sobolev inequality.
To summarize above, we can find a sequence of functions in the class of products with the above property only in cases $n=1,2$. 

Any suggestions on potential examples
  of functions that can do the higher
  dimentional cases?

REPLY [4 votes]: This is my first visit to MO, so I have to apologize for making this an "answer": it is just a comment on Nik Weaver's suggestion.
I don't think that NW's conjecture is much different from the original question.  Indeed, you comment that it is different because f should work against all permutations g.  But in fact you restrict to those g that have C_g bounded by a universal constant (say, 2).  This is a very small set, it is even compact in the natural topology relevant to the action on Z^d (which is the action underlying the whole question).  Therefore, it is not so different from considering finitely many g's and thus from the original question.
Anyway, this is my intuition, it is not a mathematical statement.
Nicolas Monod
[Edit after reading NW's argument]: at first sight, it seems that your reduction to your conjecture is indeed just exploiting that the set C_g less than a constant is compact (approximation argument).<|endoftext|>
TITLE: Computing the correlation between two vectors without divulging them
QUESTION [6 upvotes]: Alice and Bob respectively know a vector of $N$ real numbers $u$ and $v$. They would both like to know $\rho = \langle u,v \rangle/N$ but Alice does not want Bob to gain anymore information about $u$ than given by $\rho$ and vice versa.
The easy solution is that they go to a trusted third party that calculated $\rho$ for them and hands them the solution. That's not very interesting though.
Given that homomorphic encryption exists, we know there's a way to do this without a trusted third party, there may exist a simpler and more elegant solution to the problem though.
Let's relax the problem a little bit and allow for asymptotic solutions in three respects

The extra information can be made arbitrarily small by the protocol
The extra information can be made arbitrarily small by taking $N$ to infinity 
$\rho$ can be calculated to an arbitrary degree of precision with an arbitrarily high probability

One road is for Alice and Bob to generate vectors $\epsilon$ and $\eta$ where every entry is a random normal number with a very large variance $\nu$.
Alice sends $u+\epsilon$ to Bob who computes and publishes $u.v + \epsilon.v$
Bob sends $v+\eta$ to Alice who computes and publishes $u.v + \eta.v$
They also compute $(u+\epsilon).(v+\eta)$.
With all that, they can compute $(u.v - \epsilon.\eta)/N$.
There remains an error term, $\langle \epsilon.\eta \rangle/N$ which has standard deviation about $\nu/\sqrt{N}$
The problem is that there is a tradeoff here. If the $\nu$ is too small, too much about the vector is divulged, if it's too big there is too much noise in the result.
Another idea would be for Alice and Bob to agree on a random base of $\mathbf{R}^N$. Alice could then pick $\epsilon$ as a random linear combination of the first $N/2$ vectors, and Bob could pick $\eta$ as a linear combination of the last $N/2$ base vectors. This guarantees $\epsilon$ and $\eta$ will be orthogonal, but now Bob would know the projection of $u$ over a very large subspace. Not that good.
Thougths?

REPLY [4 votes]: cis.syr.edu/~wedu/Research/paper/duthesis.pdf 
My favorite protocol is the 2nd one. One variation of it would be:
Alice breaks down u as $u_1 + \ldots + u_p$ and transmits set $( \{u_1^0, u_1^1\}, \{u_2^0, u_2^1\} \ldots, \{u_p^0, u_p^1\} )$ to Bob where $u_i^1$ are just a random vectors. Bob computes the dot product of each vector with a $v$ and adds random number $\epsilon_i$. Using the oblivious transfer protocol Alice gets back $(u_1^0.v +\epsilon_1, \ldots, u_p^0.v + \epsilon_p)$ and thus can compute $u.v + \sum \epsilon_i$. Bob can then divulge  $\sum \epsilon_i$ and $u.v$ is known.
I'm a little disappointed there isn't an answer that doesn't rely on modular arithmetic. I was hoping for a purely geometric protocol.<|endoftext|>
TITLE: Bounded linear functionals and representations
QUESTION [5 upvotes]: Suppose that $A$ is a unital C$^*$-algebra and that $\varphi: A \to \mathbb{C}$ is a bounded linear functional. Then there exists a Hilbert space $H$, a representation $\pi: A \to B(H)$ and vectors $\psi, \eta \in H$ such that $$\varphi(a) = \langle \pi(a)\psi, \eta \rangle$$
for all $a \in A$ (this can be proved by decomposing the functional as a linear combination of four states and considering the direct sum of the representation spaces associated to the GNS-construction for each state).  
My question is: 

Assuming further that $\| \varphi \| \leq 1$, can we choose $H$, $\pi$ and $\psi, \eta$ as above, satisfying the additional requirement that $\| \psi \| \leq 1$ and $\| \eta \| \leq 1$, such that (again) $\varphi(a) = \langle \pi(a) \psi, \eta \rangle$ for all $a \in A$?

Note that this is clearly true for a positive functional - simply write $\varphi(a) = \langle \pi(a)\xi, \xi \rangle$ (using the GNS-construction) and note that $$ 1 \geq \| \varphi \| = \sup_{\| a \| \leq 1} |\langle \pi(a) \xi, \xi \rangle | \geq |\langle \pi(1) \xi, \xi \rangle| = \| \xi \|^2$$

REPLY [4 votes]: Another point of view is that this result can be seen as a special case of its "completely contractive" version (in the same way that Stinespring's dilation theorem for completely positive maps generalizes the GNS representation of states). In particular, the following is true: if $\Phi:A\to B(H)$ is any completely contractive map, then there exists another Hilbert space $K$, a representation $\pi:A\to B(K)$, and operators $V_1, V_2:H\to K$ with $\|V_i\|\leq 1$ such that 
$$
\Phi(a) =V_2^*\pi(a)V_1
$$
for all $a\in A$. This result may be found e.g. in Chapter 8 of Paulsen's Completely Bounded Maps and Operator Algebras.  Essentially the idea is to use Arveson's extension theorem to embed the original $\Phi$ as the "upper right corner" of a completely positive map into the $2\times 2$ matrices over $B(H)$, and then use the Stinespring theorem.<|endoftext|>
TITLE: Cayley Graphs of Z/nZ with invertible adjacency matrices
QUESTION [8 upvotes]: Let $G = \mathbb Z/n\mathbb Z$ and let $\emptyset\neq S\subseteq G$.  Then the Cayley digraph of $G$ with respect to $S$ has vertex set $G$ and directed edges of the form $g\rightarrow gs$ with $s\in S$.  I don't assume $S$ is symmetric or that $S$ generates $G$.  Let $A_S$ be the adjacency matrix of the corresponding Cayley digraph.  Let us say $S$ is good if $A_S$ is invertible.
Define $p_S(x)=\sum_{k\in S}x^k$ where I identify $G$ with $\lbrace 0,\ldots, n-1\rbrace$ in the usual way.  Discrete Fourier analysis says $S$ is good iff $p_S$ has no root which is an $n^{th}$-root of unity, i.e., $p_S$ is relatively prime to $x^n-1$.  Alternatively, $S$ is good if and only if $\sum_{s\in S}s$ is an invertible element of the group ring $\mathbb CG= \mathbb C[x]/(x^n-1)$.
Obviously if $n$ is prime, then all proper non-empty subsets are good since the cyclotomic polynomial is $p_G$.  On the other hand, if $S$ is a proper subgroup of $G$, then it is easy to see that $S$ is not good.  Let us say that for $S,T\subseteq G$ the sum $S+T$ is unambiguous if each element of $S+T$ can uniquely be expressed as a sum of an element of $S$ with an element of $T$. For example, a coset $g+H$ is unambiguous. Clearly, if $S+T$ is unambiguous, then $p_{S+T}=p_Sp_T\bmod (x^n-1)$ and so if either $S$ or $T$ is bad, then so is $S+T$.


Question. Is there some nice characterization of bad subsets of $G$ as, say, built from proper subgroups via unambiguous sums and perhaps some other operations?

REPLY [2 votes]: I don't think there is any nice characterization for this problem.
$1+x+x^{7}+x^{13}+x^{19}+x^{20}$ isn't relatively prime to $x^{30}-1$ despite not being characterized by any sort of reasonable construction.
It comes from adding two cosets and subtracting one, like so: $(1+x^{10}+x^{20})+(x+x^7+x^{13}+x^{19}+x^{25})-(x^{10}+x^{25})$. You can't get it by just combining cosets. Choosing elements from cosets doesn't help because that would be the wrong elementary cyclotomic polynomial.
This is equivalent to the problem: When does a (multi)set of roots of unity sum to $0$? You shouldn't expect a particularly good answer. Compare it to the problem "when does a set of complex numbers of norm $1$ sum to $0$", which just asks you to find polygons of side length $1$. There are quite a lot of these.
This problem is more tractable when you can mandate that the size of the set is small. Then you can give an explicit construction for all bad sets. But as far as several graduate students who worked on the problem in the second form can tell, there is not any construction that works in general that is better than "take multiples of an elementary cyclotomic polynomial that have only $0$ and $1$ for coefficients."<|endoftext|>
TITLE: Geometric applications of Ekeland's variational principle
QUESTION [11 upvotes]: I'm looking for geometric applications of Ekeland's variational principle in order to see it at work in a context I'm familiar with. Let me recall the principle itself:
Definition. Let $(X,d)$ be a metric space, let $f$ be a real-valued function on $X$, and let $\epsilon > 0$. A point $m \in X$ is an $\epsilon$-minimizer if (1) $f(m) \leq \inf f + \epsilon$ and (2) $m$ is the unique minimizer of the perturbed function $x \mapsto f(x) + \epsilon d(m,x)$. 
Ekeland's weak principle. If $f$ is lower semi-continuous and $(X,d)$ is complete, there is an $\epsilon$-minimizer for every value of $\epsilon > 0$. 
Ekeland's strong principle. Assume $f$ is lower semi-continuous and $(X,d)$ is complete. If for a given $\epsilon > 0$ the point $y \in X$ satisfies $f(y) \leq \inf f + \epsilon$, then there is an $\epsilon$-minimizer $m \in X$ such that $f(m) \leq f(y)$ and which is at distance $\leq 1$ from the point $y$.
There are some applications in Ekeland's paper, but I'd like to see something more geometric using, for example, the Hausdorff distance on the space of convex sets. Actually, this principle caught my eye and I'm just curious as to what a geometer can do with it.

REPLY [5 votes]: Here is a quick application of the Ekeland's Variational Principle to Spectral Theory. Let $A$ be a bounded  linear symmetric operator on a Hilbert space $H$, and let $\mathbb{S}$ be the unit sphere of $H$. Then, an elementary result states:

$$\inf_{x\in\mathbb{S}}(Ax\cdot x)=\min \sigma(A)$$ 
  $$\sup_{x\in\mathbb{S}}(Ax\cdot x)=\max \sigma(A)$$

The standard proof is not complicated (it relies on the spectral radius formula, the identity $\|A^2\|=\|A\|^2$ for symmetric operators, plus some translation argument). Here is a completely different proof via the  Ekeland's principle, that also gives a nice geometrical insight.
A well-known basic fact is that  $(\lambda,x)\in \mathbb{R}\times\mathbb{S}$ is a pair eigenvalue-eigenvector for $A$ iff it is a pair critical value-critical point for the quadratic form of $A$ restricted on the unit sphere, namely the bounded and smooth function $q: \mathbb{S}\ni x\mapsto (Ax\cdot x)$. This just because $\nabla_{\mathbb{S}}q(x)=2(Ax-q(x)x)$. On the same lines, $\lambda\in\mathbb{R}$ is a spectral value iff it is a Palais-Smale level of $q$, that is, there exists a sequence $(x_j)_{j\ge0}\subset \mathbb{S}$ with $\nabla_{\mathbb{S}}q(x_j)=o(1)$ and $q(x_j)=\lambda+o(1)$ (Indeed, this is equivalent to $(A-\lambda)x_j=o(1)$ for a sequence of norm-one vectors, which exactly means that the symmetric operator $A-\lambda$ is not invertible). The proof via the EVP is now clear: take a minimizing resp. maximizing sequence for $q$ on $\mathbb{S}$. By the EVP, one can assume it is a Palais-Smale sequence, ending the proof.<|endoftext|>
TITLE: Unicity of a vector space frame's dual frame
QUESTION [6 upvotes]: The Wikipedia page on vector space frames gives a construction to find a dual frame for a given frame. Specifically, given a set of vectors $\{ e_k \}$ in a Hilbert space $\mathcal{H}$ such that for all $v\in\mathcal{H}$ and some $ A \leq B<\infty$ we have
$$A||v||^2\leq\sum_k|\langle e_k|v\rangle|^2\leq B ||v||^2,$$
the construction gives (among other things) a set of linear functionals $\phi_k:\mathcal{H}\rightarrow\mathbb{C}$ such that for all $v\in\mathcal{H}$
$$v=\sum_k e_k \cdot\phi_k(v).$$
My question is, how unique are these functionals? (or alternatively, their dual images in $\mathcal{H}$.) The construction gives a natural way to find these dual images, which depends on the inverse of the map $S(v)=\sum_k\langle v|e_k\rangle e_k$ and thus on the inner product used, but that doesn't mean the corresponding functionals will depend on the inner product.
If I have, as in one standard example, three noncollinear vectors in a two-dimensional space, then I would naively expect to have "one real degree of freedom" in choosing the coordinates. How does this show up? If there are other sets of functionals, what do their dual images in $\mathcal{H}$ look like? Is the one picked by $S$ special? if so, how? Does any of this depend on the (in)finiteness of the space dimension?

REPLY [2 votes]: You are essentially asking about the structure of the set $D_X$ of all dual frames $Y$ of a given frame $X$. As mentioned by Dustin G. Mixon, there is always one preferred dual frame, namely the canonical dual frame $X'$. Now regarding uniqueness: $D_X=(X')$ if and only if $X$ is a Riesz basis. 
If $H$ has finite dimension, then a frame $X$ is a Riesz basis if and only if $|X|=\dim H$. So you will get infinitely many duals as soon as $|X|\geq \dim H+1$, and only the canonical dual when $|X|=\dim H$.
In general, the set $D_X$ has a natural structure of affine space via the consideration of idempotents. Here is a detailed outline.
Analysis/frame/synthesis operator: for any family $(x_j , j\in J)$, we consider the so-called analysis operator $\theta_X:H\longrightarrow \mathbb{C}^J$ sending $x$ to $((x,x_j))_{j\in J}$. By definition, $X$ is a frame if $\theta_X$ is bounded above and below from $H$ to $\ell^2(J)$. That is, $\theta_X$ is bounded and the positive so-called frame operator $\theta_X^*\theta_X$ is invertible (with spectrum in $[A,B]$, the constants of the definition). Denoting by $(e_j,j\in J)$ the canonical basis of $\ell^2(J)$, note that the adjoint (called the synthesis operator) satisfies $\theta_X^*e_j=x_j$.
A useful projection: when $X$ is a frame, the projection $p_X$ onto the range of $\theta_X$ is given by $p_X=\theta_X(\theta_X^*\theta_X)^{-1}\theta_X^*$. 
Riesz basis: a frame $X$ is a Riesz basis if and only if $\theta_X^*$ is injective, if and only if $\theta_X$ is surjective, if and only if $\theta_X$ is invertible ($\theta_X^*$ yielding the equivalence between $X$ and the canonical basis of $\ell^2(J)$). That's equivalent to $p_X=1$.
Dual frame: a frame $Y$ is called a dual frame of a frame $X$ if $\theta_Y^*\theta_X=\mbox{Id}_H=1$. That is $v=\sum_j (x_j,v)y_j$ for every $v\in H$. That's equivalent to $\theta_X^*\theta_Y=\mbox{Id}_H=1$. That is $v=\sum_j (y_j,v)x_j$ for every $v\in H$. 
Canonical dual fame: given a frame $X$, the canonical dual $X'$ of $X$ is given by the formula $\theta_{X'}^*=(\theta_X^*\theta_X)^{-1}\theta_X^*$. It is indeed trivial to check that $\theta_{X'}^*\theta_X=1$.
Idempotent characterization of duality: it is easy to see that, given two frames $X,Y$, they are alternate duals of each other if and only if $\theta_X\theta_Y^*$ is idempotent in $B(\ell^2(J))$.
Affine structure: given a frame $X$, the mapping $Y\longmapsto \theta_X\theta_Y^*$ is a bijection from the set $D_X$ of all duals of $X$ to the set of all idempotents $q\in B(\ell^2(J))$ with range equal to that of $\theta_X$. The latter is simply
$$
D_X\simeq p_X+p_X B(\ell^2(J))(1-p_X)
$$
where $p_X$ corresponds to the canonical dual $X'$.
Since $p_X=0$ is impossible ($\theta_X^*p_X=\theta_X^*\neq 0$), it follows that $D_X$ is a singleton (unique dual) if and only if $p_X=1$ ($X$ Riesz basis). Otherwise, it is of course infinite and the canonical dual is optimal in the sense that it minimizes the norm of the idempotent $\theta_X\theta_Y^*$. It is indeed the only $Y$ for which the latter is $1$.<|endoftext|>
TITLE: Connected components of the boundary of an open subset
QUESTION [8 upvotes]: Hi!
Let f be a (continuous, $C^\infty$... whatever) function from $\mathbb{R}^n$ ($n \geq 2$) to $\mathbb{R}$. Assume that each connected component of $f^{-1} (0; \infty)$ and $f^{-1} (-\infty; 0)$ is unbounded. Also assume that $f$ changes sign so that neither $f^{-1} (0; \infty)$ nor $f^{-1} (-\infty; 0)$ is empty. Is it possible to prove that there exists at least one connected component of $f^{-1}(0)$ which is unbounded? Thanks.

REPLY [4 votes]: Yes, it is possible to prove that $f^{-1}(0)$ has an unbounded component. I'll give a proof of this. I see that BS already has one proof -- maybe that can also be translated into a similar argument. In fact, the result holds in rather more generality than that stated. We can replace $X=\mathbb{R}^n$ by any locally compact Hausdorff space which is connected, locally path connected, singly connected, and has one end. Saying that $X$ has one end means that $X\setminus C$ has a single unbounded component for each compact $C$. Note that $\mathbb{R}^n$ satisfies all of these properties for $n\ge2$, but has two ends for $n=1$ when the result fails. It also possible to generalize a bit further and, instead of requiring $X$ to be locally path connected and singly connected, replace this by the condition that the first Čech cohomology group vanishes.
The condition that $X$ is locally path connected and singly connected gives the following.

Lemma 1: If $U$ is a connected open subset of $X$ and $C\subseteq U$ is a closed set such that $U\setminus C$ is disconnected, then $X\setminus C$ is disconnected.

Proof: As $U\setminus C$ is disconnected, it can be written as the union $V
\cup W$ of disjoint nonempty open sets $V,W$. Given any continuous curve $f\colon[a,b]\to X$ whose endpoints $f(a),f(b)$ are not in $C$, it is possible to define an intersection number as follows. Choose disjoint closed intervals $(s_i,t_i)$ ($i=1,2,\ldots,n$) such that $f^{-1}(C)\subseteq \bigcup_i(a_i,b_i)$ and $[a_i,b_i]\subseteq f^{-1}(U)$. Set $\epsilon_i=1$ if $f(a_i)\in V,f(b_i)\in W$. Set $\epsilon_i=-1$ if $f(a_i)\in W,f(b_i)\in V$, and $\epsilon_i=0$ otherwise. Then define the intersection number as $I(f)=\sum_i\epsilon_i$. This counts the number of times that $f$ passes through $C$, with a positive count for each time it goes from $V$ to $W$ and a negative count when it goes in the opposite direction. It can be seen that if $f$ remains within $U$ then, independently of the choice of $a_i,b_i$, we have $I(f)=1$ if $f(a)\in V,f(b)\in W$ and $I(f)=-1$ if $f(a)\in W,f(b)\in V$. Otherwise, $I(f)=0$. Similarly, if $C$ is not in the image of $f$ then $I(f)=0$. By breaking $f$ down into intervals on which its image is alternately contained in $U$ and disjoint from $C$, we see that $I(f)$ is independent of the choice of $a_i,b_i$.
Next, if $(s,t)\mapsto f_s(t)$ is a continuous map from $[0,1]$ to $X$ such that $f_s(0)$ and $f_s(1)$ are not in $C$, then it can be seen that $I(f_s)$ is independent of $s$. In fact, for a given $s_0$, choosing $a_i,b_i$ as above in the definition of $I(f_{s_0})$, it can be seen that the same choice applies for calculating $I(f_s)$ for any $s$ in a small interval about $s_0$, showing that $I(f_s)$ is locally constant and, hence constant. Therefore $I(f)=I(g)$ for curves $f,g$ which are homotopic relative to their endpoints.
So, for points $P,Q\in X\setminus C$, we can define $I(P,Q)$ to be the intersection number, $I(f)$, for any curve $f$ joining $P$ to $Q$. As the space $X$ is singly connected, this is well-defined and independent of $f$.
Choose a point $P\in V$. Let $V^\prime$ be the set of points $Q\in X\setminus C$ with $I(P,Q)=0$ and $W^\prime$ be those points with $I(P,Q)\not=0$. Then, $V^\prime,W^\prime$ are disjoint open sets whose union is $X\setminus C$. If these are nonempty then it implies that $X\setminus C$ is disconnected. However, $P\in V^\prime$. Also, given any $Q\in W$ the local path connectedness implies that there is a curve joining $P$ to $Q$ in $U$, so $I(P,Q)=1$ and $Q\in W^\prime$. QED
Together with the properties that $X$ is locally compact Hausdorff and connected with one end, we can prove the result. Set $U=f^{-1}((-\infty,0))$ and $V=f^{-1}((0,\infty))$, which are disjoint open sets whose connected components are unbounded (i.e., not relatively compact). Also, set $F=X\setminus(U\cup V)$. We can show that compact connected components of $F$ only occur as isolated islands inside $U$ or $V$.
Suppose that $C$ is a compact connected component of $F$ and that $P\in C$. Then, by local compactness, there is a relatively compact open set $S$ containing $C$. From the definition of connected components, for each $Q\in\partial S$, there will be a closed subset of $F$ containing $P$ but not $Q$ and which is open in the subspace topology on $F$. Then, by compactness of $\partial S$, there is a closed subset of $F$ containing $P$ and disjoint from $\partial S$, and open in the subspace topology. Call this $K_0$. Intersecting with $S$ if necessary, we can suppose that $K_0\subseteq S$, so $K_0$ is compact. Then, by local compactness again, there is a relatively compact open set $W$ containing $K_0$ and such that $\bar W$ is disjoint from $F\setminus K_0$. In particular, $W$ is a relatively compact open set containing $P$ such that $\partial W$ is disjoint from $F$. Replacing $W$ by its connected component containing $P$, we can assume that it is connected. As $C$ is connected and intersects with $W$ but not $\partial W$, it must be contained in $W$. Suppose that $W\setminus K$ was not connected. By the lemma, $X\setminus K$ would be disconnected. Then, as $X$ has one end, $X\setminus K$ would contain at least one relatively compact connected component, say $T$. As the boundary of $T$ is contained in $W$, $T\cap W$ is nonempty. Then, $T\setminus F$ is a nonempty relatively compact connected component of $X\setminus F$, contradicting the assumptions. So, by contradiction, $W\setminus K=W\setminus F=(W\cap U)\cup(W\cap V)$ must be connected, so $W$ is disjoint from either $U$ or $V$. This implies that $\partial W$ is contained in either $U$ or $V$.
So, we have shown that every compact connected component of $F$ is contained in a relatively compact open set whose boundary is contained in either $U$ or $V$. Not all connected components of $F$ con be of this form. To show this, let $U^\prime$ be the union of all relatively compact open sets with boundary contained in $U$ and $V^\prime$ be the union of all relatively compact open sets with boundary contained in $V$. By local compactness, we have $U\subseteq U^\prime$ and $V\subseteq V^\prime$. Then, $U^\prime,V^\prime$ are disjoint open sets. Suppose not. Then, there would exist relatively compact open sets $U_0,V_0$ with boundary contained in $U,V$ respectively, and with nonempty intersection. By connectedness of $X$, there must be a point $P$ in the boundary of $U_0\cap V_0$. This will be in the boundary of either $U_0$ or $V_0$ so, wlog, take $P\in \partial U_0$. Then, $P\in U$. Also, $P\in\bar V_0$. As $P$ is not in $V$, so not on the boundary of $V_0$, this gives $P\in U\cap V_0$. So, $U\cap V_0$ is nonempty and, as its boundary is disjoint from $U$, it is closed in $U$. This means that $U\cap V_0$ contains a connected component of $U$, contradicting the conditions that $U$ has unbounded components and $V_0$ is compact.
As $X$ is connected, $X\setminus(U^\prime\cup V^\prime)$ is nonempty. Furthermore, for any $P$ in this set, the connected component of $F$ containing $P$ is not compact, by the argument above.<|endoftext|>
TITLE: Torsion in cuspidal cohomology
QUESTION [10 upvotes]: Following Lemma 2.7 from Vogtmann's Rational Homology of Bianchi Groups, I want to define cuspidal cohomology as $$H_{\mathrm{cusp}}(M)=\frac{H_1(M)}{i_*(H_1(\partial M))}$$ where $i:\partial M\to M$ is the inclusion and I take integral coefficients. The motivation is that Abelian covering spaces such that the preimage of a cusp consists of disjoint copies of the same cusp are corresponding to exactly those holonomies $\pi_1(M)\to G$ that factor through the cuspidal cohomology.
I verified for all hyperbolic cusped 3-manifolds in the SnapPea census that if $H_1(M)=\mathbb{Z}^n\oplus T$ where $T$ is torsion, then $H_\mathrm{cusp}(M)=\mathbb{Z}^{n-\mbox{number of cusps}}\oplus T$. This is to be expected rationally by half-lives-half-dies. It seems to also hold integrally for hyperbolic 3-manifolds.
Is there an argument that this holds for all hyperbolic cusped 3-manifolds? (And maybe a counterexample if you take a non-hyperbolic manifold)

REPLY [10 votes]: Consider the long exact sequence on homology (coefficients in $\mathbb{Z}$)
$$\to H_1(\partial M)\overset{i}{\to} H_1(M)\to H_1(M,\partial M) \to H_0(\partial M) \to $$
You are looking for $H_1(M)/i_\ast(H_1(\partial M)) \cong im\{ H_1(M)\to H_1(M,\partial M)\} \cong ker \{ H_1(M,\partial M)\to H_0(\partial M)\}$ by the exactness of the sequence. 
If $H_1(M)\cong \mathbb{Z}^n\oplus  T$, $T$ torsion, then $H_2(M)\cong \mathbb{Z}^{n-1}$ (the rank $n-1$ follows from Euler characteristic). By universal coefficients, we have a short exact sequence
$$0\to Ext(H_1(M),\mathbb{Z})\to H^2(M) \to Hom(H_2(M),\mathbb{Z})\to 0.$$
One computes $Ext(H_1(M),\mathbb{Z})\cong T$, $Hom(H_2(M),\mathbb{Z})\cong \mathbb{Z}^{n-1}$ (see p. 195 of Hatcher), so that $H^2(M)\cong \mathbb{Z}^{n-1} \oplus T$. By Lefschetz duality, $H^2(M)\cong H_1(M,\partial M) \cong \mathbb{Z}^{n-1}\oplus T$. So you are looking for the torsion in 
$$ker\{ H_1(M,\partial M)\to H_0(\partial M) \} \cong ker\{ \mathbb{Z}^{n-1}\oplus T \to \mathbb{Z}^c\},$$
which is clearly isomorphic to $T$.<|endoftext|>
TITLE: determinants and polynomials in matrices
QUESTION [5 upvotes]: Muirhead (1982, "Aspects of Multivariate Statistical Theory") references on page 59
a result (from MacDuffee, 1943, chap 3, "Vectors and Matrices") a book I cannot find):
" The only polynomials in the elements of a matrix satisfying $p(I)=1$ and $p(AB)=p(A)p(B)$ for all matrices, are the integer powers of det B:  $p(B) = (\det B)^k$ for some integer $k$.
Where otherwise, can I find this result and discussions of it?

REPLY [8 votes]: Francois Ziegler's answer is not massive overkill. The proof is simple. 
Suppose you have a continuous multiplicative mapping 
$P: \operatorname{Mat}_n(\mathbb R) \to (\mathbb R, \cdot)$ as you started with, then it restricts to a continuous group homomorphism 
$P:GL(n)\to (\mathbb R\setminus\{0\}, \cdot\;)$, which is analytic (using $\exp$).
Its derivative at $\mathbb I_n$ is a Lie algebra homomorphism
$P':\mathfrak g\mathfrak l(n)\to \mathbb R$ which must vanish on each commutator.
The space of all commutators is the codimension 1 Lie subalgebra $\mathfrak s\mathfrak l(n)$.
Since $P'$ is also linear, it is of the form $P'(X) = k.\operatorname{Trace}(X)$ for some $k$. 
This integrates to $P(A) = \det(A)^k$. Here $k$ must be integral if the ground field is $\mathbb C$. In the real case any $k$ works if $\det(A)$ is always $\ge 0$, and integral generally.<|endoftext|>
TITLE: Reals added after Cohen forcing
QUESTION [10 upvotes]: Let $V_1$ be a generic extension of $V\models GCH$ obtained by adding $\aleph_{\omega}-$many Cohen reals. Then we have the following:
1- In $V_1$ there are $\aleph_{\omega+1}-$many reals,
2- In $V_1$ there are only $\aleph_{\omega}-$many Cohen reals.
What can we say about the other reals in $V_1$? Are they generic for some forcing notion over $V$?

REPLY [16 votes]: In your extension $V_1$, there are actually continuum many,
that is $\aleph_{\omega+1}$ many $V$-generic Cohen reals.
In general, whenever you add even a single Cohen real, then
the extension wills have continuum many $V$-generic Cohen
reals, because if $c$ is a $V$-generic Cohen real and $x$
is any real in the ground model, then the bit-wise sum
$c\oplus x$ will be a $V$-generic Cohen real, since this
induces an automorphism of the forcing in the ground model,
and these are all different. (In your case, as Goldstern mentions in the comments, we may view the forcing as adding all but one of the Cohen reals, and then a final Cohen real, to achieve $\aleph_{\omega+1}$ many $V$-generic Cohen reals by this reasoning.)
But it is true that not every new real of $V[G]$ is a
$V$-generic Cohen real. For example, one may easily
construct reals that obey some regular pattern, repeating
their digits in pairs, for example, which prevent them from
being Cohen reals, even if they are not in the ground
model.
Nevertheless, if $V[G]$ is a forcing extension obtained by
adding any number of Cohen reals and $z$ is any real in
$V[G]$, then I claim that $z\in V[c]$ for some $V$-generic
Cohen real in $V[G]$. This is because the countable chain
condition of the forcing means that one needs only
countably much information from $G$ in order to construct
$z$, and restricting the generic sequence of Cohen reals to
any countable domain is isomorphic again to adding a single
Cohen real.
In general, if $V\subset V[G]$ is any forcing extension and
$W$ is a model in between $V\subset W\subset V[G]$, such as
$W=V[z]$ for some real $z$, then $W$ is also a forcing
extension of $V$ by a complete subalgebra of the Boolean
algebra giving rise to $G$.
In the case of adding a Cohen real, forcing which has a
countable dense set, every subalgebra of its Boolean
algebra also has a countable dense set, and all nontrivial
forcing notions with a countable dense set are isomorphic
to adding a Cohen real. In this sense, every real added in
a Cohen real forcing extension (and this includes every
real in your model by my observations above) is generic
over $V$ for forcing that is isomorphic to the forcing to
add a Cohen real.
So the answer to your final question is that yes, these extra reals are $V$-generic for some forcing notion, and that forcing notion is isomorphic to the forcing to add a single Cohen real!
Let me conclude with an interesting tidbit:
Theorem. In the forcing extension $V[c]$ obtained by
adding a single $V$-generic Cohen real, there is a family
of continuum many pairwise mutually generic Cohen reals.
Indeed, there is a perfect set $P$ in $V[c]$, all of whose
finite subsets are mutually $V$-generic Cohen reals.
Proof. Consider the forcing to add such a perfect set $P$.
We want to force to create a tree, all of whose branches
are $V$-generic Cohen reals, and such that any finitely
many branches are mutually generic Cohen reals. Let
conditions be finite binary trees, ordered by
end-extension. It is dense for the leaves to be extended
into any given dense subset of Cohen forcing. And for
finite products of Cohen forcing with itself, it is dense
to extend the tree so that all pairs (or triples etc.) of
branches are inside any given dense set in the product
forcing. Thus, this forcing will create such a tree and
hence such a perfect set.
Finally, observe that our tree forcing has only countably
many conditions, and thus it is isomorphic to adding a
single Cohen real. So the forcing extension $V[c]$ already
has such a pefect set. QED
Of course, the branches through the perfect set will not be
$V$-generic for the forcing to add continuum many Cohen
reals, since that forcing is not a subalgebra of the
forcing to add only one. The reals in the perfect set are
only mutually generic when taken finitely many at a time,
but not fully mutually generic for infinite collections.
For example, the perfect set contains reals that are the
limits of other of its elements, and this violates mutual
genericity for those infinite families.<|endoftext|>
TITLE: Quadratic reciprocity and Weil reciprocity theorem
QUESTION [12 upvotes]: I was told that Weil reciprocity theorem (one has two meromorphic function $f,g$ on a complex curve $C$, so $\prod\limits_{x\in C} g(x)^{ord_xf}=\prod\limits_{x\in C}f(x)^{ord_xg} \ $ where $ord_xf$ is the smallest degree in Taylor expansion of $f$ at $x$, product is taken only by points in divisors of $f,g$, we assume that these divisors are not intersected with each other) was introduced by Weil after thinking about quadratic reciprocity. Could you explain me the connection between them?

REPLY [16 votes]: There are already good answers by quid and by Dustin Clausen here. I thought, though, that I'd take the time to write out something more leisurely and expository. To get from Weil Reciprocity to Quadratic Reciprocity, one must make some things more general and some things less general, and there is a choice of which order to do these things in. I will first describe the less general route and will then make some comments about what happens when you make everything as general as possible.
First, we specialize from a general curve $C$ to $\mathbb{CP}^1$. Let $f$ and $g$ be polynomials in $\mathbb{C}[x]$, with roots at the disjoint sets $\alpha_1$, ..., $\alpha_a$ and $\beta_1$, ..., $\beta_b$ and with leading terms $f(x) = f_{\infty} x^{a} + \cdots$ and $g(x) = g_{\infty} x^b + \cdots$. Then Weil reciprocity says
$$\prod_{j=1}^b f(\beta_j) = \prod_{i=1}^a g(\alpha_i) \cdot (-1)^{ab}\  f_{\infty}^{b} \ g_{\infty}^{-a}. \quad (1)$$
(The extra terms on the right come because $f$ and $g$ both have poles at $\infty$; Francois discusses this in a comment above.)
Of course, $(1)$ is easy to prove directly by writing $f(x) = f_{\infty} \prod (x-\alpha_i)$ and $g(x) = g_{\infty} \prod (x-\beta_j)$. Also, this makes it clear that the above identity holds over any algebraically closed field.
We now generalize to a non-algebraically closed field $k$. Let $f$ and $g$ be relatively prime polynomials in $k[x]$. Let $f(x) = f_{\infty} r_1(x) r_2(x) \cdots r_c(x)$ be the factorization of $f$ into monic irreducibles. Similarly, let $g(x) = g_{\infty} s_1(x) \cdots s_d(x)$. Let $K_i$ be the field $k[x]/r_i(x)$ and let $L_j = k[x]/s_j(x)$. For $u$ and $v \in k[x]$, I'll write $(u \bmod v)$ for the image of $u$ in $k[x]/v$. The generalization of $(1)$ is
$$\prod_{j=1}^d N_{L_j/k}(f \bmod s_j) = \prod_{i=1}^c N_{K_i/k}(f \bmod r_i)  \cdot (-1)^{ab}\  f_{\infty}^{b} \ g_{\infty}^{-a}. \quad (2)$$
Exercise 1: See that, when $k$ is algebraically closed, $(2)$ specializes to $(1)$. Exercise 2: Deduce (2) from (1), by grouping together terms from $(1)$ over $k^{\mathrm{alg}}$.
We now specialize to the case $k = \mathbb{F}_p$. The norm map from $\mathbb{F}_{p^n}$ to $\mathbb{F}_p$ is raising to the $(p^n-1)/(p-1)$ power. So $(2)$ becomes
$$\prod_{j=1}^d (f \bmod s_j)^{\frac{p^{\deg s_j}-1}{p-1}} = \prod_{i=1}^c (f \bmod r_i)^{\frac{p^{\deg r_i}-1}{p-1}}  \cdot (-1)^{ab}\  f_{\infty}^{b} \ g_{\infty}^{-a}. \quad (3)$$
We specialize further to the case that $f$ and $g$ are irreducible to get
$$(f \bmod g)^{\frac{p^{b}-1}{p-1}} = (g \bmod f)^{\frac{p^{a}-1}{p-1}} \cdot (-1)^{ab}\  f_{\infty}^{b} \ g_{\infty}^{-a}. \quad (4)$$
Let $p$ be odd, and raise both sides to the $(p-1)/2$ to get
$$(f \bmod g)^{(p^b-1)/2} = (g \bmod f)^{(p^a-1)/2} \cdot (-1)^{ab(p-1)/2} \ f_{\infty}^{b (p-1)/2} g_{\infty}^{-a(p-1)/2}. \quad (5)$$
Recall Euler's criterion: For $u \in \mathbb{F}_p$, we have $u^{(p-1)/2} = \left( \frac{u}{p} \right)$. It has a generalization: For $u \in \mathbb{F}_{p^n}$, the power $u^{(p^n-1)/2}$ is $\pm 1$ according to whether or not $u$ is square. So, defining the quadratic residue symbol in $\mathbb{F}_p[x]$ in the obvious way, equation $(5)$ says
$$\left( \frac{f}{g} \right) = \left( \frac{g}{f} \right) \cdot (-1)^{ab(p-1)/2} \ \left( \frac{f_{\infty}}{p} \right)^b \left( \frac{g_{\infty}}{p} \right)^{-a}. \quad (6)$$
In short, $\left( \frac{f}{g} \right)$ is equal to $\left( \frac{g}{f} \right)$ up to some elementary terms, just like in quadratic reciprocity. One can rewrite the elementary correction terms to make them look more like the terms that show up in standard QR, but I'll leave this as is.
We can get more general statements by (a) not restricting ourselves to the case that $f$ and $g$ are irreducible (b) working with curves $C$ other than $\mathbb{A}^1$ (c) raising both sides of $(4)$ to the $(p-1)/g$ power for some other $g$ dividing $p-1$. I was going to write more about this, but I think it is long enough as it is.<|endoftext|>
TITLE: Is this seemingly novel irrational constant also transcendental?
QUESTION [24 upvotes]: I recently discovered a constant that is constructed as follows: 
$\chi=\sum_{n=1}^\infty (\frac{\cos{n}}{2|\cos{n}|}+\frac{1}{2}) 2^{-n}$
Furthermore I can prove that it is an irrational number whose decimal approximation is .555609809015...
I conjecture that it is also a transcendental number but have not been able to prove this. 
Anyone else want to take a crack at it? I am not a professional mathematician so if this number is not novel, then I appologize in advance.
One of the reasons I want to know this is that if this prototype number is proven transcendental than that would imply that an infinite set of similar transcendental numbers exists on the open interval (0,1) that are given by
$\chi_f=\sum_{n=1}^\infty (\frac{\cos{f(n)}}{2|\cos{f(n)}|}+\frac{1}{2}) 2^{-n}$ 
where f(n) is any algebraic function of n.

REPLY [65 votes]: Yes. It's known to be transcendental. The sequence of coefficients of your number is a variant of a Sturmian sequence. It has very low complexity. The definition of this: let the digit sequence be $a_1,a_2,a_3\ldots$ taking values in $ \lbrace 0,1\ldots,d-1 \rbrace ^{\mathbb N}$. A subword of length $k$ is a string $a_ia_{i+1}a_{i+2}\ldots a_{i+k-1}$. The complexity, $p(k)$, is a function from $\mathbb N$ to $\mathbb N$ taking $k$ to the number of subwords of the sequence of length $k$.
In a 2007 paper in the Annals of Mathematics (vol 165, p547--565), Adamczewski and Bugeaud (On the complexity of algebraic numbers. I. Expansions in integer bases) showed that if a number is algebraic, then its digit sequence in base $b$ has complexity satisfying $p(k)/k\to\infty$.
In your case, the complexity of the sequence of base 2 digits satisfies $p(k)=2k$. How to see this? 
Define a map $f$ from $[0,1)$ to $\lbrace 0,1\rbrace$ by $f(x)=1$ if $x\in [0,1/2)$ and 0 otherwise.
The $n$th term of your sequence is $f(\alpha n\bmod 1)$, where $\alpha=1/(2\pi)$. Write $T$ for the transformation from $[0,1)$ to itself given by $T(x)=x+\alpha\bmod 1$. Then the $n$th term is just $f(T^n0)$. The sub-block of the digit sequence of length $k$ starting at the $j$th term is $f(T^j0)\ldots f(T^{j+k-1}0)$. Since the $T^j0$ are dense in $[0,1)$, we need to ask how many blocks $f(x)\ldots f(T^{k-1}x)$ are possible.
Consider taking $x=0$ and moving it around the circle (=$[0,1)$) once. As you move it, the $T^ix$ also each move around the circle one time. The sequence changes each time one of the $(T^ix)_{0\le i< k}$ crosses 0 or 1/2. This is a total of $2k$ changes. Hence the sequence takes on $2k$ values as $x$ moves around the circle, hence the estimate for the complexity.<|endoftext|>
TITLE: Does this "jumping-ahead" ordinal function exist?
QUESTION [9 upvotes]: While working on a project in operator algebras with a collaborator (and fellow MO user), we are able to successfully complete a transfinite induction assuming that the following has an affirmative answer.

Question:  Let $\delta$ be a cardinal, considered as an initial
  ordinal, so that it is equal to the
  set of all ordinals of cardinality
  strictly less than $\delta$.  Does
  there exist a nondecreasing function
  $\phi \colon \delta \to \delta$ such
  that, for all ordinals $1 \leq \lambda < \delta$,
  there exists $\gamma < \lambda$ such
  that $\phi(\gamma) \geq \lambda$?

If it happens to matter, we are only concerned with the case where $\delta$ is a limit cardinal.  Any reasonably (transfinitely) constructive approach to writing down such $\phi$ seems to quickly run into issues of ordinal notation that are beyond our expertise.  If an abstract existence argument is available, we will certainly still be happy.  On the other hand, we fear that this may somehow depend on large cardinal issues.

REPLY [15 votes]: Not unless $\delta$ has countable cofinality (e.g., $\delta = \aleph_\omega$). This will fail for $\delta = \aleph_1$, for example. Let $\phi: \delta \to \delta$ be any increasing function and recursively define $\lambda_0 = 0$ and $\lambda_{n+1} = \phi(\lambda_n)+1$. Since $\phi$ is increasing, the sequence $(\lambda_n)$ is increasing, and since $\delta$ has uncountable cofinality we have $\lambda = \sup \lambda_n < \delta$. However, for any $\gamma < \lambda$ we must have $\gamma < \lambda_n$ for some $n$,
so that $\phi(\gamma) \leq \lambda_{n+1} < \lambda$.
(If $\delta$ has countable cofinality it's easy. For instance, if $\delta = \aleph_\omega$ then we define $\phi$ by letting $\phi(\lambda) = \aleph_{n+1} + \lambda$ for $\aleph_n \leq \lambda < \aleph_{n+1}$.)<|endoftext|>
TITLE: Crepant resolutions of ODP's on a 3-fold 
QUESTION [5 upvotes]: It seems to be a well-known fact that if we have a 3-fold $X$ with only ODP singularities (ordinary double point) and a smooth Weil divisor $E$ passing through them, then by blowing-up $X$ along $E$ we get a small crepant resolution of $X$.  I was wondering if anyone knew of a quick proof of this and/or a reference where it is proved.  Thanks

REPLY [3 votes]: The more general phenomenon behind this is that $E$ is not a Cartier divisor at the ODPs. (A Cartier divisor could not be smooth, because that would imply that $X$ is smooth!)
The pre-image of this divisor will be Cartier and hence the blow-up is not an isomorphism (as it is when one blows up a Cartier divisor). On the other hand the divisor is locally defined by $2$ equations near the ODPs, so the fibers over these points can be at most dimension $1$. Hence they are exactly dimension $1$. In other words, the blow up is a small morphism. 
Then you need to see that the blow up is indeed smooth. You may do this by verifying that the pre-image of $E$ is smooth which  should not be too hard as its pre-image ought to be just the blow up of E at the ODP points (check this by an explicit local computation). These points are smooth on $E$, so the blow-up of $E$ remains smooth. Then its a smooth Cartier divisor, so the ambient space has to be smooth along it, but it is already smooth everywhere else.<|endoftext|>
TITLE: Can continuity of a function be checked by restricting to smooth curves?
QUESTION [22 upvotes]: Well-known example: Consider the function 
$$f(x,y)=\left\{\begin{array}{c}
\frac{x^2y}{x^4+y^2} & \text{if }(x,y)\neq(0,0)\\
0 & \text{if }(x,y)=(0,0)
\end{array}\right.$$
When restricted to any straight line through the origin, this function is continuous. However, if we approach the origin along the parabola $y=x^2$, we get a limit of $\frac 12$, so $f$ is actually discontinuous. The question is whether smooth curves can always ferret out discontinuity in this way.

Does there exist a function $f:\mathbb R^2\to \mathbb R$ which is discontinuous at a point $x$, but is continuous at $x$ when restricted to any smooth curve?

Discontinuity is witnessed by a sequence $\{x_i\}$ converging to $x$ so that $\{f(x_i)\}$ does not converge to $f(x)$. Since we could take $f$ to be the characteristic function on $\{x_i\}$, the above question is equivalent to the following.

Given a convergent sequence $\{x_i\}$ in $\mathbb R^2$, must there be a smooth curve passing through infinitely many of the $x_i$?

REPLY [13 votes]: Yes you can do this. Suppose you have a sequence of points $c_n$ which converges very fast to zero, so that $n^k c_n$ is bounded for all $k$. Let $h$ be a smooth function that is equal to $0$ for $x\le -1$  and $1$ for $x\geq 0$. Let $t_n=\frac{1}{(n+1)^2}+2\sum_{k=1}^n \frac{1}{k^2}$ and
$$g(x)=\sum_{n\geq 0}h(1+n^2(x-t_n))h(1-n^2(x-t_n))c_n$$
then $g(t_n)=c_n$ and g is smooth because the fast convergence of the $c_n$ implies that the derivatives of all orders of the terms in the summation are uniformly bounded. This proof works in greater generality and the result is called "the general curve lemma". See 12.2 in Kriegl-Michor, "The convenient setting of global analysis".

REPLY [12 votes]: Usually, by a "smooth curve through $0\in\mathbb R^2$", one means a $C^k$ mapping $f:[-a,a]\to\mathbb R^2$ where $k\ge 1$ is the desired degree of smoothness, $a>0$, $f(0)=0$, and $f'(x)\ne 0$ for all $x\in[-a,a]$ (without the last condition, you can get quite ugly images even of $C^\infty$ mappings). Equivalently, a smooth curve is a graph of a $C^k$-function near the origin after some rotation of coordinates. Now, the answer depends on $k$. If $k=1$, then there is a subsequence $x_n$ approaching $0$ from some limiting direction. Choosing this direction as the positive semiaxis, we see that we can rarefy the sequence a bit and put everything on a $C^1$ graph with $0$ derivative at the origin. On the other hand, to request any particular modulus of continuity for $f'$ is already impossible because that would require some estimate of the kind $|\mbox{arg}(x)|\le \omega(|x|)$ with some fixed function $\omega$ tending to $0$ at $0$ where $\mbox{arg}$ is measured from the limiting direction. However, we can construct a sequence of points whose absolute values tend to $0$ very fast while their arguments tend to $0$ very slowly.

REPLY [2 votes]: Just for the sake of exposition, take ${\Bbb C}$ as the model of ${\Bbb R}^2$.  Given a sequence $(z_n)$ of non-zero complex numbers which converges to 0, the compactness of $S^1$ entails the existence of a subsequence such that Arg$(z_n)$ converges. Then, passing if necessary to an even thinner subsequence $(w_n)$, one can also get the convergence od the arguments monotonic and one-sided (in some appropriate sense). Connecting the dots, even just with line segments, gives a curve smooth at 0, and a little more care will make it smooth everywhere.  (I assume "smooth" here means $C^1$...I don't think $C^n$ is much harder, but I haven't thought about $C^\infty$.)<|endoftext|>
TITLE: Maximum size of $k$-wise linearly independent set within $\lbrace 1, 2, 3, ..., u \rbrace^k$
QUESTION [5 upvotes]: Given a positive integer $u$, how many $k$-dimensional vectors whose coordinates are all in $\lbrace 1, 2, 3, ..., u\rbrace$ can you choose so that any $k$ of them are linearly independent? Equivalently, what is the size of the largest subset of $\lbrace 1, 2, 3, ... u \rbrace^k$ so that each hyperplane through the origin contains at most $k-1$ of them?
If $k=2$, two vectors are linearly dependent iff they have the same slope, so the maximum number of pairwise independent vectors is the number of distinct slopes $y/x$ with $1\le x,y \le u$, 
$$ -1 + 2\sum_{n=1}^u \phi(n),$$
since the number of slopes up to $1$ with reduced denominator $n$ is $\phi(n)$, and slopes other than $1$ come in reciprocal pairs.

REPLY [4 votes]: Tony already mentioned that the maximum size of a set of vectors that are $k$-wise linearly independent over a finite field $\mathbb F_q$ grows linearly with $q$. 
In our situation, however, this is no longer true, and the right order of asymptotics is $O(u^{k/(k-1)})$. That is, if you keep $k$ fixed, the maximum number of $k$-wise linearly independent vectors from $\lbrace 1,2,\dots, u\rbrace ^k$ is $\sim u^{k/(k-1)}$. One has the same order of magnitude for the minimal number of linear subspaces needed to cover the points $\lbrace 1,2,\dots, u\rbrace ^k$. These statements are proved in 

I. Barany, G. Harcos, J. Pach and G. Tardos, "Covering Lattice Points by Subspaces",
  Per. Math. Hung. 43, 2001, 93-103. 

Here is the arxiv link. I think that the right order of magnitude for the maximum number of such $k$-dimensional vectors so that any $r$ are linearly independent is not known for $r < k$. See this artcle for references on such generalizations.<|endoftext|>
TITLE: Categories internal to schemes and subschemes of invertible arrows
QUESTION [7 upvotes]: Let  $X = (X_1 \rightrightarrows X_0)$ be a category in schemes, such that the source and target maps are both smooth. By an argument using finite limits we can construct a subscheme $X_1^{iso} \subset X_1$ such that  $X^{iso} = (X_1^{iso} \rightrightarrows X_0)$ is a groupoid in schemes, and this is universal in the sense that any internal functor $Y\to X$ from a groupoid $Y$ factors (strictly!) through $X^{iso}$.
My question is, can we show that the source and target maps of $X^{iso}$, which are composites $X_1^{iso} \hookrightarrow X_1 \to X_0$, are also smooth? If the inclusion of the scheme of invertible arrows was an open immersion I think we would be done.
I would also be happy if the proof only went through for algebraic spaces instead of schemes (this might be easier, who knows!).
This is a vast generalisation of the question I asked at M.SE a while back, which dealt with the case that the category $X$ was a monoid. User 'Matt E' showed it was true for the case that the monoid was smooth and of finite type over a field.

REPLY [5 votes]: The answer to your question is "yes".  We need the following properties of $X_1^{iso}$:

The map $X_1^{iso} \to X_1$ given by $(f,g) \mapsto f$ on scheme-valued points is a monomorphism.  This is straightforward from uniqueness of inverses.
$s,t: X_1 \to X_0$ are formally smooth.  We use this for existence of a lift of an infinitesimal deformation.
$s,t: X_1 \to X_0$ are locally of finite presentation.  Using basic permanence properties (see e.g., Stacks 02KL with $X=S=X_1$ and $Y = X_0$), we find that $i: X_0 \to X_1$ is also locally of finite presentation.  Combining with the fiber product construction, the two maps from $X_1^{iso}$ to $X_0$ are also lfp.
Infinitesimal formal categories are formal groupoids.  This is basically the statement that tangent spaces have abelian group structure.

We will show that the monomorphism $X_1^{iso} \to X_1$ is formally étale by rephrasing Matt E's argument in a bit more generality.  
Let $j: U \to T$ be a closed immersion of affine schemes over $X_0$ defined by a square zero ideal $I$.  Let $\bar{f}$ be a $U$-point of $X_1^{iso}$, and let $f$ be a $T$-point of $X_1$, such that $f \circ j$ is equal to $\bar{f}$ followed by the monomorphism.  By assumption, $\bar{f}$ has an inverse $\bar{g}$, and by formal smoothness of both $s,t: X_1 \to X_0$, there exists some $T$-point $\tilde{g}$ of $X_1$ that restricts to the $U$-point $\bar{g}$.
We now translate by $\tilde{g}$ to bring us to an infinitesimal neighborhood of the origin.  $\bar{g}$ and $\bar{f}$ are inverses as $U$-points, so $\tilde{g} f$ and $f \tilde{g}$ are $T$-points whose restriction to $U$ factors through identity.  The square zero condition on $I$ gives an additive bijection between lifts of the identity and elements of $\operatorname{Hom}_{\mathcal{O}_U}(e_U^*\Omega_{X_1/X_0},I)$, where $e_U: U \to X_1$ is the identity section (SGAI Exp. 3 Proposition 5.1).  Because the Hom space is an abelian group, we can easily form inverses by negating.  Translating back yields $(\tilde{g} f)^{-1} \tilde{g} = \tilde{g} (f \tilde{g})^{-1}$ as the unique inverse of $f$.
To summarize, if $f$ is a $T$-point in $X_1$ whose restriction to $U$ factors through $X_1^{iso}$, then there is a unique $T$-point $(f,f^{-1})$ in $X_1^{iso}$ that maps to $f$.  Thus, $X_1^{iso} \to X_1$ is formally étale, and the composition with either source or target $X_1 \to X_0$ is formally smooth and locally of finite presentation, hence smooth.
This argument works for algebraic spaces without change (although you may need to find new references).<|endoftext|>
TITLE: Spectrum of $L^\infty(X,\mu)$
QUESTION [10 upvotes]: Suppose that $(X,\Sigma,\mu)$ is a measured set with respect to $\sigma$-algebra $\Sigma$.
Suppose that $L^\infty(X,\mu)$ is the set of all $\mu$-equal bounded $\Sigma$-measurable functions on $X$. Indeed equally, one may say that $L^\infty(X,\mu)$ is the dual of $L^1(X,\mu)$. 
What is the spectrum of $(L^\infty(X,\mu),\|\cdot\|_\infty)$ as a Banach (C^*) algebra?
Thank you very much.

REPLY [11 votes]: I'm recording here some references on the object $\tilde X = \mathrm{Spec}(L^\infty(X,\Sigma,\mu))$ (many of which were communicated to me recently by Balint Farkas), assuming for sake of simplicity that $\mu$ is a probability measure to avoid some technicalities.  As the references below show, this object has been discovered and used multiple times in the literature, but lacks a standardised name.
The earliest reference I know of that uses this space is
Halmos, Paul R., On a theorem of Dieudonne, Proc. Natl. Acad. Sci. USA 35, 38-42 (1949). ZBL0031.40701.
who calls it the "Kakutani space" of $X$.  Another early reference (referring to $\tilde X$ as a "perfect measure space") is
Segal, I. E., Equivalences of measure spaces, Am. J. Math. 73, 275-313 (1951). ZBL0042.35502.
and the space is referred to as the "Gelfand space" of $X$ and used to relate ergodic theory with topological dynamics in
Ellis, Robert, Topological dynamics and ergodic theory, Ergodic Theory Dyn. Syst. 7, 25-47 (1987). ZBL0592.28015.
An equivalent construction of the space (based on the Loomis-Sikorski theorem) also appears in
Doob, Joseph L., A ratio operator limit theorem, Z. Wahrscheinlichkeitstheor. Verw. Geb. 1, 288-294 (1963). ZBL0122.36302.
and implicitly in
Fremlin, D. H., Measure theory. Vol. 3. Measure algebras, Colchester: Torres Fremlin (ISBN 0-9538129-3-6/pbk). 693 p., 13 p. (2004). ZBL1165.28002.
The space is also discussed in Chapter 12 of the recent text
Eisner, Tanja; Farkas, Bálint; Haase, Markus; Nagel, Rainer, Operator theoretic aspects of ergodic theory, Graduate Texts in Mathematics 272. Cham: Springer (ISBN 978-3-319-16897-5/hbk; 978-3-319-16898-2/ebook). xviii, 628 p. (2015). ZBL1353.37002.
(who call it the "Stone model" of $X$) and in a recent preprint of Jamneshan and myself we refer to it as the "canonical model" and rely on it to perform various product-type constructions on abstract measure-preserving systems.  As we note in that paper, it behaves in many ways like the Stone-Cech compactification of locally compact Hausdorff spaces.
Some key properties of this space (discussed for instance in my paper with Jamneshan):

$\tilde X$ is a Stonean space (compact Hausdorff and extremally disconnected).  In particular, by a result of Gleason, it is projective in the category of compact Hausdorff spaces: given any surjective continuous map $\pi: Y \to Z$ on compact Hausdorff spaces, any continuous map from $\tilde X$ to $Z$ can be continuously lifted to $Y$.  One can also identify $\tilde X$ with the Stone space of the probability algebra of $X$ (the measurable sets modulo the null sets). In particular, by the Loomis-Sikorski theorem, the probability algebra of $X$ is isomorphic to the Baire sets of $\tilde X$ modulo the meager sets.

The map $X \mapsto \tilde X$ can be viewed as a covariant functor from the category of probability spaces (or an abstraction of this category which we call the category of opposite probability algebras) to the category of compact Hausdorff probability spaces.  In fact it is left-adjoint to the forgetful functor from compact Hausdorff probability spaces to opposite probability algebras (much as the Stone-Cech compactification is left-adjoint to the forgetful functor from compact Hausdorff spaces to locally compact Hausdorff spaces).  Related to this, $\tilde X$ is "universal" amongst all compact Hausdorff spaces whose probability algebra is isomorphic to that of $X$, in the same way that the Stone-Cech compactification is universal amongst all compactifications of a locally compact Hausdorff space.

There is a natural lift $\tilde \mu$ of $\mu$ to $\tilde X$, using the Baire $\sigma$-algebra on $\tilde X$.  A Baire set in $\tilde X$ is $\tilde \mu$-null if and only if it is meager.  If $X$ is itself a compact Hausdorff space (with the Baire $\sigma$-algebra), then $\mu$ is Radon and there is a continuous probability-preserving map from $\tilde X$ to $X$, whose image is the support of $\mu$.

One has an isomorphism $C(\tilde X) = L^\infty(\tilde X)$.  Thus $\tilde X$ enjoys a strong version of Lusin's theorem that we refer to as the "strong Lusin property": every bounded Baire-measurable function on $\tilde X$ is equal outside of a null set (or equivalently, a meager set) to a unique continuous function. Similarly, any Baire-measurable map from $X$ to a compact Hausdorff space $K$ can be uniquely identified with a continuous map from $\tilde X$ to $K$.<|endoftext|>
TITLE: High-Dimensional Analogs of Polygon Spaces
QUESTION [28 upvotes]: [Edit: I had a mistake in the numerology (took d=6,5 instead of d=5,4). Edit: I mistakenly identified my mistake, it is 6,5 but I got the indices shifted by one.]
Background: Polygon spaces
Given a polygon P in space we can consider the space of all its embeddings when the lengths of the edges are fixed and mod out by rigid motion of space. Such spaces are called polygon spaces: see, for example, the paper Polygon spaces and Grassmannians by Jean-Claude Hausmann and Allen Knutson. (See also the MO question Space of simple polygons on $n$-vertices as a set of points in $\mathbb{R}^{2n}$.)
The dimension of these spaces can be recovered by a simple numerology: The number of degrees of freedom is $3n$; from this you have to subtract $n$ constraints given by the length of edges, and the dimension 6 of rigid motions of space, and you are left with $2n-6$.
If you consider embeddings into the plane rather then into space then the dimension of your space is $2n-n-3 = n-3$. The idea that such spaces of polygons have symplectic structure is attributed in the Hausmann and Knutson's paper to Cappell. [AK adds: It seems to have first been observed by Klyachko.]
The fact that $n-3$ is half of $2n-6$ is no coincidence and indeed the space of planar embeddings correspond to Lagrangian submanifolds of the polygon spaces.
In between the simple numerology and the complicated spaces we can identify some intermediate objects: the tangent vector spaces. Those are the vector spaces of infinitesimal motions of an embedded polygon in $R^3$ or $R^2$. After modding out with infinitesimal motions arising from rigid motions of the whole space (or plane) we are left with vector spaces of dimensions $2n-6$ and $n-3$ respectively.
Summary: Talking about polygon spaces we can identify three levels:
Numerology: $2n-6$ degrees of freedom to embeddings of polygons into $R^3$, $n-3$ degrees of freedom to embeddings of polygons into $R^2$.
Linear algebra: The vector spaces of infinitesimal motions of polygons embedded in the plane and space.
Symplectic manifolds: The polygon spaces and their Lagrangian submanifolds.
The polygon spaces have various structures on them and are quite exciting.
The idea:
Polygons are triangulations of $S^1$. Extend the notion of a "polygon space" to certain embeddings of triangulated $(2k+1)$-dimensional spheres. I will mainly talk about the "next case" which are triangulations of $S^3$.

Problem: Describe an appropriate analog of "polygon spaces" for triangulations of 3-dimensional spheres.

Details: From numerology to spaces

Numerology --->>> Linear algebra (vector spaces) --->>> Varieties/spaces.

1) The Numerology:
The numerology refers to the dimension of our hypothetical analogs for polygon spaces.
Given a sequence of non-negative integers $(f_{-1}=1) ,f_0, f_1, f_2 \dots$ and
an integer $d \ge 0$ we define a new sequence $g_0[d]$, $g_1[d]$ , $g_2[d], \dots$ as follows:
$$g_0[d] = 1,$$
$$g_1[d] = f_0 - d,$$
$$g_2[d] = f_1 - (d-1) f_0 + {{d} \choose {2}},$$
$$g_3[d] = f_2 - (d-2) f_1 + {{d-1} \choose {2}} f_0 - {{d} \choose {3}},$$
etc.  (But below we worry only about g_2 and g_3)
The sequence $f_0, f_1,\dots $ usually comes as the $f$-vector of some simplicial complex and the new sequences $g_i[d]$
are important in studying the combinatorics of these complexes.
polygons
Now, if we have a polygon with n vertices we have:
$$f_0 = f_1 = n ,$$
$$g_2[3] = -(n-3)$$ and $$g_2[4] = -(2n-6),$$
which are the dimensions of the polygon spaces for embeddings in $R^2$ and $R^3$ respectively.
The next case: triangulations of $S^3$.
The next analogous case is to start with a triangulation $K$ of $S^3$ with $f_0$ vertices $f_1$ edges $f_2$ triangles and $f_3$ 3-simplices.
Here (using Euler's theorem) it is easy to verify that for every $K$ we have $$g_3[6] = 2g_3[5].$$
(Both quantities are negative.)
Thus, the suggestion is that $-g_3[6]$ should be the dimension of a "generalized polygon space" for some sort of embedding of the triangulation $K$ (possibly into $ R^5$), and $-g_3[5]$ the dimension for "embedding" of $K$ (possibly into $R^4$). Hopefully, the latter will correspond to a Lagrangian submanifold of the former.
(Remark: in the comments Misha proposed making the embedding into certain CAT (1) spaces.)
Let me just verify the identity I described: $g_3[6] = f_2-4f_1+10f_0-20$ and $g_3[5]=f_2-3f_1+6f_0-10.$ For triangulated 3-spheres, Euler's theorem asserts that $f_0-f_1+f_2-f_3=0$ and also $f_2=2f_3$ and therefore $f_2=2f_1-2f_0$. It follows that $$g_3[6]=-2f_1+8f_0-20,$$ while $$g_3[5]=-f_1+4f_0-10.$$
2) The linear algebra
The linear algebra refers to the tangent vector spaces of our hypothetical spaces.
polygons:
For a graph $G$ consider the quantity
$X$ = (the number of edges - $m$ the number of vertices + ${{m+1} \choose{2}})$.
$X$ ($=g_2$) is a lower bound (interesting of course only when $X$ is non-negative) on the space of infinitesimal stresses when the vertices of the graph are embedded in $R^m$. $-X$ is a lower bound on the number of infinitesimal flexes for such embeddings (this is interesting when $X$ is negative).
For polygons embedded in $R^2$ and $R^3$. $X$ is non-positive and $-X$ is the dimension of the infinitesimal flexes (not just a lower bound). Those are the tangents of the polygon spaces.
Triangulations of $S^3$.
For a simplicial 2-dimensional complex $K$ we can let
$$Y=f_2 - (d-2) f_1 + {{d-1} \choose {2}}f_0 - {{d} \choose {3}}.$$
(We called it $g_3[d]$ before.) We care about $d=6$ so $$Y=f_2-4f_1+10f_0-20.$$
There are some known notions of high dimensional infinitesimal flexes and stresses which are bounded by $Y$ (or $-Y$, respectively).
There are conjectures for simplicial spheres, which are proven for simplicial polytopes, that these bounds are tight.
These spaces are perhaps tangent to some hypothetical polygon-like spaces (but only at very special points).
3) How the polygon-like space may look?
This particular suggestion probably does not work, but it allows to describe what I am up to.
We embed the vertices of $k$, the triangulated $S^3$ in some fixed location. We need 4 degrees of freedom for every edge and one constraint for every triangle.
For example, we can let the edges be arcs between the corresponding vertices which are parabolas; the "triangles" will be quadratic (describing a surfaces of minimum area) which extend the arcs corresponding to edges.
The condition is that the areas of the "triangles" are prescribed.

REPLY [15 votes]: Before describing my construction, I will explain why it is natural: In Gil's setup, we are supposed to fix a certain set of vertices (points in some space $X$) and vary edges connecting them. If edges are geodesics, we then need $X$ where geodesics with fixed end-points are far from being unique. Round $n$-spheres $S^n$ provide natural examples, where antipodal points are connected by $n-1$ dimensional families of geodesics. However, in the sphere, every point is antipodal to a unique point, so $S^n$ is not good enough. Spherical buildings are a natural generalizations of round spheres where antipodes are not unique. 
Here is a general recipe for getting spaces of $d$-dimensional space of geodesics connecting "antipodal" points. 
Let $G$ be a real Lie group which is the set of real points of an algebraic semi-simple group over reals (I assume that $G$ has no compact factors). I will assume that  $r$, the (real) rank of $G$, is $>1$. Let $R$ be the root system of $G$ and $R_+$ be the set of positive roots with respect to a choice of Borel subgroup $B\subset G$. Let $W$ be the Weyl group of $G$. 
One then associates with $G$ a spherical (Tits) building $X$. One can think of $X$ combinatorially as a cell complex encoding inclusions between parabolic subgroups of $G$. I will think of $X$ as a metrized cell complex, where each cell is isometric to a cell in the Coxeter complex $(S,W)$, here $S$ is the sphere of dimension $r-1$. Given such metric, one can talk about geodesic segments in $X$, which are (globally) length-minimizing paths. 
The group $G$ acts on $X$ by isometries, preserving the combinatorial structure of $X$. Then every  geodesic in $X$ has length $\le \pi$ and is contained in an apartment in $X$, which is an isometrically embedded copy of $S$ in $X$. Chambers in $X$ are the facets, they are stabilized by the Borel subgroups of $G$.  
More generally, stabilizers $G_\sigma$ of cells $\sigma$ in $X$ are parabolic subgroups in $G$ (by the definition of $X$). I will use the notation $G_x$ for the $G$-stabilizer of a point $x$ in $X$. Then $G_x=G_\sigma$, where $\sigma\subset X$ is the smallest cel containing $x$. Two  oriented geodesics are called congruent if there exists an orientation-preserving isometry between them induced by an element of $G$. Then, given an oriented geodesic $\gamma\subset X$, the space of $\Gamma_\gamma$ of oriented geodesics in $X$ congruent to $\gamma$ identifies naturally with $G/G_\Gamma$, where $G_\gamma$ is the stabilizer of $\gamma$ in $G$. It is easy to see that $G_\gamma$ is again an algebraic subgroup of $G$, so $\Gamma_\gamma$ is a homogeneous manifold. This explains why sets of congruent geodesics are useful for defining "edges'' between points in $X$. 
Two points in $X$ are called antipodal if they are within distance $\pi$. The key property of $X$ is that it has abundance of antipodal points and geodesics between such points are far from being unique. First of all, if $x, y$ are antipodal and belong to an apartment $S\subset X$ then we have $r-2$-dimensional space of  geodesics between $x$ and $y$ contained in $S$. However, most of these will not be congruent to each other. More interesting construction of geodesics connecting $x$ to $y$ is as follows: Let $\xi$ be a germ of a geodesic in $X$ emanating from $x$. Then $\xi$ can be extended (uniquely) to a geodesic $\gamma_\xi$ connecting $x$ to $y$. One obtains congruent geodesics $\gamma_\xi$ by using germs $\xi$  which belong to a single $G_x$-orbit, where $G_x=G_\sigma$ is a parabolic subgroup of $G$ ($\sigma$ is the minimal cell in $X$ containing $x$). 
Thus, the space of congruent geodesics connecting $x$ to $y$ (with germs at $x$ of type $\xi$) is naturally  identified with the quotient 
$$
G_x/G_\xi=G_\sigma/G_\tau,
$$
where $\tau$ is the smallest cell in $X$ containing $\xi$. 
Our goal is then to find buildings $X$ and pairs of cells $(\sigma, \tau)$, with $\sigma\subset \tau$, so that:

$dim(G_\sigma/G_\tau)=4$. 
The Levi subgroup of $G_\sigma$ admits an epimorphism to a group locally isomorphic to $SL(2, {\mathbb R})$ so that $G_\tau$ is contained in the kernel. 

I will do so when $G$ is split, since computations are easier in this case. Then each parabolic subgroup $P$ of $G$ corresponds (up to conjugation) to a root subsystem $R'$ in $R$, where ${R}'$ is obtained by omitting some nodes in the Dynkin diagram of $R$. 
Then $P$ has dimension 
$$|R_+'| + |R_+| + r$$ 
Note that $|R_+| + r$ is the dimension of the Borel subgroup $B$ of $G$, the smallest possible dimension for a parabolic. In particular,
$$
d(R')=dim(P)-dim(B)= |{R'}_+|.   
$$
Thus, if we take, say, $G_\tau=B$, then I can get any value for 
$$
d(R')=dim(P)-dim(B)=dim(G_\sigma/G_\tau)
$$
(not exceeding, say, $r/2$) by taking $R'$ to be a direct sum of rank 1 root subsystems in $R$ (so that Dynkin diagram for $R'$ contains no edges). This will satisfy condition 1, of course. 
Here is a bit more interesting example. Let $G=SL(5, {\mathbb R})$, $r=4$. Take $R'$  obtained by removing the one of the middle nodes from Dynkin diagram of $R$ so that $R'\cong A_1\oplus A_2$ and the Levi factor of $P$ is locally isomorphic to $SL(2, {\mathbb R})\times SL(3, {\mathbb R})$. Then $d(R')=1+3=4$. In particular, $P$ is a maximal parabolic and, hence, $\sigma=x$ is a vertex of $X$. Alternatively, one can use $G=SL(3, {\mathbb R})\times SL(3, {\mathbb R})$, where it does not matter which node you remove, you still get $R'\cong A_1\oplus A_2$. Below, I will consider $G= SL(5, {\mathbb R})$. 
Now, the space of geodesics $\Sigma_{x,y,\xi}$  connecting $x$ to an antipodal point $y$ and having a fixed regular congruence type, is a smooth 4-dimensional manifold. (Regular congruence type corresponds to the assumption that the germ $\xi$ is not contained in any wall of $X$, so $\tau$ is a chamber in $X$ and $G_\tau$ is Borel.) This is the degree of freedom for edges in $X$ that Gil asked for. 
Now, I can explain why the condition 1 above is useful. Suppose that $H$ is a group which 
admits an epimorphism $\rho: H\to SL(2, {\mathbb R})$. Then $H$ contains a codimension 1 subgroup $H'$ obtained as the preimage of a Borel in $SL(2, {\mathbb R})$.  In the context of the pair $P=G_\sigma, G_\tau$, I get another parabolic $P'=H'$ of codimension 1 in $P=H$ and containing $G_\tau$. Now, using $P'$-orbit of the above germ $\xi$ (instead of the $P$-orbit), I will get a 3-dimension submanifold in the manifold of geodesics $\Sigma_{x,y,\xi}$. 
Thus, we got spaces of edges connecting $x$ to $y$ which have dimensions $4$ and $3$ respectively, which is what Gil asked for. 
Our next goal is to impose restrictions on 2-faces of the 3-sphere, i.e., triangles $T_i$ whose vertices are antipodal vertices $x_1, x_2, x_3$ in $X$ and whose edges are described above. Let $\xi_i, i=1,2,3$ denote the germs (at $x_i$) of the oriented edges $[x_i,x_{i+1}]$, each germ $\xi_i$ is contained in a unique chamber $\sigma_i$ in $X$. Generically, the chambers $\sigma_i$ are mutually antipodal. Let $C_3(X)$ denote the space of ordered triples of pairwise antipodal chambers in $X$. This space is birational to $U/T$, where $U$ is unipotent radical of $B$ and $T\subset G$ is maximal torus normalizing $U$ and acting on $U$ via conjugation. Hence, $C_3(X)$ has dimension $|R_+|-r$, in our example $G=SL(5, {\mathbb R})$, it is $10-4=6$. 
Thus, we have plenty of functions on $C_3(X)$ to choose from in order to impose one restriction for each triangle. Below is a somewhat random choice, based on my reading of the paper by Fock and Goncharov  "Moduli spaces of local systems and higher Teichmuller theory". A chamber in  $X$ (in the case of  $G=GL(n, {\mathbb R})$ is given by a complete flag $F=(V_0\subset V_1\subset ... \subset V_n)$ in $V={\mathbb R}^n$. If $n=3$, then $C_3(X)$ is 1-dimensional with a single (most natural) invariant of a a triple of flags $(F_1,F_2,F_3)$  given by Goncharov's triple ratio,  
$$
\rho(F_1,F_2,F_3)=\frac{\langle f_1, v_2 \rangle \langle f_2, v_3 \rangle \langle f_3, v_1 \rangle}
{\langle f_1, v_3 \rangle \langle  f_2, v_1 \rangle \langle f_3, v_2 \rangle}
$$
Here $f_k$'s are linear functions on ${\mathbb R}^3$ whose kernels are 2-planes in the flags $F_k$ and $v_k$'s are basis vectors in the lines in the flags $F_k$. 
For $n\ge 4$, given a triple of flags $(F_1,F_2,F_3)$, taking quotient of  $V$ by the subspace $V_{n-3}$ appearing in the $i$-th flag, I reduce the dimension to $3$ and, hence, can get the triple ratio $\rho_i$ of the image of my flag in the resulting 3-dimensional space. Now, I will take function $h: C_3(X)\to {\mathbb R}$ given by 
$$
h(F_1,F_2,F_3)=\rho^2_1+\rho^2_2+\rho^2_3. 
$$ 
Then, I impose one restriction $h(T_i)=t_i\in {\mathbb R}_+$ for each triangular face $T_i$ in the 3-sphere. This works for general $n$, but I will restrict to the case $n=5$.  
At the moment, I have no idea where the symplectic structure on the resulting space of maps of $S^3$ to $X$ would come from (actually, I do, I just do not know how to make it work). Instead of $G=SL(5, {\mathbb R})$ one may have to use another Lie group. The easiest way to get a symplectic structure is to use some form of symplectic reduction, which is how it was done first by Klyachko and then in my  paper with Millson "The symplectic geometry of polygons in Euclidean space", Journal of Diff. Geometry, Vol. 44 (1996) p. 479-513, 
or in my paper with Millson and Treloar "The symplectic geometry of polygons in hyperbolic 3-space", Asian Journal of Math., Vol. 4 (2000), N1, p. 123-164. The latter used Poisson Lie theory, which, somehow, looks more promising.<|endoftext|>
TITLE: Parameterization of complex analytic subvarieties
QUESTION [12 upvotes]: Let $V$ be an analytic subvariety of some open set of $\mathbb{C}^n$ (intersection of finitely many zero-locii of analytic functions). Hironaka's desingularization theorem provides a parameterization of $V$, that is: there exists a proper, onto map $M\rightarrow V$, where $M$ is a complex manifold, which is a biholomorphism outside the strict transform of the singular set of $V$. Therefore $V$ can be locally parameterized (in that sense), around anyone of its points, by an onto holomorphic mapping from an open set of some $\mathbb{C}^m$.
My question is the following: is it true that $V$ can be globally parameterized by an onto holomorphic mapping from an open set of $\mathbb{C}^m$?
The answer to this question boils down to knowing whether any complex manifold (obtained this way) can be the range of a holomorphic map from some open set of $\mathbb{C}^m$, which is a statement that I unfortunately don't know how to handle. Neither was I able to find in the litterature any specifics regarding the conformal nature of the manifold $M$ when $V$ lies in an affine space.
I thank the community in advance for any help/comment/rebuke !

REPLY [6 votes]: The answer to your question is yes, even in a much broader setting and moreover the parametrizing open set can be chosen always to be the unit polydisc!
Theorem. (Fornaess-Stout '77). If $X$ is an $n$-dimensional complex manifold, then there exists a locally biholomorphic map $\Phi$ from the open unit polydisc in $\mathbb C^n$ onto $X$ with the property that for each $x\in X$, the fiber $\Phi^{-1}(x)$ consists of no more than $(2n+1)4^n+2$ points.
Of course, if $X$ is just a complex space, you have to desingularize before, so that your map is no more finite, but still surjective of course.
Observe finally, that there is no assumption here on $X$, it works in the abstract case also, you don't need it to be an analytic subset of an open set of some complex vector space!
The paper where the theorem is has been indicated to me by T.-C. Dinh.<|endoftext|>
TITLE: Is the Brauer correspondence injective ? 
QUESTION [8 upvotes]: Let $G$ be a finite group and $k$ a field. In modular representation theory the Brauer correspondence establishes a bijection between the isomorphism classes of indecomposable $kG$-modules with trivial source and vertex $P$ and the isomorphism classes of non-zero projective indecomposable $k[N_G(P)/P]$-modules. This bijection is induced by the Brauer map 
$$M \mapsto \operatorname{Br}_P(M) := \frac{M^P}{\sum_Q tr_{Q}^P (M^Q)}$$
$(Q < P)$ where $M^P$ are the $P$-invariants and $tr^P_Q: M^Q \to M^P,\;\; m \mapsto \sum_{g \in P/Q}gm$ is the trace map. This was proved in Theorem 3.2 of the paper 
M. Broué: On Scott Modules and p-Permutation Modules: An Approach through the Brauer Morphism. Proc. Amer. Math. Soc. 93(1985), 401-408
and the statement can also be found in this paper of M. Wildon (end of p. 12). 
However, to my understanding, Broué's proof doesn't show that if $M_1,M_2$ aren't ismorphic then $\operatorname{Br}_P(M_1), \operatorname{Br}_P(M_2)$ aren't isomorphic as well (as $k[N_G(P)/P]$-modules). 
Question: Does someone know, if the Brauer correspondence $[M] \mapsto [\operatorname{Br}_P(M)]$ ($[-]$ denotes isomorphism classes) is injective ? If so, a proof or a reference which contains a proof is also welcome.

Let me just explain why I think  Broué's proof doesn't show injectivity in full strenght: He decomposes $kG \otimes_{kP}k = \bigoplus_{i=1}^n M_i$ into indecomposable $kG$-Modules whereby the ones with vertex $P$ are exactly those such that $\operatorname{Br}_P(M_i) \neq 0$. Assume this holds for $i=1,...m$. Then 
$$k[N_G(P)/P] = \operatorname{Br}_P(kG \otimes_Pk)= \bigoplus_{i=1}^m \operatorname{Br}_P(M_i)$$
and the $\operatorname{Br}_P(M_i),\;(i=1,...m)$ can be shown to be indecomposable. Since every indecomposable projective $k[N_G(P)/P]$-module is isomorphic to a direct summand of $k[N_G(P)/P]$, the Brauer correspondence is surely surjective. 
However, to my understanding, it may happen that $M_i,M_j$ aren't isomorphic, but $\operatorname{Br}_P(M_i) \cong \operatorname{Br}_P(M_j)$ (i.e. this summand has multiplicity $>1$ in $k[N_G(P)/P])$.

REPLY [4 votes]: As Geoff Robinson's answer makes clear, the Brauer map is injective when restricted to indecomposable $p$-permutation $kG$-modules with vertex $P$, because it agrees with the Green correspondence. This is stated in Theorem 3.4 of the paper by Broué cited in the question. Natalie's answer mentions Thévenaz's book, which also has a proof of this fact.
If we allow non $p$-permutation modules (but still require fixed vertex $P$) then the map is not injective. For example, take $G = P = \langle g \rangle \cong C_8$ and let $k = \mathbf{F}_2$. Let $V_k$ be the unique $k$-dimensional indecomposable representation of $G$, for $1 \le k \le 8$. Then $V_3$ has a basis $v_1,v_2,v_3$ such that $v_i(g+1) = v_{i+1}$ for $i=1,2$ and $v_3(g+1) = 0$. Clearly $V_3^P = v_3$. Since 
$$v_2g^2 = v_2 + v_2(1+g^2) = v_2 + v_2(1+g)^2 = v_2$$ 
and $\mathrm{Tr}_{\langle g^2 \rangle}^P v_2 = v_2(1+g) = v_3$, we have $\mathrm{Br}_P(V_3) = 0$. A similar argument shows that $\mathrm{Br}_P(V_5) = \mathrm{Br}_P(V_7) = 0$. But since $V_3$, $V_5$ and $V_7$ have odd dimension, they all have full vertex $P$.
There is a related example in Example 1.3.6 of the note of mine linked in the question.<|endoftext|>
TITLE: Large cardinal axioms and total recursive functions
QUESTION [11 upvotes]: Are there known relationships between large cardinal axioms (say Mahlo or Woodin cardinals)
and total recursive functions (over the natural numbers) of the type:
$ZFC$ + large cardinal axiom $\vdash$ $f$ is a total recursive function,
but:
$ZFC \not\vdash$ $f$ is a total recursive function,
like this is known for different axiom systems of reverse mathematics, e.g.:
$ATR_0 \vdash$ the Goodstein function $G$ is total recursive,
but for the weaker system $ACA_0$ we have:
$ACA_0 \not\vdash$ the Goodstein function $G$ is total recursive.
Or, more generally asked, is there a relationship of the following kind:
If $LCA_1$ (large cardinal axiom 1) is stronger (wrt. consistency strength)
than $LCA_2$, then
$ZFC + LCA_1$ proves the totality of a recursive function $f$ which
grows faster than all recursive functions having totality proofs in
$ZFC + LCA_2$?
This question arises in investigations of the relationship between
learnability and provability, where provably total recursive functions
are used to schedule the learning process. A learning systems $\Lambda(\Sigma_1)$
using $\Sigma_1$ as a background theory is stronger than a learning system
$\Lambda(\Sigma_2)$, if $\Sigma_1$ allows totality proofs of recursive functions
growing faster than all recursive functions having totality proofs in $\Sigma_2$.

REPLY [8 votes]: $\def\zfc{\mathrm{ZFC}}\def\lca{\mathrm{LCA}}$Yes. Normally, $\zfc+\lca_1$ proves not only the consistency of $\zfc+\lca_2$, but also the uniform reflection principle of $\zfc+\lca_2$ (at least for arithmetic formulas). In particular, the $\Sigma^0_1$-reflection principle is a $\Pi^0_2$ sentence provable in $\zfc+\lca_1$ but not in $\zfc+\lca_2$, and it can be expressed as the totality of the following function $f$: if $x$ is a code of a $\zfc+\lca_2$ proof of a $\Sigma^0_1$-sentence of the form $\exists u\,\theta(u)$ with $\theta\in\Delta^0_0$, let $f(x)$ be the minimal $u$ such that $\mathbb N\models\theta(u)$, otherwise $f(x)=0$.
To see that $f$ grows faster than any provably total recursive function of $\zfc+\lca_2$, let $g$ be such a function. We may assume without loss of generality that $g$ is increasing, and $g(x)=y$ has a $\Sigma^0_1$-definition $\exists w\,\lambda(x,y,w)$ which is provably total and (strictly) increasing in $\zfc+\lca_2$. Let $\theta(x,u)$ be the formula $\exists y,w\le u\,\lambda(2^x,y,w)$ (this is strictly speaking not $\Delta^0_0$, but this is easy to fix, I'll leave it like this to simplify the presentation). Then the formulas $\exists u\,\theta(\ulcorner n\urcorner,u)$ (where $\ulcorner n\urcorner$ denotes the binary numeral of $n$) have $\zfc+\lca_2$ proofs of Gödel number bounded by $p(n)$ for some polynomial $p$, but their smallest witnesses have magnitude at least $g(2^n)>g(p(n))$ for large enough $n$.
This argument shows that $f(m)>g(m)$ for infinitely many $m$. If we want this to hold for all sufficiently large $m$, it suffices to make the function increasing by using $f_2(x)=\max_{y\le x}f(y)$ instead of $f$.

REPLY [7 votes]: Since large cardinal axioms have arithmetic consequences not provable without them, the answer to your first question is yes. For example, let $f(n)=1$, provided that $n$ is not the Goedel code of a proof of a contradiction in  ZFC. Since large cardinals imply Con(ZFC), the theory ZFC+large cardinals proves that $f$ is total. But ZFC alone does not prove this (if consistent), since it is relatively consistent with ZFC that there are proofs of contradictions from ZFC.
For the more general question, suppose that the stronger theory not only proves the weaker theory, but also proves that whenever the weaker theory proves that a TM program computes a total function, then it really does. This is the situation for most large cardinals---for example the theory asserting that there is a weakly compact cardinal implies towers of smaller inaccessible cardinals, and so if a program is total inside such a $V_\kappa$, then it really is in $V$. Given this situation, let $f(n)$ be the function which first inspects all smaller $m\leq n$ to see which code proofs from the weaker theory that a certain function $g_m$ is total, and then let $f(n)$ be larger than all such $g_m(n)$. Our assumption on the stronger theory ensures that we can be confident that the computation of $g_m(n)$ converges, so $f(n)$ will be defined.<|endoftext|>
TITLE: Orbifold vs Alexandrove space vs Limit of manifolds
QUESTION [5 upvotes]: I have a question about some defitions : 
  Orbifold, Alexandrov space, limit of manifolds in Gromov-Hausdorff distance sense.
Consider following example. 
Let $r> 0$   
$L_c = \{ (x cos \theta, x sin \theta, cx) |  0 \leq x$ and $0 \leq \theta < 2\pi \}$ 
$S$ : $(z-\sqrt{2} r)^2 + x^2 + y^2 = r^2$ 
$T$ : $ (z- \sqrt{2} R)^2 + x^2 + y^2 = R^2$ 
If $R$ is sufficiently large, then we have a two dimensional sphere $U_c$ enclosed by
   $L_c$, $S$, and $T$.
First notice the following. $lim_{r \rightarrow 0} U_c$ is an 
   orbifold for some $c$ 
Question : Is any orbifold is a limit of manifolds in Gromov-Hausdorff sense ? 
If this question is wide, we can restrict to the case of nonnegatively curved orbifolds.
   : Is a nonnegatively curved $n$-orbifold a limit of positively curved $n$-manifolds ? 
Question 2: In the following paper, a space with curvature $ \geq k$ is defined.
M. Gromov Y. Burago and G. Perelman, A.d. alexandrov spaces with curvature bounded
   below, Uspekhi Mat. Nauk 47 (2) (1992), 3–51.
Is a $n$-dimensional space with curvature $ \geq k$, which is smooth except 
  finite points, is a limit of $n$-manifolds of 
  positive sectional curvature $\geq k$ ? I believe that this question is trivial and 
  it is true.    
I do not think that all orbifolds or spaces with curvature $ \geq k $ are limits of 
  manifolds. 
However I can not deny it. 
Since ${\bf R}^3={\bf R}^4 /S^1 = lim_{k \rightarrow \infty} {\bf R}^4/{\bf Z}_k$, 
  orbifolds are different from spaces with curvature $ \geq k $. But they are 
  obtained from the sequences of manifolds. 
MOTIVATION : Hsiang-Kleiner classified positively curved manifolds with $S^1$-action. 
  I want to extend this result to positively curved orbifolds with $S^1$-action. 
If orbifold is a limit of manifolds then the problem is simple. 
Accordingly I want to know the questions. Thank you for your attention.

REPLY [10 votes]: By Perelman's Stability Theorem, if a (compact) limit of $n$-dimensional Alexandrov spaces of curvature $\ge k$ has the same dimension, then the convergent spaces are eventually homeomorphic to the limit space. So in this context a limit of manifolds is always a topological manifold.
And there exist non-manifold examples of Alexandrov spaces, e.g., $\mathbb R^3/\{id,-id\}$ which is the cone over $\mathbb {RP}^2$. Or, if you want a compact example, take $S^3/\mathbb Z_2$ where $\mathbb Z_2$ acts on $S^3$ with two fixed points (think of $S^3$ lying in $\mathbb R^4$ and consider the reflection in some 1-dimensional axis). This is a compact orbifold of curvature $\ge 1$ and it is not a limit of manifolds of the same dimension with a uniform lower curvature bound.

REPLY [9 votes]: Q1. Note that one oriented orthonormal frame bundle $FO$ over a smooth orbifold $O$ is a smooth manifold.
This frame bundle admits a one-parameter family of metrics which collapse to the original manifold and its curvature can be made bounded from below.
Proof. Equip $FO$ with $SO(n)$-invariant metric, consider product space $[\varepsilon\cdot SO(n)]\times FO$ and factorize along diagonal action. (This often called Cheeger's trick, but I think it was known before Cheeger.)  
Q2. It is an open question. It is expected that cones over some positively curved manifolds can not be approximated. Vitali Kapovitch has examples of such $n$ dimesional cones which can not be approximated by $m$-dimensional manifolds with $m< n+8$.<|endoftext|>
TITLE: Generic section of a vector bundle over G(2,4)
QUESTION [8 upvotes]: I want to prove that on a smooth cubic surface in $\mathbb{P}^3$ there are exactly $27$ lines and I want to do it in a following way. First, our cubic is given by $F(x,y,z,t)=0$ and this polynomial gives a section of $Sym^3(T)^*$ where $T$ is the tautological bundle over $G(2, 4)$ (just by restricting $F$ to the point of $G(2,4)$). The zeros of this section are exactly the lines contained in our surface. 
Now I need to show that this section is generic (then use Poincare duality and finish the proof). We need to check that it's transverse to the zero section and that's where the problem is.
I started with taking local coordinates (planes spanned by $(1,0,x_2,x_3)$ and $(0, 1, y_2, y_3)$), one can write the condition for transversality in terms of the tangent spaces (inversibility of some matrix involving $\frac{\partial F}{dx_3}$ and $\frac{\partial F}{dx_4}$) but I failed to show that the tangent vectors actually span the tangent space. Is there an easy way to do that?

REPLY [11 votes]: Your question is beafully answered in Harris' paper Galois Groups of Enumerative Problems, Duke Math. J. Volume 46, Number 4 (1979), 685-724.
More precisely, in Section III Harris proves the following result:

Proposition. Let $V \subset \mathbb{P}^n$ be a smooth hypersurface and $Z(V)$ the Fano scheme of lines on $V$. For any $u \in Z(V)$ let us denote by $l_u$ the line in $V$ corresponding to $u$. Then $Z(V)$ is singular at $u$ if and only if the normal bundle $N$ of $l_u$ in $V$ is special, that is $h^1(V, N) >0$.

He works in in local coordinates, exactly as you attempted to do.   
Let now $V \subset \mathbb{P}^3$ be an irreducible cubic surface. It is not difficult to see that if $V$ contains infinitely many lines then it must be singular. Hence if $V$ is smooth the Fano scheme $Z(V)$ is zero-dimensional.  
By adjunction any line $l$ on a cubic surface $V$ is a $(-1)$-curve, hence $N=\mathcal{O}_{\mathbb{P}^1}(-1)$. Therefore $h^1(V, N)=0$, i.e. any line on $V$ correspond to a smooth point of $Z(V)$. 
This precisely means that the section of $\textrm{Sym}^3 T^*$ whose zero locus is $Z(V)$ is transverse to the zero section. Hence the number of lines on $V$coincides with the degree of the zero-dimensional scheme $Z(V)$, which is given by  $$c_4(\textrm{Sym}^3 T^*)=27.$$<|endoftext|>
TITLE: Chain conditions in quotients of power sets
QUESTION [7 upvotes]: Several days ago a friend asked me the following:

We know that in $\mathcal P(\mathbb N)$ we can find a family of size continuum that every [distinct] two intersect in a finite set. Can we do that with $\mathcal P(\mathbb R)$, that is a family of size $2^{\frak c}$ many subsets of real numbers that the intersection of any [distinct] two is finite, or at least less than $\frak c$?

The question, if so, asks about $(2^{\frak c})^+$-c.c. in the Boolean algebras $\mathcal B_\kappa=\mathcal P(\mathbb R)/\sim_\kappa$ where $\sim_\kappa$ is the equivalence relation defined as $A\sim B\iff |A\triangle B|<\kappa$. The first question asks for $\mathcal B_\omega$ and the latter asks for $\mathcal B_\mathfrak c$.

Assuming GCH (or at least that $2^{\frak c}=\aleph_2$) gives a relatively simple positive answer to the latter question:
Consider the tree $2^{<\omega_1}$, it is of size $\aleph_1$ so we can encode the nodes as a real numbers. This tree has $2^{\omega_1}=\aleph_2$ many branches, each defines a subset of $\mathbb R$ using the encoding, and every distinct two branches meet at most at countable set of points.

I consulted with several other folks from the department and I was told that most of these questions are very well known, so an answer about consistency and provability is almost certainly out there. Naive Google search got me nowhere, so I came to ask here the following:

In the particular case of the question above, can we say anything in ZFC about the chain-condition of $\mathcal B_\kappa$ for $\omega\leq\kappa\leq\frak c$? 
My partial answer above shows that with GCH we have an answer for $\cal B_\frak c$, but does that also answer $\cal B_\omega$ or do we need to assume stronger principles as $\lozenge$ for suitable cardinals?
How far does this generalized, when replacing $2^\omega$ by any infinite cardinal $\mu$, and asking the similar question about $(2^\mu)^+$-c.c. in the similar quotients?

I'd be glad to have a reference to a survey of such results, if it exists.

REPLY [5 votes]: I'll add in that Shelah has used pcf theory to investigate related questions. Typically these results are tucked away inside long papers dealing with other questions, but I know that the last section of [Sh:410] explicitly deals with ``strongly almost disjoint families", and characterizes their existence in terms of pcf.
For example, if $\aleph_0<\kappa\leq\kappa^{\aleph_0}<\lambda$, then  the existence of a family of $\lambda^+$ sets in $[\lambda]^{\kappa}$ with pairwise finite intersection is equivalent to a ``pcf statement''.
I'm not sure which version of the paper to link to, as the published version has been reworked a few times. I THINK that the most recent version is here:
Sh:410<|endoftext|>
TITLE: Generalizations and relative applications of Fekete's subadditive lemma
QUESTION [9 upvotes]: Fekete's (subadditive) lemma takes its name from a 1923 paper by the Hungarian mathematician Michael Fekete [1]. A historical overview and references to (a couple of) generalizations and applications of the result are found in Steele's book on probability and combinatorial optimization [2, Section 1.10], where a special mention is made to the work of Pólya and Szegő on the structure of real sequences and series [3, Ch. 3, Sect. 1] and that of Hammersley [4], motivated by percolation theory, on subadditive functions, the
continuous analogue of subadditive sequences, whose systematic study was initiated, as far as I know, by Hille and Phillips in the 1957 edition of their beautiful monograph on functional analysis and semigroups [5, Ch. VII]. The same Steele acknowledges that his own 1989 proof of Kingman's subadditive ergodic theorem [6], of which Birkoff's celebrated theorem is a corollary, was eventually inspired by Fekete's lemma. Now, my question is:

Can you point out further generalizations (and corresponding
  (interesting) applications) of Fekete's lemma?

Added later. Fekete's lemma can be used to prove that the limit occurring in the spectral radius formula does actually exist. And this counts (to me) as an (interesting) application.
Bibliography.
[1] M. Fekete (1923), Über die Verteilung der Wurzeln bei gewissen algebraischen Gleichungen mit. ganzzahligen Koeffizienten, Math. Zeit., Vol. 17, pp. 228-249.
[2] M.J. Steele, Probability theory and combinatorial optimization, SIAM, Philadelphia, 1997.
[3] G. Pólya and G. Szegő, Problems and Theorems in Analysis, Vol. I, Springer-Verlag, Berlin, 1998 (reprint of the 1978 Edition).
[4] J.M. Hammersley (1962), Generalization of the fundamental theorem of subadditive functions,  Proc. Cambridge Philos. Soc.,
Vol. 58, pp. 235-238.
[5] E. Hille and R.S. Phillips, Functional analysis and
semi-groups, American Math. Soc., 1996 (revised edition).
[6] J.M. Steele (1989), Kingman's subadditive ergodic theorem, Annales de l'I.H.P., Section B, Vol. 25, No. 1, pp. 93-98.

REPLY [5 votes]: Since you mentioned Kingman's subadditive ergodic theorem, you may find interesting the following semi-uniform subadditive ergodic theorem:
Let $T \colon X \to X$ be a continuous map of a compact metric space $X$.
If $f_n \colon X \to [-\infty,+\infty)$ is a subadditive sequence ($f_{n+m} \le f_n + f_m \circ T^n$) of upper semicontinuous functions then:
$$
\sup_{\mu} \lim_{n \to \infty} \frac{1}{n} \int_X f_n d\mu = 
\lim_{n \to \infty} \frac{1}{n} \sup_{x \in X} f_n(x)  ,
$$
where the first $\sup$ is taken over all $T$-invariant probability measures.
References:

Schreiber. J. Diff. Eq. 148 (1998), 334--350. 
Sturman, Stark. Nonlinearity 13 (2000), 113--143.
Morris. Proc. London Math. Soc. (3) 107 (2013) 121–150. See Appendix A.<|endoftext|>
TITLE: Rooks in three dimensions
QUESTION [36 upvotes]: Given is an infinite 3-dim chess board and a black king. What is the minimum number of white rooks necessary that can guarantee a checkmate in a finite number of moves?
(In 3-dimensional chess rooks move in straight lines, entering each cube through a face and departing through the opposite face.  Kings can move to any cube which shares a face, edge, or corner with the cube that the king starts in.
See Wikipedia).
Update: Comments (below the line) give interesting information on this problem including a connection to Conway's angel problem with 2-angel (Zare) and interesting comments towards positive answer and connection with "kinggo" (Elkies). Also a link to an identical SE question is provided MSE Question 155777 (Snyder).

REPLY [7 votes]: The following is for a finite board (the question actually assumes an infinite board). 
For an $N\times N \times N$ board, wouldn't $2N$ rooks suffice? The idea comes from adapting the checkmate with two rooks for the two dimensional case in which the two rooks alternate rows and force the king to the last rank.
For the three dimensional case, for each $y$ with $1\leq y \leq N$, place one rook at $(1,y,k)$ and another at $(2,y,k+1)$. Then, for $y$ going from $1$ to $N$, move the rook at $(1,y,k)$ to the square $(1,y,k+2)$. Again, for $y$ going from $1$ to $N$, move the rook at $(2,y,k+1)$ to $(2,y,k+3)$. Each for loop over $y$ involves moving, alternately, the rooks with $x$ coordinate $1$ by increasing their $z$ coordinate by $2$ units, or the rooks with $x$ coordinate $2$ by increasing their $z$ coordinate by $2$ units. We alternate, so that a for loop in which the rooks with $x$ coordinate $1$ are moved is followed by a for loop in which the rooks with $x$ coordinate $2$ are moved, and vice versa. Eventually, either the rooks with $x$ coordinate $1$ or the rooks with $x$ coordinate $2$ will have $z$ coordinate $N$.
The effect of this is that a subset of the squares guarded by the rooks form a "floor" of two layers that keeps moving upward. So if the black king is between this "floor" and the top of the $N\times N\times N$ cube, it gets pushed to the top face. The "floor" must be moved upward in such a way that it never becomes disconnected, so that the black king can never escape to beneath the "floor" through some gap.
For an infinite board, @Noam Elkies has already mentioned $5N$, so the following is not an improvement: the "ceiling" (and the other walls of the $N\times N \times N$ cube) can be formed by placing $N$ more rooks at $(N,y,N)$, for each $y$ with $1\leq y\leq N$, and $N-1$ more rooks at $(x,1,N)$, for each $x$ with $1 \leq x \leq N-1$, and $N-1$ more rooks at $(x,N,N)$ for each $x$ with $1 \leq x \leq N-1$.<|endoftext|>
TITLE: Smooth Models of Hyperelliptic Curves (+ a concrete question)
QUESTION [5 upvotes]: My general question is : given a hyperelliptic curve $y^2 = f(x)$ with disc$f(x) \ne 0$, is there a general formula for finding a smooth complete model of the curve?
Specifically, I want a smooth, complete model of the curve $C_0 : y^2 = x^5 + 1$ over some field $k$ of characteristic 0.
Silverman (Exercise 2.14, the Arithmetic of Elliptic Curves) seems to think that I can find such a model by considering the closure of the image of $C_0$ in $\mathbb{P}^4$ under the map $[1,x,x^2,x^3,y]$.
It seems to me that the closure of the image of this curve is the scheme $\text{Proj }k[X_0,X_1,X_2,X_3,X_4]/(X_2X_0 - X_1^2, X_3X_0^2 - X_1^3, X_3X_0 - X_1X_2, X_4^2-X_2X_3-X_0^2)$
But, you can quickly check that there are infinitely many points at infinity (ie $X_0 = 0$), namely the points $(0,0,1,a^2,a)$. However, this is impossible since the points at infinity are closed, and hence finite since the affine piece $C_0$ is embedded in $D_+(X_0)\subseteq\text{Proj } k[X_0,\ldots,X_4]$.
thanks

REPLY [11 votes]: An alternative, and I think easier, way to get a smooth model for $y^2=f(x)$ is to glue together two non-singular affine curves. If $f(x)$ has degree $2d$ or $2d-1$, then glue $C:y^2=f(x)$ and $C':v^2=u^{2d}f(1/u)$ via $x=1/u$ and $y=v/u^d$. Note that the affine curves $C$ and $C'$ are both smooth. Also, the "points at infinity" on $C$ are the points where $u=0$, so there are two such points if $f$ has even degree and one such point if $f$ has odd degree.
I want to reiterate what Dalawat says. You generally should not answer your own question. Instead, edit the question if there are further things that you want to say.

REPLY [3 votes]: Anyway, an interesting tidbit I just learned, which I'm going to explain here mostly for my own benefit (though I'd appreciate any clarifying comments), is why the closure of the image of the curve $C_1 : y^2 = x^6 + 1$ in $\mathbb{P}^4$ has two points at infinity, whereas the curve $C_0 : y^2 = x^5 + 1$ in $\mathbb{P}^4$ has only one point at infinity.
From here on let $C_i$ denote the affine curve given by the equations above over $K$. Let $C_i'$ be the smooth complete model of $C_i$, ie the closure of the image of the map to $\mathbb{P}^4$.
If we consider the projection onto $x$ map $x : C_i \rightarrow \mathbb{P}^1$, we get an inclusion of function fields $K(x)\subseteq K(x,y)$ of index 2. The smooth complete model also has the same function field $K(x,y)$, so the number of points it has at infinity is just the number of preimages of $\infty$ under the projection $x$, extended to $C_i'$. Since $x$ is a double cover, this is either 1 or 2.
Now, the trick here is to note that at least for nonsingular curves (for example the ring of integers of a number field), points correspond to valuation rings of their function field, so here, we are looking at valuations of $K(x,y) = K(C_i')$ lying above the valuation ring $K(1/x)$ of $K(x) = K(\mathbb{P}^1)$. Now, noting that
$$K(x,y)\otimes_{K(x)} K((1/x)) = \bigoplus_{v\mid \frac{1}{x}} K(x,y)_v$$
we see that there are two valuations if and only if $K((1/x))(y) = K((1/x))$.
Indeed, in our case $y = \sqrt{x^d + 1}$, and the square root exists in $K((1/x))$ if and only if $d$ is even.
This is pretty awesome, especially since you can't otherwise detect the behavior of the curve $C_i$ at infinity without first finding a normalization.<|endoftext|>
TITLE: Local structure of rational varieties
QUESTION [47 upvotes]: I've been asked this question by a colleague who's not an algebraic geometer; we both feel that the answer should be "no", but I don't have a clue how to prove it.
Here's the question:
let $X$ be a smooth rational variety (over the complex numbers, say). Is it true that every point of $X$ has a Zariski open neighbourhood that is isomorphic to an open subset of ${\mathbb P}^n$?

REPLY [8 votes]: Here is a preprint of Ilya Karzhemanov constructing counterexamples in dimensions $n\geq 4$:
http://research.ipmu.jp/ipmu/sysimg/ipmu/1588.pdf<|endoftext|>
TITLE: Zero divisors in group algebras of torsion-free groups of length 3
QUESTION [5 upvotes]: It is easy to see that there are no group divisors of length 2 in a group algebra of a torsion-free group. I saw somewhere mentioned that it is possible to do it for length 3. How?

REPLY [6 votes]: It is open for the group algebra over the field with $2$ elements, see 
http://arxiv.org/abs/1202.6645 and http://arxiv.org/abs/1112.1790
and references there. I saw on the website of Mikhailov an unpublished paper, where he  contributes this question to I. Rips.<|endoftext|>
TITLE: Can a nontrivial spectrum smash to zero with $K$-theory? 
QUESTION [6 upvotes]: Let $E $ be a (possibly nonconnective) spectrum. Suppose $E \wedge K = 0$ (where $K$ is complex $K$-theory). Does it follow that $E = 0$?

REPLY [10 votes]: Another type of example: Let $E$ be the mod $p$ homology spectrum. The integral homology groups of the periodic $K$-theory spectrum are rational vector spaces, i.e. the integral homology groups of the mod $p$ spectrum are zero.

REPLY [8 votes]: Sure.  Smashing a based space with a spectrum is equivalent to smashing
its suspension spectrum with that spectrum.  So it suffices to give a 
nontrivial space whose reduced $K$-homology is trivial.  A classical 
example due to Luke Hodgkin is $Coker J$.  See Hodgkin, Luke; Snaith, Victor.
The K-theory of some more well-known spaces. 
Illinois J. Math. 22 (1978), no. 2, 270–278.<|endoftext|>
TITLE: construction of matrices verifying an identity
QUESTION [5 upvotes]: Let $A,B\in\mathbb{R}^{n\times n}$. Suppose that B is nonsingular and $AB\neq BA$. Can we always find real numbers $t_1,⋯,t_p$ such that 
$$B\left(\displaystyle\prod_{i=1}^{p}(A+t_{i}B)\right)A=A\left(\displaystyle\prod_{i=1}^{p}(A+t_{i}B)\right)B?$$
N.B :   $p$ is not fixed.

REPLY [12 votes]: No. A counter-example is
$$A = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0  & 0 \\ 0 & 0  & 2 & 0 \\ 0 &0  &0  & 2 \end{pmatrix} \quad 
B = \begin{pmatrix} 0 & 1 & 1 & 0 \\ -1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & -2 & 0 \end{pmatrix}.$$
Observe that both matrices live in the image of the embeding $GL_2(\mathbb{C}) \to GL_4(\mathbb{R})$. In these terms, the matrices are
$$A = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \quad
B = \begin{pmatrix} i & 1 \\ 0 & 2i \end{pmatrix}.$$
For any real $t$, the element $A+tB$ is of the form $\left( \begin{smallmatrix} u & v \\ 0 & 2u \end{smallmatrix} \right)$ for $u$ an nonzero complex number. So any product $\prod_{i=1}^p (A+t_i B)$ is of the form $\left( \begin{smallmatrix} u & v \\ 0 & 2^p u \end{smallmatrix} \right)$ for $u$ a nonzero complex number. Now
$$A \begin{pmatrix} u & v \\ 0 & 2^p u \end{pmatrix}  B = 
\begin{pmatrix} u & u + 2iv  \\ 0 & 2^{p+1} i u \end{pmatrix}$$
$$B \begin{pmatrix} u & v \\ 0 & 2^p u \end{pmatrix}  A =
\begin{pmatrix} u & 2^{p+1} u + 2iv  \\ 0 & 2^{p+1} i u \end{pmatrix}$$
So $2^{p+1} u = u$ and, as $u \neq 0$, we have a contradiction.

A vague attempt to explain how I found this. Let $C$ be the centralizer of $A^{-1}B $. The set of elements $M$ such that $AMB=BMA$ is $C A^{-1}$. The curve $A+tB$ is in $AC$ and it seems hard to find any properties of it which aren't true of all of $AC$. So we basically want to know whether $(AC)^p$ meets $CA^{-1}$ for sufficiently large $p$. I flirt with Hecke operators (because we are basically multiplying double cosets) and ergodic theory (because we want to know if a small subset of a group spreads out under multiplication). 
Eventually I decide that the answer is probably "no". If so, then the most likely way to prove it is to find a function $f:GL_n(\mathbb{R}) \to \mathbb{R}$ which is more or less multipliciative, large for positive powers of $A$, small for negative powers of $A$ and bounded on $C$. A character would be great. But the only character of $GL_n$ is determinant and it doesn't work. Eventually, I realize that I can arrange for $A+tB$ to lie in a subgroup $H$ of $GL_n$ which has additional characters; let $T$ be the abelianization of $H$. It also seems like a good idea to make all of the eigenvalues of $A^{-1} B$ complex, as then $A+tB$ is always invertible. In other words, $T$ should have rank at least $2$, and should have lots of $\mathbb{C}^{\ast}$ factors. So I decide to try $H$
$$\begin{pmatrix} a & b & \ast & \ast \\ -b & a & \ast & \ast \\ 0 & 0 & c & -d \\ 0 & 0 & d & c \end{pmatrix}$$
with character $f=\frac{c^2+d^2}{a^2+b^2}$. By taking $f(A) = f(B)>1$, I ensure that $f(C)=1$, so $f$ is bounded on $C$ and large on $A$ as desired. At this point I stop trying to be systematic and just try some examples until one works.<|endoftext|>
TITLE: Hyperplane arrangements and covering numbers
QUESTION [5 upvotes]: Let $H$ be a set of $(d-1)$-dimensional hyperplanes in $\mathbb{R}^d$.  For each hyperplane $h \in H$ let $D(h)$ and $\bar{D}(h)$ be the corresponding half spaces of $\mathbb{R}^d$.  For a point $x \in \mathbb{R}^d$, define the function
$$
b(h,x) = \begin{cases} 
1 & x \in D(h) \\\
0 & \text{otherwise}.
\end{cases}
$$
For points $x,y \in \mathbb{R}^d$, define the psuedometric,
$$
\sigma(x,y) = \sum_{h \in H} |b(h,x) - b(h,y)|.
$$
So $\sigma(x,y)$ counts the number of hyperplanes in $H$ that pass between the points $x$ and $y$. I believe that the following lemma is true, but have not yet found a proof.
Lemma
Let $N$ be the number of hyperplanes in $H$.  There exists a polynomial $p$ (depending perhaps on $d$, but not on $N$) such that, for every positive integer $k$, there is a discrete subset $T(k) \subset \mathbb{R}^d$ containing not more than $p(k)$ elements with the property that for every $x$ in $\mathbb{R}^d$ there is a point $y \in T(k)$ with $\sigma(x,y) < \frac{N}{k}.$
Another way to state the Lemma is:  We require at most $p(k)$ `balls' of radius $\frac{N}{k}$ (with respect to the metric $\sigma$) to cover $\mathbb{R}^d$.
The Lemma is related to what is called Vapnik–Chervonenkis dimension, and it arises when trying to prove results about certain empirical processes.  I suspect this result (or something equivalent) to be in the literature, but I have not yet been able to dig it out.
Has anybody seen this result (or something equivalent) before?  Alternatively, does anybody know a nice proof?

REPLY [3 votes]: The Lemma follows from the Cutting lemma:
J. Matousek, Lectures on Discrete Geometry, Theorem 6.5.3.
This gives a partition of $\mathbb{R}^d$ into $O(k^d)$ generalized simplices (intersections of $d$ hyperplanes), each intersected by at most $n/k$ hyperplanes of $H$. To obtain the set $T(k)$, pick one point from each of the simplices.<|endoftext|>
TITLE: Sum of the reciprocal of perfect numbers
QUESTION [8 upvotes]: It is well known that the sum of the reciprocal of prime numbers is $+\infty$. This proves that there infinitely many prime numbers. On the other hand it is also known that the series of the reciprocal of twin prime numbers converges, so nothing can be said about the finiteness of the set of twin prime numbers. This is indeed an open problem. A similar open problem is the existence of infinitely many perfect numbers. So this is the question:
let $\mathcal{Pe}$ be the set of perfect numbers. What is known about the series
$$
\sum_{n \in \mathcal{Pe}} \frac{1}{n} 
$$
This is just curiosity, and could be a trivial question. Note that if there are no odd perfect numbers (as it seems to be the case), the series converges. Indeed any $n$ perfect is of the form $2^{p-1}(2^p-1)$.

REPLY [3 votes]: Jonathan Bayless and Dominic Klyve, Reciprocal sums as a knowledge metric, Amer Math Monthly 120 (November, 2013) 822-831, give the first 150 decimal places of the sum of the reciprocals of the perfect numbers. They calculate the sum for the known perfects to that many places, and rigorously bound the contribution from the unknown perfects (if there are any).<|endoftext|>
TITLE: "Modern" proof for the Baker-Campbell-Hausdorff formula
QUESTION [30 upvotes]: Does someone has a reference to a modern proof of the Baker-Campbell-Hausdorff formula?
All proofs I have ever seen are related only to matrix Lie groups / Lie algebras and
are not at all geometric (i.e. depend on indices, bases ect.)
By a 'modern' proof I'm thinking of a proof entirely in terms of differential geometry, i.e. 
in terms of the tangent bundle on the Lie group manifold or even better in terms of jets.
I'll keep the formulation vague on purpose, to higher my chances to get a good reference.
I think the question is pretty clear anyway.
Beyond the plain BCH-equation I would like to get a deeper understanding WHY the commutator 
(and the linear structure of the Lie algebra) is enough to define the group product locally
and what is geometrically going on.

REPLY [3 votes]: Loring Tu (of Bott & Tu fame) wrote a very nice paper on the subject: Une courte démonstration de la formule de Campbell-Hausdorff, J. Lie Theory 14 (2004) 501–508, in French but freely available here.<|endoftext|>
TITLE: Preissmann and Byers Theorems
QUESTION [6 upvotes]: I'm starting to study at the elementary level the relationship between topology and geometry of a Riemannian manifold of negative curvature. The first two theorems, simple and interesting in this direction are:
$\bf{Preissmann~~ Theorem}:$ Let  be $M$ a Riemannian manifold with sectional curvature $K<0$, then every non trivial Abelian subgroup of the fundamental group $\pi_1(M)$ is  cyclic infinite.
$\bf{Byers~~ Theorem}:$  Let  be $M$ a Riemannian manifold with sectional curvature $K<0$, then every no trivial solvable subgroup of the fundamental group $\pi_1(M)$ is  cyclic infinite, and $\pi_1(M)$ have no cyclic subgroup of finite index.
$\bf{My ~~Question}:$ I'm looking for nontrivial examples (counterexamples) for  Byers  Theorem, i.e, non trivial  examples of a Riemannian manifold that:

Has a no solvable fundamental group.

EDIT:
2  Has a cyclic $\bf{infinite}$ subgroup of finite index of the fundamental group.(In this case the trivial examples are welcome.)

REPLY [3 votes]: 1) To construct a manifold with no solvable fundamental group, take for example a finite unsolvable group $G$ and embed it into $SU(n)$ as a discrete subgroup. This embedding is obtained by realizing $G$ as a subgroup of the permutations $S(G)$ of the set $G$,
then note $S(G)$ is isomorphic to a subgroup of $SU(n)$ via monomial matrices ($n=|G|$).
The quotient $SU(n)/G$ carries a natural manifold structure with fundamental group $G$. Note these examples are compact. Taking product with $R$, one obtains non-compact examples.
2) For manifolds with cyclic subgroups of finite index, consider products of real projective spaces $RP^n$ or lens spaces $S^{2n+1}/Z_p$ with manifolds with finite fundamental group.
To combine 1) and 2), take products.<|endoftext|>
TITLE: Inner? Automorphisms
QUESTION [17 upvotes]: It is a fairly easy result of Edna Grossman's that any automorphism of a (finitely generated) free group which acts by conjugation on every primitive element ( primitive element in$F_n$ is one which is a member of a generating set of order $n$) is in fact inner. The question is: what is the right generalization of this fact? For surface groups, one can replace primitive elements by simple curves, but I a, not sure if that the"canonical" thing to do...
UPDATE Looking up citations of Henry's (@HW's) great reference, I found that the paper he cites is not quite "the last work". A more recent (and very relevant) word seems to be Bogopolski-Ventura (On endomorphisms of torsion-free hyperbolic groups)  from which I quote the abstract:
Let $H$ be a torsion-free $\delta$-hyperbolic group with respect to a finite generating set $S$. Let $a_1,..., a_n$ and $a_{1*},..., a_{n*}$ be elements of $H$ such that $a_{i*}$ is conjugate to $a_i$ for each $i=1,..., n$. Then, there is a uniform conjugator if and only if $W(a_{1*},..., a_{n*})$ is conjugate to $W(a_1,..., a_n)$ for every word $W$ in $n$ variables and length up to a computable constant depending only on $\delta$, $\sharp{S}$ and $\sum_{i=1}^n |a_i|$. 
As a corollary, we deduce that there exists a computable constant $\mathcal{C}=\mathcal{C}(\delta, \sharp S)$ such that, for any endomorphism $\phi$ of $H$, if $\phi(h)$ is conjugate to $h$ for every element $h\in H$ of length up to $\mathcal {C}$, then $\phi$ is an inner automorphism. 
Another corollary is the following: if $H$ is a torsion-free conjugacy separable hyperbolic group, then $\mbox{Out}(H)$ is residually finite. 
When particularizing the main result to the case of free groups, we obtain a solution for a mixed version of the classical Whitehead's algorithm.

REPLY [10 votes]: Have you looked at the following paper of Minasyan and Osin, who generalize Edna Grossman's Theorem to the relatively hyperbolic case?
Minasyan, A.(4-SHMP-SM); Osin, D.(1-VDB)
Normal automorphisms of relatively hyperbolic groups.
Trans. Amer. Math. Soc. 362 (2010), no. 11, 6079–6103.
It's the state of the art for these questions, I believe.<|endoftext|>
TITLE: Equidistibution of horocycles through Hecke eigenvalues of Maass cusp forms
QUESTION [5 upvotes]: At the end of this very nice post:
http://blogs.ethz.ch/kowalski/2012/05/21/who-needled-buffon/
E. Kowalski talks about the equidistribution of the points $\frac{j+i}{N}$ when $j=1,\dots,N$ and $N$ tends to $\infty$ in the fundamental domain of $SL_2(R)/SL_2(Z)$. 
He says that it "follows here most easily from non-trivial bounds on Hecke eigenvalues of Maass cusp forms".
Can anyone fill in the details of give a good reference? any related thoughts will also be appreciated!

REPLY [4 votes]: Weyl's criterion is that to check equidistribution it is enough to check against a basis of test functions.  Ignoring the non-compactness of the space for the moment, a natural basis is given by the eigenfunctions of the Laplacian, that is by the Maass eigenforms.  These have a Fourier expansion:
$$f(z) = \sum_{n\neq 0} a_n \sqrt{|n|y} K_{ir}(2\pi|n|y)e(nx)\,.$$
where $K_{ir}$ is the $K$-Bessel function and $e(w)=\exp(2\pi iw)$.  It is now easy to compute the average above in those terms: set $y=1/N$ and average $x$ over $\frac{1}{N}\mathbb{Z}/\mathbb{Z}$ -- the average of $e(nx)$ being zero unless $n$ is divisible by $N$, when it is one. Letting $\mu_N(f)$ denote the average of $f$ over the given set, and changing variables via $n=Nm$, we have
$$ \mu_N(f) = \sum_{m\neq 0} a_{Nm} \sqrt{m} K_{ir}(2\pi|m|)\,.$$
The goal is to show that this tends to zero with N.  Since the $K$-Bessel function decays exponentially, by the Bounded Convergence Theorem it is enough to know that the $a_{n}$ decay with $n$.
It now happens that if $f$ is also a Hecke eigenform then the $a_n$ are basically the corresponding Hecke eigenvalues, and then bounds toward the Ramanujan Conjecture show that the Hecke eigenvalues, hence the Fourier coefficients, decay.  It is believed that $a_n$ decay roughly like $n^{-1/2}$, and bounds of the form $n^{-\theta}$ are known.
Returning to the non-compactness, there is also continuous spectrum, but it is explicit (as is its Fourier expansion) so the analogous calculation for it is not hard.<|endoftext|>
TITLE: On the Geometric Interpretation of the Local Criterion for Flatness
QUESTION [8 upvotes]: The local criterion for flatness goes this way:
Let $\phi : (A,m)\rightarrow (B,m')$ be a local morphism of local Noetherian rings, and $M$ a finitely generated $B$-module. If $x\in m$ is a non zero-divisor on $M$ then $M$ is flat over $A$ iff $M/xM$ is flat over $A/xA$. 
One usual geometric interpretation (see for instance Eisenbud, Commutative Algebra with a View towards Algebraic Geometry, chapter 6.4) is:
If we have a morphism of affine varieties $X\rightarrow Y$ over $\mathbb{A}^1$ such that the maps to $\mathbb{A}^1$ are flat and dominant, for any point $p$ in $\mathbb{A}^1$ choose a point $p'$ in $Y$ above $p$ and a point $p''$ in $X$ above $p'$. If the map of fibers $X_{p}\rightarrow Y_{p}$ is flat in a neighborhood of $p''$ in $X_{p}$, then the map $X\rightarrow Y$ is also flat in a neighborhood of $p''$ in $X$.
I fail to see the obviousness of this interpretation: does this mean that if $R$ and $S$ are the respective affine rings defining the affine varieties $Y$ and $X$ over the field $k$, if $P'$ and $P''$ are the maximal ideals defining the points $p'$ and $p''$, if we have $S_{P''}$ flat over $R_{P'}$, there exist an element $f''$ of $S$ not contained in $P''$ such that $S_{f''}$ is flat above $R$? 
I mean that using the local criterion for flatness I see how I can get the flatness of the rings localized at maximal ideals coming from the flatness on the fibers, but how to extend it to a neighborhood of each points ? 
Edit: after re-reading the clear answer from Akhil Mathew, I cannot help but wondering if there is a way to get the geometric interpretation of Eisenbud without using the result on the open locus for flat maps which is above the level of chapter 6 of Eisenbud classical book. Can somebody enlighten me here?
Edit2: an interesting thread on Math.SE which gave me all the answers I needed using generic flatness instead https://math.stackexchange.com/a/2321347/14860

REPLY [7 votes]: The flat locus is open in any event under reasonable hypotheses (EGA IV-3, Th. 11.1.1), so being flat "at a point" and "in a neighborhood of a point" are equivalent. 
Also, this particular result is true more generally: if $f: X \to Y$ is a morphism of finite type $S$-schemes  ($S$ being noetherian), and if $X, Y$ are flat, then $f$ is flat if and only if each of the maps $f_s: X_s \to Y_s$ are flat. This follows from the fact that one direction of the local criterion of flatness is true under more generality: that is, if $B \to C$ is a morphism of local noetherian rings both local and flat over the local noetherian ring $(A, \mathfrak{m})$, then $B \to C$ is flat if and only if the fiber $B/\mathfrak{m} B \to C/\mathfrak{m}C$ is flat. See for instance Proposition 4.10 of this document.<|endoftext|>
TITLE: A product identity for partitions
QUESTION [8 upvotes]: For a partition $\lambda=(\lambda_1\ge \lambda_2\ge \dots)$, let 
$m_\lambda=\prod_i (\lambda_i-\lambda_{i+1})!$ be the product of factorials of consecutive differences and let $v_\lambda=\prod_{i | \lambda_i\neq 0} \lambda_i$ be the product of its nonzero terms. Then the following identity follows from representation theory of $S_n.$ $$\prod_\lambda m_\lambda=\prod_\lambda v_\lambda,$$ where both products are over partitions of an integer $n$. In fact for the examples I checked, it seems that if you write each side as a double product, the collection of terms on both sides is the same. However I can't see a combinatorial explanation for this fact. Does anyone know of one?

REPLY [6 votes]: Let me add a few remarks to Igor Pak's nice answer. First, notice that Hoare proves something stronger than the product identity, namely the observation that every number appears the same number of times on each side of the equation. In other words:

Theorem: (Stanley 1972, Elder 1984) The total number of occurrences of an integer k among all unordered partitions of n is equal to the number of occasions that a part occurs k or more times in a partition.

The case of $k=1$ is more popular, and I've seen it stated as an exercise in several occasions. Some people call the case $k=1$ Stanley's theorem and the general case Elder's theorem. Hoare gave a bijective proof, another independent proof was given by M.S Kirdar and T.H.R. Skyrme, "On an Identity Related to Partitions and Repetitions of Parts." Canad. J. Math. 34, 194-195, 1982 (Their proof uses generating functions).
One funny curiosity is that Stanley sent the general result above to the problem section of American Mathematical Monthly in 1972, but it was rejected as "a bit on the easy side, and using only a standard argument". Notice that Hoare's bijection was presented in the AMM in 1986.
One further generalization says that the number of occurrences of an integer k among all partitions of n is equal to the number of boxes among partitions of n whose hook-type is $(j,k-j-1)$. Setting $j=0$ recovers the theorem above. This was proved in C. Bessenrodt, "On hooks of Young diagrams, Ann. of Comb., 2 (1998), pp. 103–110 and in R. Bacher, L. Manivel "Hooks and Powers of Parts in Partitions",
Sem. Lothar. Combin., vol. 47, article B47d, 2001.<|endoftext|>
TITLE: What information can one recover from the induced map on homology?
QUESTION [7 upvotes]: The following question came up while constructing delay embeddings of time series data.
Consider an unknown topological space $X$ and an unknown continuous function $f:X \to X$. We are given a combinatorial representation $S_X$ of $X$ via a finite simplicial complex and an unknown isomorphism $\gamma_* $ from the simplicial homology $\text{H}^\Delta_*(S_X)$ (which is computable) to the singular homology $\text{H}_*(X)$. 
Similarly, instead of any direct knowledge of $f$, we have a simplicial map $\phi:S_X \to S_X$ and the commuting relation 
$$f_* \circ \gamma_* \equiv \gamma_* \circ \phi_*$$
where the star subscripts indicate maps induced on homology groups.
So from the matrix representations of $\phi_*$ one can deduce how $f$ maps cycles in $X$. Perhaps more non-trivially, one could compute the Lefschetz number of $f$ via the alternating sum of traces formula and deduce the existence of fixed points. Here's my question:

What else can one infer about $f$ from the $\phi_*$ matrices?

Do the determinants, characteristic polynomials and other matrix invariants of the $\phi_*$'s also carry useful information about $f$?
Update: Since a clarification has been requested, here are the details of how one might construct $S_X$ from $X$. To begin with, $X \subset \mathbb{R}^n$ is a $k$-dimensional Riemannian submanifold of Euclidean space. One assumes the existence of a finite point set $P \subset \mathbb{R}^n$. 
It is easy to see that if $P$ is sufficiently dense in $X$ then there is a radius $\epsilon > 0$ so that the union $U_\epsilon(P)$ of $n$-dimensional $\epsilon$-balls around points in $P$ covers $X$; in addition, if $\epsilon$ is small relative to the curvature of $X$, the map sending any point in $U_\epsilon(P)$ to its nearest point in $X$ is a strong deformation retraction, and so there is a homotopy equivalence between $U_\epsilon(P)$ and $X$.  
In the question, $S_X$ is the Cech nerve of the obvious cover $\lbrace B_\epsilon(p)~|~p \in P\rbrace$ of $U_\epsilon(P)$. The map $\gamma_*$ comes from the fact that $S_X$ has the homology isomorphic to that of $U_\epsilon(P)$ by the nerve theorem, and that $U_\epsilon(P)$ in turn has the same homology as $X$ via the retraction outlined above. So in particular, there is no need to worry about which ring the homology coefficients come from: any PID will suffice.

REPLY [2 votes]: When $X$ is a surface, then the moduli of the eigenvalues of $\phi_{\star}$ give lower bounds on the log of the dilatation of $f$, and therefore on the entropy of $f$. If we know additionally that the train track associated to $f$ is orientable, then the larger eigenvalue is the log of the dilatation. A more precise statement can be found in the following paper by Birman, Brinkman and Kawamuro http://arxiv.org/abs/1001.5094 where they associate several polynomials to $f$, one of them being actually associated to $\phi_\star$.<|endoftext|>
TITLE: Left adjoint of totalization?
QUESTION [9 upvotes]: There is a functor from bicomplexes to chain complexes sending a bicomplex to its associated total chain complex.  Does this functor have a left adjoint, and if so, what is it?

REPLY [4 votes]: It's not to hard to work out by hand what such an object would need to be if it existed.  By considering the case that $D$ is a bicomplex supported in degree $(m,n)$, the adjunction condition
$\operatorname{Hom}_{bicomplex}(L(C), D) = \operatorname{Hom}_{complex}(C, \operatorname{Tot}(D))$
imposes the condition that $L(C)$ have copies of $C_k$ in all degrees $(m,k-m)$.  Similarly, by examining the case where $D$ has a single nonzero arrow, we find that up to signs, the differential $L(C)_{m,n} \to L(C)_{m-1,n}$ is given by $C_{m+n} \to C_{m+n-1}$, and the same holds for the vertical arrows. 
As I understand it, there is a direct sum totalization, and a direct product totalization, and they have different adjunction properties.  In particular, if $D$ has large support, then the left side of the adjunction condition involves a countably infinite product in each degree.  While the direct product totalization yields an adjunction, the direct sum totalization will give you a dimension mismatch.
If you work with finite periodic complexes, the totalizations coincide, and the adjunction holds.
Since chain complexes are (up to some details about signs) representations of the $x \mapsto ax+b$ group $\mathbb{G}_a \rtimes \mathbb{G}_m$, the adjunction amounts to Frobenius reciprocity, and the problem here is quite similar to the problem of finding a well-behaved induction functor for passing from representations of this group to representations of its direct product with itself.  Since the group has positive dimension, the usual induction ends up being some kind of direct integral.  The direct product totalization seems to be some kind of completed restriction, but it is rather unclear to me.<|endoftext|>
TITLE: Are there infinite primes among powerful order terms of Dirichlet arithmetic progressions?
QUESTION [5 upvotes]: I am interested in primes and some of their varying characteristics which make them 
“qualitatively” different such as whether each prime generated by a certain type of Dirichlet arithmetic progressions belongs or not to the class of primes that are obtained from powerful order terms of that progression. Particularly, I wonder whether one can go beyond Dirichlet Theorem and establish that there is an infinite supply of this sort of primes in these progressions. 
The case I am particularly interested is the case when d=1 and a=2k with k positive integer, but I would like to formulate the following Conjecture for the general case  
For any two coprime positive integers a and d, there are infinite primes of the form $an^2m^3+d$ with n and m integers >=0
If true this would strengthen Dirichlet theorem on arithmetic progressions selecting only the terms of the progressions whose order is a powerful number (a number that if prime p divides it, $p^2$ also divides it).
I wonder about the likeliness of the conjecture (which intuitively seems high) and whether there is any proof I am still not aware of. I welcome estimates on the toughness of the proof, as well as any hints towards its development.

REPLY [3 votes]: I'm sorry this isn't much of an answer, but you might be interested in this paper of De Koninck, Kátai and Subbarao.  Your conjecture is almost strong enough to mean that for any squarefree $a$, there are infinitely many primes $p$ where the squarefree part of $p-d$ is exactly $a$ (the difference being that you do not require $(a,mn)=1$).
Section 4 in the cited paper establishes a sort of converse formulation: for any powerful $K$, there are infinitely many primes $p$ where the powerful part of $p-1$ is exactly $K$ (with the expected asymptotics).  Of course this is a far denser set in which to look for primes, so this is certainly a much easier problem.  Presumably no significant complications arise if we shift this by $d$ rather than $1$.<|endoftext|>
TITLE: Is a finite CW complex minus a point still homotopy equivalent to a finite CW complex?
QUESTION [17 upvotes]: Let $X$ be a finite CW complex and $x_0$ a point in $X$.
My question is then just:
Is $X-\{x_0\}$ still homotopy equivalent to a finite CW complex?

REPLY [45 votes]: The answer to your question is no. Here is a counterexample.
Let $X$ be the CW-complex obtained by attaching a 2-cell to the space $[-1,1]$ via the attaching map
$S^1\cong [-1,1]/(-1\sim 1) \longrightarrow [-1,1]$ given by (the continuous extension of) $x\mapsto x\sin(1/x)$.
Then $X\setminus \{0\}$ is homotopy equivalent to an infinite wedge of $S^1$'s.<|endoftext|>
TITLE: Existence of points on varieties which avoid a given number field.
QUESTION [13 upvotes]: Let C be a geometrically integral curve over a number field K and let K' be a number field containing K. Does there exist a number field L containing K such that

$L \cap K' = K$, and
$C(L) \neq \emptyset$?

Note that the hypotheses on C are necessary -- the curve x^2 + y^2 = 0, with the origin removed, is not geometrically integral, but gives a counterexample for K = Q and K' = Q(i).
Also, I can prove that this is true when C has prime gonality. It would be odd, though, for this to be a necessary hypothesis.

REPLY [8 votes]: I think this is a consequence of (variants of) Hilbert's irreducibility theorem. Let me explain why. Suppose that $C$ is a geometrically integral curve defined over a number field $K$. Let $K'/K$ be a normal finite extension. We will show that infinitely many points of $C$ are defined over a field disjoint from $K'$.
Since both the hypotheses and the conclusion are birational invariants, we may suppose that $C$ is a closed subset of $\mathbb{A}_K^2$ (take an affine open of $C$, embed it in $\mathbb{A}^N$ for some $N$ and take a generic projection to $\mathbb{A}^2$). 
Choose a generic projection $p:C\to\mathbb{A}^1_K$. The curve $C$ is described by an equation $F(t,x)=0$, where $t$ is the coordinate of $\mathbb{A}^1_K$, and $F$ is an irreducible polynomial.
Now, since $C$ is geometrically integral, $F_{K'}$ is still irreducible. By [Serre, Topics in Galois theory, Proposition 3.3.1], $x\mapsto F_{K'}(\lambda',x)$ is irreducible for every $\lambda'\in K'$ outside of a thin set. Hence, by [Serre, Topics in Galois theory, Proposition 3.2.1], $x\mapsto F_{K'}(\lambda,x)$ is irreducible for every $\lambda\in K$ outside of a thin set. Since $K$ is Hilbertian, this holds for infinitely many $\lambda\in K$.
Let us fix such a $\lambda$. We denote by $q$ and $q'$, the points of $\mathbb{A}^1_K$ and $\mathbb{A}^1_{K'}$ with coordinate $\lambda$. By choice of $\lambda$, $x\mapsto F_{K'}(\lambda,x)$ hence also $x\mapsto F(\lambda,x)$ are irreducible polynomials. Hence
$p_{K'}^{-1}(q')\subset C_{K'}$ (resp. $p^{-1}(q)\subset C$) consists of a unique (reduced) point
$p'\in C_{K'}$ (resp. $p\in C$). Let $L$ and $L'$ be the residual fields of $p$ and $p'$.
By construction, $p'=p\times_{q} q'$ so that $L'=L\otimes_K K'$. This implies that $L$ is disjoint from $K'$.<|endoftext|>
TITLE: Distribution of the maximum of the norm of k-averages of n i.i.d. d-dimensional random vectors
QUESTION [10 upvotes]: Suppose $X_1, ... X_n$ are i.i.d. random vectors in $d$-dimensional space (i.e., $R^d$) with continuous centrally symmetric density function $f(\cdot)$ (i.e., symmetric with respect to the origin). For concretenes's sake, assume $X_i$ is just a multivariate normal distribution with covariance matrix equal to the identity.
What is the expectation (as a function of $d$, $k$ and $n$) of the maximum value of
$||X_{i_1}+...+X_{i_k}||^2$ where the maximum is taken over all possible $1 \leq i_{1} < i_{2} < \ldots < i_{k} \leq n$ and $||\cdot||$ denotes the Euclidean norm? 
Are large deviation bounds known? In general, what is the relation between $f(\cdot)$ and the expectation of the maximum?

REPLY [2 votes]: The problem can be solved if the distribution $f(\cdot)$ is in a Levy stable distribution family. In your concrete example, since the $d\dim$ normal distribution $N(\mu,I_d)$ is "additive", the exact distribution of the $k$-sum $X_{i_1}+\cdots+X_{i_k}$ is $N(k\mu,kI_d)$. The choice of these $k$ random variables $X_{i_1},\cdots,X_{i_k}$ does not affect the mean nor covariance since the $X_i$'s are i.i.d. So it suffices to derive the distribution of $\|Y\|^2$ for a $d$-dimensional $Y\sim N(k\mu,kI_d)$, and there are $\left(\begin{array}{c}
n\\k\end{array}\right)$ possible combinations of one realization value of $\|Y\|^2$ if $Y=X_{i_1}+\cdots+X_{i_k}$ is the sum of an i.i.d. sample of size $k$ therefore the probability $Pr(max\|X_{i_1}+\cdots+X_{i_k}\|^2=y)$ will be $\left(\begin{array}{c}
n\\k\end{array}\right)\cdot f_{\|Y\|^2}(y)\cdot F_{\|Y'\|^2}(y)$. The maximum is taken over all possible $1 \leq i_{1} < i_{2} < \ldots < i_{k} \leq n$, same below. The $Y'=\sum_{j\notin\{i_1,\cdots,i_k\} }X_j$ follows a similar derivation but $F$ is the c.d.f.
To calculate the latter $f_{\|Y\|^2}(y)$, we need to notice that $\frac{Y}{\sqrt{k}}\sim N_d(\sqrt{k}\mu,I_d)$, we can compute its $L^2$ norm $\frac{1}{k}\sum_iY^2_i=:\|\frac{Y}{\sqrt{k}}\|^2\sim\chi^2(d)$. Thus the p.d.f. $f_{\|Y\|^2}(y)=f_{\chi^2(d)}(ky)$
To draw a conclusion, $Pr(max\|X_{i_1}+\cdots+X_{i_k}\|^2=y)=\left(\begin{array}{c}
n\\k\end{array}\right)\cdot f_{\chi^2(d)}(ky)\cdot F_{\chi^2(d)}((n-k)y)$ in case of $d\dim$ multivariate normal as you indicated in OP. For other families of $f(\cdot )$ closed under addition, we can proceed the same way, and the resulting density may not be of closed form if we do not know one under square transformation (in contrast that we know $\chi^2(d)$ in above.).<|endoftext|>
TITLE: Projective modules over non-rational group rings
QUESTION [6 upvotes]: Let $G$ be a finite group. We know that the $K$-group $K_0(QG)$ of the rational group ring $QG$ is a free abelian group generated by the irreducible representations of $G$ over $Q$. Now let $R$ be a subring of the rationals where the order $|G|$ is invertible. What is the relation between $K_0(RG)$ and $K_0(QG)$?

REPLY [4 votes]: I would guess that the map on $K_0$ is an isomorphism, butI could only show the surjectivity right now:
The inclusion of rings $RG\rightarrow QG$ induces a map on $K_0$. Given a projective $RG$-module - say it is a submodule of $RG^n$ - to its $Q$-span. It is a projective submodule of $QG^n$.
So let us first show that this map is surjective, ie. every projective $QG$ module arises this way. Given any such $P'$ the obvious candidate for a preimage would be $RG^n\cap P'$.
First note that it is as a $R$-module a direct summand of $RG^n$. $R$ is a PID and hence one just has to verify that the quotient is $R$-torsionfree. But $RG^n/(RG^n\cap P')$ embeds into the $Q$-vectorspace $QG^n/P'$ and hence it is $R$-torsionfree.
So we have a section of $R$-modules $s:RG^n/(RG^n\cap P')\rightarrow RG^n$. It need not be a $RG$-map. So let us make it equivariant by setting $s'(x):=\frac{1}{|G|}\sum_{g\in G}gs(g^{-1}x)$. Note that it is still a section (project down again; it is a $RG$ map).
So we have found a $RG$-complement of $RG^n\cap P'$; hence $RG^n\cap P'$ is a projective $RG$-module. So the map $K_0(RG)\rightarrow K_0(QG)$ is surjective.<|endoftext|>
TITLE: $2$-categorical structure in Grothendieck's Galois Theory
QUESTION [18 upvotes]: Grothendieck's Galois Theory, as developed in SGA I, V.4, or very gently in Lenstra's notes, establishes an equivalence between profinite groups and Galois categories. We can put this into the following more general framework: Let $\mathsf{sets}$ denote the category of finite sets; for a topological group $\pi$ let $\pi\mathsf{-sets}$ denote the category of finite sets on which $\pi$ acts continuously. There are functors
$\mathsf{Cat}/\mathsf{sets} \leftrightarrow \mathsf{TopGrp}^{\mathrm{op}}$
given by $(F : C \to \mathsf{sets}) \mapsto \mathrm{Aut}(F)$ (which is topologized as a closed subgroup of $\prod_{X \in C} \mathrm{Aut}(F(X))$) and in the other direction by $\pi \mapsto \pi\mathsf{-sets}$. In order to avoid set-theoretic problems, $C$ and $\pi$ should be essentially small. There are canonical maps $\eta_C : C \to \mathrm{Aut}(F)\mathsf{-sets}$ and $\varepsilon_{\pi} : \pi \to \mathrm{Aut}(\pi\mathsf{-sets})$ which satisfy the triangular identities; hence we have an adjunction! And every adjunction restricts to an equivalence of categories between its fixed points. Now it's fairly easy to recognize profinite groups as those fixed points on the right hand side, but Grothendieck's nontrivial insight is the classification of the fixed points on the left hand side, which he calls Galois categories (in short: $C$ has finite limits and colimits, which $F$ preserves, there are mono-epi decompositions, monos split off, and $F$ is conservative).
Question 1. Does this point of view of Grothendieck's Galois theory have already appeared somewhere?
My real question is the following: Actually $\mathsf{Cat}/\mathsf{sets}$ is a $2$-category. A morphism $(C,F) \to (C',F')$ is a functor $P : C \to C'$ together with a chosen isomorphism $F \cong F' P$. A $2$-morphism between morphisms $P,Q : C \to C'$ is a natural transformation of the underyling functors, which is base-point preservering in the obvious sense. 
Question 2. How can we endow $\mathsf{TopGrp}^{\mathrm{op}}$ with the structure of a $2$-category in such a way that the adjunction above becomes an $2$-adjunction?
- - Edit - - The comments + answers make me believe that it is just the "trivial" $2$-categorical enrichment with identities as $2$-morphisms. And since there is no reaction to Q1, I expect that the answer is "no, this is new"?

REPLY [6 votes]: EDIT: answer is expanded on OP's request.
The category $\pi Set_{fin}$ of finite, continuous $\pi$-sets is just the functor category $TopCat(\mathbf{B}\pi,finSet)$ where $\mathbf{B}$ gives the one-object topological groupoid associated with a topological group, and $finSet$ is the category of finite sets, viewed as a topologically discrete category. The groupoid $\mathbf{B}\pi$ is canonically pointed, so induces the functor $\pi Set_{fin} \to finSet$ i.e. an object of $Cat/finSet$. If you replace $TopGrp$ with the equivalent category $TopGpd_{1obj}$ of one-object topological groupoids, then this naturally comes with a notion of 2-arrow, and it is, as John Baez points out, conjugation by an element of the codomain. Then 2-arrows in $\pi Set_{fin}$ are natural transformations of functors to $finSet$. 
Now really you are working with $TopGrp^{op}$, so you have two choices as to the direction of the 2-arrows, if you are taking 2-arrows as specified above. This should fall out of the definitions. Also, since $TopGpd_{1obj}$ is a (2,1)-category, you need to restrict attention to the (2,1)-category $Cat/finSet_{(2,1)}$ underlying $Cat/finSet$. Here $Cat/finSet_{(2,1)}$ is the isocomma category: objects are categories over $finSet$, arrows are triangles commuting up to a natural isomorphism and 2-arrows are natural isomorphisms in $Cat$ that are compatible with the 2-arrows in the triangles. 
In more detail: given two objects $F\colon C\to finSet$ and $G\colon D\to finSet$, an arrow $F\to G$ is a pair consisting of a functor $f\colon C\to D$ and a natural isomorphism $c\colon G\circ f \Rightarrow F$. Given two arrows, $(f,c),(g,d)\colon F\to G$, a 2-arrow between them is a natural isomorphism $a\colon f\Rightarrow g$ such that the obvious 3-dimensional diagram commutes. Since all 2-arrows are invertible, from this commuting 3-d diagram we can write down an invertible endo-arrow $c+(G\circ a)+d^{-1}$ of $F$ (here $+$ is vertical composition of natural transformations--from right to left--$G$ really denotes the identity transformation on the functor $G$ and $\circ$ is horizontal composition).
I'm fairly confident that this is a 2-arrow in $TopGpd_{1obj}$. If one doesn't restrict to the (2,1)-category $Cat/Set_{(2,1)}$ then this doesn't work. (So really you should be thinking about $Gpd$-enriched categories, not so much general 2-categories.)
EDIT: Actually, it is easy to see that this is a 2-arrow in $TopGpd_{1obj}$, because the homomorphism induced by $(f,c)$ is conjugation by $c$ (with whiskering): $\alpha \mapsto c+(f\circ \alpha)+c^{-1}$.
Thus we have a 2-functor $Cat/finSet_{(2,1)}\to TopGpd_{1obj}^{op}$.
Now in going the other way, I think we actually need to use not just $TopGpd_{1obj}$, but the isococomma category $\ast/TopGpd_{1obj}$, where a functor between topological groupoids respects the basepoint up to a 2-arrow (which is an automorphism of the canonical basepoint). I haven't checked but this looks like the description of the functor in the preceeding paragraphs works better. Then the functor $\ast/TopGpd_{1obj}^{op} \to Cat/finSet_{(2,1)}$ is just exponentiation with $finSet$, and we don't have to think too hard about what the 2-arrows etc do.
Then you need to worry about the adjunction. But if you already have a 1-adjunction, half the work is done.<|endoftext|>
TITLE: Klein's Protocols: A window into our mathematical past
QUESTION [26 upvotes]: Klein's Protocols in over 8,000 pages recording seminars organized from 1872 to 1913 by Felix Klein and given by Klein, his colleagues, students and other invited speakers, including luminaries such as Hilbert and Minkowski, provide a unique window into our mathematical past. Eugene Chislenko and Yuri Tschinkel have presented a beautiful introduction to the corpus, noting how the breadth of the topics reflect the broad interests and knowledge of Klein (the authors claim that Klein was one of the last three grand mathemagicians able to soar over the full realm of the mathematics of their times, the other two being Hilbert and Poincare).
Klein 's Protocols is huge and handwritten in German, so I thought it would be helpful for aspiring and established mathematicians with an interest in the history of ideas if a listing were available on Math Overflow of some gems in the corpus. Can you make a contribution? 
Please provide keywords and page references for any entry. E.g., personally, I would love to have a copy of the figure titled "On regular solids in 4-dimensional space" by W. I. Stringham in Vol. II on pg. 59 (Monday, November 29, 1880).
See also these articles (pages 16-21).

REPLY [18 votes]: I compiled an index of the Protokolle which can be found here:
http://page.mi.fu-berlin.de/moritz/klein/
I transcribed all the titels, names, dates and page numbers. You can do a case insensitive search. You can also view the scanned version there.
I hope this makes it easier to find the "gems in the corpus".<|endoftext|>
TITLE: Important open questions in the field of Tropical geometry
QUESTION [26 upvotes]: What are some of the important unanswered questions in the field of tropical geometry?

REPLY [8 votes]: Three notes:
(1) There was an earlier MO question on "Learning Tropical Geometry."
(2) Bernd Sturmfels is giving a series of three lectures on "Algebraic Geometry: Tropical, Convex, and Applied" at the MathFest conference next week.
(3) Most directly to your question, see the engaging and succinct 2006 2-page manifesto, "Ten Questions in Tropical Geometry," PDF download:

Question 1. Is it possibly true that every tropical variety is shellable?
Question 2. Give a good formulation for the moduli space of curves of degree
  $d$ and genus $g$ lying in $\mathbb{TP}^2$.
Question 3. Investigate matroid subdivisions.
Question 4. Classify all “root system polytopes”,...
Question 5. Compute the (positive) tropical flag variety $GL_4/B$ in its Plücker embedding.
Question 6. What do the face lattices of tropical polytopes look like?
Question 7. When does tropicalization commute with intersection?
Question 8. What is the best axiomatization of tropical oriented matroids?
Question 9. Is the tropical discriminant of a defective point configuration a
  subfan of the secondary fan?
Question 10. How can you tell if something is a tropical variety?<|endoftext|>
TITLE: Is $M_g$ finitely covered by a scheme over the integers?
QUESTION [27 upvotes]: This question was prompted by my almost-answer to the question Does smooth and proper over $\mathbb Z$ imply rational? , but I never got around to asking this until now.
It is well known that $M_g$, the moduli space of curves, is a smooth stack. In fact it is globally a quotient of a smooth variety by the action of a finite group: if we take curves with level $n \geq 3$ structure, then we get a fine moduli space, and the quotient by $\mathrm{Sp}(2g,\mathbf{Z}/n)$ recovers $M_g$. However, we know that $M_g$ is in fact smooth over $\mathrm{Spec}(\mathbf Z)$, but this construction does not work over the integers: we need $n$ not divisible by the characteristic. This leads to the first question:
Question (a): Could it be true that $M_g = [X/G]$ where $X$ is a scheme, smooth over the integers, and $G$ is a finite group?
It is very possible that this question is too naive. Maybe there is a standard argument for why this can not be true, involving structure of group schemes and wild inertia and whatnot. Nevertheless one can ask for something weaker:
Question (b): Is there a scheme $X$, smooth over the integers, with a finite map $X \to M_g$? What if we only ask it to be proper and generically finite?
A (presumably harder) question is what happens over the boundary, i.e. if one replaces $M_g$ with $\overline M_g$ or $\overline M_{g,n}$. As I mentioned in my answer to the question linked above, it is known that over certain $\mathrm{Spec}(\mathbf Z[\frac 1 d])$ one can write $\overline M_g$ and $\overline{M}_{g,n}$ as global quotients by actions of finite groups, now using non-abelian level structures.

REPLY [3 votes]: Here's a suggestion in the genus $g=2$ case. If one considers a genus 2 Riemann surface, then it is well-known that it is hyperelliptic, and the fixed points of the hyperelliptic involution are 6 Weierstrauss points. If one quotients by the hyperelliptic involution, the quotient is a Riemann sphere with 6 distinguished points which are the images of the Weierstrauss points. So moduli space is isomorphic to the space of 6 points on a sphere, up to conformal equivalence. There is a finite cover of this, which is the space of 6 marked points, up to conformal equivalence. Thus, moduli space is 
$$ S_6 \backslash ((\mathbb{CP}^1)^6- \Delta)\ /\ PSL(2,\mathbb{C}),$$
where $\Delta$ denotes the large diagonal, and is cut out by $\Delta = \{ (z_1,\ldots, z_6) | z_i \neq z_j\ for\ i\neq j\}$, and $S_6$ acts by permuting coordinates and $PSL(2,\mathbb{C})$ acts coordinatewise . Since $PSL(2,\mathbb{C})$ acts faithfully on ordered triples, we may normalize the last three coordinates to be $0, 1, \infty$, so that $$ ((\mathbb{CP}^1)^6- \Delta) / PSL(2,\mathbb{C}) \cong  \{ (z_1,z_2,z_3)\in \mathbb{C}-\{0,1\} | z_i\neq z_j, i\neq j\}.$$
This is isomorphic to an affine variety defined over $\mathbb{Z}$ by the usual trick of introducing three new coordinates and equations $(z_i-z_j)t_{ij}=1$, $i\neq j \in \{1,2,3\}$ and has a quotient by $S_6$ which is isomorphic to $M_2$. The group elements of $S_6$ act as integral fractional transformations. 
So the suggestion is to take the affine scheme defined by the spectrum of the coordinate ring of this affine variety defined over $\mathbb{Z}$. But I don't know enough about schemes or stacks to know if this works for what you want to do.<|endoftext|>
TITLE: The anticanonical bundle on a flag variety is ample
QUESTION [9 upvotes]: Hello,
I would like to get references or answers, for the following. How do I show that the anti-canonical line bundle (i.e. dual to top wedge power of cotangent bundle) on a flag variety (of a semisimple algebraic group in char. 0) is ample?
Thank you,
Sasha

REPLY [4 votes]: Let me try to add to Jim Humphreys's comprehensive answer by pointing out that the (very) ampleness of line bundles coming from regular dominant weights was already observed by Borel and Weil, and essentially appears as Thm.4 in Serre's Bourbaki report

Représentations linéaires et espaces homogènes kählériens des groupes de Lie compacts (d'après Armand Borel et André Weil). Séminaire Bourbaki, Vol. 2, Exp. No. 100, 447–454, Soc. Math. France, Paris, 1995.

(See the example following Thm.4 on p.453.)
Although Borel and Weil work over $\mathbb C$ their argument is probably worth summarizing here. They begin by observing that the line bundle $L_\lambda \to G/B$ coming from a weight $\lambda$ is spanned by global sections iff $\lambda$ is dominant. In this case the image of the map $G/B \to \mathbb P (H^0(G/B,L_\lambda)^\ast)$ defined by sections is of the form $G/P$ where $P\supset B$ is the parabolic subgroup defined by the simple roots perpendicular to $\lambda$. Thus if $L_\lambda$ is not very ample, so that the aforementioned fibration $G/B \to G/P$ is not an embedding, one concludes that $P$ must properly contain $B$ and consequently that $\lambda$ must be perpendicular to some simple root, hence is not regular.<|endoftext|>
TITLE: Are any two Dirichlet domains for a Fuchsian group "comparable"?
QUESTION [9 upvotes]: Let $\Gamma$ be a [EDIT: finitely generated] Fuchsian group of the first kind (i.e. a discrete subgroup of $PSL_2(\mathbf{R})$ acting on the upper half-plane admitting a fundamental domain of finite hyperbolic area). Let's say two fundamental domains $D$ are comparable if each one is contained in a finite union of $\Gamma$-translates of the other.
I was quite shocked to learn that two different fundamental domains needn't be comparable: you can take the standard fundamental domain for $PSL_2(\mathbf{Z})$ and give it an infinite sequence of longer and longer triangular "teeth" sticking out sideways, with corresponding indentations on the other side -- then this won't be comparable with the usual domain.
There's a standard "nice" class of fundamental domains, though. For any $x_0$ that's not an elliptic point, there is the Dirichlet domain with centre $x_0$, given by the set of points closer to $x_0$ than to any other $\Gamma$-translate of $x_0$.
Is it true that any two Dirichlet domains are comparable?

REPLY [6 votes]: Here are some details of the solution patched together in the comments. There is an implicit assumption that $\Gamma$ is finitely generated, else it does not have a finite sided fundamental domain. Pick a $\Gamma$-equivariant system of pairwise disjoint horoball neighborhoods $B_\xi$ of the cusps $\xi \in \partial \mathbb{H}^2$. 
Suppose that $D$ is a finite-sided fundamental domain for $\Gamma$ (bounded by geodesic paths), for example a Dirichlet domain or a Ford domain. Consider a cusp $\xi$ on which $D$ accumulates. Since $D$ is finite sided, the only way it can accumulate on $\xi$ is for there to exist a concentric horoball $B' \subset B_\xi$ such that $D \cap B'$ is the region of $B'$ between two rays (taking this concentric horoball is necessary to avoid the parts where $D$ mucks around inside $B_\xi$ doing unpleasant things close to the boundary of $B_\xi$). After equivariantly shrinking the horoballs, we can assume that $D$ hits each horoball in this standard manner.
Now suppose that $D_1,D_2$ are two such fundamental domains. After further equivariant shrinking of the horoballs, we can assume that both $D_1$ and $D_2$ hit each horoball in the standard manner. It's now easy to prove that $D_1,D_2$ are comparable: their portions outside the horoballs each have compact closure and so are comparable; and if we pick a representative cusp $\xi$ of each $\Gamma$-orbit of cusps, their portions inside $B_{\xi'}$ for cusps $\xi'$ in the $\Gamma$-orbit of $\xi$ translate to a finite union of standard intersections with $B_{\xi'}$.<|endoftext|>
TITLE: Generating the graded S_n-module associated to an operad
QUESTION [8 upvotes]: Suppose I have a symmetric operad $\mathcal{P}$ defined over $\text{Vect}_{\mathbb{K}}$ with generators and relations in degrees at most $l$.  Now suppose I already know $\mathcal{P}(k)$ as an $S_k$-representation for $0 \leq k \leq l$.

How do I determine $\mathcal{P}(l+1)$ from $\mathcal{P}(0), \mathcal{P}(1), \ldots, \mathcal{P}(l)$ ?  More generally, how do I determine $\mathcal{P}(k)$ for some large $k$?

My current technique is to find vector space generators for each grade of the operadic ideal of relations by building trees in which some relation appears.  Then I use linear algebra and Schur-Weyl duality to determine each grade of the ideal as a representation.  This technique is reliable for small cases, but the number of trees grows large and this generating set is usually quite redundant.
I am hoping (perhaps naively) for some combinatorial rule in terms of Young diagrams which can answer the question explicitly.

Edit: Vladimir Dotsenko suggests that I limit the scope, so here's a (more reasonable) question which is also interesting:

Assume further that $\mathcal{P}$ is generated by a single associative binary operation.  Now can we determine $\mathcal{P}(k)$ from an initial segment using the combinatorics of Young diagrams?

It seems like the Littlewood-Richardson rule might be of some help here.  There also seems to be some plethysm going on.  Perhaps these tools are sufficient?

REPLY [7 votes]: Operads that are quotients of the associative operad (which I believe your updated question is aimed at) are, in characteristic zero, fairly extensively studied since the 1950s or so, under the somewhat misleading for geometers name "varieties of associative algebras". (One of the first papers that sort of created this field of study is, to the best of my knowledge, Specht's  "Gesetze in Ringen", http://www.ams.org/mathscinet-getitem?mr=35274.) The main bit of knowledge one needs to translate papers on varieties of algebras into the operadic language are 
"variety of associative algebras" = "quotient of the associative operad" 
and
"codimensions of a variety"="dimensions of components of an operad".
Armed with this knowledge, have a look at recent surveys, for example the book of Giambruno and Zaicev (Google Books link: http://tinyurl.com/giambrunozaicev), and you will see many hints on how to approach this kind of questions. In all fairness, the main question people in this area are interested in is asymptotic growth rather than exact formulas, but on the way lots of things are done, including some representation-theoretic computations, so you might find what you need. 
P.S. Feel free to e-mail me if you are interested in some particular operads, maybe I happen to know precisely what you want to find out.<|endoftext|>
TITLE: First nonzero eigenvalue of the Laplacian on the submanifold
QUESTION [6 upvotes]: Consider a compact, connected $n$ dimensional Riemmanian manifold $\mathcal{N}$ and its $m$ dimensional closed submanifold $\mathcal{M}$ (with the metric coming from from the one defined on $\mathcal{N}$).
Question:
Is there any way to give a lower bond for the first nonzero eigenvalue of Laplace- Beltrami operator  ($\Delta f = - div(grad f)$) defined on  $\mathcal{M}$? In particular, is such lower bound related to the first nonzero eigenvalue of the Laplacian on $\mathcal{N}$? 
I am especially interested in in the case when $\mathcal{M}$ is of "small" codimension and is given by the intersection of level sets of some smooth functions defined on $\mathcal{N}$. The original motivation for this question comes from the problem of "incheritance" of measure concentration by $\mathcal{M}$, when it is known that it occours for $\mathcal{N}$.

REPLY [10 votes]: In the general case (especially for high codimension) there will not be such a relation: Every compact Riemannian manifold can be isometrically embedded into the Euclidean space $R^N$. As the image of the submanifold is compact, you can change the metric "near infinity" such that it defines a metric on the sphere $S^N.$
But in special cases, there might be a relation, see for example Yau's conjecture: The first non-zero eigenvalue of a compact embedded minimal hypersurface in (the round) $S^n$ is n.
You should take a look into the paper of S. Montiel & A. Ros, Minimal immersions of surfaces by the first eigenfunctions and conformal area, Invent Math.<|endoftext|>
TITLE: Journals and other sources with "easy reading" papers?
QUESTION [43 upvotes]: Some time ago the journal "Algebra and Analysis" (English translation is published in 
"St. Petersburg Mathematical Journal") had a special section which was called "easy readings for professional mathematicians", which tried to present in accessible and interesting way some ideas. It seems now this section is discontinued unfortunately.
So my question is the following: are there some journals which publish "easy readings"? Or may be there are some collections somewhere else on web? 
PS
May be some recommendations for concrete "easy reading" papers are also valuable.
PSPS
As a kind of bigger and vaguer ingredient of the question, let me add the following:
do you feel some over-representation of some hard-to-read papers over some beautiful
and short papers? If mathematics is "mainly about follow-ups" it seems it makes progress more difficult, can something be changed or nothing ?  One of reasons is probably the gap between "thinking and explaining", however as Gil Kalai answered he does not feel much gap, in part due to "Part of the reasons is being involved in semi-formal Internet acticities like blogs, polymath projects, MO, etc., ", this I think gives some hope that 
giving a bigger weight to user-friendly ways of communication may improve readability of papers...

REPLY [3 votes]: There are prizes for mathematical exposition, scoped to a single work, which I think provide possible sources of easily-readable material.
For example:

The AMS's Steele prize for exposition
The MAA's Chauvenet prize

REPLY [2 votes]: The Newsletter of the EMS (as well as the Gazette des mathématiciens, usually in French) often contains good exposition. The EMS also recently launched a journal for survey articles (but not always as broadly readable than your target, I think).<|endoftext|>
TITLE: Computational complexity of calculating the nth root of a real number
QUESTION [7 upvotes]: Several sources state that the computational or time complexity of square rooting is the same as that of multiplication (or division). See for example:

Jean-Michel Muller, "Elementary Functions: Algorithms and Implementation" (Birkhäuser, Boston, USA, 2006, 2nd edn.) on p.93 where the Newton-Raphson method is considered. It is stated that "...the complexity of square root evaluation or division is the same as that of multiplication..."
H. Alt, "Square rooting is as difficult as multiplication", Computing, 1979, 21, (3), pp.221-232. It is stated on p.231 "...that we have shown that square rooting is of the same complexity as the other arithmetical operations mentioned. The proof can be extended to higher roots, as well."

Although there are several different approaches to implement computation of the elementary operations (+,-,x,/) it is therefore possible to implement square rooting such that the complexity is equivalent to an implementation of multiplication (a single multiplication of 2 real numbers).
What does "the proof can be extended to higher roots" mean? Is it possible to implement the calculation of the nth root with complexity equivalent to a single multiplication or to the multiplication of $n$ real numbers? If not, what is the complexity of calculating the nth root of a real number?

REPLY [12 votes]: First, note that the asymptotic complexity of arithmetic operations stated in the common literature concerns operations on numbers with arbitrary precision, and the running time is expressed as a function of the desired number of digits. From the standpoint of asymptotic complexity it makes no sense to ask for operations with constant precision (e.g., single floats, as you mentioned in the comments): there are only $O(1)$ such numbers, hence the operation can be evaluated in time $O(1)$ (e.g., by a look-up table).
Let me thus denote the desired precision as $m$ (since you use the customary $n$ for something else). The result you quote is that $\sqrt a$ can be computed in time $O(M(m))$, where $M(m)$ is any function (satisfying some mild regularity conditions) such that multiplication of two $m$-bit integers can be performed in time $M(m)$. (The currently known asymptotically fastest multiplication algorithm has $M(m)=m\log m\,2^{O(\log^*m)}$.) The algorithm uses Newton iteration $x\mapsto x-\frac{x^2-a}{2x}$. This iteration has a quadratic rate of convergence, hence $O(\log m)$ iterations suffice, and each step takes $O(1)$ multiplications and divisions, leading to the estimate $O(M(m)\log m)$ on the total running time. The extra factor of $\log m$ can be removed by the following observation: since the number of correct digits is roughly doubled by each iteration, we do not have to perform all operations with precision $m$, it suffices to use precision sufficient to accomodate the correct digits. Thus only the last iteration is performed with precision $m$, the last but one has precision $m/2$, the one before that $m/4$, and so on. Then the running time is $O(M(m)+M(m/2)+M(m/4)+\cdots)$. Since $M$ is essentially linear, this is can be bounded by a geometric series, whose sum is $O(M(m))$. (Note by the way that the fact that division can be done in time $O(M(m))$ also uses a similar Newton iteration argument.)
Now, what about $n$th roots in general? You can use Newton iteration again, as Denis suggests. The analysis is similar to the square root case, but since each step takes $O(\log n)$ multiplications, you get a bound $O(M(m)\log n)$. Note that if $n$ is given in binary, $\log n$ is the length of the input, hence this is an algorithm with worse than a quadratic running time. Another approach is to compute $\sqrt[n]a$ as $\exp((\log a)/n)$. Using binary splitting, the Taylor series for $\exp$ and $\log$ can be evaluated in time $O(M(m)(\log m)^2)$; using algorithms based on the arithmetic-geometric mean this can be reduced to $O(M(m)\log m)$, leading to $n$th root computation with the same time bound. This also has an extra $\log$ factor, but it is independent of $n$. I don’t know how to compute $\sqrt[n]a$ in time $O(M(m))$, and I am somewhat skeptical that such a thing is known. It might well be that the comment in Alt’s paper is only intended to cover the case of constant $n$.<|endoftext|>
TITLE: connections between Grothendieck's and Serre's duality
QUESTION [5 upvotes]: Hi, 
I would like to show that 
if $f: X \rightarrow Y=Spec \, \mathbb{C}$, where $X$ is a nonsingular complex projective variety, is the projection to a point, then the complex $f^! \mathcal{O}_Y$, appearing in Grothendieck's duality, is the dualizing sheaf for $X$. Let $F$ be a coherent sheaf on $X$. Starting from
$ Hom_{\mathcal{O}_X}(F, f^! \mathcal{O}_Y) \simeq  Hom_{{O}_Y}(Rf_* F, \mathcal{O}_Y) $
applying the cohomology functor $ H^i $ we obtain
$Ext^i(F, f^! \mathcal{O}_Y) \simeq Ext^i(Rf_* F, \mathcal{O}_Y).$
Using Yoneda's Formula, the right term becomes
$Hom^i_{D(Y)}(Rf_* F, \mathcal{O}_Y) \cong Hom_{D(Y)}(Rf_* F, \mathcal{O}_Y[i]) \cong Hom(\widetilde{H^1(X,F)}[-i], \mathcal{O}_Y),$ where, for the last isomorphism, I use Theorem 8.5 from Hartshorne'a Algebraic Geometry, p.251.
Now, by Corollary 5.5 pag.151 from Hartshorne, and remembering that $\Gamma(Y, \mathcal{O}_Y)= \mathbb{C}$, the last term is equal to $Hom(H^1(X,F)[-i], \mathbb{C}) = H^{1-i}(X,F)'$. 
Now, we have $Hom^i_{D(X)}(F, f^! \mathcal{O}_Y) \cong H^{1-i}(X,F)'$ and, shifting by $(-n+1-i)$, $Hom(F, f^! \mathcal{O_Y}[-n+1]) \cong H^n(X,F)'$. So, $f^!(\mathcal{O}_Y)[-n+1]$ is a dualizing sheaf for $X$. But we know that, for a nonsingular projective variety the (unique) dualizing sheaf is the canonical sheaf $\omega$. Thus, we must have $f^!\mathcal{O}_Y[-n+1] \cong \omega$, then $f^! \mathcal{O}_Y = \omega[n-1]$.
I should obtain $f^! \mathcal{O}_Y = \omega[n]$... what's wrong?
Thank you

REPLY [5 votes]: The proof that if $f \colon X \to Y$ is proper and smooth then $f^!\mathcal{O}_Y \cong \Omega^n_F[n]$ ($n=$ dim. rel.$(f)$) assuming only the existence of $f^!$ follows from theorem 3 in
Verdier, Jean-Louis 
Base change for twisted inverse image of coherent sheaves. 
1969 Algebraic Geometry (Internat. Colloq., Tata Inst. Fund. Res., Bombay, 1968) pp. 393–408 Oxford Univ. Press, London.
The proof relies on the "fundamental local isomorphism" and several formal properties of $f^!$. Hope it helps.<|endoftext|>
TITLE: Calculating a second fundamental form in the space of hermitian metrics
QUESTION [9 upvotes]: Let $X$ be a compact Kahler manifold and let $\mathcal M$ denote the space of hermitian metrics on $X$. We'll identify a hermitian metric with a smooth, real and positive $(1,1)$-form $\omega$. Let $\mathcal K$ be the subspace of $\mathcal M$ defined by Kahler metrics, that is, those hermitian metrics $\omega$ such that $d \omega = 0$.
It is possible to equip $\mathcal M$ with a Riemannian metric, for example by using the Hodge $L^2$ metric defined by
$$
G(U,V)_\omega = \int_X \langle U,V \rangle dV_\omega,
$$
where the inner product inside the integral is the one defined by $\omega$ on $(1,1)$-forms on $X$. Here $U$ and $V$ are vectors tangent to $\mathcal M$, or in other words, they are smooth real $(1,1)$-forms on $X$. Despite some technical issues (the fibers of the tangent space are not complete) this metric admits a Levi-Civita connection, and a curvature tensor, and it induces a metric on $\mathcal K$ by restriction.
Q: How can we calculate the second fundamental form of $\mathcal K$ in $\mathcal M$?
It is tempting to try to do this by considering the exterior derivative as a linear map from $\mathcal M$ to the space of 3-forms on $X$, and saying that $\mathcal K$ is its fiber over $0$. If we were talking about a smooth function $f : \mathcal M \to \mathbb R$, then the second fundamental form of the fiber $f^{-1}(0) \subset \mathcal M$ (assuming smoothness) would be given by the Hessian $-\nabla^2 f$ (see Lang's "Fundamentals of differential geometry, Prop. 2.1, p. 376). Is there a similar formula when the submanifold in question is defined by a map $f : \mathcal M \to \mathcal A$ where the target space is an infinite-dimensional manifold? 
An alternative approach would be to use the orthogonal projection onto the normal bundle of $\mathcal K$ in $\mathcal M$, but this projection is expressed using the Laplacian and Green operator associated to the metric $\omega$, so this road promises to be quite bumpy if at all usable. Any references or remarks would be greatly appreciated.

REPLY [3 votes]: Maybe, the following paper can help. Almost hermitian there means that the almost complex structure can also vay, and need not be integrable. 

Olga Gil-Medrano, Peter W. Michor: Geodesics on spaces of almost hermitian structures, Israel J. Math. 88 (1994), 319--332.
(pdf)<|endoftext|>
TITLE: On holomorphic branched coverings of a domain in the plane to the unit disk
QUESTION [6 upvotes]: This question is partly motivated by my answer to this question on math.stackexchange.
Let $\Omega$ be a bounded $n$-connected domain in the plane, bounded by $n$ pairwise disjoint Jordan curves. 
It was proved by Ahlfors that by solving an extremal problem (related to so-called analytic capacity), one can obtain a function $f$ holomorphic in $\Omega$ with the following properties :

$f$ is an $n$-to-$1$ branched covering of $\Omega$ onto the unit disk $\mathbb{D}$,
$f$ extends continuously to the boundary of $\Omega$, and maps each boundary curve homeomorphically onto the unit circle.

See e.g. Krantz's Geometric function theory: explorations in complex analysis, theorem 4.5.9.
My question is :
Is there another (perhaps more intuitive) way to see that such a function necessarily exists?

REPLY [2 votes]: I now know a lot more about this question than I did when I asked it, so for the sake of completness let me add :
By applying the Riemann mapping theorem $n$ times, we can assume that $\Omega$ is bounded by analytic Jordan curves. In this case, we have the following theorem due to Bieberbach, which says that there are many proper holomorphic maps onto the unit disk.
Theorem
Let $\Omega$ be a domain in $\mathbb{C}_{\infty}$ bounded by disjoint analytic Jordan curves $E_1, E_2, \dots, E_n$, and let $p \in X$. For each $j$, fix a point $\alpha_j \in E_j$. Then there is a unique proper holomorphic map $f:\Omega \to \mathbb{D}$ such that $f(\alpha_j)=1$ for all $j$ and $f(p)$=0.
For more information about proper holomorphic mappings, see the preprint there.<|endoftext|>
TITLE: Is there a notion of "ribbon 2-category"?
QUESTION [7 upvotes]: It there some notion of ribbon 2-category, which would allow for, say, talking about the Seifert surface of links (which is a 1-morphism in some ribbon category) as a 2-morphism in the category?
Thank you! (I'm sorry this question is so vague.)

REPLY [3 votes]: I'm currently trying to work out the details. Here is how far I got:
Pivotal categories

You can do graphical calculus on $\mathbb{R}^2$ with monoidal categories.
Boxes represent morphisms, lines are objects. You're allowed isotopies that don't upset the order of composition.
If you have duals, you can change inputs to outputs and vice versa.
In the graphical calculus, you still have to be careful not to "rotate" your boxes, though.
If you have a pivotal category, you only have to differentiate between inputs and outputs of a box up to cyclic order, that is, you can arbitrarily rotate boxes. You're allowed all isotopies now.

In total, this means that you have a graphical calculus where you can embed a directed graph into $\mathbb{R}^2$, label it in a way that type checks, and you don't need to worry about what's up and down.
Spherical categories
The spherical axiom says that you are allowed to put your diagram onto $S^2 \cong \mathbb{R}^2 \cup \{\infty\}$, and you're allowed isotopies on the sphere, not just the plane. This gives you one additional move, namely isotoping across $\infty$. This implies that left and right traces are equal.
(The subtlety here is that this move may validate additional axioms in the category that don't come from isotopies on the sphere. Namely, you can "pull out" an endomorphism of the monoidal unit from within a closed diagram. This means that the graphical calculus is not complete in a naïve sense. Monoidal composition of diagrams in pivotal categories is by glueing two copies of $\mathbb{R}^2$ in an essentially unique way. But when glueing two $S^2$'s together, you have to pick a point on each to perform the connected sum, and not depending on this ambiguity validates the extra axiom.)
Ribbon categories
A ribbon category is (up to issues with too big categories such as non-fusion categories) a braided, spherical category. You can think of it as a (weak) 3-category with one 1-morphism and the appropriate duality data, or you think of it as a spherical category with extra structure, the braiding.
The subtlety we had in spherical categories now vanishes into the third dimension.
Spherical 2-categories
There exist graphical calculi in $\mathbb{R}^3$ for 3-categories. Higher notions of pivotality can be defined explicitly (https://arxiv.org/abs/1211.0529). A 2-spherical category should be a monoidal 2-category (basically a 3-category with one object) that satisfies a graphical calculus on $S^3$, i.e. it satisfies the additional move of isotoping a brane (corresponding to objects in the monoidal 2-category) around the point at infinity. (No such axiom needs to exist for morphisms or 2-morphisms.)
Ribbon 2-categories
There exist explicit notions of braided 2-categories. We could define now that ribbon 2-category is a spherical 2-category with a braiding.<|endoftext|>
TITLE: How to "see" that double suspension of homology 3-sphere is homeomorphic to a sphere?
QUESTION [24 upvotes]: Is there a good way to think about/understand the result that the double suspension of a homology 3-sphere is homeomorphic to a sphere, to get intuition for why this is true?  For instance, what sort of neighborhood must one take for one of the suspension vertices in the $S^0$ in order to see that a neighborhood of this point is homeomorphic to ${\mathbb R}^5$?  Thanks!! 
I found the following related MathOverflow question, which raises relevant points and is interesting, but doesn't seem to answer my question:
"If a manifold suspends to a sphere..."
My posting of this question today was partly inspired by thinking about the comments to the very recent MO question:
"Is a finite CW complex minus a point still homotopy equivalent to a finite CW complex?"
At some point, I asked the question I'm now posting to a topologist who is known for his incredible intuition, and he remarked that he believed it was better to think in terms of taking a join with $S^1$ rather than repeatedly taking a join with $S^0$, but he wasn't sure what else to say.
Thanks again for any help with this!

REPLY [22 votes]: To get some intuition behind the double suspension theorem, you can try three pages with a lot of pictures in Ferry's notes (starting with p.166, in Chapter 26). He gives a rough sketch of proof in the case of one particular homology sphere, the one for which the theorem was first proved by Edwards.
The double suspension theorem boils down to showing that a certain non-manifold (namely, the single suspension over a homology sphere) becomes a manifold when multiplied by $\Bbb R$.
When Milnor conjectured the double suspension theorem in early 60s, he must have been aware
of the existence of other non-manifolds with this property (found earlier by Bing). It is fortunate that they also exist in a lower dimension so it's easier to visualize what's going on. One such example is $(S^3/W)\times\Bbb R\cong S^3\times\Bbb R$, where $W$ is the Whitehead continuum, and there is a rather explicit construction of this homeomorphism in Ferry's Chapter 4 (p.15).
Added later: Another example is $(M/D)\times\Bbb R\cong M\times\Bbb R$, where $D$ is a wild copy of the $n$-disk contained in the interior of the manifold $M$. The case $n=2$ is actually used the above-mentioned proof of the double suspension theorem in Ferry's notes, but is not proved there; a proof of the case $n=1$ with some pictures can be found in the Daverman-Venema book, Section 2.6.

To address the specific question about neighborhoods, the open star (in the original double-suspension triangulation) of any vertex in the suspension circle is homeomorphic to $\Bbb R^5$ $-$ at least in the case of one particular homology sphere,  the boundary of the Mazur manifold $W$. Indeed, the closed star of this vertex is the suspension over $cone(\partial W)$. As explained in Ferry's notes, $cone(\partial W)\times\Bbb R$ is homeomorphic to $W\times\Bbb R$. Hence the open star of the vertex is homeomorphic to $(W\setminus\partial W)\times\Bbb R$. The latter can be identified with the interior of $W\times I$. But it is easy to see that $W\times I$ is homeomorphic to the $5$-ball. So its interior is homeomorphic to $\Bbb R^5$.<|endoftext|>
TITLE: What else is Seiberg-Witten Theory equal to?
QUESTION [38 upvotes]: In low-dimensional topology there have been a bunch of invariants defined, and Seiberg-Witten Theory seems to make its appearance in [a lot of] them:
1) Heegaard Floer homology = SW Floer homology (Kutluhan, Lee, Taubes)
2) Embedded Contact homology = SW Floer homology (Taubes)
3) Gromov-Witten invariant = 4-dimensional SW-invariant (Taubes)
4) Turaev torsion = 3-dimensional SW-invariant (Turaev)
5) Milnor torsion (hence Alexander invariant) = 3-dimensional SW-invariant (Meng, Taubes)
6) Donaldson-Smith standard surface count = 4-dimensional SW-invariant (Usher)
7) Casson invariant (hence integral Theta divisor) = 3-dimensional SW-invariant (Lim)
8) Poincare Invariant = SW-invariant for algebraic surfaces (Okonek, et al.)
Conjectured:
8) Heegaard Floer closed 4-manifold invariant = SW-invariant (Ozsvath, Szabo)
*Analog of (1) above in dimension 4
9) Lagrangian matching invariant = SW-invariant (Perutz)
*Analog of (6) above for broken Lefschetz fibrations
10) Near-symplectic Gromov-Witten count = SW-invariant (Taubes)
*Analog of (4) above for near-symplectic manifolds, counting holomorphic curves in the complement of the degenerate circles of the near-symplectic form -- but this invariant hasn't really been defined yet
Does/should it stop there? Are there constructions out there that Seiberg-Witten Theory could possibly have a link with?

REPLY [7 votes]: Khovanov homology could probably fit into one of the two categories (although I'm not experienced enough to judge/guess whether it belongs to the "SW does it all" or to the "SW can't do it" category).
At a first glance, the two objects look quite different: Khovanov homology is a theory for knots and links in $S^3$ (and cobordisms between them), that categorifies the Jones polynomial, detects the unknot. It can be defined combinatorially in terms of knot diagrams, but I think that the original definition had a flavour of category theory and representation theory. However, Witten has recently proposed a gauge-theoretic approach to it, so maybe there's a deep connection after all.
Moreover, it's been known for some time that Khovanov homology and Heegaard Floer homology are related. Take a knot $K\subset S^3$: there's a spectral sequence, whose $E_2$-page is defined in terms of (a suitable variant of) $KhH(K)$, that converges to $\widehat{HF}(\Sigma(K))$, the Floer homology of the double cover of $S^3$, branched over $K$.<|endoftext|>
TITLE: calculating Möbius function
QUESTION [16 upvotes]: I wonder if there is any efficient way to calculate Möbius function for a array of number 1:1000000 
http://en.wikipedia.org/wiki/M%C3%B6bius_function

REPLY [3 votes]: I also like thisone, $\mathcal{O}(N \ln N)$ additions being not so different to $\mathcal{O}(N \ln \ln N)$ multiplications
% N(1+ln N) additions/affectations. an array of 2 bits elements would be enough

% matlab 
function mu = computeMu(N)
    mu = zeros(1,N); mu(1) = 1; 
    % Will compute the Dirichlet inverse of any sequence starting with 1
    for n = 1:N
        mu(n+n:n:end) = mu(n+n:n:end) - mu(n);
    end
end<|endoftext|>
TITLE: on the approximation by holomorphic functions
QUESTION [5 upvotes]: Good evening,
I have a question on the approximation of holomorphic functions on a space of cartesian product type.
Question: Let $U,V$ be domains in $\mathbb{C}^n$ and $f\in \mathcal{O}(U\times V)$ a holomorphic function on $U\times V.$ Do we always have the following : $f$ can be approximated by holomorphic functions on $U\times V$ of the form $g(z,w) = \sum_{i=1}^N h_i(z)k_i(w)$ where $h_i\in \mathcal{O}(U)$ and $k_i\in\mathcal{O}(V)$ ? (N is arbitrary)
If it is not possible, can this be true if we put some conditions on $U$ and $V$? So what are the conditions?
Any help is appreciated. Thanks in advance.
Duc Anh

REPLY [5 votes]: This is the answer : the above statement is true without any conditions on $U$ and $V.$ It is the theorem 1.7.7 in the book of Narasimhan, Analysis on Real and Complex Manifolds. One of my professors has pointed it out for me.<|endoftext|>
TITLE: The "ds" which appears in an integral with respect to arclength is not a 1-form. What is it?
QUESTION [29 upvotes]: The only reasonable way to interpret "$ds$" as a functional on tangent vectors has to be that it takes a tangent vector and spits out its length, but this is not linear.  So $ds$ is not a 1-form.  It still seems like a nice sort of object to think about integrating.  Does $ds$ fit into a larger class of gadgets generalizing differential forms?  Or it there some compelling reason that I shouldn't care about $ds$?

REPLY [7 votes]: It is an example of an absolute differential form, as defined by Toby Bartels here: http://ncatlab.org/nlab/show/absolute+differential+form.<|endoftext|>
TITLE: Topological necessary and sufficient condition for tightness
QUESTION [8 upvotes]: Recall the definition of tightness for a probability measure $\mathbb P$ on the Borel $\sigma$-algebra of a metric space $(S,d)$:

For each $\varepsilon>0$, we can find a compact subset $K$ of $X$ such that $\mathbb P(K)\geq 1-\varepsilon$. 

The question is: is there a "nice" topological characterization of metric spaces such that each Borel probability measure is tight?
In Billingsley's book Convergence of probability measures, 1968, it's said that it's an open problem. I wish know whether some progress have been done so far.
Call EPT a metric space on which each Borel probability measure is tight. 
Some remarks:

By Ulam's theorem, each separable metric space topologically complete (Polish is a shorter term) is EPT. 
A necessary and sufficient condition that each probability has a separable support is that each subset of $D$ of $S$ which is discrete have a non-measurable cardinal (i.e. we can't find a probability measure $\mu$ on $2^D$ such that $\mu(\{x\})=0$ for each $x\in D$). Hence for each EPT space, each discrete subset have a non-measurable cardinal.  
If we assume the metric space separable, we have the answer from Dudley's book Real Analysis and Probability: each probability measure on $S$ is tight if and only if $S$ is universally measurable (that is, if $\widehat S$ is the metric completion of $S$, then $S$ is $\mathbb P$-measurable for each probability measure $\mathbb P$ on $\widehat S$).

REPLY [6 votes]: You run into questions of set theory: If $X$ has the discrete topology (which is metric) 
then your condition means: Every probability measure on the power set $2^X$ is discrete. 
This is related to the 'measurability' of the cardinal $card(X)$. For general metric spaces
the maximal cardinality of discrete subsets is the key.<|endoftext|>
TITLE: Combinatorial analogues of curvature
QUESTION [15 upvotes]: There appear to be many "combinatorial" definitions of curvature as applied to finite simplicial (or regular CW) complexes. For instance, we have the ideas of Cheeger, Muller and Schrader, of Forman and also some wonderful work by Luo. This list just refers to work that I've come across over the last few years and should not be considered complete...
Of course, each definition appears to have a different "domain" to which it applies, and no two are equivalent on the intersection of their domains. This is in sharp contrast to the smooth curvature theory, some of which is even canonically introduced to undergraduates in standard calculus sequences. 
Moreover, each theory mentioned above has inherent gaps in terms of establishing a good analogy with smooth curvature. Cheeger et. al work in a setting that is borderline non-combinatorial and focus on approximating curvature of smooth objects by piecewise flat ones "in-measure"; Forman does not require embedding but there is provably no version of Gauss-Bonnet for his definition of curvature. Luo's work restricts to triangulated 3 manifolds with boundary, etc. 
It is not straightforward to come up with a wish list of what one desires from a definition of combinatorial curvature, but certainly: Gauss-Bonnet, Myer's theorem, and well-behavedness under cell subdivision come to mind.
So: 

Are there combinatorial analogues of curvature that satisfy Gauss-Bonnet for abstract simplicial complexes and Myer's theorem for triangulated complete Riemannian manifolds?

If the answer is "yes", then what are they? And if "no", then:

What are the popular combinatorial definitions of curvature different from the ones that show up in the papers mentioned above?

REPLY [6 votes]: Perhaps, the most general concept of curvature (measure) is that of normal cycle.  This is an object $N^S$ naturally  associated to a reasonably nice compact subset $\newcommand{\bR}{\mathbb{R}}$  $S\subset\bR^n$. PL sets are reasonabky nice and, more generally, the  semialgebraic sets are nice. The compact smooth submanifolds of $S$ are also nice.
Briefly, $N^S$ is an $(n-1)$-dimensional  current  on $\bR^n\times S^{n-1}$.  Here is briefly its definition.  
If $S$ is a compact domain with $C^2$ boundary, then  the outer normal to the boundary $\newcommand{\bn}{\boldsymbol{n}}$ defines the Gauss map $\newcommand{\pa}{\partial}$ $\bn:\pa S\to S^{n-1}$, and then $N^S$ is defined as the current of integration along the  graph of the Gauss map.    $\newcommand{\ve}{{\varepsilon}}$ If $S$ is not such a domain, but it is sufficiently nice, then for $\ve>0$ sufficiently small the compact  set
$$S_\ve:=\bigl\lbrace x;\;\; {\rm dist}(x, S)\leq \ve\bigr\rbrace $$
is a domain with $C^2$ boundary and one can show  that   as $\ve \searrow 0$  the currents $N^{S_\ve}$ converge  weakly to a closet current  $N^S$ which is rather nice. (You can think of $N^S$  as being  the current of integration along several smooth oriented  manifolds   with some multiplicities. )  In other words $N^S$ can be thought of as the graph of the Gauss map of a thin tubular neighborhood of $S$.
Where does  curvature come in?  In the smooth case,   Gauss'  Theorem shows that the Gauss map encodes the curvature, so it is not surprising that in this case $N^S$ has something to do with the curvature.  How does it work in general?
On $S^{n-1}\times \bR^n$ there exist certain universal  $(n-1)$-forms
$$ \alpha_0,..., \alpha_{n-1} \in \Omega^{n-1}(S^{n-1}\times \bR^n) $$
The integral of $\alpha_k$ along $N^S$ is called the $k$-th curvature measure of $S$ and it is denoted by $\mu_k(S)$.     It can be expressed as an integral over $S$ of a certain measure, which  may be singular if  $S$ is. When $S$   is a smooth oriented  submanifold of $\bR^n$, the measure  describing $\mu_0$ is given by the integration of the Pfaffian of the  Riemann curvature.  For PL sets  this measure  will have singular contributions coming from the angles of the faces, their areas etc.  The general Gauss-Bonnet theorem states that if $S$ is a reasonably nice set, then $\mu_0(S)$ is the Euler characteristic of $S$.  The story is much more complicated and more beautiful than what I can squeeze in a few paragraphs, but I can give you some  links to places where you can read more about this subject.
First,   you should look  at these beautiful notes of J. Fu who is one of the pioneers in this field. On his homepage you will find other beautiful things.
You can also look  at these notes of mine where I  give a description of $N^S$ for simplicial sets. There is a very close connection between the normal cycle and Morse theory and my notes discuss this aspect.   
If you want a more elementary point of view on these curvature measure, then you should definitely have a look  at   this REU project that I supervised a few years ago. I  was very pleased  with the final result of these three fine young people. This survey is very written with many figures and a lot of intuition. The last part deals  with Morse theory and Gauss-Bonnet for two-dimensional simplicial complexes.<|endoftext|>
TITLE: Blackbox Theorems
QUESTION [83 upvotes]: By a blackbox theorem I mean a theorem that is often applied but whose proof is understood in detail by relatively few of those who use it. A prototypical example is the Classification of Finite Simple Groups (assuming the proof is complete). I think very few people really know the nuts and bolts of the proof but it is widely applied in many areas of mathematics. I would prefer not to include as a blackbox theorem exotic counterexamples because they are not usually applied in the same sense as the Classification of Finite Simple Groups. 
I am curious to compile a list of such blackbox theorems with the usual CW rules of one example per answer.  
Obviously this is not connected to my research directly so I can understand if this gets closed.

REPLY [10 votes]: Deligne's construction of Galois representations corresponding to modular forms.

REPLY [4 votes]: The Cohen-Structure theorem in commutative algebra (classifying complete local rings in some sense).<|endoftext|>
TITLE: To compute minors of Jacobian of symmetric polynomials
QUESTION [6 upvotes]: For any $n$ tuple $f_1,f_2,\dots,f_n$ in the polynomial ring $\mathbb{C}[x_1,x_2,\dots,x_n]$ 
 one has Jacobian, expressed by the $(n \times n)$-determinants:
$$
  J(f_1,\dots,f_n):=|\frac{\partial}{\partial x_i}(f_j)|_{1 \leq i,j,\leq n} 
$$  And,
one has for elementary symmetric polynomials $e_i$ of degree $i$, 
 $$
  J(e_1,e_2,\dots,e_n)=\prod_{1 \leq i,j,\leq n}(x_i-x_j)=\triangle
 $$
also for complete symmetric polynomials and power sum symmetric polynomials, one has a very nice formula for Jacobian, see the following link:
http://www.ams.org/mathscinet/search/publdoc.html?pg1=MR&s1=1955517&loc=fromreflist
Question: Let $n \geq 4$. Is the similar results also known for minors of Jacobian of symmetric polynomials (complete symmetric polynomials). 
Remark: For power sum and elementary symmetric polynomials, one can derive easily, I am looking for complete symmetric polynomials.

REPLY [2 votes]: The Jacobian of a family of complete functions, or of power sums, is
given in the article: A.Lascoux, P.Pragacz, {\it Jacobians of symmetric
polynomials}, Annals of Comb. 6(2002) 169-172. An interesting fact is that Schur
functions of "multiple" of alphabets occur in the case of complete functions.<|endoftext|>
TITLE: Proof for which primes H*G has torsion
QUESTION [14 upvotes]: In 1960 Borel proved a beautiful result:

Theorem. Let G be a simple, simply connected Lie group. Suppose that p is a prime that does not divide any of the coefficients of the highest root (expressed as a linear combination of the simple roots). Then $H^*G$ has no p-torsion. 

Here $H^*G$ refers to the integral cohomology of $G$ as a space.
(Interestingly, the converse of this statement does not hold. For example $Sp(n)$ does not have $2$-torsion, though the coefficients of the highest weight are (1,2). This counterexample is given by Borel).
You can check out his proof here if you're interested, but basically it seems like what he does is this: he computes the cohomology of each of the simple, simply connected Lie groups with coefficients that have the relevant primes inverted and proves the result by observation.
Since the statement of the theorem itself has little to do with the actual classification of simple Lie groups, and morally should only rely on the fact that we can recover a simple, simply-connected Lie algebra from its root system, it seems natural to ask:

Question: In the intervening 50+ years since Borel proved this theorem, do we know of a direct proof of the theorem that does not use the classification of simple Lie groups?

I really have no idea how to go about doing this... could one use Lie algebra cohomology? I can only seem to find a link between Lie algebra cohomology and deRham cohomology of Lie groups, which is of no help when asking questions of torsion. Maybe there's an integral version?

REPLY [8 votes]: I can't answer this completely, but I can point out some important follow-ups in the literature by Borel and others which need to be be taken into account.   Borel's Tohoku paper is reprinted in the second volume of his collected papers (Springer, 1983).   At the end of this volume is a full two page commentary by Borel on this paper, which reiterates his disappointment that he was unable to prove his result on torsion without going through the classification.   Here he assembles more detailed results on "torsion primes" which were afterwards proved by Demazure (1973 Inventiones paper) and especially by Steinberg (1975 Advances paper). Eventually all of this work yields a clearer picture with less reliance on the classification.  
Further closely related work then appeared in an AMS Memoir which Steinberg reviewed at length in Mathematical Reviews:
Borel, Armand; Friedman, Robert; Morgan, John W. Almost commuting elements in compact Lie groups.  Mem. Amer. Math. Soc. 157 (2002), no. 747, x+136 pp
ADDED: It's a natural problem in Lie theory to find classification-free proofs of results which are first observed case-by-case.   But I'm not aware of any substantial progress beyond Borel's careful commentary (in the 1983 volume), where he notes the equivalence of five statements about cohomology and the equivalence of three statements about orimes related to root systems and Weyl groups (with reference to Demazure and Steinberg).   As he observes, the first five imply the last three, but only case-by-case study shows the converse as well.    
Along the way Borel also points out an improvement over his original Nagoya formulation: while the "bad" primes (possibly 2, 3, 5) are those dividing a coefficient of the highest root, the "torsion" primes are the more limited ones dividing a coefficient of the coroot of this highest root.  For instance in type $G_2$, with respective short and long simple roots $\alpha, \beta$, the highest root is $3\alpha + 2\beta$ whereas its coroot is $\alpha^\vee + 2\beta^\vee$.  
There is a long history of study of the topology of a semisimple Lie group (equivalent to the topology of a maximal compact subgroup), in which a reduction is made to study of the root system and its Weyl group: for example, the determination of Betti numbers in terms of exponents or degrees for the Weyl group (Chevalley, ICM 1950).   But this kind of transition is rather subtle.  In the study of torsion primes, a key role is played by subgroups of maximal rank in a compact Lie group, these being correlated with certain subdiagrams of the extended Dynkin diagram.   Much of the technology recurs in the study of p-compact groups, as Jesper observes.   But not everything is well understood conceptually.<|endoftext|>
TITLE: Elevator pitch for the Virtual Fibering Theorem
QUESTION [21 upvotes]: There has been a great deal of excitement among topologists about the proof of the Virtual Haken Theorem, and in fact of the  Virtual Fibering Theorem (for closed hyperbolic 3-manifolds, but I'm guessing they will soon be proven in full generality). The proof is lucidly discussed in Danny Calegari's blog. The theorems state that every compact orientable irreducible 3-manifold with infinite fundamental group has a finite cover which is Haken or a surface bundle over a circle, correspondingly. This implies various good things for a 3-manifold with fundamental group π, including:

π is large, meaning that π has a finite index subgroup which maps onto a free group with at least 2 generators. In particular the Betti numbers of finite covers can become arbitrarily large.
π is linear over $\mathbb{Z}$, i.e. π admits a faithful representation $\pi \to \mathrm{GL}(n,\mathbb{Z})$ for some $n$. (Thurston conjectured that $n\leq 4$ is sufficient).
 
π is virtually biorderable.

Stefan Friedl, from whose comment the above list is an excerpt, summarizes the situation as follows:

It seems like every nice property of fundamental groups which one can possibly ask for either holds for π or a finite index subgroup of π.

All well and good. But how could you `sell' that to somebody who isn't a classically-oriented 3-dimensional topologist? 
An elevator pitch is defined by Wikipedia as follows:

An elevator pitch is a short summary used to quickly and simply define a product, service, or organization and its value proposition. The name "elevator pitch" reflects the idea that it should be possible to deliver the summary in the time span of an elevator ride, or approximately thirty seconds to two minutes. In The Perfect Elevator Speech, Aileen Pincus states that an elevator speech should "sum up unique aspects of your service or product in a way that excites others."

The Virtual Fibering Conjecture (or the Virtual Haken Conjecture) was the grand conjecture in 3-manifold topology following Geometrization, and thus must have/ should have/ ought to have (I believe) a compelling elevator pitch. For contrast, Geometrization is easy to `sell' because it directly applies to the Homeomorphism Problem in 3-manifold topology: Given two 3-manifolds, determine whether or not they are homeomorphic. Geometrization allows you to canonically decompose both manifolds into submanifolds with geometric structure, and then to compare geometric invariants. In terms of "The Goals of Mathematical Research" as given in the introduction to The Princeton Companion to Mathematics, this corresponds to the goal of Classifying.

Question: What is a good elevator pitch for Virtual Fibering (or for Virtual Haken), explaining the utility of these results in terms of "the fundamental goals of mathematical research" (Solving Equations, Classifying, Generalizing, Discovering Patterns, Explaining Patterns and Coincidences, Counting and Measuring, and Finding Explicit Algorithms). The target would be mathematicians who are not 3-dimensional topologists.

Everyone in the approximate vicinity of the field instinctively feels that these are historic results, but I'd like to be able to justify that feeling (in the abovementioned sense) to myself and to others.

REPLY [24 votes]: OK, I will also give it a shot.
First of all, I don't like to sell Geometrization because it helps with the homeomorphism problem. The Geometrization Theorem is an object of stunning beauty ("most 3-manifolds are hyperbolic" should be an exciting statement for anybody in an elevator who has seen the art of M.C.Escher), and beauty in mathematics is usually a sign that we are on the right track. And indeed, the beauty of Geometrization begets all kinds of results.
Regarding the new results of Agol, Wise et al. one should perhaps not jump right to "virtual Haken" or "virtually fibered" but one should look at the "real theorem",
the Virtually Compact Special Theorem which goes as follows:
If $N$ is a finite volume hyperbolic 3-manifold, then $\pi_1(N)$ is virtually compact special, i.e. $\pi_1(N)$ is virtually a quasi-convex subgroup of a Right Angled Artin Group (RAAG).
One can explain a RAAG to anybody who has seen group theory in an elevator between about 3 floors. The fact that "simple" objects like RAAGs contain all hyperbolic 3-manifold groups (up to going to a finite index subgroup) is stunning and beautiful. All the goodies, e.g. largeness, linear over $\mathbb{Z}$, virtual fibering, LERF, virtually biorderable etc come from that statement (well, together with Agol's fibering theorem, tameness etc.).
This can be seen clearly by looking at Diagram 4 in a recent survey paper on 3-manifold groups by authors whose names escape me at the moment. It is really stunning how the Virtually Compact Special Theorem answers all open questions at once. 
It is one of the great achievements of Dani Wise to have found the "right statement". 
(Note that largeness, linear over $\mathbb{Z}$, biorderable do NOT follow from virtual fibering or virtual Haken alone.) 
Back to the elevator:
The results make me think that hyperbolic 3-manifolds are like Jack in the Box.
If you take a hyperbolic integral homology sphere you look at a tiny manifold, but when you press a button (i.e. go to an appropriate finite cover), the 3-manifold suddenly becomes a grand object of beauty (e.g. has as many fibered faces in the Thurston norm ball as you could wish).
(This analogy also works with tiny seed, a bit of water, blooming flower etc. for the botanically minded elevator companion)
So to conclude, I think the Geometrization Theorem and the Virtually Compact Special Theorem of Agol-Wise are stunningly beautiful results. The fact that the statements are so beautiful made it highly plausible that they were right, even before they were proved (I can't imagine that any serious person doubted the Poincare conjecture after Thurston stated the Geometrization conjecture). And ideally it's this beauty which I would like to communicate.<|endoftext|>
TITLE: How Does My Radio Work?
QUESTION [65 upvotes]: Bear with me for a moment while I invoke the real world; the main question at the end is purely mathematical.
I live in an area with $n$ AM radio stations and $m$ FM radio stations.  
AM station number $j$ wants to send me the signal $\phi_j(t)$.  FM station number $k$ wants to send me the signal $\psi_k(t)$.
Of course if they just sent those signals, my radio would recieve their sum and have no idea how to disentangle them.  Therefore, the signals are first encoded.  My (possibly ill-informed) understanding is that (modulo a gazillion bells and whistles), AM station $j$ sends the signal $\phi_j(t)\sin(\omega_j t)$. where $\omega_j$ is some constant, and FM station $k$ sends the signal $A_k \sin(\psi_k(t))$ where $A_k$ is some constant.
My radio then receives the signal 
$$\sum_{j=1}^n\phi_j(t)\sin(\omega_j t)+\sum_{k=1}^mA_k\sin(\psi_k(t))\quad\quad(1)$$
Having received this signal, and knowing the values of the $\omega_j$ and the $A_k$, my radio is then somehow able to compute any one of the signals $\phi_j(t)$ or $\psi_k(t)$ and play it for me on request.  (In fact, I'm pretty sure it can recover $\phi_j$ on the basis of $\omega_j$ alone, without knowing the values of the other $\omega$'s.)
It's not obvious to me that this is mathematically possible, though my radio seems to have no problem doing it.
Question 1 (Pure Mathematics).  For what values of $\omega_1,\ldots,\omega_n,A_1,\ldots,A_m$ is it possible to recover the functions $\phi_1,\ldots,\phi_n,\psi_1,\ldots\psi_m$ from expression (1) alone?  And what assumptions are being made on the class of allowable functions from which the $\phi_j$ and $\psi_k$ are drawn?
Question 2 (Part Engineering, part Pure Mathematics).  If (as is not impossible), AM and/or FM works entirely differently than I think it does, thus rendering Question 1 entirely unmotivated, then how do AM and FM work, what is the correct analogue of expression (1), and what is the right answer to the corresponding new version of Question 1?
Edited to add: I'm aware that there are all sorts of issues with distorted transmissions, error-correcting, etc.  I want to abstract away from all of these and understand the basics.

REPLY [11 votes]: I've been trained as an engineer, and I can tell you that engineers have a somewhat simplified view of the matter. (But, not only a simplified view, of course.) The other answers fill in some detail, but I think a higher-level view is useful.
There is no such thing as perfect recovery of the transmitted signal. The best you can hope is to bound the error.
For most modulation techniques the basic idea is that the spectrum $X$ of a signal $x$ is nearly 0 outside a narrow band: if $|f-f_0|\gt B$ then $X(f)\approx0$. Both AM and FM are essentially means of transforming a spectrum centered around $0$ into one centered around $f_0$. So, in order to recover a signal, the main concern is to make sure that the spectrums $X_1$, $X_2$, …, $X_n$ do not overlap. This is achieved in a rather uninteresting way: regulation. Then you can extract one signal by shifting $f_0$ to $0$ (convolution with a Dirac impulse in frequency domain, meaning multiplication with a harmonic signal in the time domain), and then applying a low pass filter (multiplication with a rectangular function in the frequency domain, meaning convolution with a sinc in the time domain). See also this related question.
There are broad-spectrum modulation techniques, which are used for example in fourth generation mobile-phone networks, that do not rely on the assumption that the signal covers a narrow band. The two main ones are frequency hoping (use some narrow-band modulation technique but change $f_0$ often in some pseudorandom sequence) and spread spectrum (multiply the signal with a pseudorandom sequence before using a narrow-band modulation technique). The signals obtained thru such methods have a wide band, but are bounded $|X(f)| < c$ for some $c$ for all $f$. This way they behave as background noise as far as demodulating any narrow-band signal is concerned.<|endoftext|>
TITLE: Tweetable way to see that Willmore energy is Möbius invariant?
QUESTION [8 upvotes]: Consider a compact orientable Riemannian manifold $M$ (without boundary) isometrically immersed into $\mathbb{R}^3$.  The Willmore energy of $M$ is the functional
$$\mathcal{W} = \int_M H^2 dA$$
where $H$ is the induced mean curvature.  I am looking for a short, sweet, and utterly convincing (though not necessarily utterly formal) way of demonstrating that the Willmore energy is Möbius invariant.  Of course, $H$ itself is rigid motion invariant, so the only thing that really needs to be explained is scale invariance and invariance w.r.t. sphere inversions.
My oddball way of seeing scale invariance is that if you think of the surface as a conformal immersion $f:M \rightarrow \mathbb{R}^3$ then the Willmore energy is simply the (squared) $L^2$ norm of the so-called mean curvature half-density $H|df|$, where $|df|: TM \rightarrow \mathbb{R}; X \mapsto |df(X)|$ can be thought of as the (isotropic) length element.  And since mean curvature times length is scale invariant, so is the Willmore energy.  But that's an oddball way of seeing things, and certainly not the simplest explanation.
As for sphere inversions, my only thought is that inversions are reflections in hyperbolic geometry.  And reflections are isometries... So I have a sneaking suspicion that hyperbolic space provides a cute explanation -- perhaps for Möbius invariance on the whole -- but such an explanation eludes me.
Other perspectives are, of course, very welcome!
Update: it is tempting to try to show that the Willmore energy is more generally conformally invariant, but this statement is not true -- in two dimensions conformal structure is much more flexible than in dimensions three or higher.  In particular, given a smooth surface $M$ equipped with a conformal structure there are many immersions $f: M \rightarrow \mathbb{R}^3$ such that the induced metric is compatible with the conformal structure, and not all of these immersions will have the same Willmore energy.  A concrete example is the Dirac spheres, which are conformal immersions of $S^2$ with progressively larger constant mean curvature-half density, hence progressively larger Willmore energy (some pictures here, unfortunately low-resolution).  But since there is only one conformal structure on $S^2$, $\mathcal{W}$ cannot be conformally invariant.

REPLY [9 votes]: The Willmore energy $\mathcal{W} = \int_M H^2 dA$ differs from the functional $$\widetilde{\mathcal{W}} = \int_M (H^2-K) dA$$ just by a constant as one can see from the Gauss - Bonnet theorem ($K$ here is the Gaussian curvature of $M$).
The expression $H^2-K$ in $\widetilde{\mathcal{W}}$ is the half of the square of the length of the trace-free part of the second fundamental form which is a (pointwise) conformally invariant density of conformal weight $-2$, while "dA" can be seen as a density with conformal weight $2$, so the entire integrand $(H^2-K) dA$ is independent of a choice of a metric.
Thus $\widetilde{\mathcal{W}}$ is manifestly conformally, and in particular, Möbius invariant. So is $\mathcal{W}$.
(A Liouville's theorem ensures that conformal maps of $\mathbb{R}^n$, $n\ge 3$, are restrictions of Möbius transformations.)
Edit. The above is an attempt to address the original request for a "tweetable" argument.
Of course, the precise statement is that the Willmore energy is conformally invariant with respect to the conformal transformations of the ambient space. (Otherwise we would not be able to invoke the Liouville's theorem).
The correct definition of the Willmore energy involves an immersion $f\colon M \rightarrow \mathbb{R}^3$ and the induced conformal structure on the immersed manifold. The Dirac spheres show, in particular, that the Willmore energy does depend on the immersion.<|endoftext|>
TITLE: Terminology: Banach spaces equipped with continuous associative product?
QUESTION [6 upvotes]: This is admittedly a low-interest question mathematically, and is arguably a question I could resolve if I had time over the next few days to go and look through a large number of the Banach algebra/functional analysis books on my shelves and in the library. However, it strikes me that this is easily crowd-sourceable and that people may know of texts I am less familiar with. My reason for asking on MO rather than MSE is that I think it will get better answers here.
So: the usual definition of a Banach algebra is that it is a (complex) algebra equipped with a complete vector-space norm, such that $\Vert ab\Vert\leq \Vert a\Vert \Vert b\Vert$ for all elements $a,b$.
Now suppose we have a (complex) algebra $A$ equipped with a complete vector-space norm $\Vert\cdot\Vert$ and a constant $K>0$ such that $\Vert ab\Vert\leq K\Vert a\Vert \Vert b\Vert$ for all $a,b$. These are much rarer in the literature, most likely for the following reason: a standard exercise doled out to students is to show that there is an equivalent norm on $A$ for which multiplication is contractive, i.e. rendering $A$ (in this new norm) a Banach algebra in the usual sense. In this sense "one has nothing new".
However, in some joint work I am writing up, I am toying with the idea of working in this greater generality, in order to let certain technical functorial constructions have more natural formulations.
(In a bit more detail, it is to do with certain homologically flavoured constructions for Banach algebras and Banach bimodules more naturally living in a world where multiplication need not be contractive.)
So my question is this:

do these kinds of algebra have a standard name, and where are the established sources for such terminology?

I have a dim recollection that they are given a name of their own in Zelazko's old book on Banach algebras, but I don't recall what the name was, and I can't find anything in Bonsall & Duncan.
Note: I am not after arguments as to what terminology should or should not be, or observations about one definition being a "semigroup object in Ban$_1$" while the other is a "semigroup object in Ban". Rather, I need some idea of whether one choice of terminology is standard, and hence least likely to cause confusion/irritation to the intended audience, should I decide to pursue this course.

REPLY [4 votes]: Yemon, I have used the term "weak Banach algebra" for such things. I don't think there is a standard term, though. I vaguely recall seeing people simply call them Banach algebras (probably in some older papers when the terminology in the subject hadn't really stabilized).
(I ran into this issue when dealing with the Lipschitz algebra $Lip_0(X)$ for $X$ a complete finite diameter metric space. You really want to use Lipschitz number as the norm, even though this only makes it a weak Banach algebra. There's no real penalty for doing this, and the advantage is that it allows you to identify $X$ isometrically with the normal spectrum of $Lip_0(X)$.)
Edit: I've just realized that this is what Gelfand meant by "normed ring". E.g., on the first page of his book Commutative Normed Rings (1960) he writes:
"A normed ring is a complex Banach space in which an associative multiplication is defined that is permutable with the multiplication by complex numbers, distributive with respect to addition, and continuous in each factor."
and there is a footnote which says "In another terminology, a Banach algebra."
A few pages in he proves that you can always achieve $\|xy\| \leq \|x\|\|y\|$ by renorming.<|endoftext|>
TITLE: From convex polytopes to toric varieties: the constructions of Davis and Januszkiewicz
QUESTION [20 upvotes]: One of the most useful tools in the study of convex polytopes is to move from polytopes (through their fans) to toric varieties and see how properties of the associated toric variety reflects back on the combinatorics of the polytopes. This construction requires that the polytope is rational which is a real restriction when the polytope is general (neither simple nor simplicial). Often we would like to consider general polytopes and even polyhedral spheres (and more general objects) where the toric variety construction does not work.
I am aware of very general constructions by M. Davis, and T. Januszkiewicz,
(one relevant paper might be Convex polytopes, Coxeter orbifolds and torus actions, Duke Math. J. 62 (1991) and several subsequent papers). Perhaps these constructions allow you to start with arbitrary polyhedral spheres and perhaps even in greater generality.
I ask about an explanation of the scope of these constructions, and, in simple terms as possible, how does the construction go?

REPLY [2 votes]: In the definition of moment-angle manifold in Davis-Januszkiewicz's 1991 paper, the orbit space is a simple convex polytope $P$ which is contractible. We can replace $P$ by an arbitrary smooth nice manifold with corners $Q$ and define the generalized moment-angle manifold $\mathcal{Z}_Q$ in the similar way as usual moment-angle manifold (see https://arxiv.org/abs/2011.10366). Let $F_1,\cdots,F_m$ be all the facets of $Q$ and $\lambda: \{ F_1,\cdots,F_m \} \rightarrow
    \mathbb{Z}^m$ be a map such that $\{\lambda(F_1),\cdots,\lambda(F_m)\}$ is a unimodular basis of $\mathbb{Z}^m\subset \mathbb{R}^m=T_e(S^1)^m$.
\begin{equation}
       \mathcal{Z}_{Q} = Q\times (S^1)^m / \sim
     \end{equation}
where $(x,g) \sim (x',g')$ if and only if $x=x'$ and $g^{-1}g' \in \mathbb{T}^{\lambda}_x$
where $\mathbb{T}^{\lambda}_x$ is the subtorus of $(S^1)^m$ determined by
the linear subspace of $\mathbb{R}^m$ spanned by the set $\{ \lambda(F_j) \, |\, x\in F_j \}$.
The free quotient of $\mathcal{Z}_Q$ under the action of some subtorus in $(S^1)^m$ of rank $m-n$ gives analogues of toric manifolds in this setting (where $n$ is the dimension of $Q$). Such a space can also be defined from a (non-degenerate) characteristic function on the facets of $Q$. The equivariant cohomology ring of $\mathcal{Z}_Q$ with $\mathbf{k}$-coefficients is isomorphic to a ring $\mathbf{k}\langle Q\rangle$ (called the topology face ring of $Q$) that is determined not only by the face poset of $Q$ but also by the $\mathbf{k}$-cohomology rings of all the faces of $Q$. The definition of topology face ring is a direct generalization of the face ring of a simple convex polytope.
Furthermore, we can replace $Q$ by an arbitrary finite CW-complex $X$ with a panel structure $\mathcal{P}$ and define moment-angle complex $(D^2,S^1)^{(X,\mathcal{P})}$ and do the similar calculations for its cohomology and equivariant cohomology (see https://arxiv.org/abs/2103.04281). The definition of panel structure
is due to M. Davis's 1983 paper "Groups generated by reflections and aspherical manifolds not covered by
Euclidean space" in Ann. of Math.
Roughly speaking, a panel structure on $X$ defines "abstract faces" on $X$ which allows us to do the similar construction as $\mathcal{Z}_Q$. But it is more convenient to think of
$(D^2,S^1)^{(X,\mathcal{P})}$ as the colimit of a diagram of CW-complexes of the form
$ f\times \underset{j\in I_f}{\prod} D^2_{(j)} \times
 \underset{j\in [m]\backslash I_f}{\prod} S^1_{(j)}$ where $f$ ranges over all
the "abstract faces" of $(X,\mathcal{P})$ and $I_f$ denotes all indices of the panels that contain $f$. Note that "panels" plays the role of facets here.
In general, $(D^2,S^1)^{(X,\mathcal{P})}$ may not be a manifold. The free quotient of $(D^2,S^1)^{(X,\mathcal{P})}$ under some torus action can also be considered as a far-reaching generalization of manifolds with locally standard torus actions. In addition, the topology face ring of the panel structure $(X,\mathcal{P})$ also makes perfect sense, which is isomorphic to the equivariant cohomology ring of $(D^2,S^1)^{(X,\mathcal{P})}$.
In particular, when $X$ is the cone of the barycentric subdivision of a simplicial complex $K$ (with a canonical panel structure $\mathcal{P}_K$), the topological face ring of $(X,\mathcal{P}_K)$ is nothing but the face ring (Stanley-Reisner ring) of $K$ (see Section 5 of that paper).<|endoftext|>
TITLE: $W^{1,1}$ simplicial approximation
QUESTION [7 upvotes]: Let $f$ be a continuous real-valued function defined on an $n$ dimesional simplex $\Sigma\subset \mathbb{R}^n $. The classical simplicial approximation scheme provides a sequence $f_k$ of piecewise affine functions converging uniformly to $f$: namely, on each simplex $S$ of the $k$-th baricentric subdivision of $\Sigma$, one takes ${f_k}  _ {|S}$ to be the affine interpolation of the data of $f$ on the vertices of $S$. Precisely, if $\omega$ is a modulus of continuity for $f$, it follows easily  that $$ \big\| f - f_k \big\|_{\infty,\Sigma}\,  \le\, \omega\bigg( \Big(\frac{n}{n+1}\Big )^k \operatorname{diam}\Sigma \bigg)\, .$$
Notice that $\nabla f_k(x) $ is  defined and locally constant at any point $x\in\Sigma$ not in the  $(n-1)$-skeleton of the $k$-th barycentric subdivisions; in particular, a.e. in $\Sigma$. However, understanding the behaviour of this sequence is not as simple as before, even for smooth $f$. For instance,  the norm of $\nabla f_k(x) $  is in general not bounded by the uniform norm of $\nabla f $.
Questions. Assuming   that $f$ has bounded and continuous first order derivatives, can we conclude that $\nabla f_k $ converges to $\nabla f $ in $L^1(\Sigma)$? How to dominate pointwise $|\nabla f_k (x) |$ (independently of $k$), and how to bound $\|\nabla f_k -\nabla f\|_{1,\Sigma} $ ?

REPLY [3 votes]: You can represent on every simplex $\sigma$ of your triangulation the derivative $D f_k$ by
$$
  D f_k = \frac{1}{\lvert \sigma \rvert} \sum_{i = 0}^n \int_{\sigma} \Bigl(\frac{1}{(1 - \beta_i (x))^n} - 1\Bigr)\, Df(x)[a_i - x]\,dx D\beta_i,
$$
where $\beta_i$ are the maps that give barycentric coordinates 
$$
  \beta_0 (x) + \dots + \beta_n (x) = 1
  \quad \text{ and } \quad 
  \beta_0 (x)a_0 + \dots + \beta_n (x)a_n = x.
$$
For a proof, see J. Van Schaftingen, Approximation in Sobolev spaces by piecewise affine interpolation, [arXiv:1312.5986]).
The kernel appearing in this integral representation is absolutely integrable ($a_i - x$ controls $\beta_i (x)$), uniformly with respect to simplices that do not degenerate. This should imply in fact an $L^\infty$ approximation property under your assumptions. (If the function $f$ is merely in a Sobolev space, one can improve the convergence by translating the triangulation points.)<|endoftext|>
TITLE: Does a small contraction occur between smooth varieties?
QUESTION [7 upvotes]: Let $V, \tilde{V}$ be smooth algebraic varieties over $\mathbb{C}$ and $f \colon \tilde{V} \rightarrow V$  a projective (or proper) birational morphism. Assume that the exceptional locus $E \subset \tilde{V}$ has codimension $\ge 2$. 
Question Is $f$ an isomorphism?

REPLY [5 votes]: Yes. Suppose $f$ contracts a curve $C$. Then for any ample divisor $D$, we have $D\cdot C>0$. 
But $D=f^*f_*D$ by your hypotheses on the exceptional locus, and so $D\cdot C=f_*D\cdot f_*C=0$, a contradiction.<|endoftext|>
TITLE: A natural center of a convex weakly compact set in Banach space
QUESTION [16 upvotes]: Question: Let $S$ be a convex weakly compact set in Banach space $H$. Propose a natural way to define the unique center $O \in S$.
Motivation: A lot! For example, in game theory $S$ can be a set of possible (fair) allocations, and we need to suggest a natural method to choose one.
Discussion. If S consists of 2 points, or line in $\mathbb R^2$, we have no natural way to select a center, thus $S$ should be convex and weakly compact. To define these properties, we need vector space and topology, thus the natural setting is Bahach space. 
If $H$ is $\mathbb R^n$, the natural choice is centroid (center mass), but to define it for general case, we need a natural notion of "uniform density" in a Banach space. Is this someting standard which I do not know?
My main example is $H = L^1$, space of all intergable functions $[0,1]\to \mathbb R$. In this case if $S$ consists of all functions with range in $[a,b]$, the center should naturally be a constant function $f(x)\equiv (a+b)/2$. Also, $O$ should be tractable to compute, at least for such a simple examples of $H$ and $S$. A good axiomatic foundation ($O$ is the unique point saisfying axioms A1, A2, and A3) would be a plus.

REPLY [2 votes]: I believe that, at least in the case of a weakly compact convex set $K$ in a uniformly convex Banach space $X$, the circumcenter could be a possible good candidate. By circumcenter I mean a point which is a solution to the following minimization problem:
$$
\bar r := \inf_{x\in K} \min\{ r\geq0 \mid K\subseteq\bar B(x,r) \},
$$
where $\bar B(x,r)$ denotes the closed ball.
It is immediate to verify that the problem has a solution. Indeed, if $(x_n)_{n\in\mathbb N}\subset K$ is a minimizing sequence, that is, there exist $r_n\to\bar r$ such that $K\subseteq\bar B(x_n,r_n)$, possibly extracting a subsequence we can assume that $x_n\rightharpoonup\bar x$, then we have
$$
|y-\bar x| \leq \liminf_{n\to\infty} |y-x_n| \leq
\liminf_{n\to\infty} r_n = \bar r
\qquad \text{for all $y\in K$},
$$
which implies that $K\subseteq\bar B(\bar x,\bar r)$.
The remaining problem is the uniqueness. Assume that $x_1$ and $x_2$ are two circumcenters and set $\varepsilon=|x_1-x_2|$. By the uniform convexity, there exists $\delta>0$ such that $|u|\leq\bar r$, $|v|\leq\bar r$ and $|u-v|\geq\varepsilon$ imply $\left|\frac{u+v}2\right|\leq\bar r-\delta$. Now, if we take $y\in K$, we have $|x_1-y|\leq\bar r$, $|x_2-y|\leq\bar r$ and $|(x_1-y)-(x_2-y)|=\varepsilon$, therefore $\left|\frac{x_1+x_2}2-y\right|\leq\bar r-\delta$. But this contradicts the minimality of $\bar r$, because we have found $K\subseteq\bar B\left(\frac{x_1+x_2}2,\bar r-\delta\right)$.
To tackle the non uniformly convex case, I just have an idea, but I'm not sure if it's going to work. Maybe someone else can think about it. Given a weakly compact convex set $K_0$, we define $r_0$ as the optimal radius above and we consider the set $K_1=\{ x\in K \mid K\subseteq\bar B(x,r_0) \}$. The set $K_1$ is non-empty, convex (thanks to the convexity of the norm) and closed (I believe, by the same reasoning used for the existence). Thus, we have found another weakly compact convex set $K_1\subseteq K_0$. By iterating this argument, we find a nested family of compact sets $K_{n+1}\subseteq K_n$. Then the intersection $K_\infty=\bigcap_{n\in\mathbb N}K_n$ is non-empty. The hope is that one could get that $K_\infty$ is a singleton by proving something like $r_{n+1}\leq r_n/2$, or $\mathop{\mathrm{diam}}(K_{n+1})\leq\mathop{\mathrm{diam}}(K_n)/2$.
Note to future self for improvement
We have $\mathop{\mathrm{diam}}(K_{n+1}) \leq r_n \leq \mathop{\mathrm{diam}}(K_n) \leq 2r_n$.

The first inequality is sharp.
Example: $K_0=[0,1]\times[-1,1]\subset\mathbb R^2$ with the
$\lvert\,\cdot\,\rvert_\infty$ norm. $r_0=1$ and $K_1=[0,1]\times\{0\}$.
The third inequality is simply useless.
The second inequality is the one to be improved.

Existing literature
The improvement of the second inequality is related to the Jung's constant of a metric space $(X,d)$, given by
$$
J(X) = \sup\left\{\frac{2\mathrm{rad}(A)}{\mathrm{diam}(A)} \bigg\vert A\subseteq X\right\}
$$
where
$$
\mathrm{rad}(A) = \inf_{x\in X}\sup_{y\in A} d(x,y), \qquad\qquad
\mathrm{diam}(A) =\sup_{x,y\in A} d(x,y).
$$
We have already noted that $1\leq J(X)\leq 2$. If $J(X)<2$, then we get
$$
\mathrm{diam}(K_{n+1})\leq r_n\leq J(X)/2 \mathrm{diam}(K_n),
$$
therefore
$\mathrm{diam}(K_n) \leq \left(\frac{J(X)}2\right)^n \mathrm{diam}(K_0)$
converges geometrically to $0$.
This gives us another criterion for the uniqueness of the center: it is sufficient that $J(X)<2$. For example, $J(L^\infty)=1$ [Pichugov 1988] and we have the uniqueness of the center, even though $L^\infty$ is not uniformly convex.
See also this paper and this.
Failure in $L^1$
It is known that $J(L^1)=2$. I have found a nice example that proves this and that proves also that, unfortunately, the iterative procedure suggested above fails in $L^1$.
Consider in $L^1([0,1],\mathrm{leb}^1)$ the family $W=(w_n)_{n\in\mathbb N}$ of the wavelet functions
$$
w_n(x) = \begin{cases}
1  & \frac i{2^n}\leq x<\frac{i+1}{2^n},\ \text{$i$ even}, \\
-1 & \frac i{2^n}\leq x<\frac{i+1}{2^n},\ \text{$i$ odd},
\end{cases}
$$
It is immediate to compute $\lVert w_m-w_n\rVert_1=1$ for every $m\neq n$.
Therefore $\mathrm{diam}(W)=1$. We want to show that also $\mathrm{rad}(W)=1$. Given a function $f\in L^1([0,1],\mathrm{leb}^1)$ and $\varepsilon>0$ we can find $n\in\mathbb N$ and a function
$$
f_n = \sum_{i=0}^{2^n-1} a_i \chi_{\left[\frac i{2^n},\frac{i+1}{2^n}\right)}
$$
such that $\lVert f-f_n\rVert_1\leq\varepsilon$. (This can be done in many different ways: by hand, or using the fact that $W$ is complete in $L^2([0,1],\mathrm{leb}^1)$).
Then
$$
\lVert f_n-w_{n+1}\rVert_1 =
\sum_{i=0}^{2^n-1}
 \int_{\frac i{2^n}}^{\frac{i+1}{2^n}} \frac{|a_i-1|+|a_i+1|}2 dx \geq
\sum_{i=0}^{2^n-1}
 \int_{\frac i{2^n}}^{\frac{i+1}{2^n}} 1\,dx = 1,
$$
hence $\lVert f-w_{n+1}\rVert_1 \geq \lVert f-f_n\rVert_1 - \lVert f_n-w_{n+1}\rVert_1 > 1-\varepsilon$.
This proves that a ball containing $W$ must have radius at least $1$.
Furthermore, if we take $K=\mathrm{co}(W)$, then the ball of radius $1$ covering $K$ can be centered at any point of $K$, which means that $H=K$ and the procedure described above doesn't converge to a single point.<|endoftext|>
TITLE: Koszulness of the cohomology ring of moduli of stable genus zero curves
QUESTION [14 upvotes]: Let $n \geq 3$. The ring $H^\bullet(\overline{M}_{0,n},\mathbf Q)$ was determined by Sean Keel. It is generated by the cohomology classes of boundary divisors $D_{A,B}$ corresponding to partitions $A \sqcup B = \{1,\ldots,n\}$ of the marked points with $|A|, |B| \geq 2$, and where $D_{A,B} = D_{B,A}$. All relations are given by: (i) demanding that the product $D_{A,B} \cdot D_{A',B'}$ vanishes if the two divisors are disjoint (i.e. if there are no containments between the four sets $A,A',B$ and $B'$); (ii) the relation $$ \sum_{\substack{\{i,j\} \subseteq A \\ \{k,l\} \subseteq B}} D_{A,B} = \sum_{\substack{\{i,k\} \subseteq A \\ \{j,l\} \subseteq B}} D_{A,B} $$ which follows by pulling back the WDVV relation on $\overline{M}_{0,4}$ to $\overline M_{0,n}$. 
It follows in particular that $H^{2\bullet}(\overline{M}_{0,n},\mathbf Q)$ is a quadratic algebra. I was asked during a seminar today whether this algebra is Koszul, but I had no idea what to answer. So, is it Koszul? If so, is its Koszul dual interesting?

REPLY [11 votes]: It is: https://arxiv.org/abs/1902.06318 - this paper also explains how to use the Koszul dual algebra for something, where something is estimating Betti numbers of the free loop spaces of $\overline{M}_{0,n}$; those, by Gromov & Ballmann-Ziller, allow one to estimate the number of closed geodesics of bounded length.<|endoftext|>
TITLE: Is the intersection of boundaries of convex bodies a topological sphere?
QUESTION [5 upvotes]: Let $K_1, K_2, \ldots K_n$ be convex bodies in $R^d$. Assume that for any index set $I$, $\cap_{i \in I} K_i$ is not empty and is not properly contained in any body $K_i$ for $i\in I$.
Is it true that $\cap_{i \in I} \partial K_i$ is a disjoint union of topological (or even better PL) spheres?

REPLY [5 votes]: To get a "YES" answer, you have to assume that 
at any point of $p\in\partial K_i\cap \partial K_j$
any two supporting hyperplanes to $K_i$ and $K_j$ 
have angle $> \tfrac{\pi}2$.
The proof is by induction on $n$.
WLOG we may assume that all $\partial K_i$ are smooth.
Assume $S_{n-1}=\partial K_1\cap \partial K_2\cap \dots\cap K_{n-1}$ is a sphere.
Note that $f=\mathop{\rm dist}_{\partial K_n}$ is a concave function on $S_{n-1}\cap K_n$,
Perturb $f$ so it become smooth.
The function has one maximum point and by Morse Lemma the level set 
$S_n=f^{-1}(0)=\partial K_1\cap \partial K_2\cap \dots\cap K_{n}$ is a sphere.<|endoftext|>
TITLE: Existence of dg realization for 6 functors
QUESTION [26 upvotes]: Is there a way  to lift 6 functors on constructable sheaves to the dg world?

REPLY [4 votes]: This answer only goes half way, but I think it is worth pointing out that the inner workings of the six functor formalism have been studied a lot in the motivic literature. 
Motivic six functor formalisms: The definite reference would be the thesis of Ayoub which you can find on his homepage. The formalism of cross functors (based on unpublished works of Voevodsky and Deligne) gives a general framework how to set up the whole six functor formalism. Another relevant reference is the work of Cisinski-Déglise on triangulated categories of mixed motives. They develop the framework further, with the notions of motivic categories fibered over the category of schemes.
My point is that the input to these frameworks can be things more general than triangulated categories, these frameworks work with stable model categories, $\infty$-categories or dg-categories (as long as you feed in the right data). In the motivic setting, the framework is usually applied to yield six functors for categories of motives. But if you check the validity of the axioms for étale sheaves or $\ell$-adic sheaves, the framework would give you dg-versions of the corresponding derived categories. You should also look at the mixed Weil cohomologies paper of Cisinski-Déglise: plugging in $\ell$-adic cohomology in their constructions gives you dg-versions of $\ell$-adic derived categories. 
A grain of salt: The frameworks above do not (explicitly) produce dg-enhancements of the six functors. Four of the functors are easy to deal with because they are derived functors of functors on the level of abelian categories - so they have dg-enhancements. Note, however, the exceptional functors are only constructed on the triangulated level in the abovementioned references. But I think that these can be upgraded to dg-versions: e.g. the construction of $f_!$ via a colimit over the category of compactifications of $f$ and then using the adjoint $i_!$ for an open immersion and $p_\ast$ for a proper map should work in the corresponding dg-setting. Similarly, for the definition of $f^!$ one could apply a version of the $\infty$-categorial adjoint functor theorem, together with suitable compact generation properties. If it's possible, it's only going to be a matter of time before an $\infty$-version of these formalisms will be available in the motivic world.
Constructibility conditions: The frameworks above usually work with fairly big categories. However, in the paper of Cisinski-Déglise, you can find the description of compact objects in these big categories - they agree with the classically defined constructible objects. Moreover, under rather weak assumptions, the six functors also preserve compact objects - the framework above then gives you dg-versions of $\ell$-adic constructible sheaves.
Another grain of salt: Ok, this is all for the algebraic setting, working over schemes and such. If you are more interested in the locally compact topological setting, the literature mentioned above probably does not apply directly. However, all the techniques are there, and I am fairly sure that the framework can be adapted to this setting as well.<|endoftext|>
TITLE: Refining open covers in locally path connected spaces
QUESTION [8 upvotes]: Suppose $X$ is a locally path connected topological space and $\mathcal{U}$ is an open cover of $X$ (consisting of path connected sets if we want). 
One often wants the intersection $A\cap B$ of pairs of elements $A,B\in \mathcal{U}$ to be path connected, or perhaps stronger, that the intersection of finitely many elements of $\mathcal{U}$ be path connected. This, for instance, is the case in some groupoid versions of the van Kampen theorem (like the one in Peter May's A Concise Course in Algebraic Topology). Having covers with this property also simplifies life a bit in the study of shape invariants constructed via nerves of covers. If we don't start with a cover this nice perhaps we can at least get to one by refinement.
Question: Is it always possible to find an open cover $\mathcal{V}$ of $X$ refining $\mathcal{U}$ such that the intersection of every pair of elements in $\mathcal{V}$ is path connected (or empty)? Can we do even better and find $\mathcal{V}$ such that the intersection of finitely many elements of $\mathcal{V}$ is path connected?
I'm less confident such refinement is possible for general locally path connected $X$. I'd be perfectly content to assume $X$ is paracompact Hausdorff.

REPLY [6 votes]: If $X$ is a smooth manifold, there is always a refinement to a given open cover such that any finite intersection is connected. Namely, choose a complete Riemannian metric on $X$ and choose a refinement by geodesically convex set. 
One can surely push this property to somewhat more general spaces like metric spaces which locally admit unique paths realizing the distance. Maybe, simplicial complexes can be equipped with a metric with this property.<|endoftext|>
TITLE: blow-ups of secant varieties
QUESTION [8 upvotes]: Let $C$ be a curve embedded in ${\mathbb P}^n$ by a full linear system  (I am particularly interested in the case of an elliptic curve but it seems natural to ask the question in more generality). Let $S_k$ be the $k$-th secant variety, i.e. the union of all $k$ planes 
intersecting $C$ at (at least) $k+1$ points. Construct $X_k$ inductively:
$X_1$ is the blow-up of ${\mathbb P}^n$ along $C$, $X_2$ is the blow-up of $X_1$ along the proper preimage of $S_1$ in $X_1$ etc. Is there a reasonable moduli interpretation for $X_i$, especially for the last one $X_d$, $d=[(n-3)/2]$?
For a particular value of $n$ Bertram identified $X_d$ (or something similar) with the 
moduli space of semi-stable rank 2 vector bundles on $C$; but I am interested in the case
of an arbitrary $n$.

REPLY [2 votes]: You can probaly describe your space as some kind of fibration in $\mathcal{M}_{0,m}$ over the dual linear system $\mathbb{P}^{n*}$, via Kapranov's blow-up construction of $\overline{\mathcal{M}_{0,m}}$ . This is done in
http://arxiv.org/abs/0903.5515
for some of the Bertram's cases but it should hold in general. Your space may have an interpretation in terms of moduli of vector bunldes (or sheaves).<|endoftext|>
TITLE: Deciding the convexity of semialgebraic sets
QUESTION [6 upvotes]: Given a basic closed semialgebraic set, $S \subset \mathbb{R}^n$, defined by
$S = \{ x \in \mathbb{R}^n \mid g_1 (x) \geq 0 \land \dots \land g_m (x) \geq 0\}$
where $m \in \mathbb{N}$ and $g_1, \dots, g_m \in \mathbb{R}[x]$, how can one decide if $S$ is convex? Some context: this question arose while reading Schweighofer's slides on LMI representations of convex semialgebraic sets [pdf].
Let us introduce the predicate $p (x) = \bigwedge_{i=1}^m g_i (x) \geq 0$, so that we can write $S$ in the more parsimonious form $S = \{ x \in \mathbb{R}^n \mid p (x)\}$. From Boyd & Vandenberghe, we have the following definition:

A set $C$ is convex if the line
  segment between any two points in $C$
  lies in $C$, i.e., if for any $x_1,
> x_2 \in C$ and any $\theta$ with $0 \leq \theta \leq
> 1$, we have $\theta x_1 + (1-\theta)
> x_2 \in C$.

Hence, $S$ is convex if and only if the following universally quantified formula
$\forall x_1 \, \forall x_2 \, \forall \theta \, \left[\, p(x_1) \land p(x_2) \land (\theta \geq 0 \land \theta \leq 1) \implies p (\theta x_1 + (1-\theta) x_2) \, \right]$
where $x_1, x_2$ range over $\mathbb{R}^n$ and $\theta$ ranges over $\mathbb{R}$, evaluates to true. The formula above can be decided using a quantifier elimination package like QEPCAD or REDLOG. 
Question: other than quantifier elimination, is there any procedure that would allow one to decide the convexity of a given basic closed semialgebraic set?

REPLY [6 votes]: In this paper by Ahmadi et al the authors show that a very special case of this question (whether a polynomial is convex, so the set is the epigraph) is NP hard in many cases, so no really easy algorithm is likely to exist.<|endoftext|>
TITLE: Beautiful descriptions of exceptional groups
QUESTION [51 upvotes]: I'm curious about the beautiful descriptions of exceptional simple complex Lie groups and algebras (and maybe their compact forms). By beautiful I mean: simple (not complicated - it means that we need not so many words to describe this).
For $G_2$ we know the automorphisms of octonions and the rolling distribution (and also the intersection of three $Spin_7$-s in $Spin_8$).
For $F_4$ we know the automorphisms of Jordan algebra $H_3(\mathbb O)$ and Lie algebra of commutators of right multiplications in this algebra (see Chevalley-Schafer's paper for details).
For $E_6$ we know the automorphisms of determinant in $H_3(\mathbb O)$ and Lie algebra linearly spanned by right multiplications and $\mathfrak f_4$.
For $\mathfrak f_4$, $\mathfrak e_6$, $\mathfrak e_7$, $\mathfrak e_8$ we know the Vinberg-Freudenthal Magic Square.
What do we know (expressing in a simple form) about $E_7$ and $E_8$?

REPLY [10 votes]: Here is a description that is new and you can judge whether it is beautiful.  Given any simple complex Lie group $G$ and almost any irreducible representation $V$, the stabilizer of almost any $G$-invariant polynomial $f$ on $V$ has identity component $G$.
Cartan's examples 

$G = E_6$, $V$ of dimension 27, and $f$ cubic; or
$G = E_7$, $V$ of dimension 56, and $f$ quartic

are very special cases of this general principle.  (They are very special because in these cases the ring of $G$-invariant polynomials on $V$ is generated by $f$.)
In the case of the group $E_8$, you can take $V$ to be the Lie algebra $\mathfrak{e}_8$.  Then the ring of invariant polynomial functions is a polynomial ring with generators of degree 2 (the Killing quadratic form), 8, 12, 14, 18, 20, 24, 30.  The new result says: If you take $f$ to be any of the generators besides the Killing form, then $E_8$ is the identity component of the stabilizer of $f$.
This is a very concrete description of $E_8$, because an explicit formula for the degree 8 polynomial is already in the literature (Cederwall and Palmkvist - The octic $E_8$ invariant (arXiv)).
Alternatively, there is a commutative, nonassociative, and $E_8$-invariant product on its 3875-dimensional irreducible representation, and the automorphism group of this nonassociative ring is $E_8$.
There is also a variation on the result I mentioned at the beginning that may be worth mentioning: you can also realize each simple complex Lie group $G$, up to isogeny, as the stabilizer of a cubic form on some representation.  For $E_8$, you can take the cubic form to be the one defining the multiplication on the 3875-dimensional representation.
The new results mentioned here are from Garibaldi and Guralnick - Simple groups stabilizing polynomials (MSN, arXiv).<|endoftext|>
TITLE: Clique weight-optimal matchings on n-partite graphs.
QUESTION [8 upvotes]: I am trying to analyze the results of a physical experiment consisting of $n$ "runs" of measurements, each of which generates a set of $k$ points in Euclidean space. The following problem came up when trying to quantify the difference between these point sets. I would be extremely surprised if the problem is not already solved, but I was unable to find a solution in graph theory textbooks and on google.
I will outline the problem for $n=3$ because it is the first non-trivial case. Hopefully the solution does not fail for higher $n$. Consider a $3$-partite graph $G$ with vertex bins that we will call Red, Blue and Green. Assume that each bin has the same number of vertices, say $k>0$. Assume now that this graph is complete in the standard sense, i.e., there exists a (positively) weighted edge between any two differently colored vertices. Completeness puts us in a different situation from the one discussed here.
By a matching $m$ I mean a partition of the vertices into $k$ $3$-cliques of the type (red, blue, green). Each clique $C$ is assigned a weight $w(C)$ which equals the maximum of the weights of the edges in that clique. So a clique consisting of vertices R,B and G is assigned the max weight of the three edges RB, RG and BG. The weight of the matching is defined to be $$W(m) = \max \lbrace w(C)~|~C \text{ is a clique of } m\rbrace.$$
Here is the question: 

Is there an efficient algorithm to compute $$\inf \lbrace W(m)~|~m \text{ is a matching of }G\rbrace ?$$

Note that the trivial algorithm which consists of constructing all possible matchings is known to me but happens to be a computational nightmare. If this is a known / non-research problem, I apologize. Please just point me to the correct references or search terms in this case.

REPLY [5 votes]: The problem is NP-complete which follows by reduction from the problem if a given 3-partite graph $G=(A\cup B\cup C,E)\ $ with $\lvert A\rvert=\lvert B\rvert=\lvert C\rvert=k\ $ can be partitioned into $k$ disjoint triangles. Here this problem is shown to be NP-hard by reduction from 3-dimensional matching.
The triangle partition problem can be reduced to the problem from the question by assigning weight $1$ to all edges and weight $2$ to all non-edges. Then the triangle partition exists if and only if the optimal matching (in the sense of the question) has weight 1.<|endoftext|>
TITLE: Convexity of a specific semialgebraic set
QUESTION [5 upvotes]: I have an engineering problem that may be solved using semidefinite programming. I would like to know whether a given set is convex.
Let $m \in \mathbb{R}^+$ be a positive real scalar, $l \in \mathbb{R}^3$ be a real vector, and $L \succ 0$ be a size $3 \times 3$ real positive definite matrix. Does the following inequality
$$L - m\ S\left(\frac{l}{m}\right)^T S\left(\frac{l}{m}\right) \succ 0$$
where $S(\cdot)$ denotes the skew-symmetric matrix operator and $\succ 0$ denotes positive definiteness, define a convex set?

I.e.:
Being
$$m \in \mathbb{R}$$
$$l \equiv \left[l_x\ l_y\ l_z\right]^T$$

    $$
L \equiv \left[\begin{matrix}
      L_{xx} & L_{xy} & L_{xz} \\
      L_{xy} & L_{yy} & L_{yz} \\
      L_{xz} & L_{yz} & L_{zz} \\
    \end{matrix}\right]
$$

The variables $m,l_x,l_y,l_z,L_{xx},L_{xy},L_{xz},L_{yy},L_{yz},L_{zz}$ define a $\mathbb{R}^{10}$ space.
The constraints

    $$
\left\{
\begin{matrix}
  m &> 0 \\
  L &\succ 0\\
  L - m\ S\left(\frac{l}{m}\right)^T S\left(\frac{l}{m}\right) &\succ 0
\end{matrix}
\right.
$$

which, since $m>0$, are equivalent to

    $$
\left\{
\begin{matrix}
  m &> 0 \\
  L &\succ 0\\
  m L - S\left(l\right)^T S\left(l\right) &\succ 0
\end{matrix}
\right.
$$

define a semialgebraic set on the $\mathbb{R}^{10}$ variables space.
Here, $\succ 0$ means that the left argument is a positive-definite matrix, and,

    $$
S(x) = \left[\begin{smallmatrix}
  0 & -x_3 & x_2 \\
  x_3 & 0 & -x_1 \\
  -x_2 & x_1 & 0
\end{smallmatrix}\right]\quad\text{with}\quad
x = \left[x_1\ x_2\ x_3\right]^T
$$


I did some, manipulation and rewrote the last constraint as a polynomial inequalities system:
being

    $$
mI = m L - S\left(l\right)^T S\left(l\right) =
\left[\begin{smallmatrix}
  L_{1xx} m_{1} - l_{1y}^{2} - l_{1z}^{2} & L_{1xy} m_{1} + l_{1x} l_{1y} & L_{1xz} m_{1} + l_{1x} l_{1z} \\
  L_{1xy} m_{1} + l_{1x} l_{1y} & L_{1yy} m_{1} - l_{1x}^{2} - l_{1z}^{2} & L_{1yz} m_{1} + l_{1y} l_{1z} \\
  L_{1xz} m_{1} + l_{1x} l_{1z} & L_{1yz} m_{1} + l_{1y} l_{1z} & L_{1zz} m_{1} - l_{1x}^{2} - l_{1y}^{2}
\end{smallmatrix}\right]
$$

then, through Sylvester's criterion,
  
    $$
mI \succ 0 \Leftrightarrow
\left\{
\begin{matrix}
\det\left(mI_{1,1}\right) = L_{1xx} m_{1} - l_{1y}^{2} - l_{1z}^{2} &> 0\\
\det\left(mI_{1:2,1:2}\right) &> 0\\
  \det\left(mI\right) &>0
\end{matrix}
\right.
$$

It would be sufficient that the polynomials were concave to guarantee set convexity, however they are not concave.
Although not being concave, it does not imply that set is not convex; for example, the first polynomial $L_{1xx} m_{1} - l_{1y}^{2} - l_{1z}^{2}$ is not concave itself but defines a convex set if constraint $m>0$ is taken into account.
(This representation also gave me some suspicions that maybe the $L \succ 0$ constraint is implicit on the other.)
I also tried to write the set as a linear matrix inequality (LMI), but I couldn't (my knowledge in this area is really short). 

Update:
I was able to check that this set is close under positive scalar multiplication, since
$$ (\gamma\ m) (\gamma\ L) - S\left(\gamma\ l\right)^T S\left(\gamma\ l\right) = \gamma^2 \ \left( m L - S\left(l\right)^T S\left(l\right)\right) \succ 0 \quad \text{for} \quad \gamma > 0$$
then it is a cone. If one can prove the set is close under addition then it will be proven to be a convex cone.


Now, the questions are:
Which methods can I use to check if the defined set is convex?
If so, is it possible to represent it as an LMI?

REPLY [3 votes]: Your set is indeed a convex cone.
Since it is a cone, it suffices to show that the $m=1$ section is convex. 
But this is equivalent to show that the (symmetric matrix valued) "function" $u\mapsto P(u)=S(u)^*S(u)$ is "convex", i.e. $P((u+v)/2)\prec (P(u)+P(v))/2$, because the set is basically the "epigraph" $(u,L)$ : $L\succ P(u)$.
Now it is easily checked that the quadratic function $P$ satisties the parallelogram identity $P((u+v)/2)+P((u-v)/2)=(P(u)+P(v))/2$, which does the job since $P$ is positive.
EDIT : in fact any set of $(x,y)$ defined by an inequality $S(y)-A(x)^*A(x) \succ 0$, with $S(y)$ $n\times n$ symmetric and linear in $y$, and $A(x)$ $n\times d$ and linear in $x$, is convex and moreover defined by a Linear Matrix Inequality.
Indeed this is equivalent to 
$$\left(\begin{matrix}
          S(y) & A(x)^*  \\\
          A(x) & I  \end{matrix}\right) \succ 0$$
as easily seen by row and column operations. In fact substituting $I$ by $\lambda I$, you can "re-homogenize" the problem.<|endoftext|>
TITLE: Largest graphs of girth at least 6
QUESTION [21 upvotes]: Let $e_6(n)$ be the greatest number of edges in a simple graph with $n$ vertices and girth at least 6.
Let $G_6(n)$ be the set of simple graphs of order $n$ with girth at least 6 and $e_6(n)$ edges.
My question: Is there any $n$ for which none of the graphs in $G_6(n)$ is bipartite?
From computer experiments, I have found that the only values of $n\le 50$ for which $G_6(n)$ has any non-bipartite graphs at all are 7 (7 edges), 9 (10 edges), 15 (22 edges), 27 (53 edges), and 43 (106 edges). However, in all those cases $G_6(n)$ includes bipartite graphs as well.
A table (needs checking, please don't cite yet): "[n=44,e=108,g=12]" means $e_6(44)=108$ and there are 12 graphs. All the graphs are bipartite unless the notation is like "[n=15,e=22,g=2+1]" which means there are two bipartite graphs and one non-bipartite graph.
[n=5,e=4,g=3], [n=6,e=6,g=1], [n=7,e=7,g=1+1], [n=8,e=9,g=1], [n=9,e=10,g=3+1], [n=10,e=12,g=3], [n=11,e=14,g=1], [n=12,e=16,g=1], [n=13,e=18,g=1], [n=14,e=21,g=1], [n=15,e=22,g=2+1], [n=16,e=24,g=4], [n=17,e=26,g=4], [n=18,e=29,g=1], [n=19,e=31,g=1], [n=20,e=34,g=1], [n=21,e=36,g=3], [n=22,e=39,g=2], [n=23,e=42,g=1], [n=24,e=45,g=1], [n=25,e=48,g=1], [n=26,e=52,g=1], [n=27,e=53,g=2+2], [n=28,e=56,g=1], [n=29,e=58,g=1], [n=30,e=61,g=1], [n=31,e=64,g=1], [n=32,e=67,g=5], [n=33,e=70,g=3], [n=34,e=74,g=1], [n=35,e=77,g=1], [n=36,e=81,g=1], [n=37,e=84,g=3], [n=38,e=88,g=2], [n=39,e=92,g=1], [n=40,e=96,g=1], [n=41,e=100,g=1], [n=42,e=105,g=1], [n=43,e=106,g=2+3], [n=44,e=108,g=12], [n=45,e=110,g=183], [n=46,e=115,g=1], [n=47,e=118,g=1], [n=48,e=122,g=1], [n=47,e=118,g=1], [n=48,e=122,g=1], [n=49,e=126,g=1], [n=50,e=130,g=1].
Update Nov 2015: For $51\le n\le 63$, all extremal graphs are bipartite except for $n=63$, where there are 3 bipartite extremal graphs and 4 non-bipartite extremal graphs (187 edges).

REPLY [2 votes]: I am collecting some varied thoughts on the problem, in the hopes that it will inspire someone to finish the problem.
I suggested earlier that the graphs in $G_{n+1}$ could be built incrementally from graphs in $G_n$ by adding one vertex and thee appropriate number of edges.  Brendan McKay assured me that this would not be possible for $n=44$ as "that graph had too many edges", to reinterpret his assurance.  Even so, it might be useful to consider the subgraph relation on the union of the $G$'s and see if most of them can be built up incrementally, and to characterize the ones that aren't and are primitive in some sense.
It is clear that removing one vertex and its adjacent edges from an example in $G_{n+1}$ does not reduce the minimum girth, and that adding a vertex and single edge also does not reduce the girth, so that the function $e(n)$ is increasing in $n$ for $n>4$ and further increases by no more than the minimum degree taken over all the vertices of all the members of $G_{n+1}$.
There likely is a nice argument bounding the maximum degree among all members of $G_n$, but I don't see it.  I can build a graph on an even number of vertices by gluing a number of length 3 paths together at their endpoints, but this gives an average degree of slightly less than 3 and a max degree of slightly less than n/2, so this is useful more for providing a lower bound for $e(n)$ than anything else.
Another construction giving a bipartite involves associating each point in a set L with a small subset (of size 3, say) of another set R in a way that no two subsets of R so chosen intersect in more than one point.  The result has girth 6 or more and if both L and R have 7 points, a maximal example resembles a BIBD (or for me, an adjacency matrix of 0's and 1's with order 7 and absolute determinant value of 24) which I believe corresponds to Brendan's example for $n=14$.  Perhaps BIBD's contribute more examples?   They might be a significant subclass of the primitive graphs in the subgraph relation I mention above.
Also, why so many graphs for $n=45$?  It makes me think of the combinatorial explosion of equivalence classes of Hadamard matrices, although it might be better to think of equivalence classes (under row and column permutations and perhaps under switching as well) of 0-1 matrices having maximal determinant values.  Are there combinatorial analogues in the literature which might suggest such a brief plethora of examples?
Gerhard "Binary Matrices On My Mind" Paseman, 2012.06.28<|endoftext|>
TITLE: On Grothendieck's period relations
QUESTION [9 upvotes]: Let $V$ be a smooth projective variety defined over $\mathbf{Q}$ and denote by
$$
\omega: H_{dR}^*(V,\mathbf{Q})){\otimes_{\mathbf{Q}}}\mathbf{C}\rightarrow H_{B}^*(V,\mathbf{Q})\otimes_{\mathbf{Q}}\mathbf{C},
$$
Grothendieck's comparison isomorphism between algebraic De Rham cohomology and Betti cohomology. Choosing $\mathbf{Q}$-rational basis on both sides we may think of $\omega$ as given by a square matrix. 
In many places in the literature it is said that algebraic cycles (defined over $\mathbf{Q}$) on the $n$-iterated product of $V$, namely $V^n$, give rise to polynomial relations in the entries of the matrix $\omega$.
Q: How does one obtain such polynomial relations?

REPLY [12 votes]: Here is an example to flesh out Keerthi's nice answer in a very simple case.
Take $V$ to be the $0$-dimensional variety $V=\operatorname{Spec} F $ where $F/\mathbf{Q}$ is a finite Galois extension of degree $d$. The left hand side is given by $H^0_{dR}(V) = F$, while $V(\mathbf{C})$ is the finite set $\Sigma=\operatorname{Hom}(F,\mathbf{C})$, so that $H^0_B(V(\mathbf{C}),\mathbf{Q}) \cong \mathbf{Q}^{\Sigma}$. The comparison isomorphism is then just the usual isomorphism $\omega : F \otimes \mathbf{C} \xrightarrow{\cong} \mathbf{C}^\Sigma$.
Choosing a basis $(a_1,\ldots,a_d)$ of $F$ over $\mathbf{Q}$ and the canonical basis of $\mathbf{Q}^{\Sigma}$, the entries of the matrix of $\omega$ are just $\sigma(a_i)$ with $\sigma \in \Sigma$ and $1 \leq i \leq d$. Each time you have a polynomial relation $P(a_1,\ldots,a_d)=0$ with $P \in \mathbf{Q}[X_1,\ldots,X_d]$ (which you can interpret as an algebraic cycle on $V^d =\operatorname{Spec}  F^{\otimes d}$), you get a corresponding relation on the $\sigma(a_i)$'s. As you have probably guessed, the motivic Galois group in this case is just $\operatorname{Gal}(F/\mathbf{Q})$ seen as a finite algebraic subgroup of $\mathrm{GL}(H^0_B) \cong \mathrm{GL}_{d/\mathbf{Q}}$. (NB : this "definition" of the Galois group is actually closer in spirit to Galois's original definition.)
Another example is given by an elliptic curve $E/\mathbf{Q}$ which has CM by an order in an imaginary quadratic field $K$. In this case $\mathrm{GL}(H^1_B(E)) \cong \mathrm{GL}_{2/\mathbf{Q}}$. The complex multiplication provides an algebraic cycle on $E \times E$, and the motivic Galois group is the normalizer of $K^\times$ seen as a subgroup of $\mathrm{GL}_{2/\mathbf{Q}}$.<|endoftext|>
TITLE: Generating All Permutations Without Repetition Using Two Generators
QUESTION [12 upvotes]: It is well known that all symmetric group can be generated using two generators
The two generators are:
1) $(1,2)$
2) $(1,2,3,\dots ,n)$

Question: Is there a deterministic algorithm to generate all permutations without repetition using only these two generators?
  (Bonus 1: The algorithm generates the permutations in a cycle. Bonus 2: Not requiring the inverse of generator 2)
  Edit: As point out by John, this is equivalent to a Hamiltonian path in the Cayley graph of $S_n$ with these two generators.

It is easy to generate all of them without repetition using $n-1$ generators, by the
Steinhaus-Johnson-Trotter algorithm.
It is easy to generate all of them, with repetition, using two generators.
However I was unable to find a way to generate all without repetition and using only two generators.
As this approach seems natural, I suspect someone should have worked on it but I was unable to find any references online.  
Does anyone knows the status of this problem?

REPLY [3 votes]: There has been a recent paper by Sawada and Williams where the problem is solved for the harder variant where you always shift in the same direction, i.e., you never require the inverse of generator 2 (Bonus 2 problem).
The paper is available here:
https://epubs.siam.org/doi/abs/10.1137/1.9781611975031.37
A demonstration of that algorithm can be run on the Combinatorial Object Server website:
http://page.math.tu-berlin.de/~muetze/cos/
Go to "Permutations", then select "Prefix swaps and rotations (Sawada-Williams)".<|endoftext|>
TITLE: Strictly Positive Measures on Countable Boolean Algebras
QUESTION [7 upvotes]: Let $B$ be a Boolean Algebra. 
A strictly positive measure on $B$ is a function $m$ from $B$ to $[0,1]$ such that (i) $m(b)=0$ iff $b=0$,  (ii) $m(1)=1$, and (iii) $m(a+b)=m(a)+m(b)$ whenever $a$ and $b$ are disjoint.
Is there a strictly positive measure on every countable Boolean Algebra?

REPLY [8 votes]: Yes. Let $M$ be the space of all measures on $B$. This is a compact space when endowed with with the pointwise convergence topology since it is a closed subspace of $[0,1]^B$. If $b$ is a nonzero element of $B$, then the set $U_b = \lbrace m \in M : m(b) \gt 0 \rbrace$ is open and dense in $M$. By the Baire Category Theorem, the intersection of all these sets is nonempty.

For an explicit construction, let $b_1,b_2,\ldots$ enumerate $B\setminus\lbrace0\rbrace$ and for each $n$ let $m_n$ be a measure on $B$ such that $m_n(b_n) \gt 0$ (e.g. let $m_n$ be the characteristic function of an ultrafilter containing $b_n$). Then $m = \sum_{n=1}^\infty 2^{-n}m_n$ is as required.<|endoftext|>
TITLE: Probability that random weights on $K_n$ satisfy triangle inequality
QUESTION [20 upvotes]: Given $K_n$, if a random real weight between $[0, 1]$ is chosen for every edge, what is the probability that the graph satisfies the triangle inequality? How about the discrete version, where the weights are integers in $[0, k]$?
It is easy to see that if $n = 3$ the probability is $1/2$ and (empirically) that the probability approaches zero as n goes to $\infty$. Has anyone studied the problem before? Any exact or asymptotic results are appreciated.
Notes:

This question was posted on math.SE here. I got no answers and it seems pretty inactive at the moment.
This is my first question on mathoverflow, so I am sorry if this is not research-level, but it seems to me that it is.

REPLY [5 votes]: Based on a conversation with Dan Romik, here is a generalization of Doug's bounds on volume.
Let $H(n,t)$ be the hypergraph of all $t$-tuples of a set with $n$ elements, and let $n > k > t$ be another integer.  Suppose each hyperedge $T$ is colored by an i.i.d. random variable $x_T$ that takes values in some measure space $X$.  Suppose furthermore that for each subset of $k$-subset $K$, there is some non-trivial symmetric, measurable restriction $R \subset X^{\binom{k}{t}}$ on the colors $x_T$ for $T \subseteq K$.  Let $P(n)$ be the probability that all of the restrictions on the coloring of $H(n,t)$ are satisfied.
Theorem:  For every $m \ge k$,
$$\limsup_{n \to \infty} \frac{\log P(n)}{n^t} \le \frac{\log P(m)}{m^t}.$$
Corollary: The limit 
$$\alpha = \lim_{n \to \infty} \frac{\log P(n)}{n^t}$$
exists, and one obtains better and better bounds on $\alpha$ by computing $P(m)$ for specific values of $m$, beginning with the case $m=k$ which implies that $\alpha < 0$.  In general one obtains $\alpha \in [-\infty,0)$.
Proof.  The theorem is a corollary of Rödl's theorem that there exists a packing of blocks of size $k$ which are disjoint on hyperedges of $H(n,t)$, and which cover a fraction of the $t$-tuples that converges to 1 as $n \to \infty$.
Theorem: (1) If the condition $R$ contains a cube $I^{\binom{k}{t}}$, where $I \subset X$ is some event with positive measure, then $\alpha > -\infty$.  (2) If there is a finite partition $\{I_i\}$ of $X$ such that $R$ is disjoint from each $I_i^{\binom{k}{t}}$, then $\alpha = -\infty$ because $P(n) = 0$ when $n$ is large enough.
For instance, suppose that $X$ is a compact Riemannian manifold with Riemannian measure.  Then condition (1) is satisfied if the interior of $R$ contains at least one point on the diagonal.  Condition (2) is satisfied if the closure of $R$ is disjoint from the diagonal.
Proof. Case (1) is just the remark that the probability $P(n)$ is at least the probability of landing in $I^{\binom{n}{t}}$.  Case (2) follows from Ramsey's theorem.<|endoftext|>
TITLE: Lifting isomorphisms between derived categories
QUESTION [7 upvotes]: (Remark: I first asked this question at math.stackexchange. As it received no answer, I'm posting it here).
Suppose $A$ and $B$ are commutative rings. Let $A\to B$ be a surjective ring homomorphism. I will denote by $D(A)$ and $D(B)$ the derived categories of unbounded complexes over $A$ and $B$.
Suppose $M,N \in D(B)$ are two complexes over $B$. Let $F:D(B)\to D(A)$ be the forgetfull functor.
Suppose that we know that $F(M) \cong F(N)$. Does it follows that $M\cong N$ in $D(B)$?
If we had a quasi-isomorphism $F(M) \to F(N)$, then it will of course lift to $D(B)$, because since $A\to B$ is surjective, an $A$-linear map of complexes over $B$ will automatically be $B$-linear.
However, isomorphisms in the derived category might pass through a third object $K$, which might not be defined over $B$. Thus, I suspect the answer to my question is no, but I have no idea how to find a counterexample.
Thank you for any idea!

REPLY [13 votes]: Let $A=k[x]$ and $B=k[x]/(x^2)$, let $X$ be the complex $\hskip{.1in}\dots\to 0 \to B\stackrel{x}{\to} B\to 0\to \dots$, and let $Y$ be $\hskip{.1in}\dots\to 0\to k\stackrel{0}{\to}k\to 0\to\dots$. Then $X$ and $Y$ are isomorphic in $D(A)$, but not in $D(B)$.
The point is that $X$ is isomorphic to the third object in a triangle containing a map $\zeta:k\to k[2]$, where $\zeta$ represents an element in the kernel of $\operatorname{Ext}^2_B(k,k)\to\operatorname{Ext}^2_A(k,k)$.<|endoftext|>
TITLE: Total ring of fractions vs. Localization
QUESTION [18 upvotes]: Let $R$ be a commutative ring and denote by $K(R)$ its total ring of fractions, the localization of $R$ with respect to $R_{\mathrm{reg}}$. For every multiplicative subset $U \subseteq R$ there is a canonical map of $R$-algebras
$$H : K(R)[U^{-1}] \to K(R[U^{-1}]).$$
This may fail to be surjective; see Kleiman's article about Misconceptions about $K_X$, available here. At least it is true when $R$ is reduced and $\mathrm{Spec}(R)$ has finitely many irreducible components.
Question. Is $H$ an isomorphism for every reduced commutative ring $R$?
Actually this is Exercise 3.15a in Eisenbud's book on commutative algebra (page 113), but my gut feeling is that the answer is no in general and that this is another misconception about $K(R)$ or Eisenbud forgot to mention the noetherian hypothesis. There is a hint on page 716 which says that $K(R)$ is zero-dimensional; well again this is only clear when $R$ is noetherian.
What I have done so far: The map $H$ is well-defined by the universal properties. One checks that $H$ is injective. It is easy to see that $H$ is surjective iff the following property holds, where $\tau : R \to R[U^{-1}]$ denotes the canonical localization map:
$(*)$ If $r \in R$ such that $\tau(r) \in R[U^{-1}]$ is regular, then there is some regular $s \in R$ such that $\tau(r) | \tau(s)$.
The general case can be easily reduced to $K(R)$; we arrive at the
Equivalent question. Is the class of reduced rings with the property that every regular element is a unit (sometimes called "classical ring", see MO/42647) stable under localization?
If $R$ is zero-dimensional and reduced, this means that every element is associated to some idempotent element (see Gilmer's Zero dimensional rings); the same then must be true for localizations. Hence nothing goes wrong for zero-dimensional rings. This applies, in particular, to artinian rings, and therefore also to finite rings. But I strongly suspect that an infinite product of finite nontrivial rings has a chance to be a counterexample, for example $\prod_{n} \mathbb{Z}/p^n$. This is is a reduced (EDIT: no!) classical ring of positive dimension (thus contradicting Eisenbud's hint). Is the localization at $p$ also classical?
More general. What can be said about the structure of reduced classical rings?

REPLY [14 votes]: A counterexample from one of my other MO answers seems to work again. Let $k$ be an algebraically closed field. Let $R$ be the ring of functions $f: k^2 \to k$ such that there exists a polynomial $\overline{f} \in k[x,y]$ with
1. $f(x,y) = \overline{f}(x,y)$ for all but finitely many $(x,y) \in k^2$ and
2. $f(0,0) = \overline{f}(0,0)$.
$R$ is reduced: If $f(x,y) \neq 0$, then $f(x,y)^n \neq 0$ for all $n$. $\square$
Every element $f$ of $R$ is either a unit or a zero divisor: 
Case 1: $f$ is nowhere zero. In this case, $\overline{f}$ must lie in $k^{\ast}$, as otherwise $\overline{f}$ vanishes at infinitely many points and $f=\overline{f}$ at all but finitely many of them. So $f^{-1}$ is equal to the nonzero constant $\overline{f}^{-1}$ at all but finitely many points, and $\overline{f}^{-1} \in R$. So, in this case, $f$ is a unit.
Case 2: $f(x_0,y_0)=0$. Without loss of generality, we may assume that $(x_0,y_0) \neq (0,0)$. This is because, if $f(0,0)=0$ then $\overline{f}(0,0)=0$, implying that $\overline{f}$ vanishes at infinitely many points and $f=\overline{f}$ at all but finitely many of them, so we can find some other $(x_0,y_0)$ at which $f$ also vanishes. Let $\delta(x,y)$ be $1$ if $(x,y) = (x_0, y_0)$ and $0$ otherwise. Then $f \delta=0$ and $\delta \neq 0$, showing that $f$ is a zero divisor. $\square$
The set of functions vanishing at $(0,0)$ is clearly a maximal ideal of $R$; which we will denote $(0,0)$. We claim that $R_{(0,0)} \cong k[x,y]_{(0,0)}$. Proof sketch: We claim that $f = \overline{f}$ in the localization. To see this, let $g$ vanish at the finitely many points where $f \neq \overline{f}$, but $g(0,0) \neq 0$. Then $fg=\overline{f} g$, and $g$ is invertible in the localization. This shows that $f s^{-1} = \overline{f} \overline{s}^{-1}$ for any $f$ and $s$. $\square$.
Clearly, $k[x,y]_{(0,0)}$ is not classical. 
I think this construction can clearly be generalized to make classical rings which have any local ring of dimension $\geq 2$ as a localization.<|endoftext|>
TITLE: Embedding algebraic surfaces in projective space
QUESTION [9 upvotes]: Suppose I have a smooth complete surface $S$ embedded in $\mathbf{P}^n$. Is there a known or conjectured lower bound (preferably sharp) on the degree of $S$ in terms of its numerical invariants?
In the case of curves this is the classical Castelnuovo bound (http://en.wikipedia.org/wiki/Castelnuovo_bound).

REPLY [6 votes]: This is not really an answer, but it's too long for a comment: since Alex was asking for bound in terms of numerical invariants, I'll try to give another point of view (which will be certainly less precise of Harris' result explained by jvp) which could perhaps be of some interest.
Let $S$ be a smooth surface and suppose $A\to X$ is an ample line bundle. Then some multiple $m_0 A$ of $A$ will give an embedding of $S$ in some projective space $\mathbb P^N$. After the embedding, the very ample line bundle $m_0 A$ "becomes" isomorphic tautologically to the restriction $\mathcal O_X(1)$ of $\mathcal O(1)$ to $X$, and the degree of $S$ relative to this embedding is just $\mathcal O_X(1)^2=m_0^2 A^2$.
Now, given an ample line bundle $A$ on a $n$-dimensional smooth projective manifold, an important problem of algebraic geometry is to find effective bounds $m_0$ such that multiples $mA$ of $A$ become very ample for $m \ge m_0$ . From a theoretical point of view, this problem has been solved by Matsusaka and Kollár-Matsusaka. Their result is that there is a bound $m_0 = m_0 (n, A^n, A^{n−1}\cdot K_X )$ depending only on the dimension and on the first two coefficients $A^n$ and $A^{n−1}\cdot K_X$ in the Hilbert polynomial of $A$.
Later on, Siu gave an effective version of the Big Mastusaka theorem, which reads as follow:
Theorem. (Siu '93). Let $A$ be an ample holomorphic line bundle over a compact complex manifold $X$ of complex dimension $n$ with canonical line bundle $K_X$. Then $mA$ is very ample for 
$$
m\ge m_0:=\frac{(2^{3^{n-1}}5n)^{4^{n-1}}(3(3n-2)^nA^n+K_X\cdot L^{n-1})^{4^{n-1}3n}}{(6(3n-2)^n-2n-2)^{4^{n-1}n-\frac 23}(A^n)^{4^{n-1}3(n-1)}}.
$$
Thus, if you fix any polarization on your (abstract) manifold, then you get a lower bound on the degree of the embedding with respect to this polarization only in numerical terms.<|endoftext|>
TITLE: Rank of $x (x^2 - 1) = c (c^2 - 1) y^2 $ over  $\mathbb{Q}$ for given rational values of $c$
QUESTION [5 upvotes]: Can anything be said in general about the rank etc over $\mathbb{Q}$ of the family of Weierstrass equations (in slightly non-standard form) $x (x^2 - 1) = c (c^2 - 1) y^2$ for various given rational values of $c$ ? I have a good reason for asking, so this isn't idle curiosity.
Naturally, it would be simplest if the Weierstrass equation has the same behaviour for each of these values of $c$, or least with a manageable (finite) amount of variation. Obviously there is always a solution $x, |y| = c, 1$; but that might be a trivial solution of a rank 0 case.

REPLY [2 votes]: Not an answer, just an image of Weierstrass curves for several values of $c \in (1,3]$:

      

(I was curious to see how the curve varies with $c$.)<|endoftext|>
TITLE: Criteria for Positivity of Pseudoddifferential Operators on Manifolds
QUESTION [7 upvotes]: Let $(M,g)$ be a Riemannian Manifold and $L^2$ the Hilbert space given by the volume form associated to the metric.  Let $L_0^2$ be the subspace which is orthogonal to the constant functions.  When is a pseudodifferential operator on $M$ a positive operator on $L^2_0$?
For second order operators the Laplacian $\Delta$ is the main example.
For order zero, the obvious examples are multiplication by $f$ where $f \in C^\infty(M)$ is a smooth function and $f > 0$.  Conversely if $f < 0$ anywhere then it is clear that the multiplication operator is not positive.
If $A$ is positive on $L^2_0$ then
$(\Delta^{p/2} A \Delta^{p/2} v, v) = (A \Delta^{p/2} v, \Delta^{p/2} v) > 0$
for $v \in L^2_0$ non-zero.  So we can use the Laplacian as a sort of natural way to change the order of a given positive operator.  Note that the principle symbol of such an operator is $||\xi||^{p}\sigma(A)(x,\xi)$.
How else can I construct more positive pseudodifferential operators? So far I can only come up with operators whose symbols in a fiber look like $||\xi||^{p}f(x)$.  I am looking for "more interesting" symbols, such as those whose restriction to the co-sphere at a point is non-constant.
Ideally of course I would just like a global criterion for a symbol to quantize to a positive operator, but something tells me that this is a hard problem. If it is any easier, I would also be interested in specific examples, like the sphere with the round metric.

REPLY [2 votes]: Let $A$ be a selfadjoint (pseudo)differential operator of order 2 on $(M,g)$ with a nonnegative symbol. It is a consequence of the Fefferman-Phong inequality that $A$ is semi-bounded from below, i.e. $A+C\ge 0$, where $C$ is a constant.
Now you could object that the total symbol is not invariantly defined: true but considering that $A$ acts on half-densities (identified with functions on a Riemannian manifold), you get that
$$
a_2+ \Re a_1
$$
is indeed invariantly defined. Here the symbol of $A$ is
$a_2+a_1+r_0$, where $a_j$ is of order $j$, $r_0$ of order 0, $a_2\ge 0,\quad a_2+\Re a_1\ge 0$.<|endoftext|>
TITLE: Finding the Nash Equilibrium of $0-1$ poker with one betting round
QUESTION [5 upvotes]: While working on a hobby project I encountered a difficult math problem. Or at least, difficult for me. Here is the problem:
Given an $a > 0$, find all pairs of a value $λ \in [0,1]$ and a function
$f \colon [0,1] \to [0,1]$ such that $z$ is minimal.
$
b(v) = \int_0^v f(x)\ \mathrm{d}x \\
z=\int_0^1{
\max \left(
\begin{array}{l}
(1 + 2a)λ, \\
(2 + 2a)x - λ, \\
2 a b(x) + 1, \\
(2 + 2a)x + b(1) - 2b(x) \\
\end{array}
\right)
}
\mathrm{d}x
$
That is the complete problem, but a partial solution would already help me a lot, for example:

A solution only for certain values of $a$
Finding values of $z$ without knowing $λ$ and $f$
Only one pair $(λ, f)$ for an $a$ instead of all the pairs
Are there even multiple pairs for one $a$? (considering different $f$'s that yield the same $b$ as equal)
Is it possible that all values of $a$ have at least one pair in which the $f$ has $\{0, 1\}$ as its range?

I'll give you some info on what this formula is about. The goal of the project I'm working on is to have a better understanding of poker by finding Nash equilibria of very simplified versions of poker. The game I'm now trying to find equilibria for is a two player game that goes as follows:
In the beginning both players are required to bet a certain amount, the ante, variable $a$ in the formula. Both players get a 'card', which is a uniformly distributed number between 0 and 1. Then the one and only betting round follows. Both players have only one coin with a value of 1 that they can use. Player 1 starts. There are five different ways the betting round can go:

Check > Check
Check > Bet > Fold
Check > Bet > Call
Bet > Fold
Bet > Call

After the betting round the payout is done. If the betting round ends with check or call, there will be checked who of the players has the highest card. The winner will get the ante, or the ante plus one if a bet followed by a call occurred. The goal for each player is to have as much expected profit as possible.
When player 1 uses the best response to the strategy of player 2, the expected value of the game will be $z - 1 - a$. A positive value indicates profit for player 1, a negative value profit for player 2.
If player 1 checks, then player 2 will check with probability $f(c)$ and bet otherwise, where $c$ is the card player 2 has. If player 1 bets, then player 2 will call if his card is at least $λ$.
If I have the solution to this problem and didn't make any mistakes in making the formula, then I have the optimal strategy for player 2 and the expected value in the Nash equilibrium. And then I only have to do something similar for player 1.
Any ideas on minimizing $z$?

REPLY [7 votes]: $0-1$ poker games have been studied by Borel, Von Neumann and Morgenstern, and others. The most detailed treatment I know is by Bill Chen and Jerod Ankenman, who wrote a series of posts in rec.gambling.poker in 2003, "The [0,1] game: Part 1-14." Some of this was included in their 2006 book, The Mathematics of Poker.
The game you mention is "[0,1] Game #9" on pages 198-203.
Before going into the solution, note that it is a common misconception that mixed strategies are ubiquitous. In fact, for games with finitely many options, mixed strategies are typically required for finitely many boundary cases. In a game with finitely many states like rock-paper-scissors that may be everything, but in a game like this with no atoms in the distribution of cards, mixed strategies are not required at all. Because of the hidden information, when you bet, your opponent still will not know whether you are betting for value (with a strong hand which is ahead on average when called) or with a bluffing hand (which is behind your entire calling range). At the Nash equilibrium, you may be indifferent between betting or checking with a range of hands, when checking will never win and betting as a bluff gives the same value as giving up. So, instead of choosing a function which gives you a probability of betting when dealt that card, you might as well partition the possibilities into sets for each action.
The following excerpts show the Nash equilibrium:

X's strategy consists of the following thresholds:

$x_0$ between check-folding and bluffing.
$x_1^*$ between check-folding and check-calling.
$x_1$ between value betting and check-calling.

Y's strategy likewise consists of three thresholds:

$y_0$ between checking and bluffing if X checks.
$y_1^*$ between calling X's bet and folding to X's bet (if X bets)
$y_1$ between value-betting and checking if $X$ checks.

... 
$0 \lt x_1 \lt y_1 \lt x_1^*, y_1^* \lt y_0 \lt x_0 \lt 1$
[This follows the convention that lower cards are stronger, and corrects a typo.]
...
Solution:
$y_1 = (1-\alpha)/(1+\alpha)$
$ x_1 = (1-\alpha)^2/(1+\alpha)$
$y_0 = (1+\alpha^2)/(1+\alpha)$
$x_0 = 1 - \alpha(1-\alpha)^2/(1+\alpha)$
$y_1^* = 1 - \alpha$
$x_1^*= 1-\alpha$

where $\alpha$ is the probability with which you fold a bluff-catcher with a bet of that size relative to the pot to make someone indifferent to bluffing. Since you have a bet of $1$ into a pot of size $2a$, $\alpha = 2a / (1+2a)$.<|endoftext|>
TITLE: The space of lattices and modular forms of weight 1/2
QUESTION [12 upvotes]: Suppose my favorite way of thinking about modular forms is as functions on the space of (real, 2D) lattices.  One can identify this space with $SL_2(\mathbb{Z}) \backslash GL_2(\mathbb{R})$, i.e. bases for the lattice up to reparameterization.
A function $f : SL_2(\mathbb{Z}) \backslash GL_2(\mathbb{R}) \rightarrow \mathbb{C}$ is a modular form of weight $k$ if it satisfies the "scaling relation":

$f(a\ R_\theta\ T) = a^{-k} e^{- i k \theta} f(T)$

where $R_\theta$ denotes the appropriate rotation matrix, and $a$ a positive real.
[More precisely $f$ only has to be defined on one connected component of the space.]
(I think) the standard definition is equivalent to this, by considering the value of $f$ on canonical lattices $\langle 1,\ z \rangle$ for $z$ in the complex upper half-plane.  Note it's entirely clear in this language that e.g. $G_4$ is a modular form of weight 4.

So, my question is: is it possible to make sense of modular forms of half-integral weight (for concreteness, say $\vartheta$) in a similar way?
I'm aware that a necessary step is to pass to some kind of double cover such as $Mp_2$, to make the scaling relation make sense for $k=1/2$; but I am having trouble making this sufficient.  In particular, as sub-questions:

What group plays the role of $SL_2(\mathbb{Z})$?
While I could extend $\vartheta$ by "brute force" to a function on e.g. $Mp_2$ (by callously applying the scaling relation), is there a natural way to define $\vartheta$ on the larger space, similar to the "obvious" definition:
$G_4(\Lambda) = \sum_{w \in \Lambda \setminus 0} w^{-4}$
?


Apologies in advance if this is standard -- I've been unable to locate a satisfactory answer in the literature.
Thanks,
Freddie

REPLY [7 votes]: You need to replace $SL_2(\mathbb{Z})$ with a discrete subgroup of $Mp_2$, and you want this subgroup not to contain the kernel of $Mp_2 \to SL_2$, since otherwise there are trivially no half-integer-weight forms. If you take a small enough finite-index subgroup of $SL_2(\mathbb{Z})$, then it will admit a lifting to $Mp_2$.<|endoftext|>
TITLE: Can you cover the Boolean cube {0,1}^n with O(1) Hamming-balls each of radius n/2-c*sqrt(n)?
QUESTION [16 upvotes]: (where c>0 and the balls need not be disjoint?)
This is an embarrassingly simple question, yet somehow I couldn't find an answer (not even, "this is a well-known open problem") after spending some time googling the literature on covering codes.  
A simple probabilistic argument shows that you can cover the Boolean cube with O(n) Hamming balls of radius $n/2-c\sqrt n$ each, for any $c>0$.  My guess would be that you can't do it with (much) fewer---O(1) Hamming balls seems aggressively optimistic---but I don't know if it's known how to prove that.
(In the language of coding theory, I want to know whether $K_2(n,n/2-c\sqrt n)$, the minimum size of a binary covering code with radius $n/2-c\sqrt n$, can be upper-bounded by a constant depending only on c, not on n, at least for some c>0.)

REPLY [8 votes]: Spencer's famous "Six Standard Deviations Suffice" is essentially equivalent to what you're asking (for a statement and proof see, e.g., the Alon-Spencer book or the  new proof by Lovett and Meka). It shows, e.g., that with balls of radius $n/2-6\sqrt{n}$ you need n balls to cover the cube. It also gives $\Omega(n)$ for any radius $n/2-c\sqrt{n}$.<|endoftext|>
TITLE: Computation of KO characteristic classes/numbers
QUESTION [5 upvotes]: How to compute KO characteristic classes/numbers?
They were introduced by Anderson/Brown/Peterson to study the structure of the spin cobordism ring. I looked through the literature but I did not find a nice example of computation. For instance, I would like to know how to determine $\pi^J(\mathbb{H}P^2)$. Many thanks for any comments in advance.

REPLY [5 votes]: Chapter II (in particular sections II.7 and II.9) of the book
Conner, P. E.; Floyd, E. E.
The relation of cobordism to K-theories. 
Lecture Notes in Mathematics, No. 28 Springer-Verlag, Berlin-New York 1966 v+112 pp. 
gives a nice construction of $KO$-characteristic classes for symplectic bundles, in particular showing that they satisfy the usual axioms. From this you may be able to perform computations. I'm not sure what the notation $\pi^J$ means (cokernel of the J-homomorphism?).<|endoftext|>
TITLE: Almost Complex Structure approach to Deformation of Compact Complex Manifolds
QUESTION [29 upvotes]: I don't know much about the deformation of compact complex manifolds, I've only read chapter 6 of Huybrechts' book Complex Geometry: An Introduction. There are two parts to this chapter. The second goes through the standard approach, that is, considering a family of compact complex manifolds as a proper holomorphic submersion between two connected complex manifolds. My question is about the approach taken in the first section, which I will briefly outline.
One can instead consider a deformation of complex structures on a fixed smooth manifold, as opposed to deformations of complex manifolds – by Ehresmann's result, a deformation over a connected base is nothing but a deformation of complex structure on a fixed smooth manifold. This point of view is difficult to work with because a complex structure is a complicated object, so we instead consider almost complex structures – by the Newlander-Niremberg Theorem, complex structures correspond to integrable almost complex structures.
Fix a smooth even-dimensional manifold $M$. Now Huybrechts considers a continuous family of almost complex structures $I(t)$. He does not say where $t$ comes from, but I have interpreted it to be an open neighbourhood of $0$ in $\mathbb{C}$. Now, let $I(0) = I$. The complexified tangent bundle to $M$ splits with respect to $I$. That is, $TM\otimes_{\mathbb{R}}\mathbb{C} = T^{1,0}M\oplus T^{0,1}M$. But this is true of each almost complex structure $I(t)$. Denote the corresponding decompositions by $TM\otimes_{\mathbb{R}}\mathbb{C} = T^{1,0}M_t\oplus T^{0,1}M_t$ – this is deliberately suggestive notation; we can consider the compact (soon-to-be) complex manifold $(M, I(t))$ as the fibre of a complex family over a point $t$ in the base.
For small $t$, we can encode the given information by a map $\phi(t) : T^{0,1}M \to T^{1,0}M$ where, for $v \in T^{0,1}M$, $v + \phi(t)v \in T^{0,1}M_t$. Huybrechts then says:

Explicitly, one has $\phi(t) = -\text{pr}_{T^{1,0}M_t}\circ j$, where $j : T^{0,1}M \subset TM\otimes_{\mathbb{R}}\mathbb{C}$ and $\text{pr}_{T^{1,0}M_t} : TM\otimes_{\mathbb{R}}\mathbb{C} \to T^{1,0}M_t$ are the natural inclusion respectively projection.

According to this, the codomain of $\phi(t)$ is $T^{1,0}M_t$, not $T^{1,0}M$. Is this a typo or am I missing something? Added later: As Peter Dalakov points out in his answer, it is a typo.
Anyway, Huybrechts continues with this approach. Enforcing the integrability condition $[T^{0,1}M_t, T^{0,1}M_t] \subset T^{0,1}M_t$ ensures that each almost complex structure is induced by a complex structure. Under the assumption that $I$ is integrable, $[T^{0,1}M_t, T^{0,1}M_t] \subset T^{0,1}M_t$ is equivalent to the Maurer-Cartan equation $\bar{\partial}\phi(t) + [\phi(t), \phi(t)] = 0$, where $\bar{\partial}$ is the natural operator on the holomorphic vector bundle $T^{1,0}M$, and $[\bullet, \bullet]$ is an extension of the Lie bracket.
I like this approach because if you take a power series $\sum_{t=0}^{\infty}\phi_it^i$ of $\phi(t)$ you can deduce:

$\phi_1$ defines the Kodaira-Spencer class of the deformation;
all the obstructions to finding the coefficients $\phi_i$ lie in $H^2(M, T^{1,0}M)$.

Does anyone know of some other places where I would be able to learn about this approach, or is there some reason why this approach is not that common?
Just for the record, I have looked at Kodaira's Complex Manifolds and Deformation of Complex Structures, but I haven't been able to find anything resembling the above.

REPLY [24 votes]: This approach to deformations is taken, for instance, in all of the original papers of Kodaira-Spencer and Nirenberg.   You can have a look at 
 On the existence of deformations of complex analytic structures, Annals, Vol.68, No.2, 1958
http://www.jstor.org/discover/10.2307/1970256?uid=3737608&uid=2129&uid=2&uid=70&uid=4&sid=47699092130607
but there are many other papers by the same authors. 
For a nice and compact exposition, you can look at these class notes of Christian Schnell:
http://www.math.sunysb.edu/~cschnell/pdf/notes/kodaira.pdf
Of course, the Maurer-Cartan equation and  deformations (of various structures) via dgla's have been used by many other people since the late 1950-ies: Goldman & Millson, Gerstenhaber, Stasheff, Deligne, Quillen, Kontsevich.
Regarding the formula: that's a typo, indeed. You have  two eigen-bundle decompositions,
for $I$ and $I_t$: 
$$ T_{M, \mathbb{C}} = T^{1,0}\oplus T^{0,1}\simeq T^{1,0}_t\oplus T^{0,1}_t  $$
and you   write $T^{0,1}_{t}=\textrm{graph }\phi$, where $\phi: T^{0,1}_M\to T^{1,0}_M$. So actually
$$\phi = \textrm{pr}^{1,0}\circ \left.\left(\textrm{pr}^{0,1}\right)\right|_{T^{1,0}_t}^{-1}.$$
In local coordinates, 
$$ \phi = \sum_{j,k=1}^{\dim_{\mathbb{C}} M}h_{jk}(t,z)d\overline{z}_j\otimes \frac{\partial}{\partial z_k},  $$
and  $T^{0,1}_t$ is generated (over the smooth functions) by
$$\frac{\partial}{\partial \overline{z_j}} + \sum_{k=1}^{\dim_{\mathbb{C}}M}h_{jk}\frac{\partial}{\partial z_k}. $$
Regarding the question "where does  $t$ come from?", the answer is "From Ehresmann's Theorem": given a proper holomorphic submersion $\pi:\mathcal{X}\to \Delta$, you can choose a holomorphically transverse trivialisation $\mathcal{X}\simeq X\times \Delta$,
$X=\pi^{-1}(0)= (M,I)$. In this way you get yourself two (almost) complex structures
on $X\times \Delta$, which you can compare.
ADDENDUM
I  second YangMills' suggestion to have a look at Chapter 2 of Gross-Huybrechts-Joyce.
You can also try  Chapter 1 of K. Fukaya's book "Deformation Theory, Homological algebra, and Mirror Symmetry", as well as the Appendix to  Homotopy invariance of the Kuranishi Space by Goldman and Millson (Illinois J. of Math, vol.34, No.2, 1990). In particular, you'll see how one uses formal Kuranishi theory to avoid dealing with  the convergence of the power series for $\phi(t)$. For deformations of compact complex manifolds, the convergence was proved by Kodaira-Nirenberg-Spencer. Fukaya says a little bit about the convergence of this series in general, i.e., for other deformation problems.<|endoftext|>
TITLE: Mathematics of quasicrystals
QUESTION [15 upvotes]: I want to study quasicrystals from mathematical point of view, but I'm having hard time finding materials about it. If you could suggest me some books, articles or papers, I would be glad.

REPLY [13 votes]: The trouble with the quasicrystals is that the literature in this area is dominated by non-mathematical or pseudo-mathematical papers and books. In particular, just extracting a mathematical definition of a quasicrystal from this literature is not so easy. This situation is well-illustrated by the wikipedia article on quasicrystals and the MO discussion of this topic at
What is the relation between Quasicrystals, Riemann Hypothesis, and PV numbers?
The two papers that I found most enlightening and mathematical, address this problem head-on (more on this below): 
[1]. A. Hof, "On diffraction by aperiodic structures", Commun. Math. Phys., 169 (1995), p. 25-43. 
[2]. J-B. Gouere, "Quasicrystals and almost periodicity", Commun. Math. Phys., 255 (2005), p. 655-681.
Both papers prove some nontrivial mathematical theorems, on the basis of these theorems one can then form two, somewhat different, mathematical definitions of a (quasi)crystal. Senchal's 2-page long survey paper (see Mahmud's answer) or, better, her book (see Joseph's answer) is a good introduction, the trouble is that she does not prove anything in the book and that she could be sloppy with her definitions, for instance, she conflates functions, measures and distributions, which are needed for defining crystals.  
If you look in Senchal's book, you first get the following physical definitions of crystals: 
"A crystal is any solid with essentially discrete diffraction diagram." (This includes both traditional crystals and quasicrystals.) The word "essentially" will be the difference between different mathematical definitions, which one derives from [1] and [2]. 
A (mathematical) quasicrystal is a tiling $T$ of ${\mathbb R}^n$ by convex polytopes satisfying certain properties:  

Since general tilings are hard to work with, pretty much everybody assumes that $T$ is a Voronoi tiling of ${\mathbb R}^n$ based on a certain discrete subset $N\subset {\mathbb R}^n$, which a geometer would call a separated net. 
Unfortunately, just having a separated net is not enough in order to deal with the "diffraction" issue. There is a disagreement which tilings are allowed as crystals and which are not. There is a general agreement that at least all periodic tilings (the ones where $N$ is a finite union of orbits of a discrete group of translations) and certain tilings constructed by Penrose and others via projection method, should be counted as crystals and quasicrystals respectively. I will call these two classes of tilings as "standard." 

a. Here is Hof's definition of a crystal (Senchal's definition is taken from Hof's paper). Hof in [1] takes the auto-correlation function (actually, a distribution) $\gamma$ of $N$ and computes the (appropriately defined) Fourier transform $\hat\gamma$ of $\gamma$ (this is a mathematical interpretation of the "diffraction diagram"). Then $\hat\gamma$ is a measure $\mu$ which, in general, splits as a sum of two measures $\mu_d+\mu_c$: Discrete part $\mu_d$, which is supported on a certain countable subset of ${\mathbb R}^n$ and continuous  part $\mu_c$. He proposes that "essentially discrete diffraction" means that $\mu_d$ is nonzero. Hof then proves that "standard" tilings indeed have nontrivial $\mu_d$ (According to [2], Hof even proves that $\mu_c=0$ in this case, but I did not check this). The trouble with this definition is that, as far as I can tell, there is no known purely geometric interpretation of the condition $\mu_d\ne 0$ in terms of the next $N$ itself (at least, none existed 6 years ago). 
b.  Gouere [2] (his work is an extension of Hof's approach and of a work by Lagarias) works with a slightly different definition, i.e., that $\mu_c=0$ (such sets $N$ are called Patterson sets). His main result is a purely geometric interpretation (actually, several interpretations) of this condition, see Theorem 1.1 in [1]: Patterson sets 
are the sets which are almost periodic with respect to Besikovitch's metric. 
Remark 1. As far as I can tell from reading Freeman Dyson's paper 
here, definition of a quasicrystal that Dyson proposes is the one with $\mu_c=0$. 
Remark 2. Gouere does not propose that a Patterson set is the right definition of a crystal, this is just my take on his paper. Condition $\mu_c=0$ is more limited, but, in view of [2], is geometric and also covers "standard" examples, while the condition $\mu_d\ne 0$ is more general, but is nongeometric.<|endoftext|>
TITLE: Why is this theorem attributed to J.-P. Serre?
QUESTION [5 upvotes]: Page $117$ of Atiyah, MacDonald's Introduction to Commutative Algebra text has the following theorem. Let $P(M,t)$ denote the Poincare- series of $M$.

$\textbf{Theorem.}$ $\bigl(\mathsf{Hilbert-Serre}\bigr)$. $P(M,t)$ is a rational function in $t$ of the form $f(t)/\prod_{i=1}^{s} (1-t^{k_i})$, where $f(t) \in \mathbf{Z}[t]$.

This theorem appears in the section of the book called Hilbert-Functions (page 116), so one understands that it could have possibly been discovered by Hilbert. 

But why is the above theorem attributed to J.-P. Serre?  References about when Serre was credited to the above theorem would be helpful.

REPLY [7 votes]: I don't know what Atiyah-Macdonald were thinking, but I can tell you a theorem which is attributed to Serre (correctly, I think), and is relevant to this question.
Let $M$ be a finitely-generated graded $k[x_0, x_1, \ldots, x_n]$ module. Let $H^0(M)$, $H^1(M)$, ..., $H^n(M)$ be the local cohomology modules of $M$ with respect to the maximal ideal $\langle x_0,\ldots, x_n \rangle$. These are graded modules which satisfy the following properties:
Theorem: For all integers $d$, the function
$$\dim M_d - \sum_{r=0}^n (-1)^r \dim H^r(M)_d$$
is polynomial in $d$.
Theorem (Serre vanishing) For $d$ sufficiently large, $H^r(M)_d=0$.
So Serre vanishing separates Hilbert's theorem into two parts: A certain function is a polynomial for all $d$, and that function is equal to the Hilbert function for large $d$.
I'm presenting this using the language of commutative algebra, which I don't think is the language Serre used. In sheaf cohomology language, let $\mathcal{M}$ be the sheaf on $\mathbb{P}^{n-1}$ corresponding to $M$ and let $\mathcal{H}^r(M) = \bigoplus_{d=-\infty}^{\infty} H^r(\mathbb{P}^{n-1}, \mathcal{M} \otimes \mathcal{O}(-d))$. Then the relation between sheaf cohomology and local cohomology is that
$$\mathcal{H}^r(M) \cong H^{r+1}(M)$$
for $r \geq 1$ and there is a short exact sequence
$$0 \to H^0(M) \to M \to \mathcal{H}^0(M) \to H^1(M) \to 0.$$
In this language, Serre vanishing says that, for $d$ large, $\mathcal{H}^r(M)_d=0$ for $r>0$ and $M_d \cong \mathcal{H}^0(M)_d$; this is how the result is usually stated.
The first theorem in this language is that $\dim \sum_{r=0}^{n-1} (-1)^r \mathcal{H}^r(M)_d$ is a polynomial in $d$.<|endoftext|>
TITLE: Is the $\ell$-adic cohomology of a non-proper variety unramified at good primes?
QUESTION [8 upvotes]: Let $X$ be a smooth variety of finite type over a number field $k$. Let $\overline{X} = X \times_{k} \overline{k}$, and let $\ell$ be a prime. It's well known that if $X$ is proper, then the étale cohomology groups $H^i_{et}(\overline{X}, \mathbb{Z}_{\ell})$ are unramified at any prime $\mathfrak{p} \nmid \ell$ at which $X$ has good reduction (and in fact are isomorphic as representations of $\operatorname{Gal}(\overline{K}_\mathfrak{p} / K_\mathfrak{p})$ to the étale cohomology groups of the special fibre).
Does this statement also hold if $X$ is not assumed to be proper? (I'm interested in the case of smooth affine varieties.) What about the weaker statement that $H^i_{et}(\overline{X}, \mathbb{Z}_{\ell})$ is unramified almost everywhere?
(I don't know a reference for the proof of the "well known" statement -- I couldn't find it in Milne's books or in SGA 4.5.)

REPLY [10 votes]: I think the weaker statement should be true.  Here's a sketch of an argument: by compactification theorems and resolution of singularities, there is a smooth proper
scheme $Y$ over $k$ containing $X$ as an open subscheme, such that $Y \setminus X$ is a
divisor $D$ with simple normal crossings.  Let $D_1, \dots, D_r$ be the irreducible components of $D$.  Then any $p$-fold intersection of the $D_i$'s is smooth and proper
over $k$.
There should be a spectral sequence, in terms of the etale cohomology of $\overline{Y}$ and that of the intersections of the $D_i$'s, that abuts to the etale cohomology of $\overline{X}$.  Thus the etale cohomology of $\overline{X}$ should be unramified at any prime of good reduction for $\overline{Y}$ and all of the intersections of the $D_i$'s.  I imagine you could also use this to show that at such primes the cohomology of $\overline{X}$
was isomorphic to the cohomology of the reduction.
David<|endoftext|>
TITLE: AJ conjecture for links
QUESTION [8 upvotes]: Garoufalidis proposed a conjecture on $q$-difference equations for the colored Jones polynomials of knots.
\begin{equation}
\hat{A}_K(\hat{l},\hat{m};q)J_n(K;q)=0
\end{equation}
where the actions of the operators $\hat l$, $\hat m$ are defined by 
\begin{equation}
\hat{l}J_n(K;q)=J_{n+1}(K;q) , \quad \hat{m}J_n(K;q)=q^{n/2}J_n(K;q) \ .
\end{equation}
Are there analogous $\hat{A}$-polynomials which define $q$-difference equations for the colored Jones polynomials of links? If there are, they should be polynomials with $2n$ variables $\hat{l}_i, \hat{m}_i$ ($i=1,\cdots,n$) for a link with $n$ components. Are their actions to the colored Jones polynomials of links known?
More simply, is there a paper which expresses the classical $A$-polynomial $A(l,m)$ or the character variety for the Hopf link in $S^3$?

REPLY [9 votes]: I think Ian's answer is right on.  This is a comment that is too long to be a comment.

Let $L\subset S^3$ be a link of $k$-components.  We use $X(L)$ to denote the $SL(2,\mathbb{C})$ character variety of the fundamental group of $S^3-L$.  Let $X(T^2)$ denote the $SL(2,\mathbb{C})$ character variety of the torus.  There is a restriction map
$$X(K)\rightarrow X(T^2)^k $$  the image is an algebraic set, with ideal $A(L)$.  You can lift it to a $2^k$ fold branched cover by $\mathbb{C}^{2k}$.  Which corresponds to extending an ideal to get $\tilde{A}(L)\subset \mathbb{C}[l_i^{\pm 1},m_i^{\pm 1}]$.  The ideal $\tilde{A}(L)$ is not necessarily principle, so no $A$-polynomial, just an $A$-ideal. By Poincare-Lefschetz duality applied to cohomology with coefficents in the adjoint representation, away from the singular points the image of $X(L)$ in the character variety of the torus is complex Lagrangian with respect to the complex symplectic form $2\sum_i dl_i\wedge dm_i$.
The procedure outlined by Ian yields an ideal in the exponentiated Weyl algebra $W_k$ quantizing $\mathbb{C}[l_i,m_i]$ which would be the noncommutative $A$-ideal.
Define a function 
$$ J(L): \mathbb{N}^k \rightarrow \mathbb{Z}[t_i^{\pm 1}]$$ 
which assigns to each tuple of integers the Kauffman bracket of the framed link $L$ colored by the Jones-Wenzl idempotents.
By the argument in FGL that Ian mentions there is a pairing so that $J(L)$ annihilates $\tilde{A}(L)$.
On the other hand the exponetiated Weyl algebra $W_k$ acts on the space of functions $f:\mathbb{N}^k\rightarrow \mathbb{Z}[t_i^{\pm 1}]$ as explained in Garoufalidis and Le's first paper on the subject, and their proof shows $J(L)$ is holonomic.  Hence it is annihilated by an ideal that acts like the quantization of the ideal of a complex Lagrangian subvariety of the cotangent bundle of $\mathbb{C}^k$.

Modulo normalization the noncommutative $A$-ideal lives in the annihilator.  The content of the AJ conjecture is, how close does the noncommutative $A$-ideal come to being all of the annhilator of $J(L)$?<|endoftext|>
TITLE: Cartesian product of graphs
QUESTION [5 upvotes]: Let $G, H$ be two infinite connected graphs. Suppose that we can color $G$ and $H$ in $m$ and $n$ colors respectively so that monochromatic clusters (i.e. monochromatic connected components) are of uniformly bounded diameters. Is it possible to color the Cartesian product $G\times H$ in $m+n-1$ colors so that the diameters of monochromatic clusters are bounded. Here $G \times H$ is the graph with vertices $(x,y), x\in G, y\in H$ and $(x,y)$ connected with $(p,q)$ iff either $x=p$ and $y,q$ are adjacent in $H$ or $y=q$ and $x,p$ are adjacent in $G$. 
Example. Let $Z_3$ be the graph with vertices $i\in \mathbb{Z}$ and edges $(i,j)$ where $|i-j|\le 3$. It can be colored in 2 colors as follows: color intervals $[10k+1, 10k+5]$, $k\in \mathbb{Z}$, in black and intervals $[10k+6,10(k+1)]$ in white. All monochromatic clusters have diameters 2. Then $Z_3\times Z_3$ can be colored in 3=2+2-1 colors so that all monochromatic clusters have uniformly bounded diameters. The clusters are bricks. 
 Update  The question above turned out to be too easy and not what I wanted to ask. Here is the real question. 
Let $G, H$ be two infinite connected graphs. Pick a number $\lambda\ge 1$. Suppose that we can color $G$ and $H$ in $m$ and $n$ colors respectively so that monochromatic $\lambda$-clusters (i.e. maximal subsets $X$ where  any two vertices $u,v$ are connected by a monochromatic sequence $u=x_0, x_1, ..., x_k=v$ where the distance between $x_i, x_{i+1}$ is at most $\lambda$) are of uniformly bounded diameters. Is it possible to color the Cartesian product $G\times H$ in $m+n-1$ colors so that the diameters of monochromatic $\lambda$-clusters are bounded. Here $G \times H$ is the graph with vertices $(x,y), x\in G, y\in H$ and $(x,y)$ connected with $(p,q)$ iff either $x=p$ and $y,q$ are adjacent in $H$ or $y=q$ and $x,p$ are adjacent in $G$. 
In the first version of the question $\lambda=1$. For an example, consider the graph $\mathbb{Z}$ (a line), and $\lambda=3$.

REPLY [5 votes]: Yes. In fact you only need max$(m, n)$ colors. Let's assume $m \leq n$ (switching $G$ and $H$ if necessary.) Number the colors from 0 to $m - 1$ and 0 to $n - 1$. If $x$ has color $i$ and $y$ has color $j$, then give $(x, y)$ color $i + j$ mod $n$. Then the monochromatic connected components of the cartesian product are the cartesian products of the monochromatic connected components.<|endoftext|>
TITLE: Classroom platonism 
QUESTION [14 upvotes]: I'd like to know whether any form a certain hypothesis about the
learning of higher mathematics has entered the mathematical or
educational literature.  I'll frame the hypothesis here but not defend
it since this is not a blog-in-disguise; likewise I'm not soliciting
debate.
This hypothesis opposes to some degree the common shibboleth which
holds that "mastering abstraction" constitutes the single major
plateau which undergraduate mathematics students must, but often do
not, scale.
For the sake of making the distinction,  I'll first flesh out what I
mean by "mastering abstraction."  Generally speaking, abstraction means
reducing to essentials. So mastering mathematical abstraction breaks
into two major challenges: learning modelling and learning formal
work.  Modelers must first know how to decide what they may safely
ignore and then how to select or construct formal systems that
adequately capture what remains.  With a formal system in hand,
getting answers requires skill with its internal operation, sometimes
despite the loss of intuition that arises from distance to the
original situation.  Of course a feedback cycle often arises --
"answers" from formal work can demand systemic revision of the
formalism.
On the the current hypothesis, namely that something else constitutes the major glass ceiling for advance mathematics students.  I'll call that something else cognitive platonization. (If someone else has already coined a better name I'd like to know!)  So cognitive platonization occurs when mathematicians confer objecthood on the collection of some or all configurations of a known object.  Examples abound: taking all solutions of certain differential equations as elements of a vector space, forming (iterated) power sets and cumulative hierarchies in set theory, studying state spaces in dynamical systems, moduli spaces in geometry, homology and cohomology groups or Stone-Cech compactifications in topology. Like abstraction, cognitive platonization often induces a loss of intuition due to distance from the original situation, but I contend a different sort of distance.  Abstraction involves reasoning away from a picture you may feel afraid to lose; cognitive platonization involves reasoning on the way to a picture you may fear will never congeal.
As an aside, I chose the name because some radical philosophers
challenge the very "existence" of just these sort of things I see
students struggling to comprehend.
I'd like to know several things:
1) Does the challenge of teaching cognitive platonization (known by
whatever name) have a theoretical literature?
2) Does cognitive platonization have a practical literature, meaning
materials aimed directly at students, perhaps at the (American)
college sophomore level?
3) Do any books from the popular science genre frame this issue and do
a good job at communicating its essentials to a wide-audience?
4) What testable implications of the hypothesis can anyone suggest?
Might success or failure with, say, abstract algebra or measure theory
correlate with a student's response to tasks, otherwise unrelated to
that subject matter, that indicate their ability or willingness to
embrace this process of conferring objecthood?  If so, what sort of
tasks?
Final note: I'm asking here because most mathematics education research looks at K-12 teaching and learning, or perhaps calculus.  Almost all writing about teaching higher mathematics comes from practicing mathematicians.

REPLY [6 votes]: The platonization you are looking at seems related to the idea of reification that appears in mathematics education literature (roughly the compression of a mathematical process to a mathematical object). Note: This is clearly distinct from what you are asking about, but shares the feature of bestowing objecthood on something in a way that requires the shift to (what is at first) a radically different viewpoint.
A few things on this that may be helpful can be found here.
Particularly, here are a couple of papers on reification.<|endoftext|>
TITLE: Adjoint functor theorem for infinity categories
QUESTION [15 upvotes]: In HTT, a version of the adjoint functor theorem for (locally) presentable infinity categories is proven (Corollary 5.5.2.9). Is there a more refined version of this somewhere, which more closely resembles Freyd’s original version? I.e., is there a version for infinity categories which are not necessarily (locally) presentable, but requires a solution set condition? If so, a reference would be fantastic. Thanks!

REPLY [9 votes]: The general adjoint functor theorem didn't exist in the literature when this question was originally asked, but now it does: Nguyen, Raptis, and Schrade.<|endoftext|>
TITLE: Cosheaf homology and a theorem of Beilinson (in a paper on Mixed Tate Motives)
QUESTION [14 upvotes]: I'm trying to understand the proof of Theorem 4.1 in the paper Multiple Polylogarithms and Mixed Tate Motives by AB Goncharov (http://arxiv.org/pdf/math/0103059v4.pdf). In it, the author uses cosheaf homology.
As far as I can tell, the global sections functor for cosheaves is right exact, so homology should be given by the left derived functors of the global sections functor. Similarly, the higher direct image functors should be the left derived functors of the standard direct image. The Grothendieck spectral sequence should be a homological spectral sequence. 
I have four questions:

Is this correct?
I understand that (see e.g. Hartshorne, Chapter III, Proposition 8.1) the cosheaf sending an open set $U$ to $H_q(p^{-1}(U))$ should be the $q$th direct image of the constant sheaf $\mathbb Z$. However, when on p.46 we are defining the cosheaf $\mathcal{R}_c$ and we are taking relative homology, what cosheaf are we taking the direct image of?
Where does the exact sequence (110) come from? Do we always get an exact sequence from the joining of two complexes?
Between (109) and (110), I assume that $R_i$ means left derived functor, since, as I mentioned, higher direct images of cosheaves are left, not right, derived functors. But what on Earth does he mean by the higher direct image of a subvariety (or complex of subvarieties)?

I'm guessing that the $q$th direct image of the complex of varieties should be interpreted as the cosheaf corresponding to the homology relative to the union of the varieties associated to that complex? Assuming that's the case, I'm a little unsure how to deal with the higher direct images of the truncation (maybe it corresponds to the hyperhomology of the complex on the inverse image of $U$? Then the special case of the whole complex makes sense since the hyperhomology of that complex is the relative homology. But if that's so, I don't know what to make of the hyperhomology of the truncation...)

REPLY [2 votes]: Let me begin with a reference: Marta Bunge and Jonathon Funk, Singular coverings of toposes. What follows is a very concise digest of very few of the features of that book.
Seems like the correct way of dualizing the correspondence between sheaves and local homeomorphisms is the following.
A continuous map $f:E\to X$ is a local homeomorphism iff the canonical map
$$
\operatorname{stalk}_x(f):=\varinjlim_{U\ni x}\Gamma\left(\left.f\right|_U\right)\to f^{-1}(x)
$$
given by
$$
(\text{germ at $x$ of a section $\sigma$})\mapsto\sigma(x)
$$
is bijective for any $x\in X$, where $\Gamma\left(\left.f\right|_U\right)$ denotes the set of all global sections of the map $\left.f\right|_U:f^{-1}(U)\to U$.
Now considering that $\Gamma$ is right adjoint to the inverse image, it is most natural to try the left adjoint to the inverse image. It does not always exist but in good cases (related to local connectedness) is given by connected components. So in case there are appropriate adjunctions $\pi_0\dashv f^*\dashv\Gamma$, it makes sense to consider the following condition on $f$ as a first approximation to the dual notion: it is a "co-local homeomorphism" if the canonical maps
$$
f^{-1}(x)\to\varprojlim_{U\ni x}\pi_0(f^{-1}(U))=:\operatorname{costalk}_x(f)
$$
given by
$$
f^{-1}(x)\ni e\mapsto\left(\text{connected component of $e$ in $f^{-1}(U)$}\right)_{U\ni x}
$$
are bijective. In this context then, it is natural to call matching families of connected components of inverse images of neighborhoods of a point cogerms (of cosections?) at that point. So every point in the inverse image of $x$ determines a cogerm at $x$, and the condition is that every cogerm is of this kind for a unique point in the inverse image.
This approach shows that cosheaves are "more special" than sheaves: to begin with, the $\pi_0$'s involved might be quite nasty. Even if they are nice, usually they are non-discrete. And even if they are discrete, still in general the inverse limit  carries a natural nondiscrete inverse limit topology, so it is natural to require of the above map to be not just a bijection but a homeomorphism. In this way one more or less gets complete spreads introduced by R. H. Fox in 1957 and studied by several people ever since. Their total spaces behave much better than those of local homeomorphisms (which might be non-Hausdorff even over very nice spaces). More or less, complete spreads are branched coverings.
There are several other subtleties to take into account, but, at any rate, Bunge and Funk build a theory dual to that of sheaves based on the notion of complete spread, and obtain a duality with several nice properties. In particular they deepen further the insight of of Lawvere that sheaves are "like functions" while cosheaves are "like distributions".<|endoftext|>
TITLE: Examples of calculations of Turaev-Reshetikhin TQFT of cobordisms with boundaries have genera greater than 1
QUESTION [6 upvotes]: I am studying Turaev-Reshetikhin TQFT. I describe the definition of the invariant $\tau(M)$ of a cobordism $(M, \partial_{-}M, \partial_{+}M)$ in the previous question breifly. Framings in the definition of Reshetikhin-Turaev TQFT
In many papaers or book, the only example of this invariant is just the case of a cylinder over a torus $M=T\times [0,1]$, where $T$ is a torus. In this case a cobordism is$(T \times [0, 1], T, T)$ and the top boundary is parametrized by the identity $\mathbb{id}:T \to T$ and the bottom boundary is parametrized by any homeomorphism $f: T \to T$.
I would like to know more examples, especially when $M$ is a cylinder over a closed orientable surface of genus greater than $1$. In this case, what I found difficult is that to find a special ribbon graph and show that it gives a cobordism $(M, \partial_{-}M, \partial_{+}M)$ as a result of surgery and check if the parametrizations of the boundaries are correct.
Could you give me examples of such calculations? (ex. $M$ is a cylinder over a genus 2 surface with identity parametrization on top boundary and "non trivial" parametrization on the bottom boundary.)
Thank you in advance.

REPLY [7 votes]: I guess this is it for my MO lurking. So anyway, you're interested in seeing example calculations, similar to Turaev IV.5.4, of the action introduced in Turaev IV.5.1, right?
This action is also referred to known as the quantum representation of the mapping class group and has been considered from numerous viewpoints including of course TQFTs via quantum groups, the skein theory of the Kauffman bracket, conformal field theory, and geometric quantization. Calculations in the higher genus case grow messy quickly, but the skein theory approach provides an algorithm (cf. Masbaum-Vogel, 3-valent graphs and the Kauffman bracket) for calculating explicitly the representation in a particular natural basis of the vector spaces associated to the boundary surface. This algorithm has been implemented by A'Campo and Masbaum and should be available here -- the site seems to be down at the moment, but leave me an e-mail if you want a copy of the (usually freely available) program. How it works and some example calculations have been explained by A'Campo.
Note that all calculations of this program are performed using the skein theory approach, as described in Turaev Ch. XII. Off the top of my head, the only explicit non-torus and non-computer assisted calculations of the quantum representations I remember seeing are for the four-holed sphere (cf. Masbaum, An element of infinite order in TQFT-representations of mapping class groups, Andersen-Masbaum-Ueno, Topological quantum field theory and the Nielsen-Thurston classification of $M(0,4)$, Laszlo, Pauly, Sorger - On the monodromy of the Hitchin connection). By the factorization properties of the TQFT/the quantum representations, this then gives example calculations for all surfaces of genus $g \geq 2$, as the four-holed sphere embeds in such.
Now, a different family of examples is provided by complements of links in $S^3$, i.e. 3-manifolds having boundary a disjoint union of tori. Here, in the modular category corresponding to, say $U_q(\mathfrak{sl}_N)$ or the one arising from the HOMFLY polynomial (cf. Blanchet - Hecke algebras, modular categories and 3-manifold quantum invariants), knowing the vector in the vector space associated to (a disjoint union of) tori corresponding to the knot complement more or less boils down to understanding the coloured HOMFLY invariants of the link in question, as the vector space of the torus has a basis of handlebodies containing longitudinal (coloured) links, and these link invariants have been considered by several people (in particular in the case $N = 2$).<|endoftext|>
TITLE: Atiyah-Bott-Shapiro Orientation
QUESTION [9 upvotes]: Dear community,
there are so-called orientation maps $a:MSpin\to ko$ and $b:MSpin^c \to k$, "defined" in ABS's paper "Clifford modules". Unfortunately I am not familiar with representation theory.
Let $c:MSpin \to MSpin^c$ resp. $d:ko \to k$ denote the obvious maps given by considering a spin- as a spin$^c$-manifold resp. complexification.
Is it true that $b\circ c=d\circ a$?

REPLY [14 votes]: A point in the $n$th space of $MSpin$ is an $n$-dimensional manifold equipped with a spin structure. In other words, it is a manifold equipped with a bundle of bimodules between the Clifford algebra of $\mathbb R^n$ and the Clifford algebra of $TM$. That bundle of bimodules is called the spinor bundle of $M$.
Similarly, a point in the $n$th space of $MSpin^c$ is an $n$-dimensional manifold equipped with a bundle of bimodules between the Clifford algebra of $\mathbb C^n$ and the Clifford algebra of $TM\otimes_{\mathbb R}\mathbb C$.
The map $MSpin\to MSpin^c$ is given by complexifying the bimodule.

A point in $n$th space of $KO$ (allow me to take non-connective $K$-theory - the result for connective $K$-theory then follows readily) is given by a real Hilbert space equipped with an action of $Cliff(\mathbb R^n)$, and an odd skew-adjoint clifford-linear Fredholm operator.
Similarly, a point in the $n$th space of $K$ is given by a complex Hilbert space with an action of $Cliff(\mathbb C^n)$, and an odd skew-adjoint clifford-linear Fredholm operator.
The map $KO\to K$ is again complexification.

The ABS orientation (which is not constructed in ABS)
sends a spin manifold $M$, now also equipped with a metric, to the Hilbert space of $L^2$ sections of the spinor bundle, equipped the the obvious $Cliff(\mathbb R^n)$-action. The Fredholm operator is the Dirac operator constructed from (the connection associated to) the metric and the $Cliff(TM)$-action.
The spin-c version is identical. It is then obvious by construction that the diagram you asked about is commutative.

I should say that the above argument is completely hand-wavy...
I actually don't know in which paper/textbook the ABS orientation is defined as a map of spectra (and I would like to know -- so if someone knows, please tell me). My guess is that, regardless of the approach taken, once you see the definition, it is completely obvious that the diagram is commutative.<|endoftext|>
TITLE: Short time existence on nonlinear parabolic PDE
QUESTION [12 upvotes]: I saw several papers that without proof accept the fact "Short time existence on nonlinear parabolic PDE" is there any affirmative proof of this fact?
in which book we have this fact, the number of page and number of theorem ?

REPLY [12 votes]: This is kind of meta-theorem. It has several version, and if someone was willing to write one theorem containing all the situations, it would be unreadible. 
The fully non-linear case (example $\partial_t u=\det(\nabla^2u)$ with $u(t=0,\cdot)$ convex), is really involved. In the quasilinear case (example $\partial_t u={\rm div}(|\nabla u|^{p-1}\nabla u)$, it is still complicated, but several books treat it extensively. 
The semi-linear case $\partial_tu+Lu=f(u,\nabla u)$, where $L$ is an elliptic linear operator (like the Laplacian $-\Delta$), has a rather simple philosophy. One follows the guidelines of the Cauchy-Lipschitz theory for ODEs. If $a\in X$ is the initial data, one rewrites the Cauchy problem as an integral equation
$$u(t)=e^{-tL}a+\int_0^t e^{(s-t)L}f(u(s),\nabla u(s))ds=:Nu(t).$$
When $T>0$ is small enough and the Banach space $X$ is appropriate, one proves that $N$ is a contraction in some ball $B(a;r)$ of $C(0,T;X)$. Then Picard's theorem tells that there is a unique fixed point $u$ ; this is the local solution.
By appropriate, I mean that the operators $S_t:=e^{-tL}$ form a strongly continuous semi-group over $X$. In particular $S_t\in{\mathcal L}(X)$ and $$\|S_t\|_{{\mathcal L}(X)}\le Ce^{\omega t},\qquad\forall t>0$$
for suitable constants $C$ and $\omega$. In practice, we may deal with a scale of Banach spaces $X_s$, like the Sobolev spaces $H^s$, and we have
$$\|S_t\|_{{\mathcal L}(X_s,X_r)}\le Ct^{\alpha(s-r)}e^{\omega t},\qquad\forall t>0$$
for some $\alpha>0$, which is related to the order of $L$.<|endoftext|>
TITLE:  Are there simple conditions on a category C which guaranty that Ind(C) is a Grothendieck topos?
QUESTION [12 upvotes]: The category of finite sets is not a Grothendieck topos, but its Ind category
Ind(Finite-Sets) = Sets
is a Grothendieck topos. Similarly, given a pro-finite group G, the Grothendieck topos of discrete G-sets is the Ind of the category of finite G-sets.
Question: Are there simple conditions on a category C which guaranty that Ind(C) is a Grothendieck topos?
The examples which interests me is of the following kind: I have an infinite sequence of finite groups $(G_i)$ and a corresponding sequence of quotient groups $G_i \to H_i$. I consider the category of sequences $(X_i)$ of finite sets such that each $X_i$ carries a $G_i$-action which factors through $H_i$ for almost all $i$. Even for extreme cases (i.e. $H_i=G_i$, $H_i=1$, or even $G_i=H_1=1$) I don't know if the corresponding Ind-category is a topos. Another possible variant  is to require a uniform bound on the size of the $X_i$'s.

REPLY [7 votes]: In Di Liberti–Ramos González's Gabriel-Ulmer duality for topoi and its relation with site presentations, they state (Theorem 3.17) that $\mathbf{Ind}(\mathscr C)$ is a Grothendieck topos if and only if $\mathscr C$ is extensive and pro-exact (Definition 3.14 therein).
They attribute this result to Carboni–Pedicchio–Rosický's Syntactic characterizations of various classes of locally presentable categories, though it does not appear as an explicit theorem there, so it not such a convenient reference.<|endoftext|>
TITLE: Smooth bases of matroids
QUESTION [15 upvotes]: Motivated by algebraic geometry, I've come up with a purely
combinatorial definition within the theory of matroids. 
The question is: is this concept known?
If you like matroids but not algebraic geometry, skip to the definition below.
Let $n\choose k$ denote the collection of all $k$-element subsets of
$[1,n]$ (rather than the number thereof). We can and will identify
this collection both with the set of $T$-fixed points on the
$k$-Grassmannian $Gr(k,n)$, where $T$ is the $n$-torus that acts
(unfaithfully), and also with the set of Plücker coordinates.
Let $C \subseteq {n\choose k}$ be a subcollection. Then, following 
Neil White, we can define a subscheme $\Pi_C$ of $Gr(k,n)$ by killing
all Plücker coordinates $p_S, S \notin C$. This subscheme is $T$-invariant,
and its $T$-fixed points are exactly $C$.
Easy fact: if $\Pi_C$ is irreducible, then $C$ is a matroid. The
non-Pappus matroid shows the converse is false. (This is my own motivation
for matroids -- they serve as combinatorial stand-ins for subvarieties
of Grassmannians.)
I'm interested in the smooth points of $\Pi_M$, where $M$ is a matroid.
Perhaps the most efficient way to describe $M$ is by listing
its connected flats $F$, and for each, giving the rank. (Saying that
$rank(F) \leq r$ means that for each $S$ that intersects $F$ too much,
$p_F = 0$. I'm pretty sure that the connected flats gives the 
shortest list of $F$s to give all the $S$.)

If $M \subseteq {n\choose k}$ is a matroid, call $\lambda \in M$
  a smooth base if for any connected flat $F$, $rank(F) = |\lambda \cap F|$.

Note that $\geq$ is required for $\lambda$ to be a base at all.
It's pretty easy to prove that $\lambda$ is a smooth point of $\Pi_M$
iff $\lambda$ is a smooth base of $M$ in the sense above.

Is this concept known to matroid theorists? Is this characterization
  of smooth points known to anybody?

Example: let $M$ be the Schubert matroid for a $\lambda \in {n\choose k}$,
i.e. for each $i \notin \lambda, i+1 \in \lambda$, we have a connected flat
$[1,i]$ with rank $|[1,i] \cap \lambda|$. Then $\lambda$ is a smooth
base of $M$. And indeed, $\lambda$ is the point in the Bruhat cell whose
closure is the Schubert variety $\Pi_M$.

REPLY [7 votes]: I hope this question gets a good answer. In the mean time I'll mention a concept I've seen which seems somewhat related to your condition of smoothness.
When you have a matroid $\mathcal M$ with a base $B$ with the property that all cyclic flats $F$ are spanned by $F\cap B$, this is called a fundamental transversal matroid. The base $B$ is called a fundamental base. 
Schubert matroids are a special case of fundamental transversal matroids, and the base $\lambda$ is a fundamental base in the sense above. At some point I was convinced that your definition implies at least that the matroid is transversal, but now I'm not so sure anymore.
Some further characterizations in terms of some rank inequalities, or affine representations on simplices are proved in "Characterizations of transversal and fundamental transversal matroids" by J.E. Bonin, J.P.S. Kung and A. de Mier. See also these slides for pictures. I should also point out that I haven't seen the geometric connection which you explain being mentioned in the literature on fundamental transversal matroids.<|endoftext|>
TITLE: Interesting mathematical documentaries
QUESTION [174 upvotes]: I am looking for mathematical documentaries, both technical and non-technical. They should be "interesting" in that they present either actual mathematics, mathematicians or history of mathematics. I am in charge of nourishing our departmental math library (PUCV) and I would like to add this kind of material in order to attract undergraduates toward mathematics. For this reason, I am not looking for videos of conferences or seminar talks, but rather for introductory or "wide public" material.
Here are some good examples. 

"Dimensions", by Leys, Ghys & Alvarez, http://www.dimensions-math.org/ which explains actual maths and is excellent.
"Julia Robinson and Hilbert's tenth problem", https://www.vismath.eu/en/films/julia-robinson, about the life of some great mathematicians.
BBC documentary on "Fermat's last theorem" (by the way, any information about how to purchase it would be welcome, it does not seem to be possible to do it from the BBC site http://www.bbc.co.uk/iplayer/episode/b0074rxx/horizon-19951996-fermats-last-theorem. Maybe http://vimeo.com/18216532 ?). 

Are there more examples? Thanks, Ricardo.

REPLY [4 votes]: Recently a documentary on Maryam Mirzakhani was released called Secrets of the Surface: The Mathematical Vision of Maryam Mirzakhani. It both described her personal life/journey through mathematics, as well as an introduction to part of her research aimed at a general audience.<|endoftext|>
TITLE: Can every $\mathcal{L}_{\omega_1,\omega}$ formula be expressed as a type? What about canonical forms?
QUESTION [6 upvotes]: If $\mathcal{L}$ is a countable, first-order language, it is easy to see that every $n$-type $p$ (over $\emptyset$) can be expressed as an $\mathcal{L}_{\omega_1,\omega}$-formula, namely $\bigwedge_{\psi\in p} \psi$. Does the converse hold? Namely, given an $\mathcal{L}_{\omega_1,\omega}$ formula $\psi(v_1,\ldots,v_n)$ is there an $n$-type $p$ such that $\mathfrak{A}\models\psi(\bar{a})$ if and only if $\bar{a}$ realizes $p$ in $\mathfrak{A}$?
On a related note, are there any nice canonical forms for such infinitary formulas?

REPLY [8 votes]: As noted in the other answers, not every $L_{\omega_1,\omega}$ formula is expressible as a type. 
Nevertheless, there is some sense in which $L_{\omega_1,\omega}$ is equivalent to omitting types:
Theorem: For every $L_{\omega_1,\omega}$ sentence $\phi$, there exists a countable first-order theory $T$ and a countable set of types $\{p_n:n<\omega\}$ such that any model $A$ satisfies $\phi$ if and only if $A$ can be expanded into a model of $T$ which omits every $p_n$.
One can construct $T$ and the $p_n$ as follows: introduce a new relation symbol $R_\psi(x_1,\dots,x_n)$ for every subformula $\psi(x_1,\dots,x_n)$ of $\phi$. If $\psi$ is atomic or constructed by the usual first-order operations from other subformulas, include in $T$ a corresponding axiom: for example,
$$R_{\exists x_{n+1}\,\chi(x_1,\dots,x_{n+1})}(x_1,\dots,x_n)\leftrightarrow\exists x_{n+1}\,R_{\chi(x_1,\dots,x_{n+1})}(x_1,\dots,x_{n+1}).$$
The only problem is to deal with countable conjunctions and disjunctions. If for instance $\psi(\vec x)=\bigwedge_{n<\omega}\psi_n(\vec x)$, we include in $T$ the axioms $R_\psi(\vec x)\to R_{\psi_n}(\vec x)$ for all $n$, and we include the type
$$p_\psi(\vec x)=\{R_{\psi_n}(\vec x):n<\omega\}\cup\{\neg R_\psi(\vec x)\}$$
as one of the $p_n$'s. Notice that $A$ omits $p$ iff it validates the implication
$$\bigwedge_{n<\omega}R_{\psi_n}(\vec x)\to R_\psi(\vec x).$$
The rest of the proof is easy.
If we work only with infinite models, a single type $p$ is sufficient instead of countably many. This can be seen as follows. Introduce a new predicate $N(x)$ and constants $\{c_n:n<\omega\}$, and consider the type
$$p(x)=\{N(x)\}\cup\{x\ne c_n:n<\omega\}.$$
Then for each $\psi(\vec x)=\bigwedge_{n<\omega}\psi_n(\vec x)$ introduce a new predicate $S_\psi(u,\vec x)$ together with the axioms
$$\begin{gather*}
R_{\psi_n}(\vec x)\to S_\psi(c_n,\vec x),\\
\forall u\,(N(u)\to S_\psi(u,\vec x))\to R_\psi(\vec x),
\end{gather*}$$
which will serve together with $p$ as a replacement for $p_\psi$.
This result has various interesting consequences: for example, Hanf numbers of $L_{\omega_1,\omega}$ and of FO with omitting types are the same. (The Hanf number of a logic $L$ is the smallest cardinal $\kappa$ such that for every $L$-sentence $\phi$, if $\phi$ has a model of size at least $\kappa$, then it has models of arbitrarily large cardinality. The axiom of replacement implies that every logic whose formulas do not form a proper class has a Hanf number.) As a matter of fact, both Hanf numbers equal $\beth_{\omega_1}$, but this beautiful result of Morley has a considerably more difficult proof.<|endoftext|>
TITLE: Open immersions of open manifolds
QUESTION [31 upvotes]: For concreteness, I will work in the category of smooth manifolds, but my question makes sense in topological and PL category as well. Recall that a manifold $M$ is called open if every connected component of $M$ is non-compact. 
Question. Is it true that for every $n$ there exists a compact $n$-dimensional manifold $N^n$ so that every open $n$-dimensional manifold $M^n$ admits an immersion in $N^n$? (In this context an immersion is just a local diffeomorphism.) 
I think that the answer is positive and that the manifolds $N$ are connected sums of products of projective spaces of various dimensions. 
Edit: Igor noted that real-projective spaces are not enough, one should include complex-projective spaces as factors in the products. (The reason I think that products real and complex projective spaces are the right thing to use is that real and complex projective spaces generate rings of unoriented and oriented cobordisms.) 
Some background: As we know very well, not every manifold admits an open embedding in a compact manifold. For instance, the infinite connected sum of 2-dimensional tori does not. However, it is easy to prove that every open surface admits an immersion in ${\mathbb R}P^2$. Whitehead proved that every open oriented 3-dimensional manifold admits an immersion in ${\mathbb R}^3$  (and, hence, to any 3-dimensional manifold). I also convinced myself (although I do not have a complete proof) that every open non-orinentable  3-manifold admits an immersion in ${\mathbb R}P^2\times S^1$.  More generally, every open paralelizable $n$-manifold admits an immersion in ${\mathbb R}^n$. This is a special case of the Hirsch-Smale immersion theory, which reduces existence of immersion from an open manifold to a homotopy-theoretic question about maps of tangent bundles.

REPLY [8 votes]: This is the special case announced in the comment. We assume $M$ to be an oriented 4-manifold with the homotopy type of a 2-complex. We prove that $M$ immerses in $\mathbb{C}P^2$.
In the case of compact 4-manifolds with boundary I have a more direct and elementary proof in my paper which does not depend on Phillips' theorem.
Anyway, a bundle map $TM \to T\mathbb{C}P^2$ (needed to apply the theorem of Phillips) can be constructed in this way. Firstly, endow $M$ with an almost-complex structure (which exists by our assumptions). There is a nowhere vanishing vector field on $M$, hence $TM$ splits as a Whitney sum of complex rank 1 bundles $TM = \xi \oplus \varepsilon^1$ with $\varepsilon^1$ trivial and $\xi$ a pullback of the complex universal bundle. By cellular approximation, $\xi$ is a pullback of the canonical bundle $\gamma^1_1$ on $\mathbb{C}P^1$, so $TM$ is the pullback of $\gamma^1_1 \oplus \varepsilon^1$. Now, rank 4 real oriented vector bundles over $\mathbb{C}P^1$ are classified by $\Bbb Z_2$ (via the second Stiefel-Whitney class), hence $T\mathbb{CP}^2_{|CP^1} = \gamma^1_1 \oplus \varepsilon^1$ (as real vector bundles). So $TM$ is a pullback of $T\mathbb{C}P^2$ by a suitable map $M \to \mathbb{C}P^2$ (with values in $\mathbb{C}P^1$).<|endoftext|>
TITLE: Constant Mean Curvature hypersurfaces "condensing" onto a minimal submanifold
QUESTION [7 upvotes]: Let $M$ be Riemannian manifold and $S\subset M$ a minimal submanifold, with $\dim S<\dim M-1$. According to a few references (e.g., Mahmoudi, Mazzeo & Pacard), it should not be hard to see that:

''The closer a constant mean curvature (CMC) hypersurface of $M$ is to $S$ (in the Hausdorff metric), the larger its mean curvature must be.''

I was wondering if this claim is indeed not hard to see, given that I am still unable to find a simple/elementary proof. Any suggestions?
Moreover, I was wondering if anything is known about the speed in which the value of the mean curvature $H(t)$ of a family $N_t$ of CMC hypersurfaces of $M$ "condensing" (i.e., collapsing) onto $S$ diverges to $+\infty$. For instance, is it always the case that both $H$ and $H'$ diverge to $+\infty$ as those hypersurfaces collapse onto the minimal limit submanifold? Perhaps, in some special situation (e.g., if the sequence of CMC hypersurfaces collapsing is a solution to the Mean Curvature Flow (MCF)), this is implied by some property of the MCF?

REPLY [8 votes]: Suppose that N is the CMC hypersurface which is $\epsilon$ close to S.
Take the tube of radius $\epsilon$ around S. One can check that this has
(typically variable) mean curvature on the order of $1/\epsilon$. If one
chooses this tube correctly, it is tangent to N and this gives a lower bound 
for the mean curvature of N at the point of contact. Since the mean
curvature of N is constant, this means that it is everywhere bigger than
$1/\epsilon$.
The situation is now a bit better understood than when that paper was written.
First, Rosenberg proved a theorem that if N is a surface in a 3-manifold and has mean curvature H and H is very large, then one of the two regions N bounds in the ambient
manifold has inradius less than C/h.
Second, it is only true that the set on which the $N_t$ collapse must be minimal
if the norm of the second fundamental form of N is comparable to the mean curvature.
If this fails -- and a typical scenario is a sequence of spheres joined by very small
necks, then it seems to be the case that the collapsing set is not necessarily minimal.
Finally, regarding your question -- by the same comparison result (using geodesic tubes) it is certainly the case that H is increasing as $N_t$ collapses, so this gives some sort of
average increase of H'(t).  There maybe could be some exotic examples where the mean curvature wobbles and H' is not necessarily tending to $+\infty$.  
There are lots of other good questions here -- Harold Rosenberg had several questions which remain unanswered. For example, if  T is a  `geodesic triangle' in a 3-manifold,  then is there a CMC tube which condenses to T? The problem is whether it is possible for the CMC
surfaces to turn the corners at the vertices of T.<|endoftext|>
TITLE: Covering a (hyper)cube with lines
QUESTION [7 upvotes]: Let $K_n$ be the sets of vectors $x \in  \mathbb{Z}^d $ with each coordinates $x_i$ between $1$ and $n$. For any subset $A$ of $K_n$, let $S(A)$ be the set of points $x \in K_n$ which are on some line containing at least two points of $A$ (in other words, $S(A)$ is the union of the lines passing through - at least - two points of $A$).
Such a set $A$ is said to generate $K_n$ if $S(A) = K_n$. Now let $r_d(n)$ be the smallest size of a generating subset of $K_n$.
Question : What are the best known bounds on $r_d(n)$ ? (The first non trivial case is $d=2$)
This problem may be "well-known" ; I'm almost sure this question has already been studied, but I didn't find any reference, and Google gives nothing.
The trivial bound is $r_d(n) \gg_d n^{\frac{d}{2}- \frac{1}{2}} $ : taking a generating subset of size $r_d(n)$, there are at most $O(r_d(n)^2)$ lines to consider, each one intersecting $K_n$ in at most $n$ points, so that $|K_n| \ll r_d(n)^2 \times n$.
A refinement of this argument (a typical line contains much less than $n$ points of $K_n$) gives a lower bound $r_d(n) \gg_d n^{\frac{d}{2}- \frac{1}{4} - \frac{1}{4(2d-1)} } $.

REPLY [9 votes]: I think the probabilistic method gives an A of size $O_d(n^{d/2}\sqrt\log n)$. (Formula updated according to js's comment.)
Put every point to A independently with probability p. What is the probability that a point x will be in S(A)? For any x, we can find $\Omega(n^d)$ pairs of points that are all different such that x lies on the line of any pair. (This is true because e.g. for d=2 if x is in the bottom-left part of $K_n$, then we can take the $n/4\times n/4$ grid upper-right from it, contract the $n/8 \times n/8$ bottom-left of this grid, and double each point from x to get its pair.)
The probability that both points of a fixed pair are in A is $p^2$, the probability that no such pair exists is $(1-p^2)^{n^d}$. So if $n^d(1-p^2)^{n^d}<1$, then we are done using the union bound. Unless I am mistaken this is true if $p>\Omega_d(n^{-d/2}\sqrt\log n)$.
Now of course we cannot be sure about how big A is. But if we replace the above $<1$ with a $<1/2$, then we can even add the condition that A should be at most $pn^d$, for which the probability is $\ge 1/2$. So we get $O_d(n^{d/2}\sqrt\log n)$ points. Maybe this can be further improved with some more advanced probabilistic methods.<|endoftext|>
TITLE: RS to RSK correspondence
QUESTION [5 upvotes]: The RS correspondence is a correspondence which associates to each permutation a pair of standard Young tableaux of the same shape.
The RSK correspondence associates to each integer matrix (with non-negative entries) a pair of semistandard Young tableaux of the same shape.
Given an integer matrix, replace it by a permutation matrix whose rows and columns, when partitioned according to the row and column sums of the original matrix, have block sums equal to the entries of the original matrix. There is a unique such permutation matrix with the property that there are no descents within any of the blocks (each block is a partial permutation).
For example, if
$A=\begin{pmatrix} 2 & 1\\ 1 & 0\end{pmatrix}$
then the corresponding permutation matrix is 
$\tilde A  =\begin{pmatrix} 1 & 0  & 0 & 0\\
                             0 & 1  & 0 & 0\\
                             0 & 0  & 0 & 1\\
                             0 & 0 & 1 & 0\end{pmatrix}$
Here the row and column partitions are both $(3,1)$.
It seems to be well-known (for example, it is implicit in Fulton's matrix ball construction) that to obtain the SSYT's for $A$, one may substitute for each entry in the SYT's for $\tilde A$ the integers corresponding to the blocks the rows and columns corresponding to these entries belong.
In the above example, the SYT's associated to $\tilde A$ are 
$P = Q = \begin{array}{cc} 1 & 2 & 3 \\ 4 & &\end{array}$
into which we would saubstitute $1$ for $1,2,3$ and $2$ for $4$ to get the SSYT's for $A$:
$P = Q = \begin{array}{cc} 1 & 1 & 1 \\ 2 & & \end{array}$.
Is there a nice reference for this result?

REPLY [14 votes]: [Responding particularly to Bruce...]  You may want to take a look at my thesis, which was the first place that the Knuth versions of RSK were "Fominized".  There are lots of examples, which others have told me they've found helpful in understanding this material.  (I'm sure Fomin already understood that this could be done, but it doesn't appear in his papers before 1991.)  I put a scan on the web at:
http://www.math.uconn.edu/~troby/research.html
Scroll down to: 
Applications and Extensions of Fomin's Generalization of the Robinson-Schensted Correspondence to Differential Posets, Ph.D. Thesis, Massachusetts Institute of Technology, 1991. 
The key idea is just that RS commutes with "standardization" of words or SSYT, where one adds subscripts from Left to Right in the word and corresponding tableaux. See EC2, Lemma 7.11.6.  
Thanks to Tricia Hersh for mentioning this thread at the SIAM DM conference.  This is my first posting to MathOverflow, so I'm not allowed to comment.  
Hope this helps!
Tom<|endoftext|>
TITLE: Are there infinite groups which have only a finite number of irreducible representations?
QUESTION [9 upvotes]: If I know that a certain group has only finitely many irreducible representations (let's say over the complex numbers), is that group necessarily finite?
In the following cases, you can see that there will be an infinite number of irreducible representations(irreps).

If the group is a direct product of infinitely many (non-trivial)groups. 
If the group is finitely generated, infinite, and abelian.
If there exists a normal subgroup N such that G/N satisfies conditions 1 or 2.

I'm stuck. Any ideas? Given an arbitrary infinite group, can you construct infinitely many irreps?

REPLY [3 votes]: This is not an answer, but it's a bit too long for a comment.
If I'm not mistaken, any group $G$ which embeds densely into a compact Hausdorff group $H$ has the property that it is finite if and only if it has finitely many irreducible representations. This includes the residually finite groups, which embed densely into their profinite completions. To see this, use the fact that by Peter-Weyl $L^2(H)$ decomposes as a direct sum $\bigoplus_i n_i V_i$ where $V_i$ are the finite-dimensional unitary irreducible representations of $H$ and $n_i = \dim V_i$. It follows that there are infinitely many $V_i$ if and only if $H$ is infinite if and only if $G$ is infinite; moreover, by density the $V_i$ are irreducible representations of $G$, any two which are inequivalent as $H$-representations are inequivalent as $G$-representations. 
So to find a counterexample we should look for groups that are not residually finite. Infinite simple groups certainly have this property, but I don't know any of them well enough to analyze their irreducible representations.<|endoftext|>
TITLE: Bitopological spaces and algebraic topology
QUESTION [5 upvotes]: Is it possible to introduce the concept of bitopological spaces such as $(X,T_1,T_2)$ (introduced by J.C.Kelly see Proc. London Math. Soc. (3) 13 (1963) 71–89 MR0143169, J.C. Kelly) in the homotopy theory or homology theory?

REPLY [4 votes]: In some sense, the (relatively new) field of directed algebraic topology represents an attempt to do this. This article by Marco Grandis includes a discussion (on page 8) of why bitopological spaces are too general to admit a good (directed) homotopy theory.<|endoftext|>
TITLE: What do we mean by contractible for simplicial objects in a category?
QUESTION [5 upvotes]: EDIT: removed cruft from this question.
Recall that extra degeneracies for an augmented simplicial set $X$ are maps $s_0\colon X_n \to X_{n+1}$ for $n=-1,0,1,2,\ldots$ which satisfy the usual simplicial identities with respect to the existing $d_i$, $s_i$. This definition clearly works for simplicial objects in pretty much an arbitrary category.
A simplicial set, augmented by $X_{-1} = \ast$, with extra degeneracies is contractible. In fact, given enough structure on an ambient category $C$, one can sensibly talk about homotopy of simplicial objects in $C$ (for example, one can say when $sC$ is a category with cofibrant objects, and has a notion of homotopy of maps). [EDIT: By this I mean there is some sort of model structure around relative to which we can talk about homotopy]
So my question is this: is it reasonable to think of simplicial objects in $C$ with extra degeneracies as being contractible for any category $C$ with terminal object? Certainly, ignoring size issues, we can think of such things as being contractible after we embed them in the category of simplicial sets in $Pre(C)$, if not some smaller (co)completion category.
Secondarily, can I get away with saying a simplicial object with extra degeneracies "is a contractible simplicial object?"

REPLY [3 votes]: Actually the issue is not that simple. Here is a talk by Mike Barr on showing there are three different notions of 'contractible' for simplicial objects in a category that do not coincide even for simplicial sets. They even give explicit examples.

Michael Barr (joint with John F. Kennison, R. Raphael), Contractible simplicial objects, talk 9 October 2018 in the Logic and Categories seminar, McGill University, abstract, slides.

They find "Strong extra degeneracies" implies "Extra degeneracies" implies "Homotopic to a constant simplicial object", and these are strict implications. See the slides for precise definitions. Beware that the 'extra degeneracies' given in my question are not necessarily the same as Barr et al's definition.<|endoftext|>
TITLE: Elementary Equivalence =?  Homotopy Equivalence
QUESTION [12 upvotes]: One of the most interesting novelties in recent foundational studies is Voevodsky's Homotopical Type Theory project (see here). 
Finally homotopy theory ideas have entered in a royal fashion the foundational arena! 
I wonder if other areas of logic and foundational studies can be tackled from an homotopical standpoint. For instance, in model theory, one encounters the central notion of elementary equivalence:  

two structures M and N of the same
  signature $\sigma$ are called
  elementarily equivalent if they
  satisfy the same first-order
  σ-sentences.σ-sentences.

The question:

take elementary embedding as a notion
  of weak equivalence, what kind of
  structure has the associated  homotopy
  category?  Perhaps dreaming a little,
  can one even manage to identify a
  Quillen model structure on the
  category of $\sigma$ -structures? 

NOTE: Andreas Blass has (rightly) asked why I mention elementary embedding in my question, whereas the title talks about homotopy equivalences.
 Point well taken: I should reformulate the question in a broader form, as: can you choose some maps in the category as weak equivalences, so that we can have a homotopy category, possibly with a good amount of homotopy limits and colimits to do some real computations?
PS: The dream behind this question is that perhaps one could think of an elementary substructure as a homotopical retract of the ambient structure , and more generally one could come up with a notion of "continuous deformation" of structures, just like in the topological category

REPLY [3 votes]: The dream behind this question is that perhaps one could think of an elementary substructure as a homotopical retract of the ambient structure
I have tried and failed to do something similar for models of a sufficiently nice theory, say a first order categorical theory such as ACF; here is what my thoughts were. My desired weak equivalences were: add a finite tuple to a model M and get a model Ma prime(primary) and minimal over $M\cup a$ (acyclic cofibration); represent a model $N$ as the union $\cup M_i$ of an increasing chain of elementary submodels of strictly smaller cardinality (acyclic fibration). 
The first one is an elementary equivalence between a model and its substructure, and thus fits your suggested definition. However, the second one requires one to extend your category of models and consider the category of families of models; from the categorical point of view, you formally add new limits ignoring those limits you already have in your category. 
It is then easy to define a model "pre"structure that satisfies some of the axioms of a model category; Lowenheim-Skolem theorem then means that every morphism can be decomposed as a cofibration and an acyclic fibration, and existence of prime and minimal models (that holds for sufficiently nice theories) means that every morphism decomposes as an anyclic cofibration and a fibration (axiom M2 of Quillen). But that's all: I was not able to construct a model category for an interesting theory, say non-locally modular or even ACF itself. 
We were only able to construct a model category for a trivial theory T with the empty language; then this all becomes set theory and is described in here and here<|endoftext|>
TITLE: What is the first interesting matric Toda bracket in the stable homotopy of the sphere?
QUESTION [12 upvotes]: Feel free to gloss ‘interesting’ as you see fit.  One way:

1. What is the lowest degree matric Toda bracket in $\pi_\ast(S)$ that doesn't contain zero?

By ‘degree’ I mean total homotopical degree, i.e. the $\ast$ in $\pi_\ast$.  By ‘matric’ I mean to exclude ordinary Toda brackets, that is matric Toda brackets all of whose entries are one-by-one matrices, and also to exclude brackets that are trivially determined by ordinary Toda brackets.
I'm also interested in the title question with ‘first’ replaced by ‘simplest’.  For instance:

2. What is the lowest order matric Toda bracket in $\pi_\ast(S)$, all of whose matrix entries are sums of products of Hopf elements, that doesn't contain zero?

By ‘order’ I mean the number of entries in the bracket, i.e. whether it is 3-fold or 4-fold or 5-fold or...
Now I'd like to know the same, but with the proviso that the bracket be detectable in the classical Adams spectral sequence:

3. What is the first or simplest, interesting matric Toda bracket in $\pi_\ast(S)$ that is detectable in the $\mathrm{E}_2$ term of the classical Adams spectral sequence?

Some remarks:

I believe the matric Massey product $\langle h_2^2, h_0, \left(\begin{array}{cc} h_1 h_3 & h_2^2 \end{array} \right), \left(\begin{array}{c} h_1 \\ h_2 \end{array}\right)\rangle $ in the $\mathrm{E}_2$ term of the Adams spectral sequence is the class $e_0$, but that is not a permanent cycle.
Kochman lists the matric Toda bracket $\langle \sigma, \left(\begin{array}{cc} A[31] & \nu\end{array}\right),\left(\begin{array}{cc} \eta & 0 \\ 0 & \eta \end{array}\right),\left(\begin{array}{c} \nu \\ \eta A[30] \end{array}\right)\rangle$ in degree 44 as containing an element of order 8, where $A[30]$ and $A[31]$ refer to certain elements of order 2 in those degrees.

REPLY [16 votes]: I know this post is quite old, but in case you are still interested, or anyone else is, I thought about sharing my recent thoughts about the topic. After all, this is the second result on "matric toda bracket" on google, and it frustrated me several times that this is still unanswered.
The bracket André suggested sadly uses relations which are only valid in $tmf$: in $\pi(S)$, we have $\nu^3 + \eta \epsilon = \eta^2 \sigma \neq 0$.
We want to construct a nontrivial matric Toda bracket, so we have to use a relation which cannot be written as a single product. The first one of those is given by $4\nu + \eta^3 = 0$.
Multiplying this by two relates this in some sense to the easier relation $8\nu = 0$, so we could wonder about this enabling us to exhibit the Toda bracket $\langle \nu, 8, \nu\rangle$ as twice the matric Toda bracket 
$$\left\langle \left(\begin{array}{cc} \nu & \eta\end{array}\right), \left(\begin{array}{cc} 4 & \eta^2\\ \eta^2 & 0\end{array}\right), \left(\begin{array}{c} \nu \\ \eta\end{array}\right)\right\rangle$$
This yields a degree $7$ class of indeterminacy $\nu \cdot \pi_4 + \pi_4 \cdot \nu + \eta \cdot\pi_6 + \pi_6 \cdot \eta = 0$.
We want to multiply the middle term by $2$. In order to make sense of that, consider the following setting: In a good model for spectra, we can think of bimodules over a ring spectrum $E$, i.e. there are maps $E \wedge X\rightarrow X$ and $X\wedge E \rightarrow X$ plus respective commutative diagrams. Then we can talk about Toda brackets $\langle a, x, b\rangle$, where $a,b\in \pi_* E$ and $x\in \pi_* X$. Now given a bimodule map $f:X \rightarrow Y$, a defining system for the Toda bracket $\langle a, x, b\rangle$ can be pushed forward using $f$ to obtain a defining system for the Toda bracket $\langle a, f_* x, b\rangle$, so we obtain
$$
f_* \langle a, x, b \rangle \subseteq \langle a, f_* x, b\rangle
$$
This statement immediately generalizes to matric Toda brackets.
Notice that we can use this here: $S\xrightarrow{2} S$ has a representative which commutes with the right and the left action of $S$ on itself, since it can be factored naturally through the pinching map $S\rightarrow S\vee S$. 
This can be applied as above to see
$$
2\cdot\left\langle \left(\begin{array}{cc} \nu & \eta\end{array}\right), \left(\begin{array}{cc} 4 & \eta^2\\ \eta^2 & 0\end{array}\right), \left(\begin{array}{c} \nu \\ \eta\end{array}\right)\right\rangle \subseteq \left\langle \left(\begin{array}{cc} \nu & \eta\end{array}\right), \left(\begin{array}{cc} 8 & 0\\ 0 & 0\end{array}\right), \left(\begin{array}{c} \nu \\ \eta\end{array}\right)\right\rangle
$$
This actually is an equality, since the indeterminacy is zero on both sides.
In fact, the right hand side equals the usual Toda bracket $\langle \nu, 8, \nu\rangle$.
This element is usually known as $8\cdot \sigma$, which is nonzero. We obtain
$$
\left\langle \left(\begin{array}{cc} \nu & \eta\end{array}\right), \left(\begin{array}{cc} 4 & \eta^2\\ \eta^2 & 0\end{array}\right), \left(\begin{array}{c} \nu \\ \eta\end{array}\right)\right\rangle = 4\sigma
$$
What this example tells us is that matric Toda brackets really should be expected all over the place: Whenever a simple relation (consisting of a single product) lifts to somewhere as a more complicated relation (with more than one summand), we should be able to lift Toda brackets built from the simple relation to matric Toda brackets by adding additional rows which account for the new summands, in our case the $\eta^3$. Note also that the exact way in which we did that did not matter that much, we could also have chosen something like
$$\left\langle \left(\begin{array}{cc} \nu & \eta^2\end{array}\right), \left(\begin{array}{cc} 4 & \eta\\ \eta & 0\end{array}\right), \left(\begin{array}{c} \nu \\ \eta^2\end{array}\right)\right\rangle
$$
or similar. The only way this generally might go wrong is that you are unable to do this in a way which actually reduces back to the bracket you wanted (we used $2\cdot \eta^2$ to end up with a diagonal matrix).
If you look closely into the reference André posted, there a similar thing happened, only that the lifting did not happen along an actual map, but from the $E^{\infty}$-page of a spectral sequence to the actual thing, because of nontrivial extensions.<|endoftext|>
TITLE: Second homotopy group of Cayley complex
QUESTION [5 upvotes]: Is there a good reference for information about the second homotopy group of the Cayley complex or Presentation complex of a finitely presented group, especially a hyperbolic group?  I'm looking for an argument that the second homotopy group of the Cayley complex of a hyperbolic group $G$ is finitely generated as a $G$-module in particular, but I'd welcome other interesting starting points around the second homotopy group of the Cayley complex too.

REPLY [10 votes]: If $\langle X,R \rangle$ is a finite presentation of a group $G$, then there exists an exact sequence of $\mathbb ZG$-modules
$$0 \to \pi_2(Z) \to \mathbb{Z} G^{\oplus R} \to \mathbb Z G^{\oplus X} \to \mathbb Z G \to \mathbb{Z} \to 0,$$
where $Z$ is the presentation $2$-complex of the presentation above. If one knows in addition that $G$ is of type $FP_3$, then $\pi_2(Z)$ must be finitely generated as a $\mathbb Z G$-module. It is well-known that hyperbolic groups are $FP_{\infty}$, using the Rips complex.
Any example of a finitely presented group which is not of type $FP_3$ gives a counterexample, i.e. $\pi_2$ is not finitely generated. Brady constructed a subgroup of a hyperbolic group with this property in
Brady, N. Branched Coverings of Cubical Complexes and Subgroups of Hyperbolic Groups J. London Math. Soc. (1999) 60(2): 461-480.
Much earlier, Stallings gave an example where the third homology is not finitely generated as a module over the group ring of a finitely presented group.
Stallings, J. A finitely presented group whose 3-dimensional integral homology is not finitely generated. Amer. J. Math. 85 (1963), 541–543.<|endoftext|>
TITLE: isomorphic spectral sequences => quasi-isomorphic filtered chain complexes?
QUESTION [7 upvotes]: Let $(C,\partial)$ and $(C',\partial')$ be chain complexes of $R$-modules where $R$ is a (commutative) ring. Let $F$ and $F'$ be finite filtrations of $C$ and $C'$ respectively, i.e., $$\varnothing = F_0C \subset F_1C \subset \ldots \subset F_nC = C$$ and similarly for $F'$. There exist spectral sequences associated to $F$ and $F'$, let's call them $E_{\ast,\ast}^\ast$ and ${E'}_{\ast,\ast}^\ast$. 
It is quite easy to see that an isomorphism from $F$ to $F'$ induces an isomorphism from $E$ to $E'$. What is completely unclear to me (and sorry if this is not a research level question but I couldn't find an answer) is the converse, namely:

Does an isomorphism from $E$ to $E'$ imply a (quasi-)isomorphism between $F$ and $F'$?

More precisely, if two filtrations give rise to isomorphic spectral sequences, what is the strongest statement that can be made about them? Does this statement depend on the finiteness of the filtrations, on the nature of the ring $R$, on convergence of the spectral sequences, etc?

REPLY [14 votes]: Say your two filtered chain complexes are concentrated in degree zero.  Then the spectral sequences degerate, and your questions become: If you have two filtered abelian groups and an isomorphism between the associated graded modules, can you deduce that the abelian groups are isomorphic?  The answer is no; you can take
$$
0 \subset 2 \mathbb{Z} \subset \mathbb{Z}
$$
and
$$
0 \subset \mathbb{Z} \subset \mathbb{Z} \oplus \mathbb{Z}/2.
$$
As Ralph says, you usually need something on the chain level, unless you're in very degenerate cases (where the spectral sequence determines the isomorphism type of the object in the derived category).<|endoftext|>
TITLE: Negative objects in categories
QUESTION [8 upvotes]: I seem to remember a proof that if a category $C$ has coproducts and a $0$ object, then necessarily if we had objects of $C$, say $a$ and $-a$, such that $a \oplus -a \simeq 0$, then $a\simeq 0\simeq -a$.  
But right now, I can't place this, nor am I 100% sure that that is the correct property.  I am able to show that, using the various universal properties at play, that the morphisms in such a category are necessarily quite boring, but not that it completely collapses (i.e. if I assume that all objects have a corresponding 'negative' object).

REPLY [10 votes]: There are three proofs:
1) If $C$ has infinite coproducts, you may use the Eilenberg swindle:
$$a \cong a \oplus (-a \oplus a) \oplus (-a \oplus a) \oplus \dotsc \cong (a \oplus -a) \oplus (a \oplus -a) \oplus  \dotsc \cong 0.$$
2) Here is my personal summary of Philippe's proof: (a) It is enough to prove the dual statement $a \times a^{-1} \cong 1 \Rightarrow a \cong 1$ in categories with products. (b) This may be reduced, by the Yoneda embedding which preserves products, to the case $\mathsf{Set}$. (c) In the case of $\mathsf{Set}$ it is clear. 
3) If $\mathcal{L}$ is an invertible object of a symmetric monoidal category, then $\mathcal{L} \otimes -$ is cocontinuous since $\mathcal{L}^{-1} \otimes - $ is right adjoint to it. In particular $\mathcal{L} \otimes -$ preserves initial objects. Now apply this to $(C,\oplus,0)$. Thus if $a$ is invertible w.r.t. $\oplus$, we have $a \oplus 0 \cong 0$. On the other hand, the left hand side is also $a$.<|endoftext|>
TITLE: Number of double cosets
QUESTION [7 upvotes]: If $G$ is a finite group and $H \leq G$ is a subgroup, then $|G/H| = \frac{|G|}{|H|}$.
Is there an easy way to compute $|K \backslash G / H|$, for $K \leq G$ also a subgroup?

REPLY [3 votes]: Take $K=H$ and consider the diagonal action of $G$ on $\Omega\times\Omega$, where $\Omega$ is the set of  the right cosets $H\backslash G$. Let the number of orbits of this action be $d$. Then $d$ is the number of double cosets $H\backslash G/H$:
If we denote the orbits by $\Omega_1,\Omega_2,\ldots, \Omega_d$, let $g_1,\ldots,g_d$ be representatives of the respective orbits. Then the map sending $(Hg,Hg^{\prime})$ to $Hg^{\prime}g^{-1}H$ is easily shown to be a bijection between the set of orbits and double cosets.
More details can be found under the topic of Schur bases.<|endoftext|>
TITLE: Bounding the minimum entry of an inverse matrix
QUESTION [5 upvotes]: Suppose $A$ is an $n\times n$ stochastic matrix, that is, entrywise nonnegative and row sums are all $1$. If $A$ is invertible, is it true that the minimum diagonal entry of $A^{-1}$ is no larger than $1$? 
Small matrices support this claim, but for larger ones, I don't know how to (dis)prove it. 
Edited I forgot to add the condition that the diagonal entries of $A$ are all zero.

REPLY [8 votes]: Here's a counterexample:
$$ A = \frac{1}{5}
     \left[ \matrix{ 0 & 1 & 3 & 1 \cr 2 & 0 & 1 & 2 \cr 3 & 1 & 0 & 1 \cr 0 & 3 & 2 & 0}\right], \quad
 A^{-1}=    \frac{1}{3} \left[ \matrix{ 7 & -9 & 11 & -6 \cr -8 & 6 & -4 & 9 \cr 12 & -9 & 6 & -6 \cr -13 & 21 & -14 &9} \right].
$$<|endoftext|>
TITLE: Homotopy invariance of vector bundles by parallel transport: reference needed for my students.
QUESTION [8 upvotes]: Let $M$ be a smooth manifold and $V \to [0,1] \times M$ be a smooth vector bundle. The homotopy invariance states that the restrictions $V_0$ and $V_1$ to the bottom and top of the cylinder are isomorphic.
One can prove that using parallel transport: pick a connection on $V$. For each $x \in M$, take the curve $c_x:t  \mapsto (t,x)$. Parallel transport defined by the connection along the curves $c_x$ gives an isomorphism $V_0 \cong V_1$.
This is of course an easy exercise, and my question is to find a self-contained reference that is accessible for undergraduates who know the notions of manifolds, vector bundles and integral curves of vector fields on manifolds (I wish to assign this as a seminar talk).

REPLY [2 votes]: You can check pages 33-35 of my course notes. Also you can check section 3.4 of Husemoller's book  Fiber Bundles, Graduate Texts in Math, vol. 20, Springer Verlag, 1994.  In my notes  I prove the result for arbitrary compact spaces, while Husemoller does this for paracompact spaces. In both cases no smoothness is assumed.<|endoftext|>
TITLE: Indeterminacy of long games
QUESTION [7 upvotes]: Hello, all,
Several months ago I sat in on a seminar on AD+, which was incredibly wonderful even though I could barely follow it at all. AD+ is a technical variant of AD, the axiom of determinacy, which asserts that $$ \text{For all countable } X, A\subseteq X^\omega, \text{ the game with payoff set A is determined.}$$
It is easy to show that $ZFC+AD$ is contradictory; however, assuming some large cardinals, $ZF+AD$ is consistent, and in fact has the natural model $L(\mathbb{R})$.
One of the results mentioned was the fact that a natural strengthening of AD, in which $\omega$ is replaced by an arbitrary ordinal, is inconsistent with ZF (even though assuming large cardinals, many classes of long games are determined; this I gleaned from the amazon.com preview of Itay Neeman's "Determinacy of Long Games," which I suspect has the answers to my questions, but I don't have access to it). Having lost my notes from the seminar, I have two questions:
1) What is the proof of this fact?
2) When/by whom was it proved?
Thanks to all in advance!

REPLY [11 votes]: Your statement of AD is incorrect.  $X$ should be at most $\omega$ (and at least 2), not an arbitrary set.  Specifically, for $X=\aleph_1$, determinacy of games in $X^\omega$ is inconsistent with ZF.  
Concerning your actual question, determinacy of games of length $\omega_1$, even with $X=2$, is inconsistent.  I believe that this result and the one in the first paragraph of my answer are in Mycielski's first big paper on AD, "On the axiom of determinateness" in Fundamenta Mathematicae 1963 or 1964.  For the proof, fix a coding of the well-orderings of $\omega$ as $\omega$-sequences of 0's and 1's, and consider the following game of length $\omega_1$.  Player I must, at one of his moves, play 1 (if he always plays 0, he loses).  As soon as he plays a 1, say at move $\alpha$, player II must, in his next $\omega$ moves, play a code for a well-ordering of $\omega$ of length $\alpha$.  Clearly, player I cannot have a winning strategy, since every $\alpha$ has a code.  Determinacy of this game would mean that player II has a winning strategy, but that would give a function assigning to each countable ordinal $\alpha$ one of its codes.  The existence of such a function (or of any set of reals of size $\aleph_1$) contradicts AD (which is, of course, a consequence of length-$\omega_1$ determinacy).<|endoftext|>
TITLE: Integral representation of a determinant
QUESTION [9 upvotes]: In a paper by Mathai, he uses the following integral representation of a determinant,
(or, really, what I give is a simple special case of what he gives), without any explication.
All matrices are real $p\times p$ symmetric positive definite.
\begin{equation}
    | I-U |^{-a} = \frac{1}{\Gamma_p(a)} \int_{T>0} |T|^{a-(p+1)/2} \exp(-\text{Tr}(I-U)T) \;dT
\end{equation}
where $U$ satisfies $0\lt U \lt I$ (in the cone order in the cone of positive definte matrices),
the integral is over the cone of positive definite matrices and $\Gamma_p(a)$ is the
generalized gamma function in dimension $p$, and $\Re(a) > (p-1)/2$.
Any references for this?

REPLY [12 votes]: OLDER EDIT. (Elementary derivation) I realized that my original answer was actually overkill for the question. The said integral in question follows from the definition of the multivariate Gamma function
\begin{equation*}
\Gamma_p(a) := \int_{A > 0} \exp(-\mbox{tr}(A))\det(A)^{a-(p+1)/2}(dA),
\end{equation*}
where $\Re(a)>(p-1)/2$. 
From this it follows (by a change of variables) that for a positive definite matrix $S$, 
\begin{equation*}
\int_{A > 0} \exp(-\mbox{tr}(S^{-1}A))\det(A)^{a-(p+1)/2}(dA) = \Gamma_p(a)\det(S)^a,
\end{equation*}
so that with $S=(I-U)^{-1}$ we obtain the integral in question. 
Of course, to complete the picture it may be helpful to express $\Gamma_p(a)$ in more elementary terms. Chapter 2 of Muirhead's book provides these details. I cite the result that provides this expression.

Theorem (Muirhead (1982), Thm 2.1.2) Let $\Re(a) > (p-1)/2$. Then,
  $$
\Gamma_p(a) = \pi^{p(p-1)/4}\prod_{j=1}^p \Gamma(a - (j-1)/2)
$$

(Hint: To prove the above, write the Cholesky decomposition $A=T'T$, with these change of variables, the original Gamma integral factorizes into a product of Gaussian and Gamma integrals.)

The part that I recalled below provides yet another representation that expresses the multiplicative determinantal lhs in terms of an infinite sum.
OLDER STUFF
This is actually somewhat classical knowledge. Here are two related pointers.
A partition $\tau=(t_1,\ldots,t_m)$ is a vector of nonnegative integers listed in increasing order, and $|\tau|$ denotes $t_1+\cdots+t_m$. The generalized Pochhammer symbol $(a)_\tau$ is defined as
\begin{equation*}
\newcommand{\risingf}[2]{{{#1}}^{\overline{{#2}}}}
  (a)_\tau := \frac{\Gamma_d(a+\tau)}{\Gamma_d(a)} = \prod_{l=1}^m \risingf{\bigl(a - \tfrac{1}{2}(l-1)\bigr)}{t_l}
\end{equation*}
Let $C_\tau(X)$ be the Zonal Polynomial with signature partition $\tau$. Then, the following representation exists
For a matrix $U$ satisfying $\|U\| < 1$, we have the following "binomial-theorem"
\begin{equation}
  \frac{1}{|I-U|^a} = \sum_{k\ge 0}\sum_{|\tau| = k} \frac{(a)_\tau C_\tau(U)}{k!}.
\end{equation}
Using representations for these Zonal polynomials, one can obtain the integral representation mentioned in the original post.
More directly, you can look at Chapter 7 of R. Muirhead, "Aspects of Multivariate Statistical Theory", where you'll see that actually, $|I-U|^{-a}={}_1F_0(a;U)$, a matrix argument hypergeometric function. I've to run now, if I get a chance I'll clean up my answer and fill in the details.

REPLY [2 votes]: A very similar (but different) expression for log of the determinant is given by Du and Ji in their paper "an integral representation of the determinant of a matrix and its applications". I am guessing a slight adaptation of their thing can get your formula.<|endoftext|>
TITLE: Generalized Grassmannians that parameterize the submodules of a module
QUESTION [7 upvotes]: I'm looking for something like a Grassmannian, but which parameterizes the submodules of a module rather than the subspaces of a vector space. Most specifically, I'm looking for something which parameterizes the submodules of specifically $\mathbb{Z}^n$. So another way to say it is that I'm looking for a space parameterizing for the subgroups of a free abelian group. (A moduli space?)
I've seen some references to the concept of a "Grassmannian of submodules" here and there (like the papers on the first page of https://www.google.com/search?q=%22grassmannian+of+submodules%22) but can't figure out if this handles modules like $\mathbb{Z}^n$ or not.
Does anyone know if such an object exists and if so, how to construct it? Where I can get more information on this?
EDIT: to give a bit more info, the only specific application I really care about is parameterizing the free subgroups of a free abelian group, or the "lattices" in the $\mathbb{Z}$-module $\mathbb{Z}^n$, etc. A solution which works only for that, but which doesn't handle more exotic modules would be fine for my purposes. (And if it doesn't work out for all free abelian groups in general, then having a solution for at least finitely generated free abelian groups would even be a great start.)
I framed the question in terms of the "submodules of a module" in general just because I saw some references to there being a "Grassmannian of submodules" before, so I thought such a construction might be widely known.

REPLY [6 votes]: Building on Donu Arapura's answer and Qiaochu Yuan's comment:
If the quotient is torsion then $k$ changes as a functioin of the characteristic, so no $Grass_{k,n}$ will do. If it is non-torsion then classifying the $\mathbb Z$-module is equivalent to classifying the $\mathbb Q$-subsoace it generates, so a point on the Grassmanian will be a fine enough invariant. It's also clear that every rational subspace contains a rank-k $\mathbb Z$-submodule whose quotient is torsion-free, so for the torsion-free case that is sufficient.
The next-simplest case is probably when the quotient is finite, since every other case can be viewed as a combination of this and the other one (by splitting the quotient into rank and torsion.) As Quiaochu points out, this is choosing a surjection $\mathbb Z^n\to A$, which is equivalent to choosing a set of $n$ generators of $A$. But since only the kernel of the map matters, and composing with an automorphism of $A$ doesn't change the kernel, you need to quotient out by the automorphism group of $A$. (The automorphism group of $A$ is not the image of $GL_n(\mathbb Z)$, for instance $GL_n(\mathbb Z) \to GL_n(\mathbb F_p)$ is not a surjection.)
For instance, there is only one sublattice of $\mathbb Z^n$ whose quotient is $(\mathbb Z/p)^n$: that being $p\mathbb Z^n$.
We can break this finite quotient into a product of $p$-torsion parts and consider those separately. If the $p$-torsion part is $(\mathbb Z/p)^k$, then $Grass_{k,n}(\mathbb F_P)$ will classify the choices of generators. If the torsion involves prime powers then I am not sure what to do.
But a deeper problem is that I don't think one can make a scheme over $\mathbb Z$ whose $\mathbb Z$-points have a natural bijection to the $\mathbb F_p$ points of an arbitrary scheme over $\mathbb F_p$ like $Grass_{k,n}(\mathbb F_p)$, other then silly stuff like the disjoint union of one copy of $\textrm{Spec} \mathbb Z$ for each point on the $\mathbb F_p$-scheme.<|endoftext|>
TITLE: Which immersed plane curves bound an immersed disk?
QUESTION [17 upvotes]: I am looking for a nice answer to the following question.

Which immersed plane curves bound an immersed disk?

Comments.

I am not sure what is a nice answer, but for sure I could make a stupid algorithm.
I am aware that there are plane curves which bound few "different" immersed disks. For example this Bennequin’s curve is bounding five different immersed disks. It suggests that there is no "nice" answer to the question.

REPLY [17 votes]: The answer was given by Samuel Blank in his Brandeis 1967 phd dissertation, on which Poenaru gave a Bourbaki seminar.
Then Peter Shor and C. J. Van Wyk gave a polynomial time algorithm to decide if there is an extension.
EDIT: Blank's method already gave a polynomial algorithm, but with an exponent too large to make it practical, which is needed for applications (for instance to integrated circuit design). 
The answer is in term of existence of a chain of "reductions" of a certain kind for the cyclically reduced word in the free group $F_n$ on $n$ generators determined by the immersion, where $n$ is the number of bounded components of the complement of the curve (assumed to have only transverse double points).

REPLY [3 votes]: Dear Anton,
Some time ago I was looking at this question (which is important for Chekanov's invariants of legendrian links) and the literature is rather scattered (however, look at the work of Charles Titus in MathSciNet). 
I found the following paper, but I haven't really taken a good look at it yet.
"When Does a Curve Bound a Distorted Disk?
Jack E. Graver and Gerald T. Cargo
Consider a closed curve in the plane that does not intersect itself; by the Jordan–Schoenflies theorem, it bounds a distorted disk. Now consider a closed curve that intersects itself, perhaps several times. Is it the boundary of a distorted disk that overlaps itself? If it is, is that distorted disk essentially unique? In this paper, we develop techniques for answering both of these questions for any given closed curve in the plane.
Read More: http://epubs.siam.org/doi/abs/10.1137/090767716"<|endoftext|>
TITLE: Is there a uniform Dugundji theorem
QUESTION [5 upvotes]: A theorem of Dugundji states that if $X$ is a separable metric space and $A \subseteq X$ is closed then any continuous function $f$ from $A$ to some normed linear space $L$ may be extended to a continuous map $\bar{f} \colon X \to L$. I got this formulation from van Mill's book "Infinite-Dimensional Topology". Does there exist a version of Dugundji's Theorem where we assume that $f$ is uniformly continuous an conclude the existence of a uniformly continuous extension? Since my main interest is when $L$ is a Banach space (or even a $C^*$-algebra), I would be very happy to add that assumption to get a positive answer.

REPLY [3 votes]: Note that the case of real-valued functions  is easy. A function $f:A\to\mathbb{R}$ has a uniformly continuous extension to any metric space $X\supset A$  iff it has a sub-linear modulus of continuity (that is, that verifies $\omega(t)\le a|t| + b$).
Rmk. Note that these general extensions problems may be approached as selection problems for multivalued functions. Precisely, given  $f:A\to L$ and a concave modulus of continuity $\omega$, consider the multi $F:X\to 2^L$, taking $x\in X$ to the set $F(x)\subset L$ of all admissible values for an $\omega$-continuous extension of $f$ to $A\cup \{x\}$, that is $$F(x)=\cap _ {a\in A} \bar B\big(a,\omega(d(a,x))\big)   \, .$$
Then, any extension $\tilde f$ of $f$ to $X$ has to be a selection of $F$, $f(x)\in F(x)$; note that $F(a)=\{a\}$ for $a\in A$. If $\omega$ is not too small and $L$ is compact enough, you may hope to prove that $F(x)$ is a not-empty bounded closed set, for all $x$. You may then correspondingly look for a modulus of continuity for $F$ seen as a map valued into non-empty, bounded, closed subsets of $L$ (even convex, if $L$ is a normed space) with respct to the Hausdorff distance. Finally, you may construct $\tilde f$ as a composition $c\circ F$, where $c$ is a map that picks a point $c(S)$ within every such subset $S$. This program can be achieded in a very satisfactory way in the case of the Kierszbraun theorem, where the Lipschitz constant is preserved: Hilbert spaces $L$ are a nice setting, because in this case there is a 1-Lipschitz function $c$, taking a bounded convex set $S$ to the center of the smallest closed ball containing $S$. Slightly more genreal, I guess, uniformly convex Banach spaces $L$ could work, but I guess the modulus of continuity of the extension in general will be larger. 
If $L$ is a metric space, then of course there are in general topological obstructions even for continuous extensions.<|endoftext|>
TITLE: Question on Hilbert Manifolds
QUESTION [9 upvotes]: I have a very basic question on Hilbert manifolds.
Consider the Hilbert space 
$$
\mathcal{H}:= L^2(S^1)
$$
with $S^1$ the unit circle. 
On $\mathcal{H}$ let us introduce the equivalence relation 
$$
f\sim g : \Leftrightarrow f(\cdot ) = g(\cdot + \alpha)\quad
\mbox{for some  }\alpha \in S^1.
$$
Now define the factor space 
$$
\overline{\mathcal{H}}:= \mathcal{H}/\sim.
$$
What is the structure of $\overline{\mathcal{H}}$? Is it a Hilbert manifold? If so, how to construct the smooth structure? 
I am particularly interested in computing a (Riemannian) distance between two elements of $\overline{\mathcal{H}}$.

REPLY [3 votes]: It probably is NOT a smooth manifold. I think finding a chart around the point corresponding to constants, namely, the fixed points of the action of the group of rotation, is problematic.More precisely, at a fixed point, there is not a well-defined tangent space.<|endoftext|>
TITLE: Is NBG set theory stronger than ZFC?
QUESTION [7 upvotes]: It is well known, that every statement involving only set-variables is provable in NBG set theory if and only if it is provable in ZFC. What confuses me however is that NBG has a global axiom of choice. Global choice implies that every set is ordinal definable (V=OD).
So the statement V=OD seems to be a counterexample: It only involves set-variables, it follows from NBG, but it is known to be independent of ZFC. Where am i wrong?

REPLY [18 votes]: It is not true that global choice implies $V=OD$. 
Global choice is the assertion that there is a class well-ordering of the universe. This is equivalent to the assertion that there is a global choice function, which selects from every non-empty set an element. It is not part of the axiom, however, that this class is definable. Rather, the class well-ordering of $V$ is simply one of the classes that is available in the second order part of the model, and there is no reason to suppose that all such classes are necessarily first-order definable, even from set parameters. 
The axiom $V=HOD$, on the other hand, is equivalent to the assertion that there is a definable global choice function class, or equivalently, a definable well-ordering of the universe. 
So the critical difference is in the question of whether the global well-ordering is definable or not. It might be interesting to note that there can be a global well-ordering that is definable from set parameters, and so still counts as a class in ZFC, without having $V=HOD$, simply because those parameter may be essential, and it may not be possible to define the order without them. This is what happens, for example, after adding a generic Cohen real over $L$, since in $L[c]$ we do not have $V=HOD$, and in fact $\text{HOD}^{L[c]}=L$ by the homogeneity of the forcing, but nevertheless one can use $c$ as a parameter to define in $L[c]$ a well-ordering of the universe, essentially the usual $L$-order relativized to $c$.
The proof that NGBC is conservative over ZFC is not difficult. Given any model $M$ of ZFC, one first forces global choice by class forcing:  consider the class partial order consisting of well-ordered sequences of longer and longer set length. This forcing is $\kappa$-closed for every $\kappa$, and hence adds no new sets, but the generic filter $G$ provides a class well-ordering of $M$. Now, to form the GBC model $M[G]$, one considers as classes all the classes that are definable from set parameters and this new global well-order class. One can verify that the resulting model satisfies all the GBC axioms (and this amounts essentially to the verification that this kind of class forcing works as expected). Since the sets of $M[G]$ are the same as $M[G]$, it follows that any statement that is true in every GBC model is also true in every model of ZFC, and so the extension is conservative.<|endoftext|>
TITLE: No injective groups with more than one element?
QUESTION [16 upvotes]: There are several claims in the literature that there are no injective groups (with more than one element), but I have not found a proof.  For example, Mac Lane claims in his Duality from groups paper in the 1950 Bull. A.M.S. that Baer showed him an elegant proof, but gives no hint of what it might be.  Does anyone know of an actual published proof?

REPLY [9 votes]: The argument in the slides by Maria Nogin (née Voloshina) linked to by a-fortiori in a comment was published as:
Maria Nogin, A short proof of Eilenberg and Moore’s theorem, Central European Journal Of Mathematics Volume 5, Number 1 (2007), 201–204. Also available on her homepage.

Added: The above paper was also mentioned in Jonas Meyer's answer to the same question on math.SE. As Steve D. points out in a comment there, the result  appears as Exercise 7 on page 9 of D.L. Johnson's Topics in the Theory of Group Presentations, Cambridge University Press 1980:


The argument by Eilenberg and Moore appears on pages 21/22 of Foundations of Relative Homological Algebra, Memoirs of the AMS, Volume 55 (1965). Here's a scan of the relevant passage for the convenience of the readers:<|endoftext|>
TITLE: Why is Heegaard Floer Homology defined in terms of Sym$^g\Sigma_g$ instead of Pic$^g\Sigma_g$?
QUESTION [11 upvotes]: Recall the definition of Heegaard Floer homology: $\Sigma_g$ is a closed surface, and $\{\alpha_1,\ldots,\alpha_g\}$ and $\{\beta_1,\ldots,\beta_g\}$ are sets of attaching circles.  Then Heegaard Floer homology is (more or less) the Lagrangian intersection Floer homology of $\mathbb T_\alpha=\prod_{i=1}^g\alpha_i$ and $\mathbb T_\beta=\prod_{i=1}^g\beta_i$ in $\operatorname{Sym}^g\Sigma_g$.
Now if we think of $\Sigma_g$ as a complex curve, then there is a birational map $\phi:\operatorname{Sym}^g\Sigma_g\to\operatorname{Pic}^g\Sigma_g$.

What happens if instead we consider the Lagrangian intersection Floer homology of $\phi(\mathbb T_\alpha)$ and $\phi(\mathbb T_\beta)$ inside $\operatorname{Pic}^g\Sigma_g$?  Are the resulting groups trivially the same, trivially different, or at least interesting?  (if they're not the same, then I guess there may be no good reason why they would even be invariants of the underlying three-manifold).

There is at least one concrete reason (and one philosophical reason) why one might try this definition instead of the original:

There are no holomorphic spheres in $\operatorname{Pic}^g\Sigma_g$ (because it is an abelian variety; in fact the map $\phi$ is exactly contracting all the embedded $\mathbb P^n$'s in the symmetric product).  This means we don't have to worry about some types of bubbling.
$\operatorname{Pic}^g\Sigma_g$ is a complex torus; in particular its topology is very concrete and easy to understand.  Also it is perhaps algebrogeometrically more natural than the symmetric product.
I could imagine that maybe there is some general statement whereby blowing down all the $\mathbb P^n$'s always does something understandable (perhaps nothing) to the Lagrangian intersection Floer homology.

REPLY [7 votes]: There is a tacit assumption behind this question, which I don't think is justified: that the  Abel-Jacobi images of the Heegaard tori $\mathbb{T}_{\alpha}$ and $\mathbb{T}_{\beta}$ are Lagrangian with respect to some reasonable symplectic form on the Jacobian torus. 
One can make the Heegaard tori Lagrangian by using a Kaehler form on the symmetric product that is product-like outside some neighbourhood of the diagonal. And one can probably find symplectic forms for which Abel-Jacobi is a symplectomorphism outside a neighbourhood of the theta-divisor (this is certainly true in the genus 2 case). Doing both of these things at once would suffice to make the images Lagrangian, but this might be tricky to achieve  - and it's perhaps not very natural?<|endoftext|>
TITLE: Second homotopy group of the mod 2 Moore spectrum
QUESTION [13 upvotes]: Let $S/2$ be the mod 2 Moore spectrum (i.e. the cofiber of $2: S \to S$). Then multiplication by 2 acts nontrivially on this spectrum: the homotopy groups of $S/2$ are  all $\mathbb{Z}/4$-modules by a formal argument, but not $\mathbb{Z}/2$-modules. For instance, $\pi_2(S/2) = \mathbb{Z}/4$.
The long exact sequence in homotopy groups enables one to determine the homotopy groups of $S/2$ (in the range that the stable homotopy groups of spheres are known) up to extension, but the extension problems are generally nontrivial: $\pi_2(S/2)$ is a case in point (as the aforementioned exact sequence $0 \to \mathbb{Z}/2 \to \pi_2(S/2) \to \mathbb{Z}/2 \to 0$ shows). 
Is there a general technique for resolving these kinds of extension problems? I can get it in this case using some computation in the cobar complex to get the $t-s = 1$ line of the Adams spectral sequence for $S/2$ and see that the Bockstein acts nontrivially. I'm curious about a more efficient method of doing this. (Another example I had in mind was $bo \wedge \mathbb{CP}^2$: this is $bu$ by the "theorem of Reg Wood" and this is visible in the ASS, but directly computing $KO$-groups of complex projective spaces yields nontrivial extension problems, I think.)

REPLY [5 votes]: Often these extension problem are solved using Toda brackets (as Peter May already mentioned). I will first give the general statement, not only for spectra but also for module spectra. The statement may sound a bit complicated, but is quite useful.

Let $R$ be a strictly associative ring spectrum. Let $x\in \pi_n R$ be an element in the coefficients and denote by $Cx$ the cone of $\Sigma^n R \xrightarrow{x} R$. Then we have a long exact sequence
  \begin{eqnarray*}\cdots\to\pi_* \Sigma^n R\to \pi_* R \to \pi_* Cx \to \pi_{*-1} \Sigma^n R\to \pi_{*-1}R\to\cdots\end{eqnarray*}
  which splits into short exact sequences of the form
  \begin{eqnarray*}0\to \pi_* R/x\pi_*R \xrightarrow{\alpha} \pi_* Cx \xrightarrow{\beta} \{\pi_{*-n-1}R\}_x\to 0\end{eqnarray*}
  where $\{\pi_{*-n}R\}_x$ denotes all elements which are annihilated by $x$.
Let $y\in \pi_m R$ and $z\in \pi_k R$ be elements in the coefficients of $R$ with $xy=0$ and $yz=0$. Let $\widetilde{y} \in \pi_*Cx$ be an element with $\beta(\widetilde{y}) = y$. Let $w\in \pi_*R$ be an element such that the projection of $w$ is mapped to $\widetilde{y}z$ under $\beta$. Then $w\in \langle x,y,z\rangle$.

Back to your example: We take $R = \mathbb{S}$ and $x = 2\in\pi_0 \mathbb{S}$. We have now a short exact sequence
$$ 0 \to \pi_2\mathbb{S} \to \pi_2 \mathbb{S}/2 \to \pi_1\mathbb{S} \to 0 $$
where the outer groups are isomorphic to $\mathbb{Z}/2$ and are generated by $\eta^2$ and $\eta$, respectively. Let now $y = \eta$ and $z =2$. Lift $\eta$ to an element $\widetilde{\eta} \in \pi_2 \mathbb{S}/2$. Then the statement above tells us that $2\cdot \widetilde{\eta}$ is in the image of $\langle 2, \eta, 2 \rangle = \eta^2$. In particular, $2\cdot \widetilde{\eta}$ is non-zero. This solves the extension problem.
This technique can be applied to many cases, including $KO\wedge C\eta$. I did some calculations in the category of $TMF$-modules with this.  
The problem remains how to compute Toda brackets. In general, this might be difficult, but often methods to compute $\pi_*R$ give also methods to compute the Toda brackets in $\pi_* R$. For example, Massey products in the $E^2$-term of the Adams spectral sequence converge to Toda brackets (if there are no "crossing differentials"). The Massey product $\langle 2,\eta, 2\rangle$ can, for example, be computed via cobar representatives. 
Regarding references: The statement about Toda brackets and cofiber sequences follows more or less directly from the definition of Toda brackets in the framework of triangulated categories. See for example, Section 4.6 of my Thesis -- although I was a little bit lazy there with signs. Regarding the convergence of Massey products to Toda brackets one finds a statement in Kochman's Bordism, Stable Homotopy, and Adams Spectral Sequences. Note that he uses another definition of Toda bracket, which agrees with the one in triangulated categories in the case of triple brackets (and also else if one ignores indetermenancy) as proven in some paper by Kochman.<|endoftext|>
TITLE: Can one prove complex multiplication without assuming CFT?
QUESTION [20 upvotes]: The Kronecker-Weber Theorem, stating that any abelian extension of $\mathbb Q$ is contained in a cyclotomic extension, is a fairly easy consequence of Artin reciprocity in class field theory (one just identifies the ray class groups and shows that each corresponds to a cyclotomic extension). However, one can produce a more direct and elementary proof of this fact that avoids appealing to the full generality of class field theory (see, for example, the exercises in the fourth chapter of Number Fields by Daniel Marcus). In other words, one can prove class field theory for $\mathbb Q$ using much simpler methods than for the general case.
The theory of complex multiplication is similar to the theory of cyclotomic fields (and hence the Kronecker-Weber Theorem) in that it shows that any abelian extension of a quadratic imaginary field is contained in an extension generated by the torsion points of an elliptic curve with complex multiplication by our field. To prove this, one normally assumes class field theory and then shows that the field generated by the $m$-torsion (or, more specifically, the Weber function of the $m$-torsion) is the ray class field of conductor $m$.
My question is: Can one prove that any abelian extension of an imaginary quadratic field $K$ is contained in a field generated by the torsion of an elliptic curve with complex multiplication by $K$ without resorting to the general theory of class field theory? I.e. where one directly proves class field theory for $K$ by referring to the elliptic curve. Is there a proof in the style of the exercises in Marcus's book?
Note: Obviously there is no formal formulation of what I'm asking. One way or another, you can prove complex multiplication. But the question is whether you can give a proof of complex multiplication in a certain style.

REPLY [10 votes]: (modified)
Historically, Complex Multiplication precedes Class Field Theory and many of the main theorems of CM for elliptic curves were proved directly. See Algebren (3 volumes) by Weber or Cox's book for an exposition.
Please also read Birch's article on the beginnings of Heegner points where he points this out explicitly (page three, paragraph beginning "Complex multiplication ...).
But not all. so the answer is no (unlike what I had mistakenly presumed at first and the comments below alerted me).  
The actual history is quite complicated; see Schappacher.<|endoftext|>
TITLE: Does the curvature determine the metric?
QUESTION [41 upvotes]: Hello,
I ask myself, whether the curvature determines the metric.
Concretely: Given a compact Riemannian manifold $M$, are there two metrics $g_1$ and $g_2$, which are not everywhere flat, such that they are
not isometric to one another, but that there is a diffeomorphism which preserves the curvature?
If the answer is yes:
Can we chose $M$ to be a compact 2-manifold?
Can we classify manifolds, where the curvature determines the metric?
What happens, if we also allow semi-riemannian manifolds for the above questions?
Thank you for help.

REPLY [46 votes]: It should not be surprising that, for a $2$-dimensional manifold, the Gauss curvature $K:M\to\mathbb{R}$ does not determine a unique metric $g$.  After all, the former is locally one function of $2$ variables while the latter is locally three functions of $2$ variables.  It would be remarkable if $K$ determined $g$, even up to isometry, and, of course, except for constant $K$, it does not, as the arguments of Weinstein and Kulkarni show.
Meanwhile, if one regards the curvature $\mathsf{R}(g)$ of of the Levi-Civita connection of the metric $g$  on a surface $M$ as a section of the rank $3$ bundle $\bigl(T\otimes T^\ast)_0\otimes \Lambda^2(T^\ast)={\frak{sl}}(T)\otimes \Lambda^2(T^\ast)$, then the equation $\mathsf{R}(g) = {\mathsf{R}}_0$ for a given nonzero section ${\mathsf{R}}_0$ of ${\frak{sl}}(T)\otimes \Lambda^2(T^\ast)$ is a determined second-order equation for $g$.  (Here, I am writing "$T$" for "$TM$", etc., to save space.) This then becomes a more reasonable question, one that has an easy answer:  Namely, if $M$ is compact and connected and $g_1$ and $g_2$ are metrics on $M$ such that $\mathsf{R}(g_1) = \mathsf{R}(g_2)$ and such that these curvature tensors are nowhere vanishing, then $g_2 = c g_1$ for some constant $c>0$, and conversely.  The reason for this is simple:  If ${\mathsf{R}}_0$ is a nonvanishing section of ${\frak{sl}}(T)\otimes \Lambda^2(T^\ast)$, then requiring that $\mathsf{R}(g) = {\mathsf{R}}_0$ determines the conformal class of $g$ purely algebraically (because using $g$ to 'lower the first index' of ${\mathsf{R}}_0$ has to result in a tensor that is skewsymmetric in the first two indices).  Thus, if $\mathsf{R}(g_1) = \mathsf{R}(g_2) = {\mathsf{R}}_0$, where the latter is nonvanishing, then $g_2 = e^u\ g_1$ for some function $u$ on $M$.  Plugging this back into the equation $\mathsf{R}(g_1) = \mathsf{R}(g_2)$ and computing shows that $u$ must be harmonic with respect to the conformal structure determined by $g_1$ (which is also the conformal structure determined by $g_2$).  If $M$ is compact and connected, then $u$ has to be constant.
As Misha points out, in higher dimensions, the situation is rather different and is well-described in the works of Kulkarni and Yau that he cites.  However, just qualitatively, it's worth pointing out that specifying the sectional curvature of a metric is generally a very overdetermined problem in higher dimensions, which is why the rigidity results of Kulkarni and Yau should not be surprising.
To see the nature of this, recall that the Riemann curvature tensor $\mathsf{Rm}(g)$ of a metric $g$ on $M$ is got from $\mathsf{R}(g)$ by 'lowering an index'. By the first Bianchi identity, $\mathsf{Rm}(g)$ is a section of the subbundle ${\mathsf{K}}(T^\ast)\subset {\mathsf{S}}^2\bigl(\Lambda^2(T^\ast)\bigr)$ that is the kernel of the natural map
induced by wedge product
$$
\mathsf{S}^2\bigl(\Lambda^2(T^\ast)\bigr)\to \Lambda^4(T^\ast)
$$
In fact, this map has a natural $\mathrm{GL}(T)$-equivariant right inverse, so that one has a canonical bundle splitting
$$
\mathsf{S}^2\bigl(\Lambda^2(T^\ast)\bigr) = {\mathsf{K}}(T^\ast)\oplus \Lambda^4(T^\ast)
$$
and these two subbundles are $\mathrm{GL}(T)$-irreducible.  (I'm using $\mathrm{GL}(T)$ to denote the 'gauge' group of bundle automorphisms of $T = TM$.  I don't say that it's great notation, but it will be OK for this discussion.)  An important interpretation of this splitting is the following one:  Obviously, one can regard a section of $\mathsf{S}^2\bigl(\Lambda^2(T^\ast)\bigr)$ as a quadratic form on the bundle $\Lambda^2(T)$.  Then the elements of $\Lambda^4(T^\ast)\subset \mathsf{S}^2\bigl(\Lambda^2(T^\ast)\bigr)$ are exactly the quadratic forms that vanish on all of the decomposable elements in $\Lambda^2(T)$, i.e., the elements of the form $x\wedge y$ for $x,y\in T_xM$ for some $x$.  In particular, a section $Q$ of ${\mathsf{K}}(T^\ast)$ is completely determined by its values on the 'cone bundle' $\mathsf{C}\subset \Lambda^2(T)$ that consists of the decomposable elements.
Now, any metric $g$ on $M$ determines a unique section $\Lambda^2(g)$ of ${\mathsf{K}}(T^\ast)$ with the property that 
$$
\Lambda^2(g)\bigl(x\wedge y\bigr) = |x\wedge y|^2_g = g(x,x)g(y,y)-g(x,y)^2,
$$
and the sectional curvature of $g$ is simply the function $\sigma_g:\mathsf{Gr}(2,T)\to\mathbb{R}$ defined as the ratio
$$
\sigma_g\bigl([x\wedge y]\bigr) 
= \frac{\mathsf{Rm}(g)\bigl(x\wedge y\bigr)}{\Lambda^2(g)\bigl(x\wedge y\bigr)}
\quad\text{for $x,y\in T_xM$ with $x\wedge y\not=0$},
$$
where we regard $\mathsf{Gr}(2,T)$ as the projectivization of the cone bundle $\mathsf{C}$.
The fact that the sectional curvature function is the ratio of two quadratic forms that are sections of $\mathsf{K}(T^\ast)$ shows that it is a very constrained function on $\mathsf{Gr}(2,T)$.  In fact, unless the sectional curvature is constant on $\mathsf{Gr}(2,T_x)$, the numerator and denominator of this ratio at $x\in M$ are uniquely determined up to a common multiple.  In particular, the set of such functions on $\mathsf{Gr}(2,T)$ can be regarded as the set of sections of a bundle over $M$ whose fibers are isomorphic to a space of dimension $d_n = \tfrac12 n(n{+}1) + \tfrac1{12}n^2(n^2-1) -1$ that is singular along the $1$-dimensional curve of constant functions.  
In particular, specifying a candidate sectional curvature function $\sigma:\mathsf{Gr}(2,T)\to\mathbb{R}$ that is not constant on any fiber $\mathsf{Gr}(2,T_x)$ is equivalent to specifying $d_n$ functions of $n$ variables for the $\tfrac12 n(n{+}1)$ coefficients of $g$, which is highly overdetermined when $n>2$.  This is why one has the level of rigidity for the sectional curvature that is indicated in Kulkarni's and Yau's results.  In fact, as this analysis shows, if $\sigma_{g_1} = \sigma_{g_2} = \sigma$, where $\sigma$ is not constant on any fiber $\mathsf{Gr}(2,T_x)$, then one must have $g_2 = e^u g_1$ for some function $u$ on $M$ and that function $u$ will have to satisfy $\mathsf{Rm}(e^u g_1) = e^{2u} \mathsf{Rm}(g_1)$ which is, itself, a very highly overdetermined second order equation for the single function $u$.   (It's kind of remarkable that there are any nonflat metrics $g$ that admit nonzero solutions $u$ to $\mathsf{Rm}(e^u g) = e^{2u} \mathsf{Rm}(g)$ at all, even without the compactness assumption.  That is what makes Yau's counterexample in dimension $3$ so interesting.  Indeed, though, it turns out that, up to constant scaling and diffeomorphism, there is exactly a $1$-parameter family of such exceptional metrics in dimension $3$, so they are extremely rare indeed. )
The upshot of all these observations is that specifying a nonconstant sectional curvature function $\sigma_g$ in dimensions above $2$ very nearly determines $g$ in all cases, so that, except for a very small set of degenerate cases, the sectional curvature does, indeed, determine the metric.  However, it's such an overdetermined problem that this result is not all that surprising.
In particular, specifying $\sigma_g$ is specifying a greater quantity of information about $g$ than specifying, say, $\mathsf{Rm}(g)$.  When $n>3$, even specifying $\mathsf{Rm}(g)$ is overdetermined, and, generally speaking, you'd expect $\mathsf{Rm}(g)$ to determine $g$ as well when $n>3$.  
When $n=3$, specifying $\mathsf{Rm}(g) = \mathsf{R}$ is locally $6$ second order equations for $6$ unknowns, so it's a determined system.   Back in the 80s, by using the Cartan-Kähler theorem, I showed that, when $\mathsf{R}$ is appropriately nondegenerate and real-analytic, the equation $\mathsf{Rm}(g) = \mathsf{R}$ is locally solvable for $g$, and the general solution depends on $5$ functions of $2$ variables.  Later, Dennis DeTurck and Deane Yang proved this result in the smooth category as well (see Local existence of smooth metrics with prescribed curvature, in Nonlinear problems in geometry (Mobile, Ala., 1985), 37--43, Contemp. Math., 51, Amer. Math. Soc., Providence, RI, 1986).
In higher dimensions, the natural determined curvature problem is to specify the Ricci tensor $\mathsf{Rc}(g)$ as a section of $S^2(T^\ast)$ (or some trace-modified version of it), and, there, Dennis DeTurck has the best results.<|endoftext|>
TITLE: Bayesian statistics for pure mathematicians
QUESTION [25 upvotes]: Could someone please recommend reading on Bayesian statistics presented from a pure mathematical point of view?  That is, works that start assuming a good knowledge of measure theoretic probability.  The usual reference works for statisticians gloss over the fine details, but for me this just leads to more confusion!
I already know of the following question
Statistics for mathematicians
and will follow up some of the suggestions there, but I am specifically looking for a treatment of the Bayesian approach, so if there are any recommendations that apply just to this I would appreciate hearing about them.

REPLY [3 votes]: There is a book of 2018 on the topic, that also presents some recent results of the author:

S. Watanabe, Mathematical Theory of Bayesian Statistics, Chapman and Hall/CRC, Apr 2018<|endoftext|>
TITLE: About the axiom of choice, the fundamental theorem of algebra, and real numbers
QUESTION [9 upvotes]: About fundamental theorem of algebra, there is a large collection different demonstrations.
I ask: is there some proof that avoids AC (choice axiom)?
In a general topos (with natural number object) there are the two constructions of real numbers (generalizations of the classical Dedekind and Cauchy constructions) that are different. 
In ZF theory, are the Dedekind and Cauchy constructions different? (In the "Cauchy" reals,  operates on a real number $r$ through a choice of a Cauchy sequence converging to $r$.)

REPLY [23 votes]: The fundamental theorem of algebra is, unless I miscounted quantifiers, a $\Pi^1_2$ sentence of second-order arithmetic and therefore absolute between the full universe and Gödel's constructible universe by the Shoenfield absoluteness theorem.  So, since it's provable in ZFC, it is necessarily provable in ZF as well.
I believe, though, that this metamathematical argument can be avoided, because I don't see any serious use of the axiom of choice in usual proofs by methods of analysis and (plane) topology.  
The Dedekind and Cauchy constructions of the reals are equivalent in ZF; no choice is needed.  Although AC is often involved when one wants to prove general results about sequences of reals, in the present situation one only needs to consider sequences of rationals.  So one can  fix an enumeration of the rationals and thereafter, whenever a rational needs to be chosen, taking the first suitable one in the enumeration.

REPLY [12 votes]: Regarding Cauchy and Dedekind reals. The fact that every Dedekind real has a Cauchy representation is provable in very weak systems of intuitionistic analysis. The converse fact that every Cauchy real has a Dedekind representation is equivalent to the Weak König Lemma (WKL) — that every infinite tree of $\lbrace0,1\rbrace$-sequences has an infinite path. (See note.) This very weak form of König's lemma is provable in ZF, so there is no difference between Cauchy reals and Dedekind reals in ZF.
Breaking this down further, the ability to compare a Cauchy real to a given rational number is equivalent to the Lesser Limited Principle of Omniscience (LLPO), which can be formulated as follows
$$\lnot(\exists n A(n) \land \exists n B(n)) \rightarrow (\forall n \lnot A(n) \lor \forall n \lnot B(n)),$$ where $A(n)$ and $B(n)$ are simple decidable statements (e.g., primitive recursive predicates). Note that LLPO is a consequence of the Law of Excluded Middle, and therefore true in every classical system such as ZF, but not necessarily in intuitionistic systems such as topoi. To extract a Dedekind cut from a Cauchy real, it is necessary to simultaneously compare a real with every rational number, so we need to repeat LLPO countably many times. WKL is precisely equivalent to countably many repetitions of LLPO. In set theory and topos theory, such repetitions come for free through comprehension. So the only thing that can prevent every Cauchy real to have a Dedekind cut in a topos is the failure of LLPO.

This is not quite true as I originally stated it. It is the existence of a "uniform method" for converting Cauchy reals into Dedekind reals which is equivalent to WKL. In classical systems, one can get around the uniformity requirement through clever use of the Law of Excluded Middle. In weak classical systems such as RCA0 one can prove that every Cauchy real has a Dedekind cut by providing separate methods for rational and irrational numbers, thus sweeping a lot of the computation cost in deciding whether a real is rational or not. The equivalence with WKL is recovered when one asks to simultaneously find Dedekind cuts for an infinite sequence of Cauchy reals, as demonstrated by Hirst [Representations of reals in reverse mathematics, Bull. Pol. Acad. Sci. Math. 55 (2007), 303–316; MR2369116].

REPLY [8 votes]: In classical logic, using the ZF axioms, one can show that there is an isomorphism between the Dedekind cuts in the rationals, and the equivalence classes of Cauchy sequences.   One can also (without AC) choose a representative from each equivalence class of Cauchy sequences. (EDIT: In high school one might chose decimal fractions, and represent $\pi$ by the sequence $(3, 3.1, 3.14,\ldots)$; a more sophisticated approach might choose continued fractions and represent 
$\pi$ by the sequence $(3, 22/7, 333/106,\ldots)$. 
The fundamental theorem of algebra (every non-constant polynomial with real or complex coefficients has a zero in the complex numbers) can be proved without AC. In fact, there is a quite explicit algorithm that computes, for each irreducible polynomial (with leading coefficient 1, to be on the safe side), 
the roots of this polynomial continuously from the coefficients.  
In constructive mathematics or intuitionistic logic, the story is different, as Francois Dorais explains in his answer.<|endoftext|>
TITLE: Raoul Bott's quote on Morse Theory cited by Bestvina and Kahle: where is it from?
QUESTION [17 upvotes]: I wanted to properly cite the following awesome quote:

Every mathematician has a secret weapon. Mine is Morse theory. - Raoul Bott

Now this has been attributed to Bott in precisely two places that I can find: the older of the two is Bestvina's work on PL Morse theory here and the more recent occurrence is in the 14th slide of Kahle's talk available as a rather large pdf file here.
Question:

What is the original reference for this quote?

I've combed through the likely suspect, Morse theory indomitable available here but the quote does not appear to be in that paper

REPLY [7 votes]: Here is Mladen's response to my email asking this question:

I heard him say it in a colloquium talk in 2001 (I think).

Case closed, unless Bob MacPherson has a different answer.<|endoftext|>
TITLE: elements of the pure braid group seen as Galois automorphisms
QUESTION [7 upvotes]: Let $Y_{n}$ be what I'm calling the ordered configuration space, the topological space of all ordered subsets of $\mathbb{C}$ of cardinality $n$. This can be viewed as $$\mathrm{Spec}(\mathbb{C}[t_{1}, ... , t_{n}, \{(t_{i} - t_{j})^{-1}\}_{1 \leq i < j \leq n}]).$$
The (topological) fundamental group of this is the pure braid group $P_{n}$.  So by a general version of the Riemann Existence Theorem, the elements of $P_{n}$ must be Galois automorphisms of the maximal Galois extension of the function field $\mathbb{C}(t_{1}, ... , t_{n})$ which is unramified except at the primes $(t_{i} - t_{j})$.  How may I describe algebraically the Galois automorphisms corresponding to, say, the generators of $P_{n}$?
I have a guess (something along the lines of each generator $A_{i,j}$ of $P_{n}$ sending $(t_{i} - t_{j})^{1/n} \mapsto \zeta_{n}(t_{i} - t_{j})^{1/n}$, except that seems to make the Galois group abelian) but am not sure how to prove anything systematically, and I can't find this explicitly described in any source.  Thank you for any help you can give.

REPLY [3 votes]: Your confusion comes from the fact that your description of the Galois action is, I think, correct, but you are not giving the entire Galois action because you are not having it act on the correct field. The covering generated by $(t_i-t_j)^{1/n}$ is an abelian covering (and by Kummer theory is the maximal abelian covering), so of course the Galois action on it is abelian. But there are other unramified covers whose function fields are not generated by elements of the form $(t_i-t_j)^{1/n}$).
To describe the full Galois action, you need to consider all of these covers. However, as far as I know, there is no well-behaved way to do so. For instance, each such scheme fibers over $\mathbb P^1$ minus three points: take the map $(t_1,..,t_n) \to (2t_1-t_2-t_3:-t_1+2t_2-t_3)$. It is well-defined since if both of those are $0$ then $t_1-t_2$ is zero as well, and the inequalities $t_1\neq t_2$, $t_1\neq t_3$, $t_2\neq t_3$ give three points that are not in the image. So take any unramified cover of $\mathbb P^1$ minus three points and take the fibered product with this map. You get an unramified covering of this scheme.
Writing down equations for all the umramified covers of $\mathbb P^1-\{0,1,\infty\}$ is no easy task, and those aren't even all the covers of this scheme! Thus finding an explicit representation for the Galois action is difficult.<|endoftext|>
TITLE: What's the name for the analogue of divided power algebras for x^i/i?
QUESTION [16 upvotes]: I recently came across divided power algebras here: http://amathew.wordpress.com/2012/05/27/lazards-theorem-ii/ It interests me because the free divided power algebra on one variable $x$, where $x^{(i)}$ models $x^i/i!$, seems like a good way of handling exponential and exponential-type series. (Divided power algebras that aren't generated by a single variable are more complicated to axiomatize, so I will stick to the one-variable case).
Quick summary: A system of divided powers for an element $x$ in an associative unital ring $R$ associates to each nonnegative integer $n$ a ring element $x^{(n)}$ with $x^{(0)} = 1$ and $x^{(1)} = x$ and satisfying the following condition for all $i,j \ge 0$:
$$x^{(i)}x^{(j)} = \binom{i + j}{i} x^{(i + j)}$$
As we can see, if $R$ is an algebra over the rationals, the obvious (and unique!) choice is to set $x^{(i)} = x^i/i!$ for all $i$.
Suppose I am interested in finding some $x^{[i]}$ that models $x^i/i$ instead of $x^i/i!$, i.e., I am interested in the kind of terms that appear in the expansions of logarithmic and inverse trig series. My best guess for the right analogue to consider is this: $x^{[i]}$ is defined for all positive integers $i$ with $x^{[1]} = x$ and satisfy the following condition. For each $i,j > 0$, write $ij/(i + j)$ as a reduced fraction $r/s$. Then, we must have:
$$rx^{[i]}x^{[j]} = sx^{[i + j]}$$
I'd like to know whether this structure has been studied in the past, and/or what names have been given to the structure in question. Also, I'd be interested if anybody has opinions on whether the above is a reasonable way of trying to model $x^i/i$.

REPLY [12 votes]: One possible way to generalize divided powers is to model sequences $x^n/a_n$ where $a_{n+m}/(a_na_m)$ is integral (an integer, or at least integral in some sense). You can model these in the obvious way, like divided powers. Choosing $a_n = n!$ gives you divided powers, and choosing $a_n=p^n$ gives you a sequence which, from the point of view of $p$-adic analysis, is much more "regular" in its growth. You can come up with other examples.
As you may know, one uses divided powers to construct Fontaine's ring of periods $B_{cris}$ and using $a_n=p^n$ instead of $a_n=n!$ gives a ring $B_{max}$ that is similar in nature to $B_{cris}$ but much better behaved (see III.2 of Colmez' 1998 paper "Théorie d'Iwasawa des représentations de de Rham d'un corps local").
More generally, tweaking divided powers to get various convergence conditions is something that happens often in $p$-adic Hodge theory. See for instance 5.2.3 of Fontaine's "Le corps des périodes $p$-adiques" for one of many examples of custom made convergence conditions.
In the $p$-adic setting, I would say that your log series belongs to the "space of holomorphic functions on the $p$-adic open unit disk, having order of growth $\leq 1$". Using precise analytic conditions rather than "divided powers" seems the right way to study the analytic series that interest you.<|endoftext|>
TITLE: On infinite-dimensional unitary representations of Kazhdan groups
QUESTION [7 upvotes]: Let $G$ be a finitely generated discrete group that satisfies Kazhdan's property T. Rapinchuk has proved that every unitary representation $\rho$ of $G$ on some finite-dimensional Hilbert space is locally rigid: this means that if another representation is sufficiently closed to $\rho$ (on a fixed finite set of generators) then it must be conjugated to $\rho$. The proof is nice and goes as follows:

Consider the adjoint action $\rho^{\rm ad}$ on the Lie algebra $gl_n$. As proved by André Weil, if $H^1(G, \rho^{\rm ad})=0$ then the representation $\rho$ is locally rigid.
Since $G$ is Kazhdan, every action on a Hilbert space has a fixed point. Using cocycles, this is equivalent to say that $H^1(G, \eta)=0$ for any unitary representation $\eta$ on any Hilbert space.
If $H^1(G, \rho^{\rm ad})\neq 0$ we get a contradiction, because it is easy to find an invariant positive-definite scalar product on $gl_n$ which turns the adjoint action $\rho'$ into a unitary action.

Points 1 and 3 heavily rely on the fact that the representation $\rho$ is finite-dimensional, where point 2 does not. I am interested in infinite-dimensional Hilbert spaces:

Can a Kazhdan group $G$ have a non locally rigid unitary representation into some infinite-dimensional Hilbert space?

The question might depend on which notion of local rigidity one uses in infinite-dimension, i.e. on which topology (or distance) one puts on the set of all bounded (actually, unitary) operators. I don't know if there is a standard accepted notion. I suspect that there should be plenty of non-rigid representations but I don't know any example. 
A related question on non-rigidity in infinite dimension is here.

REPLY [7 votes]: The situation in infinite dimensions is different for Kazhdan groups.
If $\Gamma$ contains a non-abelian free group, then the left-regular representation $\lambda \colon \Gamma \to U(\ell^2 \Gamma)$ admits a deformation $\lambda_t$ (for $t \in [0,1]$ say), such that $\lambda_t$ is a unitary representation, $$\sup_{g \in \Gamma} \|\lambda_t(g) - \lambda_s(g)\| \leq |s-t|,$$ $\lambda_0=\lambda$ and $\lambda_t$ not equivalent to $\lambda$ for $t \neq 0$. This is a very strong condition on a deformation; typically one does not require a uniform deformation and usually also using the strong operator topology.
An explicit construction of such a deformation for the free group itself goes back to Pytlik-Swarc in
T. Pytlik and R. Swarc, An analytic family of uniformly bounded representations of tree groups, Acta Math. 157(3-4):289-309, 1986.
The idea is then to induce this deformation to $\Gamma$ and check that the continuity is preserved (this is easy because of the strength of the assumption) and that the resulting representations remain inequivalent.
A similar behaviour is conjectured to hold for all non-amenable groups. These ideas have appeared in
Marc Burger, Narutaka Ozawa, and Andreas Thom, On Ulam stability, 
to appear in Israel J. Math., http://arxiv.org/abs/1010.0565<|endoftext|>
TITLE: Constant scalar curvature metrics in a conformal class
QUESTION [11 upvotes]: Let $(M,g)$ be a compact Riemannian manifold, then by the resolved Yamabe-problem, there exists a metric $\tilde{g}$ of constant scalar curvature in the conformal class $[g]$ of $g$. By normalizing volume, we have $s_{\tilde{g}}=Y(g)$ where $Y(g)=Y([g])$ is the Yamabe functional.
This might not be the only constant scalar curvature metric in $[g]$. So my question is: is there anything known about the set of constants $c$ where $c\equiv scal_{\tilde{g}}$ for a metric $\tilde{g}\in [g]$ of unit volume?

REPLY [12 votes]: Yes, as the OP mentions, in many cases the solution to the Yamabe problem may not be unique (we obviously ignore the effect of rescaling a metric), in the sense that there may be constant scalar metrics different from the minimizer of the total scalar curvature. Recall that the solution to the Yamabe problem was proved by finding a minimizer of the Hilbert-Einstein functional (total scalar curvature) on unit volume metrics in the conformal class $[g]$, however any critical point of this functional is a constant scalar curvature metric.
Perhaps a quick review of some uniqueness/non-uniqueness results could be of interest.
For example, in the conformal class of an Einstein metric (except the round sphere), the solution is unique. Moreover,

Anderson proved that on generic conformal classes, the solution is unique.

On the other hand, there are many non-uniqueness results:

Ambrosetti and Malchiodi (JFA, 1999) and Berti and Malchiodi (JFA, 2001) proved non-uniqueness results for conformal classes of deformations of the round metric on spheres;
Pollack (CAG, 1993) proved existence of arbitrarily small $C^0$ perturbations of any given metric (on any closed manifold!), with arbitrarily large number of solutions in its conformal class;
Schoen (1991) proved that taking the product $S^1\times S^n$ with the product of round metrics, and making the radius $r$ of $S^1$ go to $+\infty$, the number of solutions also goes to $+\infty$ as $r\to+\infty$;
Recently, de Lima, Piccione and Zedda obtained a sort of generalization, proving existence of infinitely many bifurcation points for many $1$-parameter family of product metrics, obtained by scaling one of the factors. In particular, this yields existence of a countable set of metrics in this family in which conformal class there exist at least three distinct solutions to the Yamabe problem;
More recently, in a joint paper with Piccione, we obtained a similar result (infinitely many bifurcations) for families of homogeneous metrics on spheres. Recall that the homogeneous metrics on spheres are given by scaling the round metric in the direction of the Hopf fibration, e.g., take $S^3\to S^{4n+3}\to HP^n$ and shrink the round metric on $S^{4n+3}$ in the vertical directions by a factor of $t^2$, obtaining a metric $g_t$. Then, as $t\to 0$ (i.e., as the fibers collapse), our result implies that there are infinitely many values of $t$ for which $[g_t]$ has at least 3 constant scalar curvature metrics. In a subsequent paper, we extend this result to any homogeneous fibration $K/H\to G/H\to G/K$ where $K/H$ has positive scalar curvature and either $H$ is normal in $K$ or $K$ is normal in $G$.

In most cases above, where explicit non-uniqueness of solutions to the Yamabe problem is proven, it is easy to give a qualitative description of the values of the (constant) scalar curvatures of the various solutions. For general results, however, the best available are the type of compactness results of Brendle, Khuri, Marques, Schoen etc mentioned in macbeth's answer.<|endoftext|>
TITLE: A good reference to grok hypergeometric functions?
QUESTION [15 upvotes]: When I was introduced during my degree to special functions, I made friends with a number of nice functions - Laguerre, Legendre, Hermite, Bessel, and whatnot - but I made only a passing acquaintance with the hypergeometric and confluent hypergeometric functions, mostly limited to looking them up in Arfken and recoiling in horror at the dryness of the material and the lack of physical content in the calculations.
I know, of course, that this lack of physical content is also accompanied by an astounding generality. After a while I did get the core of the idea, which I believe is "explore all special functions whose series coefficients are rational functions of $n$", and I do find it appealing, but I've not had the energy nor the motivation to follow that up and see what's interesting about the thing.
However, it appears that the long-delayed moment is here and some pretty hairy integrals (think $\int_0^\infty x^k e^{-\alpha x^2}J_m(\beta x)dx$) have pushed some ugly "${}_1 F_1$" symbols onto my page. So my question is, then: what's a good introduction to hypergeometric and confluent hypergeometric functions? I'd like one where I can get an intuitive understanding of what to expect from them in different circumstances, what nice properties they have, and generally why it really is worth it to deal with them instead of their more specific cases like Laguerre, Legendre, Hermite, Bessel, etc.

REPLY [2 votes]: I would also recommend "Special Functions and their Application" by Lebedev and Silverman (http://www.amazon.com/Special-Functions-Their-Applications-Mathematics/dp/0486606244/ref=sr_1_1/179-2267788-6339652?s=books&ie=UTF8&qid=1422720681&sr=1-1) and the series "Higher Transcendental Functions" (part of the Bateman Manuscript Project, edited by Erdelyi) which has the benefit of being free online via Caltech:
http://authors.library.caltech.edu/43491/<|endoftext|>
TITLE: Simply connected simplicial complexes
QUESTION [13 upvotes]: Let $Z$ be a simply connected, two dimensional simplicial complex.
Let $X\subset Z$ be a finite subcomplex with nontrivial $\pi_{1}$.
Must there exist a finite, simply connected subcomplex $Y\subset Z$ such that $Y\supset X$?
(Motivation: the fact that Whitehead conjecture remains unproven indicates that there are probably some very weird things that can happen in two dimensional complexes.  This question attempts to locate some of these pathologies.)

REPLY [11 votes]: The answer is negative unless you assume that $Z$ is aspherical (and even in this case I am not sure). Take Whitehead manifold $W$, which is a contractible open 3-manifold, not homeomorphic to ${\mathbb R}^3$ and let $Z$ be the 2-dimensional skeleton of a triangulation of $W$. Then $Z$ is a counter-example.
Proof.  

I claim that $W$ does not admit an exhaustion by simply-connected compact submanifolds with boundary. Indeed, if such an exhaustion $W_i$ exists, then each $W_i$ is bounded by some 2-spheres, all but one of which bounds a 3-ball in $W$. Filling in these spheres by 3-balls in $W$, we obtain an exhaustion $W_i'$ of $W$ by closed 3-balls. This would imply that $W$ is homeomorphic to the 3-space. Contradiction. 
Your question is equivalent to asking if $Z$ (provided that it is locally finite) is exhaustable by finite simply-connected subcomplexes. Note that my $Z$ is locally finite. Suppose that such an exhaustion $Z_i$ does exist, we can then enlarge each $Z_i$ by adding to it 3-simplices in $W$ whose boundary is contained in $Z_i$. This does not change $\pi_1$, of course. The resulting subcomplexes $V_i$ will exhaust $W$ and will be simply-connected. Taking $W_i$ to be a regular neighborhood of $V_i$ we obtain an exhaustion of $W$ by simply-connected compact submanifolds with boundary. This contradicts Part 1.<|endoftext|>
TITLE: Lagrangian Kleinian bottles
QUESTION [7 upvotes]: I remember some talks some time ago about proofs of nonexistence of Lagrangian Kleinian bottles in ${\mathbb C}^2$ for the standard symplectic structure, mentioning that this were the only compact surface for which the problem were open.
Question 1: Where to find references for the other surfaces?
Question 2: What is the current status for Kleinian bottles? Do there exist written proofs now?

REPLY [8 votes]: While working on my thesis, I (by accident) constructed a Lagrangian Klein bottle in $(S^2\times S^2, \omega\oplus\omega)$.  The construction works in $S^2\times D^2$ as long as the area of $D^2$ is strictly greater than half the area of $S^2$.  The method is to pick a Hamiltonian on $S^2$ that rotates $S^2$ halfway around, and realize it as the monodromy around $\partial D^2$, of a symplectic fibration over $D^2$ with fiber $S^2$. There is a circle that goes through the two fixed points of the rotation, which then traces out the Klein bottle.  See McDuff's ``Geometric Variants of the Hofer Norm'' for the precise version of how to realize such a monodromy and the constraints: the area of the base is an easy way to think of it, but it's really a ratio of volume to fiber area.
As I understand it the paper by Nemirovskii called ``homology class of a Lagrangian Klein bottle'', is flawed, because it predates the accepted proofs of non-existence as described by others.  I would love to know if its main claim is true, that a Lagrangian Klein bottle in a symplectic 4-manifold must realize a nontrivial second homology class with mod 2 coefficients. This is true for the example above. (Might need to ask this as a question.)<|endoftext|>
TITLE: The ring of continuous real-valued functions on a Stone space
QUESTION [7 upvotes]: Let $X$ be a topological space and consider the ring $C(X,\mathbb{R})$ of continuous real-valued functions on $X$ (where ring-addition and multiplication are defined in the obvious point-wise way).
It is well-known that this ring uniquely determines the topological space, if the latter is compact and Hausdorff (Gelfand-Duality).
My question is the following: What properties does the ring $C(X,\mathbb{R})$ have, if $X$ is a Stone space (that is, if $X \in Comp_2$ is also totally disconnected)? Ideally, I would like to have a statement along the following line:
$X$ is a Stone space (if and) only if the ring $C(X,\mathbb{R})$ has the property [nice property given to me by the lovely folks from MO].

REPLY [3 votes]: The answer is given by the following theorem (Davidson's $C^\*$-Algebras by Example, III, 2.5).

Let $A$ be a commutative C*-Algebra. Then the following statements are equivalent:

$A$ is an AF-algebra, i.a. a colimit of a sequence of finite-dimensional C*-algebras
$A$ is separable and the projections in $A$ generate a dense subspace.
The spectrum of $A$ is totally disconnected.


Thus, the category of locally compact totally disconnected Hausdorff spaces is equivalent to the category of commutative AF-algebras; the inverse functors are $C_0(-,\mathbb{C})$ and $\mathrm{Spm}$. Under this equivalence, compact corresponds to unital. The finite-dimensional commutative algebras are just powers of $\mathbb{C}$, corresponding to finite sets, but colimits produce interesting examples:
For example, if $X$ is the Cantor set, then it is easy to see that $C_0(X,\mathbb{C})$ is the colimit of the sequence $\mathbb{C} \to \mathbb{C}^2 \to \mathbb{C}^{2^2} \to \dotsc$, where $\mathbb{C}^{2^n} \to \mathbb{C}^{2^{n+1}}, a \mapsto (a,a)$. In general, AF-algebras are classified via so-called Bratteli-diagrams (see loc. cit).
By the way, there is a nice connection to Stone duality, which says that $C_0(-,\mathbb{F}_2)$ exhibits an equivalence of categories between the category of locally compact totally disconnected Hausdorff spaces and the category of boolean rings. In the diagram you just exchange $\mathbb{C}$ with $\mathbb{F}_2$. I would love to see a purely algebraic functor $\mathbb{F}_2 \otimes_{\mathbb{C}} (-)$, which doesn't use the spectrum as an intermediate step ...<|endoftext|>
TITLE: Weak*-closed and complemented subspaces of dual Banach spaces
QUESTION [7 upvotes]: We consider a Banach space $X$ and its dual $X^*$. 
Let $Q\colon X^\ast \to X^\ast$ be an idempotent operator. 
Question: 
Can we find an idempotent operator $P\colon X^\ast \to X^\ast$ which is weak${}^\ast$-to-weak${}^\ast$ continuous and with range isomorphic to range of $Q$ and $\mbox{im}P\subseteq \mbox{im}Q$? In fact, I am mostly interested in the case $\mbox{im }Q\cong \ell_p$ for $p\in [1,\infty)$.
Certainly, $P$ would have to be an adjoint to some idempotent on $X$. My feeling is that in general this is not the case but perhaps it might be true for some well-behaved class of Banach spaces $X$ like Banach lattices? Or Banach lattices without a complemented copy of $\ell_1$?

REPLY [6 votes]: In 
Stegall, C. Banach spaces whose duals contain $\ell_1(\Gamma)$ with applications to the study of dual $L_1(\mu)$ spaces. Trans. Amer. Math. Soc. 176 (1973), 463–477 
Stegall proved that $\ell_2$ is isometrically isomorphic to a norm one complemented subspace of $X^*$ when $X= (\sum_{n=1}^\infty \ell_2^n)_1$, yet $\ell_2$ does not embed into $X$.
It would be interesting to have an example of this phenomenon with $X^*$ separable.  
ADDED 6/25/12: 
Here is a more interesting example because the dual is separable, but the range of the projection is not $\ell_p$. Take any separable reflexive space $X$ that fails the approximation property and let $(E_n)$ be an increasing sequence  of finite dimensional subspaces of $X$ whose union is dense in $X$. Let $c(E_n)$ be the space of sequences $(x_n)$ with $x_n$ in $E_n$ and $\lim_n x_n$ existing in $X$, normed by $\|(x_n)\|= \sup \|x_n\|$.
Define  $c_0(E_n)$ to be $(\sum_{n=1}^\infty E_n)_0$. 
Consider the short exact sequence (ses)
$0\to c_0(E_n) \to c(E_n) \to X\to 0$
where the second arrow is the inclusion mapping and the third is the quotient mapping $Q$ defined by 
$(x_n)\mapsto \lim x_n$.  This sequence locally splits (with constant one), hence $Q^*$ maps $X^*$ onto a norm one complemented subspace of $c(E_n)^*$.
The ses itself does not split because $ c(E_n)$ has the approximation property (even a finite dimensional decomposition). 
This example is Proposition 2.4 in
Johnson, William B.; Oikhberg, Timur:
Separable lifting property and extensions of local reflexivity, 
Illinois J. Math. 45 (2001), no. 1, 123–137.
The construction itself is due to W. Lusky
A note on Banach spaces containing $c_0$ or $C_\infty$, 
J. Funct. Anal. 62 (1985), no. 1, 1–7.
Lusky was interested in the case that $X$ has the bounded approximation property. Then the resulting ses splits.<|endoftext|>
TITLE: Forcing as a new chapter of Galois Theory?
QUESTION [61 upvotes]: There is a (very) long essay by Grothendieck with the ominous title La Longue Marche à travers la théorie de Galois (The Long March through Galois Theory). As usual, Grothendieck knew what he was talking about: Galois Theory, far from being confined to its primary example, namely field extensions,  is pervasive throughout mathematics, and still to be fully understood. 
Ever since I first heard of forcing I was struck by its compelling analogy with field extensions: the ground model (read Q), the generic G (read a new element, say $\sqrt{2}$), the new M[G] (read 
Q[$\sqrt{2}$]), etc. 
I quote Joel Hamkins 's words here, in his sparkling paper on the multiverse:

In effect, the forcing extension has
  adjoined the “ideal” object G to V ,
  in much the same way that one might
  build a field extension such as
  Q[$\sqrt{2}$]

Of course, matters are a bit more complicated in set theory, you have to make sure the extension satisfies the axioms, that it does not "bother" the ordinals, and so on. 
Yet, one cannot really think that the analogy stops here. 
And it does not: in Galois theory, the main thing is the central theorem, establishing the Galois Connection between subfields of the extension and the subgroups of the Galois group, ie the group of the automorphisms of the extension leaving fixed the underlying field. 
No Galois connection, no Galois Theory. But wait, first we need the group. So where is it?
A hint is in the great classic result by Jech-Sochor on showing the independence of AC: by considering a group of automorphisms of P, the ordered set of forcing conditions, one can obtain a new model which is (essentially) the set of fixed points of the induced automorphisms. This is even clearer when one looks at it from the point of view of boolean valued models: each automorphism of the boolean algebra induces an automorphism of the extended universe. 

Now my question: is there some
  systematic work on classifying forcing
  extensions by their Galois group? Can one develop a full machinery which will apply to relative extensions? 

NOTE: I think this is no idle brooding: someone for instance has asked here on MO about  the 2-category of the multiverse. That is a tough question, and my sense is that before giving it a satisfactory answer some preliminary work needs to be done. Which work? Well, one needs to re-think  the classical set constructions from a structural standpoint, leaving behind its gory technical details. Now, forcing is a huge part of the multiverse, and understanding the structural algebra underpinning it would be, I trust, a huge step toward an algebraic understanding of the multiverse.

REPLY [18 votes]: It seems to me that the groups involved in producing models without AC (whether as groups of permutations of atoms or as groups of automorphisms of complete Boolean algebras) and the forcing constructions themselves are two rather separate things.  The most frequent use of forcing is to produce models of ZFC (not just ZF), and then groups are not involved.  On the other hand, the Fraenkel-Mostowski-Specker method of permutation models for the negation of AC  involves only permutation groups (and normal filters of subgroups), not forcing.  Groups and forcing come together in Cohen's method of symmetric models.  In the original (and still most common) presentation, one has a group acting on the forcing notion.  But even here, the groups and the forcing are less entangled than they seem.  Vopenka and Hajek showed (in their book "Theory of Semisets") how to get Cohen-style models by (1) starting with a permutation model, (2) forcing over it, and (3) passing to the pure (or well-founded) part.  In this presentation, the groups (and filters of subgroups) are only in step (1), and the forcing is only in step (2).  
It seems to me that, if one wants to analyze forcing from an algebraic point of view, one should begin with the simplest case, where no symmetries are involved.  The algebraic side of this is the study of complete Boolean algebras.  Afterward, one can embellish the picture by adding group actions, either acting on the Boolean algebras or producing permutation models over which to force.  The Grigorieff paper that Francois cited is an excellent place to start.  
Though the OP asked about forcing, let me also mention that the non-forcing context of permutation models might be a better place to look for Galois-like phenomena.  To begin, note that a normal filter $\mathcal F$ of subgroups of a group $G$ makes $G$ a topological group, in which $\mathcal F$ generates the neighborhood filter at the identity.  The notion of symmetry with respect to $\mathcal F$ that is used in defining the permutation model is just continuity of the standard action of $G$ on sets.<|endoftext|>
TITLE: Realiziability of hypergraphs as link (multi)sets of ordinary graphs 
QUESTION [6 upvotes]: I have a question about hypergraphs that I hope some combinatorics/graph theory experts can answer. The motivation for this question is group-theoretic and comes from the study of a certain space of measures that comes equipped with a natural affine action of the group Out(F_n). I'll skip the detailed background here, but if someone is interested, please look-up my paper with Tatiana Nagnibeda arXiv:1105.5742
Let G be a finite simple graph. For every vertex v of G the link Lk(v) is the set of vertices adjacent to v. Now form a weighted hypergraph Lk(G) whose vertex set is the same as the vertex set of G and whose hyper-edges are exactly all sets Lk(v) as v varies over the vertex set of G. Every hyperedge E in Lk(G) comes with a positive integral "weight" w(E) which is the number of vertices of G such that Lk(v)=E.
Now suppose we are given a finite weighted hypergraph H with positive integral weights on its hyperedges. 
I'd like to know if there are known necessary and sufficient condition for H so that there exists a graph G such that H=Lk(G).
If anyone knows the answer or has some suggestions regarding where to look, I'd much appreciate it.
Thanks a lot,
Ilya Kapovich.

REPLY [4 votes]: Let $A(H) = (a_{ij})$ be the incidence matrix of a weighted hypergraph $H$. This is the 0-1 matrix such that $v_i \in E_j$ if and only if $a_{ij} = 1$. (Repeated hyperedges from the integral weighting result in repeated columns of the matrix.) Then $H = Lk(G)$ for some graph $G$ if and only if the columns (or rows) can be permuted to form a symmetric matrix.
(I don't know if this is well known or not. If it's a common construction, I'm sure it is known.)
Let $G$ be a graph. Then we have:
$$v_i \in Lk(v_j) \iff v_j \in Lk(v_i)$$
As long as we label the edges so that $E_i = Lk(v_i)$, the incidence matrix is symmetric.
In the other direction, let $A'$ be a symmetric matrix resulting from permuting the columns of $A(H)$. Form $G$ by requiring that $v_i$ and $v_j$ be adjacent if and only if the $a_{ij}$ entry of $A'$ is 1. Then $E_j'$ consists of the vertices in $G$ adjacent to $v_j$. It follows that $H = Lk(G)$.
For loopless graphs we do have to add the condition that the matrix be permutable to one with 0's along the diagonal.<|endoftext|>
TITLE: isotopy inverse embeddings vs. diffeomorphisms
QUESTION [20 upvotes]: I would like to find an example, if one exists, of manifolds $M$ and $N$ with embeddings $f:M\to N$ and $g:N\to M$ such that $f\circ g$ and $g\circ f$ are both isotopic (i.e. homotopic through embeddings) to the respective identities, yet the interiors of $M$ and $N$ are not diffeomorphic. Obviously, $M$ and $N$ cannot be closed. You may assume that the manifolds have no boundary, but I would also be interested in compact examples.
By the way: in the other direction, are there conditions under which $M$ and $N$ are necessarily diffeomorphic if $f$ and $g$ as above exist?

REPLY [26 votes]: Here I'll prove that $M$ and $N$ must be diffeomorphic under your hypotheses (I'll assume that $M$ and $N$ are open, i.e. no boundary). 
Consider the direct limit (see below)  $X= M \overset{f}{\to} N \overset{g}{\to} M \overset{f}{\to} N \overset{g}{\to}\cdots$. By this, I mean the union $M \times \mathbb{N} \sqcup N\times \mathbb{N}$ modulo the identifications $(m,k)\sim (f(m),k), (n,k)\sim (g(n),k+1)$, for $m\in M, n\in N, k\in \mathbb{N}$.  Then $X$ is a smooth manifold, since it is a nested union of smooth manifolds. Similarly, we have the direct limit $Y=N \overset{g}{\to} M \overset{f}{\to} N \overset{g}{\to}\cdots$. Clearly, $X \cong Y$, since the direct limit depends only on the tail of the sequence. Also, $X$ (and therefore $Y$) is diffeomorphic to the direct limits  $ M \overset{g\circ f}{\to} M \overset{g\circ f}{\to} M \overset{g\circ f}{\to}\cdots$ and $N\overset{f\circ g}{\to} N\overset{f\circ g}{\to} N\overset{f\circ g}{\to}\cdots$. 
Claim: If $F:M\to M$ is isotopic to the identity, then the direct limit $M\overset{F}{\to} M\overset{F}{\to} M\overset{F}{\to} \cdots \cong M$. 
By the above discussion, this implies that $M\cong N$. 
To prove the claim, let $F_t:M\to M, t\in [0,1]$ be an isotopy of $F$ to the identity, so $F_1=F$, and $F_0=Id$. Take an exhaustion of $M$ by smooth compact submanifolds with boundary $K_1\subset K_2\subset K_3 \subset \cdots$, so that $F([0,1]\times K_i)\subset int(K_{i+1})$, $K_i\subset int(K_{i+1})$, and $\cup K_i=M$. The we see that $X_t = M\overset{F_t}{\to} M\overset{F_t}{\to}M\overset{F_t}{\to}\cdots$ is diffeomorphic to $Y_t= K_2 \overset{F_t}{\to} K_4 \overset{F_t}{\to} K_6 \overset{F_t}{\to} \cdots$ for all $t$ (where we really meant $F_{t| K_{2i}}$ in the maps of this direct limit and topologized by the inclusion of interiors). To see this, note that $X_t$ is a quotient of $M\times \mathbb{N}$, which is exhausted by $\{(K_k,i), k,i\in\mathbb{N}\}$. Since $Y_t\subset X_t$, we need only show that each point of $X_t$ is in $Y_t$. Suppose $x\in X_t$, then $x$ is the image of some $x'\in (K_k,i)$ for some $k,i$. If $k\leq 2i$, then $x'\in (K_{2i},i)$, so we see that $x\in Y_t$. Otherwise, $k>2i$ and we see that $(K_k,i) \overset{F_t}{\subset}(K_{k+1},i+1) \overset{F_t}{\subset} \cdots \overset{F_t}{\subset} (K_{2k-2i},k-i)$ after taking the quotient, so $x\in Y_t$. 
By the Isotopy Extension Theorem, one may see that $Y_0 \cong Y_1$. This gives the desired diffeomorphism $M\cong M\overset{F}{\to} M\overset{F}{\to} M\overset{F}{\to} \cdots $. 
We actually use the isotopy extension theorem by induction. We want to find a sequence of diffeomorphisms $G^i: K_{2i}\to K_{2i}$ so that $G^{i+1}_{|K_{2i}} = F\circ G^i$. This immediately gives the diffeomorphism 
$$\begin{matrix} Y_1= K_2 & \overset{F}{\to} & K_4 & \overset{F}{\to} &  K_6 & \overset{F}{\to} & \cdots\\
 G^1 \uparrow  &  &G^2\uparrow  &  &G^3\uparrow  & & \\ 
Y_0= K_2 & \hookrightarrow &  K_4 & \hookrightarrow & K_6 & \hookrightarrow &  \cdots
\end{matrix}$$
We will actually construct a diffeotopy $G^i_t:K_{2i}\to K_{2i}, t\in [0,1]$ such that  $G^{i+1}_{t|K_{2i}}=F_t\circ G^i_t$, where $G^i_0=Id$, and then set $G^i=G^i_1$ for the desired diffeomorphism. 
Let $G^1_t=Id$. Suppose we have constructed $G^i_t$ by induction, so that $G^i_0=Id$, $G^i_{t|K_{2i-2}}=F_t \circ G^{i-1}_t$, $t\in [0,1]$. Then $F_t\circ G^i_t: K_{2i}\to int(K_{2i+2})$ is an isotopy from $F_0\circ G^i_0=Id$ to $F_1\circ G^i_1=F\circ G^i_1$. By the isotopy extension theorem, there exists a diffeotopy $G^{i+1}_t: K_{2i+2}\to K_{2i+2}$ such that $G^{i+1}_0=Id$, and $G^{i+1}_{t|K_{2i}}=F_t\circ G^i_t$, with compact support in $int(K_{2i+2})$. This completes the proof. 
Addendum: I'll add some explanation of direct limits. Consider a sequence of smooth embeddings of open manifolds $X_i \overset{f_i}{\to} X_{i+1}$. By the direct limit $X$ of $X_1\overset{f_1}{\to} X_2 \overset{f_2}{\to} X_3 \cdots$, I mean the quotient of $X_1\sqcup X_2 \sqcup X_3 \sqcup \cdots$ with respect to the equivalence relation generated by $x_i \sim f_i(x_i)$ for all $x_i \in X_i, i\in \mathbb{N}$. Then there is a natural embedding of $X_i \hookrightarrow X$ for all $i$, in such a way that $X= \cup_i X_i$, and one gives $X$ the structure of a smooth manifold by taking the atlas of charts generated by the charts of each $X_i$. 
The space $X$ is determined by the direct limit of any subsequence $\{ X_{i_j}\}$, with maps $F_j: X_{i_j}\to X_{i_{j+1}}$ defined by $F_j=f_{i_{j+1}-1}\circ \cdots \circ f_{i_j}$. One may therefore verify that two direct limits are diffeomorphic if there are compatible diffeos. between subsequences, which justifies some of the arguments above.<|endoftext|>
TITLE: "Étalification" of a scheme
QUESTION [9 upvotes]: Let $X$ be a scheme. Does the forgetful functor
$$\mathrm{EtSch}/X \to \mathrm{Sch}/X$$
have a right adjoint $Z \mapsto \tilde{Z}$? One might call $\tilde{Z}$ the étalification of $Z$. So this is an étale $X$-scheme together with an $X$-morphism $\tilde{Z} \to Z$, which induces for every étale $X$-scheme $Y$ a bijection $\hom_X(Y,\tilde{Z}) \cong \hom_X(Y,Z)$.
If $X$ is the spectrum of a field $k$, the answer is yes, using 1) the equivalence of sites between étale $k$-schemes and continuous $G$-sets, where $G$ is the absolute Galois group of $k$, 2) the fact that every sheaf on $G$-sets is representable. Explicitly: If $Z$ is a $k$-scheme, let us denote by $Z_{\mathrm{sep}}$ the subset of all points $z \in Z$ such that $k(z)/k$ is finite and separable. Then
$$\tilde{Z} = \coprod\limits_{z \in Z_{\mathrm{sep}}} \mathrm{Spec}(k(z)).$$
For example, $\widetilde{\mathbb{A}^1}=\coprod_{\alpha \in k} \mathrm{Spec}(k)$ and $\widetilde{\mathbb{A}^2}$ is the coproduct of spectra of the form $(k[x]/(p)[T])/(q)$, where $p \in k[x]$ is irreducible and $q$ is some irreducible separable polynomial over $k[x]/(p)$.
Perhaps this construction is well-known, therefore I've put the reference request tag.

REPLY [2 votes]: Can we just do this the hard way? 
First we construct the set of points of the etalification $\bar{Z}$. A point of the etalification is a point $P$ of $X$ plus a map from the spectrum $T$ of the etale local ring at $X$ to $Z$ that forms a commutative diagram with the natural map $T \to X$. (or, if $Z \to X$ is not finite type, we need to use a finitely-generated subring of the etale local ring.)
Then for $Y\to X$ is etale with a map $Y \to Z$. We get a set-theoretic map $Y \to \bar{Z}$. One easily checks that each point is in the image of such a map. We form the finest topology such that the image of each such map is open.
Now we need a sheaf of rings.  The natural map $\bar{Z} \to Z$ given by taking the image of the special fiber is continuous. Pull back the structure sheaf of $Z$, mod out by the functions on a set $U$ that are $0$ in the images of the open balls $T$ corresponding to the points of $U$, and sheafify.
We will check that this is a scheme by finding affine neighborhoods of each point. Each point of $\bar{Z}$ is in the image of some etale $Y\to X$, and we can further choose $Y$ to be affome amd $T \to Z$ an immersion. We will check that $T \to \bar{Z}$ is an inclusion of locally-ringed spaces. Clearly it is injective. Next we check that it is continuous and open. Open subsets of $\bar{Z}$ wil come from other etale maps which agree on some etale local ring, but then these must agree on an open subset, so they will give open subsets of $Y$. Open subsets of $Y$ are just more schemes etale over $X$.
Then we check it gives an isomorphism of sheaves of rings. The sheaf of rings on $Y$ is the pullback of the sheaf of rings of $Z$, mod the functions that vanish on it. But the functions that vanish on $Y$ are exactly the functions that vanish on each etale local rings, so the two sheaves of rings are identical.
So in addition $\bar{Z}$ is etale over $X$. By construction of the sheaf of rings, $\bar{Z} \to Z$ is a morphism. Similarly the set-theoretic maps $Y \to \bar{Z}$ for all $Y\to Z$ etale over $X$ are clearly morphisms, and give the required bijections.
Obviously this is quite sketchy and there are a lot of technical details I'm missing but it seems doable.<|endoftext|>
TITLE: Does equality of Laplacians imply Kähler?
QUESTION [18 upvotes]: This question follows on from this one. 
Let $(X, \omega)$ be a Hermitian manifold and define the Laplacians $\Delta_{\partial} = \partial\partial^* + \partial^*\partial$ and $\Delta_{\bar{\partial}} = \bar{\partial}\bar{\partial}^* + \bar{\partial}^*\bar{\partial}$. 
If $(X, \omega)$ is a Kähler manifold, that is $d\omega = 0$ (or equivalently $\partial\omega = 0$ or $\bar{\partial}\omega = 0$), we have $\Delta_{\bar{\partial}} = \Delta_{\partial}$. 
More generally, on any Hermitian manifold we have $\Delta_{\bar{\partial}} = \Delta_{\partial} + [\partial, [\Lambda_{\partial\omega}, L]] - [\bar{\partial}, [\Lambda_{\bar{\partial}\omega}, L]]$ where:

 $[\bullet, \bullet]$ is the graded commutator;
 $\Lambda_{\partial\omega}$ and $\Lambda_{\bar{\partial}\omega}$ are the adjoints of wedging with the forms $\partial\omega$ and $\bar{\partial}\omega$ respectively; and
 $L$ is the Lefschetz operator, that is, wedging with $\omega$.

It is clear how the additional terms relating the Laplacians in the Hermitian case vanish if the metric is Kähler ($\partial\omega = 0$ and $\bar{\partial}\omega = 0$, so $\Lambda_{\partial\omega}$ and $\Lambda_{\bar{\partial}\omega}$ are both zero). What about the converse? That is:

If $\Delta_{\bar{\partial}} = \Delta_{\partial}$ on a Hermitian manifold $(X, \omega)$, is it necessarily Kähler?


The accepted answer in the linked question refers to balanced manifolds. These are manifolds with the property that $\Delta_{\bar{\partial}}f = \Delta_{\partial}f$ for any smooth function $f$. Not all such manifolds are Kähler. The above question is stronger as it requires equality for all smooth forms.

REPLY [3 votes]: In addition to the papers mentioned by YangMills, there is also the earlier paper by A. W. Adler which shows that if $\Delta = 2\Delta_{\bar{\partial}}$ on a hermitian manifold, then it is Kähler.<|endoftext|>
TITLE: How to Draw Complex Line Bundles
QUESTION [17 upvotes]: I am giving a presentation soon on the Classification of Complex Line Bundles and I would like to have some very "basic" visualizations to use as examples.
Background and Context
I am considering the Cech-cohomology of a principal $ \mathbb{C}^{*} $ bundle, where my sheaf $\underline{\mathbb{C}}_M^{*}$ is the sheaf of smooth $\mathbb{C}^{*}$ valued functions on the manifold $M$.  Using the exponential sequence of sheaves
$$ 0 \to \mathbb{Z}(1) \to \underline{\mathbb{C}}_M \to \underline{\mathbb{C}}_M^{*} \to 0$$
we get an isomorphism (via properties of cohomology and the connecting homomorphism)
$$H^1(M, \underline{\mathbb{C}}_M^{*}) \cong H^2(M, \mathbb{Z}(1)) $$
It turns out that $H^1(M, \underline{\mathbb{C}}_M^{*}) $ is also isomorphic to the group of isomorphism classes of principal-$\mathbb{C}^{*}$ bundles over $M$.  Since the principal- $\mathbb{C}^{*}$ bundles are in one-to-one correspondence with the complex line bundles, it should be evident how this all relates to my title. 
My Questions

(1)  Given the above information, and some knowledge of cohomology, there should be only a trivial principal- $\mathbb{C}^{*}$ bundle on the circle $S^1$.  How can we see this visually?

*See my example/analogue below.

(2)  Similarly, how can we visualize a non-trivial principal- $\mathbb{C}^{*}$ bundle on the standard 2-dimensional torus?

*Example/Analogue:
So consider a circle bundle on $S^1$, then we can consider a section of the bundle like so:
alt text http://www.cheynemiller.com/Math/Figures_files/SectionOnU.png
Now, given two sections on adjacent trivializations, 
alt text http://www.cheynemiller.com/Math/Figures_files/TransitionUaUb.png
We can imagine deforming one section into another, to get our transition functions.  Now, I can also believe that any such family of sections can be deformed into a global section, so again I want to know why this necessarily doesn't work on the Hopf bundle via pictures.

REPLY [2 votes]: To see that the Hopf bundle is not trivial, one considers its restrictions to the subspaces $N,S\subset S^2 = \mathbf{CP}^1$, with $N = \mathbb{C}^*\cup\{\infty\}$ and $S = \mathbb{C}$ where it is trivial. One should be able to write down sections given this description. Then the transition function $\mathbb{C}^* \to S^1$ can be written down. You can think of this as a map $\mathbb{C}^* \to \mathbb{C}^\ast$, and hence calculate the integral of it around $S^1 \subset \mathbb{C}^*$. This gives you the index of the transition function, which is non-zero.
You can then calculate the index of any transition function for the Hopf bundle, using the fact that it will be a Cech cocycle equivalent to the one you've written down.
Lastly, you can calculate the index of the transition function for the trivial $S^1$-bundle on $S^2$, and find this is not equal to that for the Hopf bundle.
In fancier language, because $\mathbb{C}^*$ is not simply connected, one finds that you cannot deform the Hopf bundle's transition function, which is not null-homotopic, to the transition function for the trivial bundle, which is null-homotopic.
In pictures, one has that the transition function for the Hopf bundle, restricted to the circle, loops once around the origin, but the trivial bundle's transition function is constant. If people are happy with believing that continuously deforming the transition function gives equivalent bundles (one could motivate this by writing down Cech coboundaries on $\mathbb{CP}^1$ that give the equivalence), and that transition functions which cannot be deformed to each other give inequivalent bundles (this is the important point), then this is pretty much the best picture you'll get.<|endoftext|>
TITLE: Why are ring actions much harder to find than group actions?
QUESTION [19 upvotes]: I admit freely that the following question is a bit of a fishing expedition inspired by this lovely "definition" of a module as found on Wikipedia:

A module is a ring action on an abelian group.

It takes a while for the novice to unzip this definition into the usual list of identities, but for someone with a basic understanding of (group) actions who needs a quick shorthand to remember what being a module entails, it seems hard to beat this definition.
Now, volumes have been written about group actions. This makes sense, because set theory underlies modern mathematics and for any set $X$, the automorphisms $\operatorname{Aut}(X)$ naturally have a group structure. And of course, a group $G$ acting on $X$ is just a group homomorphism $G \to \operatorname{Aut}X$. 
On the other hand — aside from this definition of a module — it is hard to come across a general theory of ring actions. In order to find interesting actions of a ring $R$, one analogously needs the set of endomorphisms $\operatorname{End}X$ to have the structure of a ring so that one may search for ring morphisms $R \to \operatorname{End}X$.

What are the most general objects $O$ for which $\operatorname{End}O$ canonically has the structure of a ring?

Certainly, $O$ at least contains abelian groups, but is there anything else?

REPLY [7 votes]: $\newcommand{\Hom}{\operatorname{Hom}}$An example for a class of rings where the ring actions are very close to group actions are Hopf algebras. As a consequence many theorems from representation theory of groups generalize to the representation theory of Hopf algebras. 
In order to give a feeling on how definitions from group actions generalize to Hopf algebras consider a group $G$ and $kG$-Modules $M,N$ ($k$ a commutative ring). Then we have the invariants $M^G := \lbrace m \in M\mid \forall g \in G:\;gm=m\rbrace$ and $G$ also acts on $\Hom_k(M,N)$ via
$(g\cdot f)(m) := gf(g^{-1}m)$. Then the $G$-linear maps can be recovered as invariants by 
$\Hom_{kG}(M,N) = \Hom_k(M,N)^G$.
Now let $H$ be a Hopf algebra over $k$ (corresponds to $kG$) with antipode $S: H \to H$ (corresponds to $g \mapsto g^{-1}$), augmentation $\epsilon: H \to k$ (corresponds to $g \mapsto 1$) and coproduct $\Delta: H \to H \otimes_k H$ (corresponds to $g \mapsto g \otimes g$). Then one defines for $H$-modules $M,N$: $$M^H := \lbrace m \in M\mid \forall h \in H: \;hm=\epsilon(h)m\rbrace$$
and if $\Delta(h)=\sum_i h_i^{(1)} \otimes h_i^{(2)}$ then $H$ acts on $\Hom_k(M,N)$ by 
$$(h\cdot f)(m):= \sum_i h_i^{(1)}f\big(S(h_i^{(2)})m\big).$$
Now, as expected, the identity $\Hom_H(M,N)=\Hom_k(M,N)^H$ holds again.<|endoftext|>
TITLE: Intersections of maximal abelian von Neumann algebras
QUESTION [5 upvotes]: Let $H$ be separable Hilbert space. Let $A$ be a maximal abelian von Neumann subalgebra of $B(H)$, and $B$ an abelian von Neumann algebra with $A\cap B={\mathbb C}I$, where $I$ is the indentity element of $B(H)$. Does there exist another maximal abelian von Neumann subalgebra of $B(H)$, say $C$, such that $C\supseteq B$ and $A\cap C={\mathbb C}I$?

REPLY [3 votes]: This is only a partial answer, but it didn't fit in the comment box.
In finite dimension, say $\dim(H)=n$, a maximal abelian von Neumann algebra $A \subseteq B(H) \cong M_n(\mathbb{C})$ just comes down to (the set of matrices that are diagonal in) a choice of basis for $H$. Similarly, $B$ consists of diagonal matrices in some (second) basis, possibly with repeated eigenvalues. So maximality forces $C$ to consist of all diagonal matrices in some (third) basis that spans the eigenspaces of the second one. The question is whether this third basis can be chosen while respecting $A \cap C=\mathbb{C}I$. If each eigenspace of $B$ has dimension an integer power of a prime number, then mutually unbiased bases are known to exist, and the answer is affirmative.<|endoftext|>
TITLE: Defining ind-coherent sheaves and their singular support
QUESTION [9 upvotes]: Q1: My first question is about defining the category $\text{IndCoh}(S)$ for a $DG$ scheme $S$. So in page $18$ of this paper, they are defined as being the ind-completion of the category $\text{Coh}(S)$. Here $\text{Coh}(S) \subset \text{QCoh}(S)$ is defined as the full subcategory of objects with bounded cohomological amplitude and coherent cohomologies. What is the ind-completion?
Q2: My second question is about the singular support of an ind-coherent sheaf. For a $DG$ scheme $Z$, the scheme of singularities $\text{Sing}(Z)$ is defined in page $19$ of this paper . In page $41$ of the same paper, given $\mathcal{F} \in \text{IndCoh}(Z)$, the singular support of $\mathcal{F}$ is defined as $\text{supp}_{A}(\mathcal{F})$
with $A=\Gamma(\text{Sing}(Z),\mathcal{O}_{\text{Sing}(Z)})$; . How should I correctly interpret and understand this definition of singular support?

REPLY [9 votes]: (Hopefully t3uji, tony pantev or Greg Stevenson will chime in with a more authoritative answer, but in the meanwhile..)
The notion of singular support of a coherent sheaf is an analog of the notion of singular support of a constructible sheaf or D-module. Let's quickly recall the latter: given a sheaf we can measure its failure to be locally constant at a particular point in a particular codirection. Namely given a covector at a point, i.e., a hyperplane in the tangent space, you ask if the sheaf behaves locally constantly moving off this hypersurface (this is a generalization of the Cauchy-Kovalevski theorem in PDE). You can measure this by taking a local function with the given covector as its differential at our point and calculating relative cohomologies (Morse groups) of its level sets near this point and "seeing if anything happens". The singular support (or microlocal support) is the collection of all covectors where our sheaf is not locally constant - ie all points and directions where "something interesting happens".
A very nice recent idea of several people (Isik, Arinkin-Gaitsgory, and others following on Orlov's work on categories of singularities --- someone who knows the history better please correct) is that one can do a very similar operation for coherent sheaves.
Recall that on a smooth variety any coherent sheaf (or bounded coherent complex) is quasiisomorphic to a perfect complex (bounded complex of vector bundles). This fails precisely at singular points of varieites by a theorem of Serre. Orlov introduced the category of singularities of a variety as the quotient of the bounded derived category by perfect complexes --- i.e., a measure of "how and where" the variety is singular (this category of singularities is supported at the singular locus). This is intimately related to the theory of matrix factorizations.  
The new notion of singular support is an "individual" version of this construction for lci schemes (schemes with cotangent complex in degrees -1,0) - i.e. one looks at a specific coherent sheaf (or bounded complex) and attaches to it its "microlocal support" --- naively speaking, the collection of points and (degree -1) codirections where the sheaf fails to be perfect (see section 0.3.7 for a way to make this intuition precise, using a description of our scheme as a local complete intersection). One can also define the singular support as an honest support for the sheaf, when considered as a module for the Hochschild cohomology sheaf of the variety (self-Ext of the identity). One can also think roughly about representing covectors as differentials of functions, and then passing to categories of matrix factorizations of this function and seeing whether our sheaf survives - again, whether the sheaf is "interesting" at a given point and codirection.
Edit: Let me add a little about the role of Hochschild cohomology. By (one) definition the Hochschild cohomology of a scheme is the self-Ext of the identity functor of the (dg) derived category. In other words, the Hochschild cohomology tautologically acts by endomorphisms of every sheaf, in a way compatible with all morphisms. One can say more -- the Hochschild cohomology can be identified with the enveloping algebra of a Lie algebra structure on the shifted tangent complex $T[-1]$, which acts on every object via the construction of Atiyah classes (see Kapranov's paper on Rozansky-Witten theory for example, Markarian's preprint on HH and many more recent papers, the latest word maybe Calaque-van den Bergh). This can be nicely interpreted in terms of derived loop spaces -- $T[-1]$ is the Lie algebra of the free loop space, and Hochschild cohomology is its "group algebra".. 
In our case we are interested in an lci scheme, and in particular the action of the top (+1) piece of the tangent complex, which after shift to $T[-1]$ lives in degree 2, or after taking enveloping algebra lives in even degree Hochschild cohomology. The singular support is then the support of the action of this commutative algebra (piece of even $HH^*$) on the sheaf. In other words, any sheaf on $X$ tautologically has a bigger action of a not quite commutative algebra, the Hochschild cohomology, but there's a commutative piece we can single out in there in the lci case and then take usual support.<|endoftext|>
TITLE: Cross sections in bundles and principal G-bundles
QUESTION [5 upvotes]: A principal $G$-bundle has a cross section iff it is trivial (e.g. Husemoller's Fibre Bundles, 3rd ed., 8.3 in chapter 4).
A principal $G$-bundle is in particular a fiber bundle with fiber $G$.
My question: does there exist a group $G$ and a non-trivial principal $G$-bundle $p:E\rightarrow B$ that does have a cross section when considered as a mere fiber-bundle?
If so, I would be glad to see a simple example. Thanks!

REPLY [7 votes]: Let $p\colon E \to B$ be a principal $G$-bundle and $s\colon B \to E$ a global section. Then the map $F\colon B\times G \to E$, $F(x,g) = s(x)\cdot g$ is a global trivialization of $E$. Here the dot denotes the right action of the group $G$. The map $F$ is surjective, because the action is transitive on fibres, and it is injective because the action is free. Continuity/smoothness is the same as the continuity/smoothness of your section $s$ and the action of the group $G$.
There's no special requierement in the definition of a section of a principal $G$-bundle, it's still the section of the underlying fibre bundle. But once you have a global section, you just use the $G$-action to trivialize the bundle.<|endoftext|>
TITLE: Should I become a Mathscinet reviewer?
QUESTION [16 upvotes]: Possible Duplicate:
When to start reviewing 

I was recently asked to become a MathSciNet reviewer, which, as far as I understand, is something that they ask almost any mathematician at a certain point in their career. 
Should I accept? What are your motivations in both directions? Is this a service to the community that I should help with, like serving as a referee for journal papers?
I'll write my own pros and cons as an answer.

REPLY [24 votes]: The question is natural but the answer is highly individual.   For what it's worth, I'll comment from my own perspective as a now-retired reviewer who produced almost 500 reviews over many decades.  In its early years the print version of Mathematical Reviews was still relatively slim but covered the most widely circulated journals and had as reviewers most of the active mathematicians of the time.  It had the great advantage of bringing together both a print database of current literature and (often) helpful commentaries by specialists.  For many decades the authors were compensated only by receiving free papers (sometimes books).
As mathematics and its offshoots proliferated in the 1960s and later, it became impossible for most people to skim all reviews.    But the evolving classification scheme helped, even though it could never meet all needs.    Until the Internet (and arXiv) developed far enough, the reviews and database played a mostly constructive role in communication.    But managing the flow of papers and editing the submitted reviews required a lot of expensive professionals, as it still does.    Some reviews were of course eccentric, such as one which simply quoted verbatim half of a two page note from a widely circulated Springer journal.   
For me personally it was a way to keep in touch with a wider range of interesting mathematics than I actually worked on at the time.   But to do the reviewing task well is time-consuming, since I always felt the need to delve into the related literature.   (At least once I discovered an earlier proof in a slightly offbeat journal of a theorem published anew in a mainstream journal by an author who hadn't been aware of the earlier proof.)    Sometimes you get correspondence (even arguments) from an author whose work you have reviewed.   All very interesting, but optional activity like refereeing.
By now MathSciNet functions mainly as an excellent database, still very expensive to maintain, and reviewers are given AMS credits for their use.  Fewer papers get full reviews, which is usually the right decision but not always.   People rely more for up-do-date stuff on other Internet sources, but the organized and flexibly searchable database is worth the cost for those institutions which can afford it.   (Not all can.)

REPLY [7 votes]: Cons:

it takes time. I should rather worry about producing good-quality research and disseminating it through slides and talks, especially now that I am relatively young. Publishing slides and expository material will "help the world" more than an AMS review.
it is not something that has a great value for the community. It might be just me, but I find the reviews of little use, considering that the papers already have abstracts. Mathscinet overall is great as a literature database and search tool, but the presence of reviews does not add much to its value.
AMS is selling access to these reviews. I am probably getting paid peanuts, or nothing at all (they did not mention payment at all on the invitation e-mail), to help building a corpus that is sold commercially. The prices are not cheap for a single user. I'd rather contribute to a public database.

Pros:

I personally do not use often the reviews, but they might be very helpful for other people. Maybe someone else relies on then more than I do (waiting for feedback here).
I am afraid to look mean and ungenerous to the eyes of the "math community" if I refuse it. For instance. refereeing for a journal is a time-consuming job, but we should not refrain from doing it because it helps the community; maybe other mathematicians see these reviews in a similar way.
it will look good in my CV as an additional "editorial service"; it is somehow a recognition that one is a competent member of the math community.

Overall, the pros do not seem too strong to me, and I realize that they are all about "how people will see me", which is probably a sign that they are flawed arguments. So, unless you guys on Mathoverflow manage to convince me of the contrary, I think I will turn down the offer, at least for now.<|endoftext|>
TITLE: The Bruss-Yor conjecture about an iterated integral
QUESTION [17 upvotes]: Is the sequence $$w_n=n! \int_0^{1/2} \int_{x_1}^{2/3} \cdots\int_{x_{n-2}}^{\frac{n-1}{n}} \int_{\frac{n}{n+1}}^1  dx_n dx_{n-1} \cdots dx_1$$ increasing for $n\ge 3$?

This is a conjecture of F. Thomas Bruss and Marc Yor in a recent paper Stochastic Processes with Proportional Increments and The Last-Arrival Problem. It is presented here with the permission of them.
Background
In the mentioned paper Bruss and Yor introduce stochastic processes with proportional increments to deal with the so called last arrival problem (l.a.p.) where an unknown number of independent random random variables $X_1,\ldots,X_N$ (which are uniformly distributed on the interval $[0,1]$) is observed and "our objective is to
stop online with exactly one stop on the very last of these points, i.e. at their
largest order statistics $X_{\langle N;N\rangle}$" (cited from Bruss and Yor).
Bruss and Yor give an interpretation of the problem (it is not so obvious what it means that $N$ is unknown) and then provide the optimal strategy. The numbers $w_n$ are then
the win probability conditioned on $N=n$. Bruss and Yor state that $w_n \to 1/e$.
Some relations
The inner integral clearly is $1/(n+1)$ and the remaining integral is thus up to the factor $n/(n+1)$ the probability that the order statistic $X_{\langle 1;n-1\rangle},\ldots,X_{\langle n-1;n-1\rangle}$ of $n-1$ independent on $[0,1]$ uniformly distributed random variables has values in $[0,1/2] \times [0,2/3] \times \cdots \times [0,\frac{n-1}{n}]$. It would be thus natural if a similar problem had appeared in some statistical context.
One can write down the latter probability using multinomial sums (which looks very
ugly). Nevertheless this suggests that some clever combinatorial arguments might help.
Finally the order statistics are closely related to Poisson processes (however, as the authors are experts in stochastic processes they have most probably checked that area).

REPLY [9 votes]: Here is a sketch of a full proof that can be (at least, in principle) verified by a human.
Claim 1. $\gamma_n=(\frac{n-1}{n})^{n-1}$ is a decreasing sequence.
This is well known and easy to prove.
Claim 2. The integral in question is just $\frac n{n+1}$ times the probability $P_n$ that $Y_k\le \frac k{k+1}$ for all $k=1,\dots,n-1$ where $Y_k$ is the $k$-th value in the list obtained by sampling $n-1$ independent random variables uniformly distributed on $[0,1]$ and putting them in the increasing order.
This has already be mentioned by the OP.
Claim 3. The joint distribution of $Y_{n-1},\dots,Y_1$ is the same as of $X_1^{\frac 1{n-1}},X_1^{\frac 1{n-1}}X_2^{\frac 1{n-2}},\dots,X_1^{\frac 1{n-1}}X_2^{\frac 1{n-2}}\cdots X_{n-1}$ where $X_j$ are i.i.d. random variables uniformly distributed on $[0,1]$.
Just condition on the maximum and use induction. From now on, we will just assume that $Y_k=X_1^{\frac 1{n-1}}\dots X_{n-k}^{\frac 1{k}}$.
Claim 4. For all $k\ge 2$, we have
$$
(\frac{k-1}{k})^{\frac{k-1}k}(\frac{k}{k+1})^{\frac 1{k(k+1)}}\le \frac{k}{k+1}
$$
Rewrite it as $(\frac{k^2-1}{k^2})^{k^2-1}\le (\frac{k}{k+1})^{k}$ and use Claim 1.
Claim 5. Let $2\le k\le n-2$. If $X_1^{\frac 1{n-2}}\dots X_{n-k}^{\frac 1{k-1}}\le \frac{k-1}{k}$ and $X_1^{\frac 1{n-2}}\dots X_{n-k-1}^{\frac 1{k}}\le \frac{k}{k+1}$, then
$X_1^{\frac 1{n-1}}\dots X_{n-k}^{\frac 1{k}}\le \frac{k}{k+1}$.
Raise the first inequality to the power $\frac{k-1}{k}$, the second to the power $\frac 1{k(k+1}$, multiply, use the fact that $X^t$ is decreasing in $t$ for $0<X<1$, and apply Claim 4.
Claim 6. 
$$
P_n\ge \frac{\gamma_n}{\gamma_{n-1}}P_{n-1}-(1-\frac 2n)^{n-1}2^{-(n-1)}
$$
Claim 5 allows one to conclude that replacing a sample $X_1,X_2,\dots$ satisfying the desired set of inequalities for $n-1$ instead of $n$ by $X_1'=\frac{\gamma_n}{\gamma_{n-1}}X_1, X_2'=X_2, X_3'=X_3,\dots$, we get a sample satisfying all the desired inequalities for $n$ except, perhaps, the very last one. Thus, the probability $Q_n$ that all inequalities except the last one are satisfied is at least $\frac{\gamma_n}{\gamma_{n-1}}P_{n-1}$ (the factor comes from the bound for the density of $X_1'$ relative to the density of $X_1$). What we need to subtract is at most the probability that the minimum of $n-1$ independent random variables uniformly distributed on $[0,1]$ is at least $1/2$ while their maximum is at most $\frac{n-1}n$. This is the last term in the inequality.
Claim 7. We have
$$
\frac{n}{n+1}P_n\ge \frac{n}{n+1}\frac{n}{n-1}\frac{\gamma_n}{\gamma_{n-1}}\frac{n-1}n P_{n-1}-\frac{n}{n+1}(1-\frac 2n)^{n-1}2^{-(n-1)}
$$  
This is just Claim 6 rewritten.
Assuming that $\frac{n-1}nP_{n-1}\ge\frac 5{16}$, we conclude that we will be able to run the induction if 
$$
[\frac{n}{n+1}(1+\frac{1}{n(n-2)})^{n-2}-1]*\frac{5}{16}\ge \frac{n}{n+1}(1-\frac 2n)^{n-1}2^{-(n-1)}
$$
Using the first three terms in the Newton's formula on the left, we see that to pass from $n-1$ to $n$, it is enough to have
$$
\frac{5}{16}\frac{n-3}{2n^2(n-2)}\ge (1-\frac 2n)^{n-1}2^{-(n-1)}
$$
Claim 8. This inequality holds for $n\ge 7$.
For $n=7$, it reduces to $8\cdot 7^4>5^6$, i.e., $(2-\frac 1{25})^2>4-\frac 78$, which is fairly obvious by Bernoulli. It remains to note that when $n$ goes up by $1$, the left hand side is multiplied by at least $(\frac 78)^2$\ge $\frac 34$ while the right hand side is multiplied by $1/2$ because of the power of $2$ and at most $(1-\frac{n-1}{n(n-2})^{-1}\le \frac{35}{29}$, which is the Bernoulli bound for the factor by which $(1-\frac 2n)^{n-1}$ differs from the decreasing sequence $(\frac{n-1}{n})^2(n-1)$. Now, $\frac 12\cdot\frac {35}{29}\le \frac{18}{29}\le \frac 34$, so we can run the induction.
The most difficult (=time and effort consuming) part of this proof is 
Claim 9. $P_1$ through $P_6$ are fine for the base of induction. 
I managed to compute $P_4$ by hand without an arithmetic error, but not $P_5$. That's why I would still prefer to see an argument that doesn't require a direct computation of any value beyond $P_3$.<|endoftext|>
TITLE: Categorifying the equality of product and coproduct of symmetric functions
QUESTION [15 upvotes]: Littlewood-Richardson coefficients are both multiplicities of $GL_n$ tensor products, and of restrictions of $GL_{m+n}$ representations to $GL_m \times GL_n$. I want to turn this equality of numbers into an equality of maps of vector spaces, and generalize it to quantum groups. In more detail:
$\def\Part{\mathrm{Part}} \def\Hom{\mathrm{Hom}}$Let $\Part$ be the abelian category whose objects are collections of finite dimensional vectors spaces $(W_{\lambda})$, one for each partition $\lambda$, such that all but finitely many of the $W_{\lambda}$ are $0$. Here 
$$\Hom_{\Part}((W_{\lambda}), (X_{\lambda})) = \bigoplus_{\lambda} \Hom_{\mathrm{Vect}}(W_{\lambda}, X_{\lambda})$$
and composition is defined in the obvious manner.
I'll write $[\lambda]$ for the element of $\Part$ which is $\mathbb{C}$ at $\lambda$ and $0$ everywhere else.
There are two potential tensor structures on $\Part$, which I will call $\otimes^T$ (for tensor) and $\otimes^R$ (for restriction).
Let $V_{\lambda}(n)$ denote the representation of $GL_n$ with highest weight $\lambda$. In $\otimes^T$,
$$\left( [\lambda] \otimes^T [\mu] \right)_{\nu} = \Hom_{GL_n}(V_{\nu}(n), V_{\lambda}(n) \otimes V_{\mu}(n)) \ \mathrm{for} \ n\gg 0.$$
Part of the claim is that, for $n$ sufficiently large, the vector spaces which we say on the right hand side are isomorphic, and this isomorphism can be chosen to commute with the obvious commutator and associator. The definition of $\otimes^T$ for other objects of $\Part$ is forced by linearity.
Similarly,
$$\left( [\lambda] \otimes^R [\mu] \right)_{\nu} = \Hom_{GL_m \times GL_n}(V_{\lambda}(m) \boxtimes V_{\mu}(n), V_{\nu}(m+n)) \ \mathrm{for}\ m,n \gg 0.$$
I believe it is not too bad to show that these two tensor structures are equivalent (including the commutator and associator) by identifying them both with the same thing in the representation theory of the symmetric group, through Schur-Weyl duality, although I haven't actually done this.

I think the same statement is true for
  $GL_n$ quantized at generic $q$. Am I right? Does
  someone know a reference for this?

Actually, what I really want is the corresponding statement in the category of crystals, so "at $q=0$". What this turns into concretely is that there is a bijection between  

$\lambda$-dominant tableaux of shape $\mu$ and content $\nu - \lambda$ and  
Yamnouchi tableaux of shape $\nu/\lambda$ and content $\mu$  

which commutes with certain jdt moves. I need certain specific cases of this, and I have now checked most of the cases I need by hand -- but I'd much rather cite a general theorem about the category of crystals than write out a lot of tableaux bijections. 
Thanks!

REPLY [5 votes]: I recently discovered "The Robinson-Schensted-Knuth" correspondence and the bijections of commutativity and associativity" by Danilov and Koshevoy (2008). It explains all the points described in the rest of this answer in far more detail and care. In summary, everything is known for $q=0$. I would still be interested to hear answers for general $q$.

A lot, perhaps all, of the crystal case is in Pak and Vallejo's papers PV1 and PV2 (full citations below):
Pak and Vallejo always work with high weight elements of the crystal, also known as Yamanouchi tableaux or tableaux with lattice reading word. I've been finding that everything is much prettier if we also work with the low weight elements. In terms of reading words, these are the ones where the reversal of the reading word has the anti-lattice property.
Let $LR_{+}(\alpha/\beta, \gamma)$ denote high-weight elements of shape $\alpha/\beta$ and content $\gamma$, and let $LR_{-}$ denote the low-weight ones. Let $\alpha \circ \beta$ denote the concatenation of $\alpha$ and $\beta$ (see Figure 1 in PV2). Note that, if $T$ is high weight of shape $\alpha \circ \beta$, for straight shapes $\alpha$ and $\beta$ then the $\alpha$ part is canonical (as PV2 uses the term), so we identify $LR_{+}(\lambda \circ \mu, \nu)$ with certain tableaux of shape $\mu$. These are the so-called $\lambda$-dominant tableaux of shape $\mu$ and content $\nu - \mu$. Similarly, $LR_{-}(\lambda \circ \mu, \nu)$ can be identified with certain tableaux of shape $\mu$.
Let $CF_{+}(\lambda, \mu, \nu)$ and $CF_{-}(\lambda, \mu, \nu)$ denote the tableaux which are in bijection with $LR_{+}(\lambda \circ \mu, \nu)$ and $LR_{-}(\lambda \circ \mu, \nu)$ as in the above paragraph. The latter is probably related somehow to $CF^{\ast}$ in PV2, but I haven't figured out how.
Let $HIVE(\lambda, \mu, \nu)$ denote hives whose boundary in cyclic order is $\lambda$, $\mu$, $-\nu^{ast}$. (Here $(\nu_1, \nu_2, \ldots, \nu_n)^{\ast} = (-\nu_n, \ldots, -\nu_2, - \nu_1)$.)
There are easy bijections (PV1, or simple extrapolation from the maps in PV1) between the following:

$HIVES(\lambda, \mu, \nu)$
$LR_{+}(\nu/\lambda, \mu)$
$LR_{-}(\nu/\mu, \lambda)$
$CF_{-}(\lambda, \mu, \nu)$
$CF_{+}(\lambda, \mu, \nu)$

One can switch $\lambda$ and $\mu$ to get five more sets in easy bijection. 
The commutor: The main result of HK is to construct a bijection between $HIVES(\lambda, \mu, \nu)$ and $HIVES(\mu, \lambda, \nu)$ and verify that it is consistent with an $\otimes^T$ commutor they call the cactus commutor. 
In fact, I think the cactus commutor also appears in PV2 as the map $\rho_2$, although they don't label it as that. Let $\omega$ be the map on a crystal that applies $w_0$ to the content and reverses the crystal operators. The definition of the cactus commutor is: $u \otimes v \mapsto \omega
(\omega(v) \otimes \omega(u))$. Now, if $u \otimes v \in LR_{+}(\lambda \circ \mu, \nu)$, then $\omega(v) \otimes \omega(u)$ will be in $LR_{-}(\mu \circ \lambda, \nu)$. When, as in this case, $\lambda$ and $\mu$ have straight shapes, $\omega$ is just the Shutzenberger operator. Now, remember that only the tableaux of shape $\mu$ is of interest in $LR_{+}(\lambda \circ \mu, \nu)$ and $LR_{-}(\mu \circ \lambda, \nu)$, so what we are looking at is the Schutzenberger operator from $CF_{+}(\lambda, \mu, \nu)$ to $CF_{-}(\mu, \lambda, \nu)$. 
I suspect that the outer application of $\omega$ is what takes us from $CF_{-}(\mu, \lambda, \nu)$ to $CF_{+}(\mu, \lambda, \nu)$
The composition
$$LR_{+}(\nu/\lambda, \mu) \to CF_{+}(\mu, \lambda, \nu) \overset{\omega}{\longrightarrow} CF_{-}(\lambda, \mu, \nu) \to LR_{+}(\nu/\mu, \lambda)$$
is what PV2 call $\rho_2$.
In the meanwhile, PV2 also have a commutor for $\otimes^R$: The map they call $\rho_1$, which goes from $LR_{+}(\nu/\lambda, \mu)$ to $LR_{+}(\nu/\mu, \lambda)$ is what I would consider the most natural possible commutor for $\otimes^R$.
According to footnote 6 in PV2, the equality of $\rho_1$ and $\rho_2$ is the main result of 
DK.
The associator PV2 also discusses associators, under the name "octahedron maps". HK show that the HIVE-based associator, introduced in KTW, is isomorphic to the $\otimes^T$ associator. 
Some experimentation suggests that the compatible thing on the $\otimes^R$ side is a map
$$\bigcup_{\rho} LR_{+}(\kappa, \lambda, \rho) \times LR_{-}(\rho, \mu, \nu) \longleftrightarrow \bigcup_{\sigma} LR_{+}(\kappa, \sigma, \nu) \times LR_{-}(\lambda, \mu, \sigma)$$
and that this map should be the map PV2 call $\zeta$ -- the "tableaux switching" map. Section 7.3 of PV2 produces a similar map using high weight tableaux everywhere, with four uses of $\zeta$. I think that three of those uses are just to switch between $LR_{+}$ and $LR_{-}$ as needed. 
Conjecture 3 of PV2 is that their associator and the HIVE based associator match. It is not clear to me whether or not their footnote 8 is claiming that HK prove conjecture 3 but, if they do, I don't see it.
PV1 Pak and Vallejo, Combinatorics and geometry of Littlewood-Richardson cones,
European J. Combin. 26 (2005), no. 6, 995–1008
PV2 Pak and Vallejo, Reductions of Young tableau bijections,
SIAM J. Discrete Math. 24 (2010), no. 1, 113–145
HK Henriques and Kamnitzer, The octahedron recurrence and $\mathfrak{gl}_n$ crystals
Adv. Math. 206 (2006), no. 1, 211–249
DK Danilov and Koshevoy Massifs and the combinatorics of Young tableaux,
Uspekhi Mat. Nauk 60 (2005), no. 2(362), 79--142; translation in
Russian Math. Surveys 60 (2005), no. 2, 269–334<|endoftext|>
TITLE: Terminology Concerning Oriented Simplicial Complexes
QUESTION [9 upvotes]: An oriented simplicial complex is a simplicial complex K equipped with a partial ordering on its vertices that restricts to a linear ordering on each simplex.  I am wondering if there is a standard name for simplicial maps $f\colon K\to K$ which preserve the partial order on vertices.  I would have liked to call them orientation preserving but I think this is usually used to just mean that the linear ordering on each simplex is preserved up to an even permutation.  I had no luck with Google.
Example: If P is a poset, then the order complex (=nerve) of the poset is naturally oriented by the original ordering on P and any order preserving map on P induces a simplicial map preserving orientation in the strong sense I described above.  In particular, after a barycentric subdivision one always can get my property.
Also an oriented simplicial complex can be viewed as a simplicial set in a natural way but not all simplicial morphisms translate into morphisms of simplicial sets.  The ones I am considering do.


Question:  Is there a standard terminology for a simplicial map on an oriented simplicial complex that preserves the ordering on the vertices?

REPLY [5 votes]: A simplicial complex with partially ordered vertices such that the vertex set of each simplex is a chain of the poset is called an ordered simplicial complex. This avoids the confusion with orientability, etc. The terminology is not new, you can find it in this paper from 1969. So, your maps are just maps of ordered simplicial complexes. Regarding the paragraph just before your question, if a map $f:K\to K$ of ordered simplicial complexes is injective on simplices, then it can be realized as a map of $\Delta$-sets. I think this is covered Friedman's wonderful notes, but it's been a while since I went through those.<|endoftext|>
TITLE: Fibrations with isomorphic Leray-Serre spectral sequences and non-isomorphic cohomology ? 
QUESTION [9 upvotes]: Are there fibrations $F_i \to X_i \to B_i$ $(i=1,2)$ with path-connected bases $B_i$ and connected fibres $F_i$ such that their corresponding Leray-Serre spectral sequences (integral coefficients) are isomorphic as abelian groups, i.e. 
$$E_r^{p,q}(1) \cong E_r^{p,q}(2)\;\;\;(2 \le r \le \infty,\;i,j \ge 0)$$
and such that their integral cohomology isn't isomorphic, i.e. there is $p \ge 0$ with 
$$H^p(X_1;\mathbb{Z}) \not\cong H^p(X_2;\mathbb{Z})$$
as abelian groups ?

REPLY [18 votes]: Such an example is given by the pair of fibrations
$$
K(\mathbb{Z}/2,1) \to K(\mathbb{Z},2) \to K(\mathbb{Z},2)
$$
(coming from the Bockstein exact sequence) and by
$$
K(\mathbb{Z}/2,1) \to K(\mathbb{Z}/2,1) \times K(\mathbb{Z},2) \to K(\mathbb{Z},2).
$$
(a trivial fibration).
In both cases, the (cohomological!) Leray-Serre spectral sequence is concentrated in even total degree.  However, the two spaces have differing $H^2$ with integral coefficients ($\mathbb{Z}$ versus $\mathbb{Z} \times \mathbb{Z}/2$).

Added (by Ralph): Here are some more details for the spectral sequences. We know 
$K(\mathbb{Z}/2,1)=\mathbb{R}P^\infty$ and $K(\mathbb{Z},2)=\mathbb{C}P^\infty$ and 
$$H^p(\mathbb{R}P^\infty;\mathbb{Z})=
\begin{cases}
\mathbb{Z} & p=0 \newline \mathbb{Z}_2 & p > 0 \text{ even }\;\;, \newline 0 & p > 0 \text{ odd }
\end{cases}
\hspace{10pt}
H^p(\mathbb{C}P^\infty;M)=
\begin{cases}
M & p> 0 \text{ even} \newline 0 & p > 0 \text{ odd } 
\end{cases}$$
where $\mathbb{Z}_2 := \mathbb{Z}/2$ and $M$ are trivial coefficients. Since $\mathbb{Z}_2$ has only two elements, the coefficient system in the LS spectral sequence of the first fibration is trivial and we obtain for $E_2^{p,q}(1)=H^P(\mathbb{C}P^\infty;H^q(\mathbb{R}P^\infty;\mathbb{Z}))$: 
$$E_2(1)=\;
\begin{array}{cccccccc}
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \newline  
\mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 & \mathbb{Z}_2 & \cdots \newline
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\newline 
\mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 & \mathbb{Z}_2 & 0 & \mathbb{Z}_2 & \cdots\newline
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\newline 
\mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & \cdots\newline
\end{array}$$
Now, for positional reasons, $E_2(1)=E_\infty(1)$. 
As the 2nd fibration is trivial, the coefficient system in its LS spectral sequence is also trivial. Hence both spectral sequences agree (in all terms), while the cohomologies differ: $H^p(\mathbb{C}P^\infty;\mathbb{Z})$ is described above and 
$$H^p(\mathbb{C}P^\infty \times \mathbb{R}P^\infty;\mathbb{Z})=
\begin{cases}\mathbb{Z} \oplus \mathbb{Z}_2^n & p= 2n \newline 0 & p \text{ odd.}  
\end{cases}$$<|endoftext|>
TITLE: Resemblance between Birman-Murakami-Wenzl algebra representations and the Lawrence-Krammer representations
QUESTION [7 upvotes]: At the end of Stephen Bigelow's paper "Braid Groups are Linear", he mentioned that there is a striking resemblance between the matrices of the Lawrence-Krammer (LK) representations and "those of a certain irreducible summand" of the Birman-Murakami-Wenzl (BMW) representations. It is also mentioned that the explicit correspondence was worked out by a paper of Zinno.
My question: is anyone aware if a computation exists for the matrices of any other irreducible summands of the BMW representations. In other words, I would actually like to see the explicit matrix for some summands of the BMW representation. 
Thanks!

REPLY [7 votes]: I have not seen this written down but this can certainly be done explicitly. The construction assumes you have explicit matrices for the representations of the symmetric group. Here you have a choice between Young's orthogonal form (which has rational entries) and Specht modules (which have integer entries).
Once you have sorted this out you can construct explicit matrices for the representations of BMW using a Morita equivalence approach. The irreducible representations of the $r$ string algebra are indexed by partitions $\lambda$ with $r-|\lambda|\ge 0$ and even. Let $V(\lambda)$ be the representation of the $|\lambda|$-string symmetric group. let $M$ be the vector space of Brauer diagrams with $r$ points on the top, $|\lambda|$ points on the bottom, with precisely $\lambda$ through strings and such that no two through strings intersect.
Then the $r$-string BMW algebra acts on $V(\lambda)\otimes M$.
More detail:
Here is the theory behind the construction. First I realised I am thinking of the Brauer category whereas OP asks about BMW. This basically means the symmetric group algebra should be replaced by the Hecke algebra.
The Brauer category has an increasing sequence of ideals. The ideal $I(k)$ is spanned by diagrams with at most $k$ through strings. The quotient $I(k)/I(k-1)$ is spanned by diagrams with precisely $k$ through strings. Then the key idea is that the category $I(k)/I(k-1)$ is Morita equivalent to the group algebra of the $k$-string symmetric group.
The functor from representations of the $k$-string symmetric group to representations of the $k+2c$ string Brauer algebra has a straightforward explicit description.
Even more detail:
The Morita equivalence I have in mind is an extension of the usual equivalence. Let $A$ be a finite dimensional unital algebra and $e\in A$ an idempotent. Let $J$ be the ideal $AeA$
and $B$ the unital algebra $eAe$. Then $J$ and $B$ are Morita equivalent since we have the bimodules $eA$ and $Ae$. Here $J$ is not unital but does have the weaker property that $JJ=J$. Then a right $J$-module $M$ is required to satisfy $MJ=M$. Something like this but without mentioning Morita theory is in Sandy Green's SLN text on Schur algebras.<|endoftext|>
TITLE: A fact about finite-dimensional manifolds I fear does not hold for Frechet manifolds (what's new?)
QUESTION [5 upvotes]: Let $M$ be a manifold equipped with a pair of surjective submersions $N_1 \stackrel{p_1}{\leftarrow} M \stackrel{p_2}{\rightarrow} N_2$ where $dim N_1 = dim N_2 = n$. Then we can find, for any point $m\in M$, a chart $U_1$ around $p_1(m)$ and a local section $s\colon U_1 \to M$ of $p_1$ such that $s(p_1(m)) = m$ and $s(U_1)$ is transverse to the fibres of $p_2$. In thinking about this we can clearly reduce to the case $N_1 = N_2 = \mathbb{R}^n$. We can even restrict attention to $M = \mathbb{R}^m$, and then it becomes a problem of linear algebra, namely finding a basis on $\mathbb{R}^m$ for which the submersions are both projections onto $n$ coordinates (we already know this separately).
I suspect that for infinite-dimensional vector spaces, and Frechet spaces in particular (really anything above Hilbert spaces in the usual hierarchy) this sort of result will not hold, and so for Frechet manifolds one cannot construct the analogous local section. 
In more detail, I'm fairly sure that given a diagram of Frechet spaces
$$V_1 \stackrel{pr_1}{\leftarrow} V_1 \times F_1 \simeq W_1 \simeq V_2 \times F_2 \stackrel{pr_1}{\to} V_2$$
where $V_1$ is known to be isomorphic to $V_2$, one cannot in general find a (nonlinear?) map $F\colon V_1 \to V_2\times F_2$ (which is a section, and passes through the origin) such that $pr_1\circ F\colon V_1 \to V_2$ is injective an isomorphism. However, I'd like to see a counterexample (or, if I'm wrong, a proof that we can do it!). (EDIT: the map might need be be non-linear in this case because the definition of submersions doesn't use tangent spaces like in the finite-dimensional case. But linear would be good.)
VERSION 2: well it turns out that I have, in addition to the $V_i$s being isomorphic, I have $F_1$ isomorphic to $F_2$. Kudos to Andrew for guessing this would be the case for the application I have in mind.

REPLY [6 votes]: (I'm strongly tempted to leave this as a comment, as it's a rather trivial counter-example and it may result in the question being changed to avoid it, but I want to take advantage of the better formatting of answers.)
Here's a trivial counter-example.  Let $V$ be a space which is isomorphic to one of its hyperplanes, in that we have an isomorphism $V \cong V \oplus \mathbb{R}$ (see shift space on the nLab).  There are lots of these around and most of the "usual suspects" for locally convex topological vector spaces have this property (especially including spaces like $C^\infty(\mathbb{R},\mathbb{R}^n)$).  We take $V_1 = V_2 = W_1 = V$, $F_1 = \lbrace 0\rbrace$, and $F_2 = \mathbb{R}$.  The isomorphisms as $W_1 = V_1$ on the left, and $W_1 = V \cong \mathbb{R} \oplus V = \mathbb{R} \oplus V_2$ on the right.  Then $\operatorname{pr}_1 \colon W_1 \to V_1$ is the identity and $\operatorname{pr}_1 \colon W_1 \to V_2$ is the composition $W_1 = V \cong \mathbb{R} \oplus V \to V$.
The only option for the section is the identity, since $W_1 \to V_1$ is an isomorphism, whence the induced map $V_1 \to V_2$ is the map $V_1 = V \cong \mathbb{R} \oplus V \to V$ which is not injective.<|endoftext|>
TITLE: For what metrics are circles solutions of the isoperimetric problem?
QUESTION [11 upvotes]: A classical result is that solutions of the isoperimetric problem on the plane, the sphere, and the hyperbolic plane are circles. Are there any other Riemannian metrics on these spaces that share this property or does this characterize metrics of constant curvature ?
I should add that by circles I mean really standard circles, not geodesic circles for the new metric.

REPLY [2 votes]: It is rather a comment and a question to the author than the  answer (but it is too long for the comment window). 
What is the   (solutions of the) isoperimetric   problem? If it is   a closed curve of a  fixed length such that it  bounds the region of a maximal volume, or  a shortest closed curve that bounds the region of certain  fixed  volume, I believe that  the solutions of isoperimetric problem are very rare. 
In order to confirm my suggestion, I would like to mention that for  metrics of constant positive curvature  the length of the circle bounding the ball of volume 1 is less than that of  for metrics of constant negative curvature.  This suggests that for the sphere  of revolution  such that the maximums of the curvature are  the poles 
the isoperimentric circles (of sufficiently small length) 
are circles (= boundary of the balls) centered at the poles. Then, in a certain coordinate system they are usual circles.   
(To see that the solutions of the isoperimetric problem are rare you could also combine two statements from the answer of Robert Bryant: curves of constant  geodesic curvature  are usually  not closed and isoperimetric curves must have constant geodesic curvature.)<|endoftext|>
TITLE: Elementary Markov Chain Question
QUESTION [5 upvotes]: Are any general conditions known on a finite transition nxn matrix that ensure that there exists at least one mth root which is also a transition matrix? It is easy to construct a 3x3 , diagonally dominant, transition matrix with distinct, real eigenvalues but no transition matrix, square root.

REPLY [5 votes]: No, as far as I know it is an open problem; there are no simple general conditions.
You can find some work on the problem, including sufficient criteria, counterexamples and some plots that give heuristics, on http://www.maths.manchester.ac.uk/~lijing/papers/hili11.pdf<|endoftext|>
TITLE: Convexity of spectral radius of Markov operators, Random walks on non-amenable groups
QUESTION [17 upvotes]: Let $P_1,P_2$ denote  stochastic transition matrices on a countable set $I$. 
Consider  $P_1,P_2$ as operators on $\ell^2(I)$ given by multiplication.
Question
Under which conditions can we show that for $t\in (0,1)$,
$$\rho(tP_1+(1-t)P_2)\le t\rho(P_1)+(1-t)\rho(P_2),$$ where $\rho(\cdot )$ denotes the spectral radius? (Of course, it would be nice to consider this problem also for more general classes of Markov operators acting on some function space.)
My hope is that the strong assumption that both matrices are stochastic, is enough to prove the inequality.
Example
My favorite example is the case where $I$ is the free group $G$ of rank $d\ge2$ and  $P_1,P_2$ are given by $P_i(g,g'):=\mu_i(  g^{-1}g'  )$, where $\mu_i$ is some measure on a set of generators of $G$.  In this case, the spectral radius of $P_i$ is strictly less than one, since the group is non-amenable (H. Kesten 1959). 
I don't know if the above inequality holds in this setting.
(Note that non-zero constant functions are not in $\ell^2(G)$.) 
An additional assumption
I'm fine with the additional assumption that $P_1$ and $P_2$ have zeros at the same places. So, regarding the previous example, the measures $\mu_i$ have the same support. A further simplification would be to assume that one of measures is equidistributed on its support. 
Related results
If $P_1,P_2$ commute, then the inequality holds. This is too restrictive.
If $P_1,P_2$ are selfadjoint, then the spectral radius can be replaced by the norm, and the inequality holds. I'm not interested in this case. I don't want to assume that $P_i$ is reversible with respect to some measure.
Finite matrices with real eigenvalues
In the following paper, Lax proved that the spectral radius on the set of $n\times n$ matrices with real eigenvalues is convex.
Kingman proved that spectral radius is log-convex on the space of finite dimensional non-negative matrices.
Cohen proved that the spectral radius is a convex function of the diagonal elements on the space of finite dimensional non-negative matrices. There are generalisations for multiplication operators by Kato.
On amenability
If the group is $G$ is amenable in the previous example, then for a large class of transition operators, the spectral radius is equal to one (e.g. uniform irreducibility and invariant measure bounded away from zero and infinity). Hence, under these conditions, all the spectral radii in the above inequality are equal to one. So, equality holds.
Example
As an example, let $G$=$\mathbb{Z}$ and consider $P_1,P_2$ corresponding to the random walks, which go one step to the left resp. right with probabilities $p_i$ resp. $q_i$, $p_i+q_i=1$, $i=1,2$.
References
LAX, P.  D.  Differential  equations, difference  equations  and matrix  theory.  Comm.  Pure Appl.  Math.  11  (1958),  175-194.
Kingman, J.F.C.: A convexity property of positive matrices. Quart. J. Math. Oxford (2) 12,
283-284 (1961)
Cohen, J.E.: Convexity of the dominant eigenvalue of an essentially nonnegative matrix. Proc. Amer. Math. Soc. 81, 657-658 (1981)
Kato: Superconvexity of the Spectral Radius, and Convexity of the Spectral Bound and the Type, Mathematische Zeitschrift, (1982)
Using commutativity
There are several papers of M.Zima about related properties for positive operators on partially ordered Banach spaces (see below). In there, some commutativity is required to prove the inequality in question.
I don't want to assume commutativity. 
M.Zima, A theorem on the spectral radius of the sum of two operators and its applications,
Bull. Austral. Math. Soc. 48 (1993), 427{434. MR 94j:47006
M.Zima, On the local spectral radius in partially ordered Banach spaces, Czechoslovak Math.
J. 49 (1999), 835{841. MR 2001m:47011

REPLY [4 votes]: I have the following counter example:
Let $I=Z_2\times G$ with $G$ a free groupe of rank $d$ and we will take $d$ large. $P_1$ is as follow 
$$P_1 \begin{cases}(1,w)\rightarrow (0,w):\forall w\in G, \text{ with probability }1 \\ (0,w)\rightarrow (1,w'): \text{ where } w' \text{ is like the randow walk on }G \text{ from }w \end{cases}$$
$$P_2 \begin{cases}(0,w)\rightarrow (1,w):\forall w\in G, \text{ with probability }1 \\ (1,w)\rightarrow (0,w'): \text{ where } w' \text{ is like the randow walk on }G \text{ from }w \end{cases}$$
It is easy to check that $P_1^2$ (resp $P_2^2 $) is the random walk on $G$ with spectral radius $\rho_d$. Therefore $\rho(P_1)=\rho(P_2)=\sqrt{\rho_d}$.
One can also check that $$\frac{1}{2}(P_1+P_2)(1_{(1,w_0)}+1_{(0,w_0)})=\frac{1}{2}(1_{(1,w_0)}+1_{(0,w_0)})+\text{a positive vector}$$
And therefore $\rho(\frac{1}{2}(P_1+P_2))\geq \frac{1}{2}$.
We can conclude taking $d\rightarrow \infty$ , $\rho_d\rightarrow 0$<|endoftext|>
TITLE: Primes and $x^2+2y^2+4z^2$
QUESTION [15 upvotes]: A few months ago, I have asked a question about primes represented by ternary quadratic forms. I got two wonderful answers, which showed me how the theory was way richer and more complex that I naively expected from the case of binary forms. I
 was also given a lot of references
to read. They were interesting but unfortunately I got submerged by the amount of theory to learn and by other things to do, and I still don't have a clear overview of how the theory works.
I hope it is not abusing the patience of the readers of this site, and in particular of the experts in that theory, to come back again with a new question, this time 
by stating a very specific result that I have obtained as a by-product of some study of modular forms, and to ask
whether this theorem can be proved using the general theory of ternary quadratic forms (which I very much believe), and if it can, how?

Proposition : a prime $p$ is represented an odd number of times by the form $x^2+2y^2+4z^2$,
  with $x,y,z$ odd positive integers, if and only if $p \equiv 7 \pmod{16}$

I hope the answer will lead me to the part of this large theory I need to learn to understand better this kind of questions (I have (infinitely) many results of the above type, and I'd like to understand how they fit in the general theory) . Also, if someone finds an elementary argument for proving the proposition, I'd be quite interested too.

REPLY [5 votes]: Here's a simpler argument. We may assume p is 7 mod 8. Let N be the number of triples of squares (r,s,u) with r+2s+4u=p. We will show that N is odd if p is 7 mod 16 and even if p is 15 mod 16. Let M be (1/64)(the number of representations of p by xx+yy+zz+tt). Jacobi's 4 square theorem (which has elementary proofs using quaternions for example) shows that M=(p+1)/8. So it suffices to show that M and N have the same parity. Now if p=xx+yy+zz+tt, then just one of x,y,z,t is even. So M=(1/16)(the number of representations of p by xx+yy+zz+4tt). In other words, M is the number of ordered quadruples of squares (r,s,t,u) with r+s+t+4u=p. Now the involution (r,s,t,u)-->(r,t,s,u) on this set of quadruples has the N fixed points (r,s,s,u), giving the result.<|endoftext|>
TITLE: ZF +  the reals are the countable union of countable sets consistent
QUESTION [9 upvotes]: Hi, i know that it is consistent with ZF without choice that the reals are the countable union of countable sets.
Is there any good reference to read a proof?
Thanks

REPLY [13 votes]: T. Jech, The Axiom of Choice. This particular proof appears in Chapter 10.
Essentially, the forcing goes through collapsing all the $\aleph_n$ (for finite $n$) to be countable, so in the full generic extension $\aleph_\omega$ of the ground model is countable too, but if we take permutations based on conditions based on finitely many collapses, then $\aleph_\omega$ of the ground model is not collapsed, and thus it becomes $\aleph_1$. 
It is not difficult to show that if the ground model satisfied GCH then every real number in this symmetric extension came from a collapse of some $\aleph_n$, and those are countable. So we have that the real numbers are a countable union of countable sets.<|endoftext|>
TITLE: Milnor patching for schemes
QUESTION [7 upvotes]: Let $R_1,R_2$, and $S$ be commutative rings with maps $R_1,R_2 \to S$ and form the fiber product $R = R_1 \times_S R_2$. A well-known theorem of Milnor says that under certain assumptions the category of finitely generated projective $R$-modules is equivalent to a category of "patching data": an object in this category consists of finitely generated projective modules $P_1$ and $P_2$ over $R_1$ and $R_2$ respectively, together with an $S$-isomorphism $P_1 \otimes_{R_1} S \cong P_2 \otimes_{R_2} S$. This may be applied, for instance, to derive a Meyer-Vietoris sequence for Picard groups of rings.
My question: does this generalize to pushouts of schemes? For instance, if I am computing the Picard group of a projective nodal cubic, which can be obtained from $\mathbb{P}^1$ by identifying two points, can I apply a Meyer-Vietoris sequence similar to the one that works in the affine case?

REPLY [5 votes]: First of all, you have to show that these pushouts exist at all. More precisely, if $A \hookrightarrow X$ is a closed immersion and $A \to Y$ is an arbitrary morphism, then the pushout $X \cup_A Y$ exists in the category of schemes. Moreover, the underlying ringed space is the pushout of the underlying ringed spaces (thus topologically it's a pushout in the usual sense, and algebraically it is given by a fiber product of structure sheaves); remark that this is not trivial at all. One can prove that the canonical map $Y \to X \cup_A Y$ is a closed immersion. You can find all that in Karl Dahlke's paper "Gluing schemes and a scheme without closed points".
Milnor's Theorem states that the functor $\mathrm{Vect}(X \cup_A Y) \to \mathrm{Vect}(X) \times_{\mathrm{Vect}(A)} \mathrm{Vect}(Y)$ is an equivalence of categories, where the right hand side is the $2$-fiber product and $\mathrm{Vect}(-)$ denotes the category of algebraic vector bundles. At least when everything is affine, but since everything in sight is local, we immediately also get it for arbitrary schemes.<|endoftext|>
TITLE: High-dimensional ribbon knots
QUESTION [8 upvotes]: Let us suppose that we have a ribbon embedding $S^n \rightarrow S^{n+2}$ for $n\geq 3$. Call this knot $K$. By a theorem of Levine (and Trotter for $n=3$ I believe) we know $K$ is unknotted if the complement is homotopy equivalent to $S^1$. The question which I wonder is, does it suffice to show that the fundamental group is $\mathbb{Z}$, given the fact that $K$ is ribbon? For $n=2$ this would be true, as Freedman mentions (that having infinite cyclic fundamental group implies being homotopy equivalent to a circle) but I am not terribly familiar with the high-dimensional case.

REPLY [3 votes]: I found a paper which discusses this issue: 
Ribbon knots and ribbon disks from Asano, Marumoto, and Yanagawa. They establish that for $n\geq 3$ a ribbon knot with infinite cyclic fundamental group is trivial.
http://ir.library.osaka-u.ac.jp/metadb/up/LIBOJMK01/ojm18_01_12.pdf<|endoftext|>
TITLE: History of provably total functions of a theory
QUESTION [5 upvotes]: Provably total functions of an arithmetical theory is one of the tools used in proof theoretic analysis of theories. 
I am looking for early history of its development. In particular,

Where was the first uses of this concept?
  Who coined the term "provably total functions"?

REPLY [2 votes]: One of the references I would recommend is P. Odifreddi's Classical Recursion Theory Volume II. On Page 324-326, it contains many references about the development of provably total functions (if you are seeking for the development within the Peano Arithmetic).<|endoftext|>
TITLE: Analog of "Spin" Chern-Simons Theory
QUESTION [9 upvotes]: 3-dimensional Chern-Simons theories, with compact gauge group $G$, are determined by $H^4(BG)$. Looking at $U(1)$, with generator $c_1^2\in H^4(BU(1))=\mathbb{Z}$ for 1st Chern class $c_1$, there are infinitely many topological theories indexed by the level $k\in\mathbb{Z}$ (for class $k\cdot c_1^2$).  However, we can define other topological theories when an additional structure is put on the manifold. For a spin structure, this new "spin" U(1) Chern-Simons theory (in 3-dimensions) happens to exist for each half-integer $k\in\frac{1}{2}\mathbb{Z}$. Main reason behind this is that for your 3-manifold $Y$, a spin 4-manifold $X$ which bounds it will have even intersection-pairing.
[[Edit]]: To expand on this, my source is Dijkgraaf/Witten. The "spin" topological theory is just taking your topological theory but requiring a choice of spin-structure. Given our functional $S=\frac{k}{4\pi^2}\int_XF^2$ modulo 1, the spin structure makes this $=k\cdot(\text{even number})$, and so we can now enlarge the levels to $\frac{1}{2}\mathbb{Z}$.
My questions are (if they make sense): Is there a structure on your manifold to have levels in $\frac{1}{n}\mathbb{Z}$ for some $n\ge 3$? Can you formulate "spin-c" Chern Simons?
From a physical-perspective I would guess no to the first one, because at least in the case of half-integers it corresponds to fermions in nature (and fermions/bosons are what we live with). But then again there is the Fractional Quantum Hall Effect...

REPLY [8 votes]: As David Roberts mentions in the comments above, indeed as one climbs up the Whitehead tower of the orthogonal group the higher Pontraygin classes become divisible by higher factors.
Notably in the next step, if you demand that also $\tfrac{1}{2}p_1$ vanishes, hence that you have not only a Spin-structure but even a String structure, then the second Pontryagian class becomes divisble by 6. 
In this case one can also divide the Lagrangian for 7-dimensional Chern-Simons theory by 6. This is discussed in 
Fiorenza, Sati, Schreiber, Multiple M5-branes, String 2-connections, and 7d nonabelian Chern-Simons theory.<|endoftext|>
TITLE: Applications of Liouville's theorem
QUESTION [9 upvotes]: I'm looking for "nice" applications of Liouville's theorem (every bounded entire map is constant) outside the area of complex analysis.
An example of what I'm not looking for : a non-constant entire function has dense image (this is essentially a corollary).
An example of the kind of thing I'm looking for : a complex matrix whose conjugacy class is bounded must be a homothety (if $A$ is such a matrix and $B$ is an other matrix, then $z \mapsto e^{-z B} A e^{z B}$ is entire and bounded hence constant, but its derivative at $0$ is $[A,B]$ : thus $[A,B]=0$).
In a similar vein : a subalgebra of $M_n (\mathbb{C})$ on which the spectral radius is submultiplicative is simultaneously triangularizable.

REPLY [9 votes]: There are many cool applications when combined with the uniformization theorem.  Not sure if you count them as "complex analysis" or not - you could really think of them as algebraic geometry.  For example:
The only meromorphic functions $f$ and $g$ satisfying $f^n+g^n = 1$ for $n>3$ are constant.  
Proof sketch:  If there were such functions they would define a map $F$ from the plane to the Riemann surface $x^n+y^n = 1$.  We can compute the genus of this guy using Hurwitz - it is $(n-1)(n-2)/2$.  So for $n>3$ the genus is bigger than two - the corresponding Riemann surface has negative curvature, and by uniformization it has the disk as its holomorphic universal cover. But then,  $F$ would factor through the disk, and so by Louville F had to be constant.
Note:
for n=1 the equation has lots of solutions
for n=2 sine and cosine work (for instance)
for n=3 the Riemann surface has genus 1, and so its universal cover is the plane.  You can find explicit solutions to the equation by using theta functions.
EDIT: another very important example of the same sort of reasoning is the little Picard theorem.  At a high level, the proof just says that the plane with 2 points deleted has a holomorphic universal covering by the disk, so any entire function which misses 2 points factors through the disk - Louville gives the contradiction.<|endoftext|>
TITLE: A graded ring $R$ is graded-local iff $R_0$ is a local ring?
QUESTION [16 upvotes]: I asked this question some months ago on math.stackexchange.com:
https://math.stackexchange.com/questions/126810/a-graded-ring-r-is-graded-local-iff-r-0-is-a-local-ring
It would be great (for me) to get it resolved.
Let $R$ be a non-zero $\bf{Z}$-graded ring (with no commutativity assumptions whatsoever). Recall that $R$ is "graded-local" if the following equivalent conditions hold:

$R$ has a unique homogeneous left ideal that is maximal among the proper homogeneous left ideals;
$R$ has a unique homogeneous right ideal that is maximal among the proper homogeneous right ideals; 
The sum of two homogeneous non-units is again a non-unit.

Note, in particular, that (3) tells us that $R_0$, the ring of degree 0 elements, is a local ring.
But it seems to me that the converse is true. That is, it seems to me that a $\bf{Z}$-graded ring $R$ is "graded-local" if and only if $R_0$ is a local ring. Since this seems a bit suspicious, I have come here to find out if this is really the case.
Here is my argument:
Let $J^g(R)$ denote the intersection of all the homogeneous left ideals that are maximal among the proper homogeneous left ideals. (This is the "graded Jacobson radical". Note that there should be at least one such "maximal homogeneous left ideal" because $R\neq 0$). As an intersection of proper homogeneous left ideals, $J^g(R$) is a proper homogeneous left ideal. I claim that it is maximal among the proper homogeneous left ideals, and hence is the unique such "maximal homogeneous left ideal".
Consider a homogeneous left ideal $I$ with $J^g(R) \subsetneq I$. Since $I$ is homogeneous there exists a homogeneous element $a\in I$ with $a \not\in J^g(R)$. Since $a$ is not in $J^g(R)$ there exists a maximal homogeneous left ideal $\frak{m}$ with $a \not \in \frak{m}$. Well, $R a + \frak{m}$ is a homogeneous left ideal, so by maximality of $\frak{m}$, $1=ra + m$ for some $r\in R$ and $m \in \frak{m}$. Taking the degree 0 components we have $1=r_0a + m_0$ where $m_0$ is a degree zero element in $\frak{m}$ and $r_0a$ is also homogeneous of degree 0. Thus, since $R_0$ is local, either $m_0$ or $r_0a$ is a unit. But $m_0$ cannot be a unit (since it would contradict the properness of $\frak{m}$). Hence $r_0a$ is a unit. In particular, $r_0a$ is left invertible, and thus $a$ is also left invertible. Thus, $I=R$. We conclude that $J^g(R)$ is a "maximal homogeneous left ideal" and hence is the unique such.
Can you see any problems with this? Is it really true that a $\bf{Z}$-graded ring $R$ is graded-local iff $R_0$ is local?
Note: It is clear that an $\bf{N}$-graded ring $R$ is "graded local" iff $R_0$ is local. (If $m_0$ is the unique maximal left ideal of $R_0$ then $m := m_0 \oplus \bigoplus_{d>0} R_d$ is the unique maximal left ideal of $R$.) But this "easy" argument for $\bf{N}$-graded rings doesn't work for $\bf{Z}$-graded rings.
If my claim is false, then a follow up question is: to show that a (not necessarily commutative) $\bf{Z}$-graded ring is graded-local, does it suffice to show that the sum of two homogeneous non-units of the same degree is again a non-unit. This is true for commutative $\bf{Z}$-graded rings, but I'm getting stuck up on the non-commutative ones.
Thank you for reading.
Update
The paper by Huishi Li mentioned below convinces me that the result is true (and that my proof is fine and can be simplified as mentioned in the comments). The paper by Huishi Li hasn't been published (beyond being put on the arXiv) so I still wonder if there is a clearly stated reference for this result in the published literature. I've checked the following three books:

Nastasescu & van Oystaeyen -- Graded and filtered rings and modules (LNM758, 1979)
Nastasescu & van Oystaeyen -- Graded ring theory (1982)
Nastasescu & van Oystaeyen -- Methods of graded rings (LNM1836, 2004)

and the result isn't stated in any of them. The second one mentions that $R$ gr-local implies that $R_0$ is local but doesn't say anything about the converse.
I'll leave the question open for a bit longer to see if anyone else chimes in, and then accept Gjergji's answer. Thanks!

REPLY [2 votes]: An elementary argument seems to work as well, even for monoids as mentioned previously.$\dagger$
Suppose $M$ and $N$ are distinct maximal homogeneous right ideals. Then $M+N=R$, and there exists $m+n=1$ with $m\in M$ and $n\in N$. Because of the grading, the grade zero parts must be such that $m_0+n_0=1$, and because $M$ and $N$ are both proper and homogeneous, neither $m_0$ nor $n_0$ can be units of $R_0$. This implies $R_0$ is not local.
By contrapositive then, we have shown if $R_0$ is local, then $R$ is graded local.
$\dagger$ I convinced myself that there are maximal proper homogenous ideals, and that the sum of homogeneous ideals is again homogeneous. I hope those lemmas were not heat induced delirium.<|endoftext|>
TITLE: What is a (generalized) BN-pair?
QUESTION [11 upvotes]: Let us consider $GL_n(K)$ over a local field $K$. It has standard subgroups $N$ and $B$. $B$ is Iwahori subgroup, $N$ consists of monomial matrices. The pair comes close to a romantic ending, i.e. forming a BN-pair, as one gets a building of affine type out of it. However, it fails axiom (BN2) asking $sBs^{-1}\not = B$. You can also observe other things are out of order: maximal parabolics of different type are conjugate, etc...
What is the set of axioms governing this generalized BN-pair that alows all the usual building geometry to take place? Is there a name and other meaningful examples?
Please, help, I cannot sleep all day because it bothers me greatly.

REPLY [4 votes]: The notion of groups with a BN-pair has been generalized to groups with a root group datum. There is a wonderful paper by Pierre-Emmanuel Caprace and Bertrand Rémy on this topic, which you can download for free:

P.-E. Caprace & B. Rémy, "Groups with a root group datum", Innov. Incidence Geom. 9 (2009), 5-77.

It has a fairly long introduction (about 6 pages) which is really worth reading, containing a lot of information about the history of the different related concepts, and explaining how they are related with the theory of buildings, and so on.<|endoftext|>
TITLE: References about the Grothendieck's way of algebraizing the notions of calculus and differential geometry
QUESTION [8 upvotes]: Hi everyone,
I'm looking for some references about the differential operators on schemes(connection, curvature, etc...).  I am reading the EGA IV 16, but EGA does not treats connection, curvature, etc.... 
Are there any articles/books that deal with the the Grothendieck's way of algebraizing the notions of calculus and differential geometry?
Thank you very much!

REPLY [7 votes]: Here are three references which were/are very helpful to me:

Berthelot, Pierre; Ogus, Arthur: Notes on crystalline cohomology. 

Chapter 2 covers much of what you are looking for.

Berthelot, Pierre: Cohomologie cristalline des schémas de caractéristique p>0. (French) 
Lecture Notes in Mathematics, Vol. 407

Here, also chapter 2 contains many things you are looking for, in a more general setup.

Grothendieck, Alexander: Crystals and the de Rham cohomology of schemes. 1968 Dix Exposés sur la Cohomologie des Schémas pp. 306–358

This doesn't contain many details, but is still very interesting (certainly not only historically).<|endoftext|>
TITLE: Doubling dimension of a Euclidean space
QUESTION [12 upvotes]: The doubling dimension of a metric space $X$ is the smallest positive integer $k$ such that every ball of $X$ can be covered by $2^k$ balls of half the radius.
It is well known that the doubling dimension $d(n)$ of the Euclidean space $\mathbb R^n$ is $O(n)$, which means that there is a constant $C$ such that for large $n$ one has $d(n)\leq Cn$. A posteriori, I can find a new constant $D$ that works for all $n$. I would like to have an explicit description of this new constant. In other words,
Question: What explicit and possibly nice and small constant $D>0$ would guarantee that $d(n)\leq Dn$, for all $n$?
Edit. As observed by Igor Rivin, $D=\log 2$ should be good for $n\geq7$, by a theorem of Verger-Gaugry. Any idea for all $n$? I have to clarify that at the moment I am not interested in the best possible constant, but in some good-looking constant, something to make aesthetically pleasant a certain formula that I found out.
Thank you in advance,
Valerio

REPLY [9 votes]: As shown in this paper,Theorem 1.2,  $D \leq \log 2.$ I remark that this paper came up in my answer to this question, and there is a bug for small $n$ ($n < 7$), but the author's interest was apparently similar to yours, so the large $n$ results should be correct. (the paper is: "Covering a Ball with Smaller Equal Balls in $\mathbb{R}^n," by Jean-Louis Verger-Gaugry)<|endoftext|>
TITLE: Forcing over set theory versus forcing over arithmetic
QUESTION [17 upvotes]: I've been trying to understand better some of the research on forcing over bounded arithmetic and its connections with lower bounds in complexity theory.  For example, Takeuti and Yasumoto have some papers where they prove results such as, for certain $M$ and certain $G$, if $M[G]$ doesn't satisfy Buss's $S_2$, then  NP ≠ co−NP.  Krajíček's book Bounded Arithmetic, Propositional Logic, and Complexity Theory also touches on this topic, showing how to recast some lower-bound proofs in complexity theory as forcing arguments.
These connections are intriguing but I don't understand them well enough yet to be able to make up my mind whether this is mostly a curiosity or whether it has the potential to lead somewhere new and interesting.  One thing that I'd like to know is the degree to which constructions in set theory carry over (or are likely to carry over) to arithmetic.  Perhaps a slightly more precise phrasing is,

Where and how does the parallel between forcing over set theory (ZFC, say) and forcing over arithmetic (or bounded arithmetic) break down?

Another way to put it might be, is there any sketch of a program (even if a wildly optimistic one) for carrying over ideas from set theory to arithmetic in such a way as to prove new lower bounds in complexity theory?  A result like the theorem by Takeuti and Yasumoto that I cited above is a start, but doesn't quite count in my mind unless we have some reason to think that insights from set theory will help us understand the properties of $M[G]$.  Can we hope for this or are there some basic obstacles that make this approach highly unpromising?

REPLY [21 votes]: Forcing in set theory is usually performed to extend a model of set theory to another model $M[G]$ of set theory in which this or that statement of set theory (such as the continuum hypothesis) changes its truth-value. On the other hand, forcing over a model of arithmetic $M$ does not produce an extension of $M$, but a rather an expansion $(M,G)$. 
A serious stumbling block for the application of forcing in $PA$ (Peano Arithmetic) to obtain independence results is a theorem of Gaifman that states that if $M$ and $N$ are models of $PA$ such that $M$ is cofinal in $N$, then $M$ is an elementary submodel of $N$, and in particular, $M$ and $N$ satisfy the same set of arithmetical sentences. Gaifman's proof, by the way, heavily relies on the so-called MRDP theorem (which states that every r.e. set is Diophantine).
Forcing has had an enormous success in clarifying definability questions in both arithmetic and set theory, but in comparsion, it has had rather limited efficacy so far in complexity theory, certainly not due to lack of effort or expertise: I know more than one expert in computational complexity whose résumé includes first rate papers on set-theoretical forcing.<|endoftext|>
TITLE: Effective way of finding generators on the curve and the rank conjecture
QUESTION [5 upvotes]: Hello everyone, 
I have never heard of a polynomial time running algorithm that finds the generators of elliptic curves efficiently. I do know that Nagell-Lutz theorem is useful in computing the torsion part in $$E(\mathbb{Q})=\mathbb{Z}^{\phi} \oplus E_{\rm{Torsion }}(Q).$$ So what about the count of $\phi$ and the effective generation of points. 
I do know 2-descent , that doesn't not work perfectly always. Regarding the 3-descent , only some curves having CM are said to pass through it. But are there any latest advancements in the area of computing the generators of the curve, if so please give some references. 
I also read the rank conjecture, for expressing the rank of the underlying abelian group $E(\mathbb{Q})$ in terms of the order of vanishing of the taylor expansion of associated $L$-function. So I again got stuck. 
Why do one go for a rank, if one has a point that has infinite order ? . 

Rank 1-curves usually mean that they have a point of infinite order on them that can be used to generate all other points by successive chord and tangent methods.
Rank 2- Curves have 2 such points of infinite order that can be used to generate all other points. 

So my question is why should  one bother about $n$ points ( of infinite order ) if we have one point of infinite order. To express an analogous statement, suppose think that you are given a secret map that will lead you to some treasure. After the hard journey, you at-last said " Eureka" and you have found a " Machine X" . Its written on the Machine X, that it will produce dollar notes, as many as you want. So its limit is infinity. You can extract infinite number of dollar notes from the machine. You took that machine and packed it and again saw the secret-map. There are in fact some other markings on the map that will lead you to a place that contains the same Machine X. 
Do you again go to those places searching for another Machine-X if you have a Machine-X that will produce infinite amount of money. I think the analogy is clear. So if we already have a point that produces infinite points why to again bother about other points that give rise to infinite points. 
So if I am right, does the task of finding $\phi$ number of generators reduce to the task of finding one generator ? . That will make the above equation look like this $$E(\mathbb{Q})=\mathbb{Z} \oplus E_{\rm{Torsion }}(Q).$$
( Which is nothing but the Rank-1 situation ).
Thank you.

REPLY [6 votes]: We do not currently know an effective algorithm to compute the rank of an elliptic curve or to compute generators for its Mordell-Weil group. One can "do descents by day and search for points by night", and in practice, the process will stop. BTW, people do 3 and 4 (and maybe even 5) descents these days, and they're not restricted to CM curves. Finally, I'll mention that if $E(\mathbb{Q})$ has rank 1, then one can use Heegner points to find a rational point of infinite order, or one can use the value of $L'(E,1)$ (sort of as described in Keller's answer) to do a very efficient search. (This last algorithm is in a paper of mine). However, neither method is practical if the conductor $N$ of the curve is too large. Roughly, for both methods one needs to compute the value of a series that doesn't begin to converge until you take $O(\sqrt{N})$ terms.<|endoftext|>
TITLE: Is the ball reducible in some high dimension?
QUESTION [18 upvotes]: Let $K$ be a bounded symmetric ($-K=K$) open convex body in $\mathbb R^n$. The critical determinant $d(K)$ of $K$ is the least possible volume $|\operatorname{det}(a_1\dots a_n)|$
of the fundamental parallelepiped of a lattice $\Lambda=\{\sum_{j=1}^n m_j a_j: m_j\in\mathbb Z\}$ such that $K\cap \Lambda=\{0\}$. Clearly, if $K'\subset K$ is another symmetric convex shape, then $d(K')\le d(K)$. The convex symmetric shape $K$ is called irreducible if the inequality is strict for every proper subset $K'$ of $K$.
It is not hard to see that the unit ball $B$ is irreducible in the dimensions 1,2,3. Moreover, if $B'\subset B$ and the radius of the largest ball contained in $B'$ is $1-\delta$, then $d(B')\le d(B)-c\delta$. However, this breaks in dimension $4$. When we cut off two opposite caps of depth $\delta$ from the unit ball in $\mathbb R^4$, we still get $d(B')<d(B)$ but the difference is now merely of order $\delta^2$. This suggests that when the dimension goes up and the number of touching points increases, we may end up in the situation when the ball is no longer irreducible. My question is whether this is really the case or whether the ball stays irreducible all the way up and just gets "less ans less" so in some sense.
The reason I'm asking is that we've just proved with Yoav Kallus that the ball in $\mathbb R^3$ is a local minimizer of the optimal lattice packing density. The corresponding statement is known to be false in $\mathbb R^2$ and I wonder if there is a trivial reason (namely, reducibility) for it to be false in some or, better, all high dimensions.
Any (relevant) ideas and/or references are welcome :).

REPLY [7 votes]: According to a talk that I found on the web, it is a theorem of Voronoi that every indecomposable root lattice is extreme.  Also the $E_8$ lattice is the union of two copies of the $D_8$ lattice with the same sphere radius.  And, of course, the rotational symmetry group of the $E_8$ is transitive on roots.  So I think that that gives it to you:  If you put two small, flat dents in a round ball in $\mathbb{R}^8$, then you cannot deform the $E_8$ lattice packing, because there is a $D_8$ lattice inside that is far away from the dents.
The same argument works for the Leech lattice, which contains a $D_{24}$ sublattice of index 8192.  It also works for $E_7$, because it contains $A_7$.  (Also $E_8$ contains $A_8$, but not in the same way, since $[E_8:A_8] = 3$ while $[E_7:A_7] = 2$.)
However, I do not think that it is known, nor even a conjecture with strong evidence, that the kissing number of the best lattice sphere packing in dimension $n \to \infty$ is more than the bare minimum $n(n+1)$.<|endoftext|>
TITLE: What makes Geometric CFT easier than CFT?
QUESTION [11 upvotes]: I've been reading:
math.stanford.edu/~conrad/249BPage/handouts/geomcft.pdf
in an attempt to shed some geometric light on class field theory. The last paragraph there reads:

In case the ground field $k$ is perfect, the essential difficulty in the proof of class field theory – proving
that the Artin map kills certain principal ideals – becomes easy to prove geometrically by means of the
interpretation of geometric points of generalized Jacobians in terms of generalized ideal class groups. (More
precisely, one has $J_m (k) = Cl_m (K)$ when $Br(k) = 1$, as happens when $k$ is finite but not when $k$ is a number
field.)

What precisely is he referring to? What particular part of class field theory ("that the Artin map kills certain principal ideals") becomes easier to prove, and why ("by means of the
interpretation of geometric points of generalized Jacobians in terms of generalized ideal class groups.")?

REPLY [5 votes]: $\def\Spec{\mathrm{Spec}\ }$Let me try to add an answer with a bit more big picture. I'm not an expert here, so I'm hoping Matt E., B. and K. Conrad and others will help improve this. I'll start by outlining the proof of geometric CFT which you'll find in Serre's Algebraic Groups and Class Fields. (There is another line of proof which you'll find in, for example, David Ben Zvi's MSRI talk, and I am still absorbing.)
Theorem 1 Let $L/K$ be a finite Galois extension (with no further conditions on the fields) with $A = \mathrm{Gal}(L/K)$. Then there is an algebraic group $G$ over $K$, an embedding of $A \times \Spec(K)$ into $G$, and maps $\Spec(L) \to G$ and $\Spec K \to G/A$, such that the diagram
$$\begin{matrix}
\Spec L & \rightarrow & G \\ \downarrow & & \downarrow \\ \Spec K & \rightarrow & G/A
\end{matrix}$$
commutes, with $\Spec L$ the preimage of $\Spec K$ within $G$, and with the two actions of $A$ on $\Spec L$ (by multipication in $G$, and by the Galois action) being equal.
This sounds very technical, but you probably have seen two special cases of it. 
Special Case 1 If $K$ contains the $m$-th roots of unity and $A = \mathbb{Z}/m$, then we can take $G$ to be the multipicative group and $A \subset G$ to be the $m$-th roots of unity; the theorem then says that $L$ is of the form $K(\alpha^{1/m})$ for some $\alpha \in K^{\ast}$. This is Kummer's thoerem. The maps $L \to G$ and $K \to G/A \cong G$ are given by $\alpha^{1/m}$ and $\alpha$.
Special Case 2 If $K$ has characteristic $p$ and $A = \mathbb{Z}/p$, then we can take $G$ to be the additive group and $A \subset G$ to be the elements of $\mathbb{F}_p$. Then $G/A \cong G$, with the map $G \to G/A \cong G$ being $y \mapsto y^p-y$. So the theorem says that there is some $\beta \in K$ such that $L = K(y)/(y^p-y-\beta)$. This is the Artin-Schrier theorem. 
If you are familiar with the proofs of Kummer and Artin-Schrier, the proof of Theorem 1 isn't much harder.
Theorem 2 Let $A$, in the above theorem, be abelian. Then we can take $G$ to be abelian as well.
In this case, $G/A$ is itself an algebraic group, which we will call $J$. 
We now specialize to the case that $K$ and $L$ are the fields of meromorphic functions on curves $X$ and $Y$ over some ground field $k$. Write $f: Y \to X$ for the covering map. We do not yet assume that $k$ is finite.
Theorem 3 We can take the group $G$ to be defined over $k$, with $A \subseteq G(k)$.
This means that we can interpret the above diagram more geometrically: We can find an open subset $X^{\circ}$ of $X$ such that, setting $Y^{\circ}:= f^{-1}(X^{\circ})$, there is a commutative diagram
$$\begin{matrix}
Y^{\circ} & \rightarrow & G \\ \downarrow & & \downarrow \\ X^{\circ} & \rightarrow & G/A
\end{matrix}$$
as before. 
Now let $k= \mathbb{F}_q$ and let $F$ denote the $q$-power Frobenius.
Theorem 4 When $A$ is abelian and $k = \mathbb{F}_q$, we can find a map $h: J \to G$ such that the composites $G \to J \to G$ and $J \to G \to J$ are $F-1$. 
Warning I suspect that I am missing a technical hypothesis, perhaps that $L$ and $\bar{k}$ be disjoint.
Special Cases In an Artin-Schrier extension, $G \cong J$ is the additive group $\mathbb{G}_a$ and the map $G \to J$ is $y \mapsto y^p-y$. If $m | q-1$, then $\mathbb{F}_q$ contains the $m$-th roots of unity. In the Kummer extension, $G \cong J \cong \mathbb{G}_m$, with $G \to J$ being $x \mapsto x^m$ and $J \to G$ being $x \mapsto x^{(q-1)/m}$. So the composite is $x \mapsto x^{q} x^{-1}$.
Generalizing the computation in my other answer shows:
Theorem 5 With the above hypotheses, if $D$ is a divisor supported on $X^{\circ}$, and $\bar{D}$ the divisor which it splits into in $X^{\circ}(\bar{k})$, then 
$$h \left( \sum_{\pi \in \bar{D}} \pi \right) = \mathrm{Art}(D)$$.
Here the sum $\sum_{\pi \in \bar{D}} \pi$ lives in $J(\bar{k})$, the map $h$ puts it into $G(\bar{k})$, and the assertion is it is equal to $\mathrm{Art}(D)$ under the embedding $A \to G(k)$.
Examples In the Kummer case, Theorem 5 says that $\mathrm{Art}(D) = \prod_{\pi \in \bar{D}} \alpha(\pi)^{(q-1)/m}$. In the Artin-Schrier case, Theorem 6 says that $\mathrm{Art}(D) = \sum_{\pi \in \bar{D}} \beta(\pi)$.  
Now, for the result which uses the most technical tools. This is basically Theorem 3.2 in Brian Conrad's notes. By the way, this works for any perfect $k$, not just a finite field.
Theorem 6 If $A$ is abelian, we can take $J$ to be a generalized Jacobian $J(\mathfrak{m}, X)$, for some conductor $\mathfrak{m}$, and $X \to J$ we can take a translation of the standard map $X \to J(\mathfrak{m}, X)^1$. Moreover, we can take the support of $\mathfrak{m}$ to be the set $S$ of critical points of $Y \to X$, and we can take $X^{\circ} = X \setminus S$. In particular, if $Y \to X$ is unrammified, we can take $J$ to be the Jacobian and $S = \emptyset$. 
Warning I might be slightly missing something in the last sentence; I would have expected to see conditions like "$Y$ is geometrically connected" and "$Y(k) \neq \emptyset$" showing up.
Let me point out the subtlety of the last sentence of Theorem 6. Let $\mathrm{char}(k) \neq 2$, and let $A = \mathbb{Z}/2$. Then $L = K(\sqrt{\alpha})$ for some $\alpha$. In our high-tech language, we can take $G$ to be the multiplicative group $\mathbb{G}_m$, with $A$ embedded as $\{ \pm 1 \}$. $X^{\circ}$ is the locus where $\alpha$ is nonzero and the map $X \to G/A \cong G$ is given by the function $\alpha$ on $X^{\circ}$. Even if $Y \to X$ is an unbranched cover, the function $\alpha$ will still have zeroes and poles (of even order). We cannot map the entire projective curve $X$ to an affine group like $\mathbb{G}_m$. Theorem 6 is telling us that, by using a very different group, the Jacobian of $X$, we can arrange for the map to $J$ to be defined everywhere on $X$.
Combining Theorems 5 and 6 and the definition of $J(\mathfrak{m}, X)$, we have
Key Theorem Let $k$ be finite and $A$ abelian. Then there is a modulus $\mathfrak{m}$, supported on the ramified primes $S$ of $Y \to X$, such that 
$\mathrm{Art}$ is trivial on principal ideals whose generators are $1 \bmod \mathfrak{m}$.

Now, what happens if we try to work with number fields? There are three problems. To me, Problem 1 feels the most serious.
Problem 1 The best attempt at a generalization of Theorem 3 is to replace $G$ and $J$ with group schemes $\mathcal{G}$ and $\mathcal{J}$ over $\mathcal{O}_K$. So the map $X^{\circ} \to J$ turns into a section of $\mathcal{J} \to \Spec \mathcal{O}_K$ defined on $X^{\circ}$. But this means that different points of $X^{\circ}$ wind up lying in different fibers of the group scheme, so a sum as in Theorem 6 makes no sense.
Problem 2 There is no global map $F$, so it is unclear what an analogue of Theorem 4 would look like.
Problem 3 Although the ray class groups of $K$ generalize the groups $J(\mathfrak{m}, X)(k)$, there is no algebraic group which generalizes $J(\mathfrak{m}, X)$, so it is not clear what an analogue of Theorem 6 would look like.<|endoftext|>
TITLE: Interpretation of universal coefficients theorem for group cohomology
QUESTION [15 upvotes]: Suppose $G$ and $A$ are abelian groups (I'm setting $G$ abelian to keep the discussion simple, though there are analogues for non-abelian $G$) with $G$ acting trivially on $A$. By the universal coefficients theorem, we have short exact sequences for all positive integers $n$:
$$0 \to \operatorname{Ext}^1(H_{n-1}(G;\mathbb{Z});A) \to H^n(G;A) \to \operatorname{Hom}(H_n(G;\mathbb{Z}),A) \to 0$$
In the case $n = 2$, this becomes:
$$0 \to \operatorname{Ext}^1(G;A) \to H^2(G;A) \to \operatorname{Hom}(H_2(G;\mathbb{Z}),A) \to 0$$
The surjection $H^2(G;A) \to \operatorname{Hom}(H_2(G;\mathbb{Z}),A)$ has a nice interpretation:
$H_2(G;\mathbb{Z})$ is the Schur multiplier of $G$, which (since $G$ is abelian) is the exterior square of $G$, so $\operatorname{Hom}(H_2(G;\mathbb{Z}),A)$ is the alternating bilinear maps from $G$ to $A$. The mapping $H^2(G;A) \to \operatorname{Hom}(H_2(G;\mathbb{Z}),A)$ is the "skew" mapping; it sends a 2-cocycle $f:G \times G \to A$ to the function:
$$(x,y) \mapsto f(x,y) - f(y,x)$$
Even without prior knowledge of the universal coefficients theorem, this short exact sequence makes sense: one can check that the skew of a 2-cocycle for an abelian group is alternating and bilinear, and the mapping descends to cohomology classes because any 2-coboundary is symmetric. The kernel of the mapping corresponds to symmetric cohomology classes, which corresponding to the abelian extension groups, given precisely a $\operatorname{Ext}^1(G;A)$. In group extension terms, the image of a given element of $H^2(G;A)$ under the skew map describes the commutator map of the extension group.
My question is: what's an analogous concrete interpretation for higher $n$?
I'm listing below some references with generalizations of the above to non-abelian $G$ and varietal generalizations:

John Burns and Graham Ellis, On the nilpotent multipliers of a group, Math. Zeitschr. 226 (1997) pp. 405–428 doi:10.1007/PL00004348.

C. R. Leedham-Green and Susan McKay, Baer-invariants, isologism, varietal laws, and homology, Acta Mathematica 137 (1976)  pp. 99–150 doi:10.1007/BF02392415 (ResearchGate copy).


Pages 135–139 (37–41 in the pdf) describe a generalization of universal coefficients to varietal laws (where the usual universal coefficients is with respect to abelian groups inside groups).

REPLY [4 votes]: This is an extended-comment, but I think no:
In homology, we know that $H_1$ is abelianization, $H_2$ is Schur multiplier, but $H_{n\ge 3}$ is ???
In cohomology, $H^1(G;A)$ is split extensions, and this fits well with $Hom(G,A)$ in the UCT. As you mention, $H^2(G,A)$ corresponds to group extensions of $G$ by $A$, and this fits well with $Hom(H_2G,A)$ in the UCT. But as we see, this all ties in to knowing $H^n(G;A)$ and $H_n(G;A)$ very well...
If we look at $H^3(G;A)$ we get crossed module extensions $0\to A\to N\to E\to G\to 0$, and these are cumbersome. We have no nice interpretation for $H^n(G;A)$ for $n>3$, except more crazy-looking exact sequences.  This is why I don't expect a "nice" map (i.e. "interpretation" of it) in the UCT to arise.<|endoftext|>
TITLE: Calabi-Yau fiber space without singular fibers implies finite quotient of product?
QUESTION [9 upvotes]: While reading this paper of Kollár, the following question came up. If $f:X\to Y$ is a fiber space (i.e. surjective holomorphic map with connected fibers) with $X,Y$ smooth projective manifolds, with $K_X\sim_{\mathbb{Q}} 0$, and with $f$ a submersion everywhere, does there exist a finite étale cover $Y'\to Y$ such that $X\times_{Y} Y'$ is isomorphic to a product family (and so $K_Y\sim_{\mathbb{Q}}0$ too)?
Note that since $f$ is a submersion, every fiber $F$ satisfies $K_F\sim_{\mathbb{Q}}0$.
The case when $\mathrm{dim}X=\mathrm{dim}Y+1$ (which includes the case when $\mathrm{dim}X=2$) is easy, since then the fibers are elliptic curves, and the $j$-invariant of each fiber gives a holomorphic function on $Y$, hence constant. It then follows from Fisher-Grauert that $f$ is a holomorphic fiber bundle, and you can apply for example Lemma 17 in this paper of Kollár-Larsen to conclude.
This same argument shows that in arbitrary dimension it is enough to show that all the fibers of $f$ are biholomorphic.
Thanks to work of Kawamata we know that $\kappa(Y)\leq 0$. We can also apply Theorem 4.4 in this paper of Fujino-Gongyo and get that $-K_Y$ is semiample.
So if $\kappa(Y)=0$ then we obtain that $K_Y\sim_{\mathbb{Q}}0$. At this point one can apply the arguments in Theorem 4.8 of this paper of Ambro and conclude.
So the question is reduced to the following problem: is it possible to have a holomorphic submersion $f:X\to Y$ with connected fibers, where $X,Y$ are smooth projective manifolds with $K_X\sim_{\mathbb{Q}} 0$ and $\kappa(Y)<0$ ?
If we assume that $Y$ is simply connected and that $X$ has $K_X\cong\mathcal{O}_X$ and $h^{p,0}(X)=0$ for $p=1,...,\mathrm{dim} X-1$ (so $X$ is Calabi-Yau in the stronger sense), then Corollary 2.5 in this paper of Zhang-Zuo implies that this is impossible.
In general I don't know the answer to this, but when $f$ is not assumed to be a submersion (so there might be singular fibers) then $\kappa(Y)<0$ can happen, for example when $X$ is an elliptically fibered $K3$ surface and $Y=\mathbb{P}^1$.

REPLY [5 votes]: I am not familiar with this topic, but I remember seeing this recent paper. It claims that if the map has no singular fiber, it is a fiber bundle (all the fibers isomorphic). Moreover, if the total space is algebraic, it can be a product family after a suitable base change, as you claimed.
Edit
I made my answer more informative, as the comments below suggested. I thought the abstract of the paper would answer to the question.<|endoftext|>
TITLE: First Table of Random Numbers
QUESTION [10 upvotes]: What was the first table of random numbers of any sort?  
The best I can do is Tippett and Pearson's Random Sampling Numbers of 1927.  
Can anybody identify an earlier table?
Thanks for any insight.  
Cheers, Scott

REPLY [4 votes]: Erastus De Forest's 1876 booklet on Interpolation and Adjustment of Series has an early table of random numbers on page 18, or in columns (5) and (6) of the below. Stephen Stigler's article on Stochastic Simulation in the 19th Century discusses this, and also work by George Darwin, Francis Galton, and William Gosset.<|endoftext|>
TITLE: Is there a concrete description of relative jet spaces that invokes neither internal Hom nor individual generators?
QUESTION [9 upvotes]: Let $X,Y$ be $S$-schemes. Define the sheaf $\underline{\hom}_S(Y,X)$ on $S$-schemes by $\underline{\hom}_S(Y,X)(T) := \hom_S(T \times_S Y,X)$. In other words, we have
$\hom_S(T,\underline{\hom}_S(Y,X))\cong \hom_S(T \times_S Y,X)$,
which explains the name: If this sheaf is representable, it is the internal hom object. It coincides with the Weil restriction $\mathfrak{R}_{Y/S}(X \times_S Y)$ (see Néron models by BLR). Grothendieck mentions in FGA 5, C.2, that $\underline{\hom}_S(Y,X)$ has only a chance to be representable when $Y$ is flat and proper over $S$. So my first question is: where Grothendieck actually proved this result?
I claim that $\underline{\hom}_S(Y,X)$ is representable whenever $Y \to S$ is finite locally free and a universal homeomorphism: one just has to look at the proof for Weil restrictions in Theorem 7.6/4 in BLR; the condition on the fibers of $X$ is not needed. In particular this holds for $Y=S[\varepsilon]/\varepsilon^{n+1}$ for some $n \geq 0$, which is the example which interests me.
For example, we have $\underline{\hom}(S,X) = X$ and $\underline{\hom}(S[\varepsilon]/\varepsilon^2,X) = T(X/S) = \mathbb{V}(\Omega^1_{X/S})$ is the tangent bundle of $X/S$. This is just a reformulation of the universal properties, but there is a sort of global description of $\Omega^1_{X/S}$, namely as $I / I^2$, where $I$ is the kernel of the multiplication $\mathcal{O}_X \otimes_{\mathcal{O}_S} \mathcal{O}_X \to \mathcal{O}_X$. My main question is: Is there any such global description for
$$T^n(X/S) := \underline{\hom}(S[\varepsilon]/\varepsilon^{n+1},X)$$? 
Probably this is what people call the $n$-th jet bundle, but I could not find any concise description in the literature. Remark that we have a seqeunce of morphisms 
$$\dotsc \to T^n(X/S) \to T^{n-1}(X/S) \to \dotsc \to T(X) \to X.$$ 
The proof of the Theorem in BLR works roughly as follows: Everything is local on $S$, to suppose it is affine. If $X$ is some affine space, you can find explicitly a representing scheme. More generally, for every quasi-coherent module $M$ on $S$ and $\mathcal{A}$ the quasi-coherent algebra on $S$ corresponding to $Y \to S$ which is finite locally free, we have $\underline{\hom}_S(Y,\mathbb{V}(M)) = \mathbb{V}(\underline{\hom}_{\mathcal{O}_S}(\mathcal{A},M))$. This implies $T^n(\mathbb{A}^m/S) \cong \mathbb{A}^{m(n+1)}$. If $X$ is affine, embed it into some affine space. If $X$ is arbitrary, glue everything. So this description is not really global and even depends on affine charts on $X$ and choices of generators and relations on the affine charts.
This can be done here even more explicit: Assume that $S=\mathrm{Spec}(R)$ and $X=\mathrm{Spec}(A)$. Let $A^{(n)}$ be the $A$-algebra generated by symbols $h_i(a)$ for every $0 \leq i \leq n$ and $a \in A$, subject to the following relations:

$h_0(a)=a \cdot 1$
$h_i(1)=0$ for $i \geq 1$
$h_i(a+b)=h_i(a)+h_i(b)$
$h_i(ab)=\sum_{p+q=i} h_p(a) h_q(b)$

Then $T^n(X/S) = \mathrm{Spec}(A^{(n)})$. The morphism $T^n(X/S) \to T^{n-1}(X/S)$ is induced by the obvious homomorphism $A^{(n-1)} \to A^{(n)}$ which inserts the generators. This description is well-known for $n=1$, where $h_1$ is just a derivation with respect to $h_0$.
I am not looking for a reference about jet spaces or alike. I would like to know if we can construct some sort of "concrete" $A$-algebra out of the $R$-algebra $A$ which turns out to be isomorphic to $A^{(n)}$. For example, when $n=1$, the well-known answer is the symmetric algebra of $I/I^2$, where $I = \ker(A \otimes_R A \to A)$, with $h_1(a)=a \otimes 1 - 1 \otimes a$.

REPLY [2 votes]: For your first question, I think the answer depends on whether you think representability means "in schemes" or "in algebraic spaces".  For schemes, you really want a projective morphism $Y \to S$ for BLR's proof to work.  For spaces, I think Olsson's "Hom stacks and restriction of scalars" gave a proof using flat proper and lfp.  (I don't see any conditions on the fibers in $X$ in your question.)
For your other questions, see Vojta.<|endoftext|>
TITLE: Geometric interpretation of families in the stable homotopy groups of spheres
QUESTION [14 upvotes]: There are  infinite families in the stable homotopy groups of spheres; many of  these can be seen by looking for "periodicity" phenomena in the Adams-Novikov spectral sequence. An example is the image of the  $J$-homomorphism. This particular infinite family can be described algebraically (at least at odd primes). If I understand correctly, these elements correspond to elements of filtration $1$ in the ANSS. That is, they are in $\mathrm{Ext}^1_{BP_*BP}(BP, BP)$ via the exact sequence of comodules
$$0 \to BP \stackrel{p}{\to} BP \to BP/p \to 0.$$
The invariant elements in $BP/p$ are the powers of $v_1$, and the coboundaries of these live in $\mathrm{Ext}^1(BP, BP)$. These correspond to multiples of the image of $J$. 
Other infinite families in the stable homotopy groups of spheres can be obtained from a similar procedure. One nice thing about the image of $J$, though, is that it has a geometric interpretation: if the stable homotopy groups of spheres are identified with the framed cobordism groups, then the image of $J$ corresponds to framed spheres. Are there similar geometric interpretations of these more exotic infinite families?

REPLY [12 votes]: This has been a burning question for quite some time, but not much is known. Surely, people believe that the next layer (i.e. the $\beta$-family) should also admit a geometric description, although as of yet nobody has been able to find one.
The geometry behind the $J$-homomorphism (homotopy of the stable orthogonal group) is very closely related to the geometry that underlies $K$-theory, so I suppose the same will be true for the $\beta$-family. There's a lot of remarkable work going on from people who try to provide a geometric construction of elliptic cohomology, and you could ask if their stuff contains the germ of a geometric construction of the $\beta$'s. Maybe some of the finite dimensional models of the String group could play a role there?<|endoftext|>
TITLE: S-matrix for the HOMFLY/Hecke category
QUESTION [12 upvotes]: This question concerns the HOMFLY-PT category, closely related to Hecke algebras.  (See here for example.)  
The minimal idempotents of this category are indexed by pairs $(\lambda_+, \lambda_-)$ of Young diagrams. (The sizes of the diagrams are arbitrary and need not be the same.  The diagram $\lambda_+$ corresponds to upward oriented strands, while $\lambda_-$ corresponds to downward oriented strands.) Consequently one can define numerical invariants of oriented links whose components are labeled by pairs of Young diagrams.  This is the "colored" HOMFLY-PT polynomial.
Of fundamental importance in this subject are the invariants $S_{\lambda_+\lambda_-,\mu_+\mu_-}$ of the Hopf link with its components labeled by pairs of Young diagrams (i.e. idempotents) $(\lambda_+, \lambda_-)$ and $(\mu_+, \mu_-)$.  In TQFT language, this is the "S-matrix" of the theory.
My Question:

Has the S-matrix for the HOMFLY-PT category been calculated and published?  If not, are partial results in this direction known?

I am aware of this paper by Morton and Lukac, which does the case where $\lambda_-$ and $\mu_-$ are both empty (i.e. all strands oriented the same direction).  This paper by Morton and Hadji is also related.  Are there other relevant papers that I have missed?
See also the BMW version of this question here.

REPLY [2 votes]: This isn't a complete answer, but it might make some partial progress. First I'll slightly restate the question. If we decompose $S^3$ into two solid tori and write $C$ and $C^{op}$ for the Homfly skein modules of these tori, then there is a pairing $$\langle -,-\rangle: C\otimes_A C^{op} \to k$$
where $k$ is the base ring (which is isomorphic to the skein module of $S^3$) and $A$ is the Homfly skein algebra of the torus. As you say, both $C$ and $C^{op}$ have bases indexed by pairs of partitions, and if I understand right, the entries of the $S$ matrix are evaluations of this pairing.
One brute-force approach for computing $\langle (\lambda, \mu), (\lambda',\mu')\rangle$ is to "move the left hand side to the right using the action of $A$ and then evaluate." An explicit presentation of $A$ is given here http://arxiv.org/abs/1410.0859. There are generators $P(a,b)$ for $a,b \in \mathbb(Z)$ (which are "the $gcd(a,b)$ power sum on slope $b/a$"), and they satisfy the commutation relations $$[P(x), P(y)] = (q^d-q^{-d}) P(x+y),\quad x,y \in \mathbb Z^2,\quad d = det[x\,\,y]$$
(If $x$ and $y$ are primitive and $d=1$ this is just the skein relation.) The paper also gives explicit formulas for the action of the generators $P(a,b)$ in the basis $(\lambda,\mu)$ that you mentioned, and these formulas actually aren't too bad (e.g. if $a,b \not= 0$ then the entries in the matrix are monomials).
One difficulty in completing the answer is that expressing the basis elements $(\lambda,\mu)$ in terms of power sums isn't too easy. (Morton does have a determinental formulas in terms of complete symmetric functions in one of his papers.) Also, I'm not sure if the evaluations $((\emptyset, \emptyset),(\lambda,\mu)\rangle$ have actually been computed.
(By the way, this algebra $A$ actually has a lot of different realization which are not obviously the same. One other realization is as (a specialization of) the Hall algebra of coherent sheaves on an elliptic curve over a finite field (this is described in the paper above). Also, if you've found these numbers since the question was asked I'd be curious to hear.)<|endoftext|>
TITLE: abelianization of adelic points of an algebraic group
QUESTION [9 upvotes]: Let $G$ be a connected reductive group defined over a number field $K$ and $G^{der}$ its derived subgroup.
Let $\mathbb{A}_K$ denote the adeles of $K$. 
Then for $G=GL_n$ we have $[GL_n(\mathbb{A}_K),GL_n(\mathbb{A}_K)]=SL_n(\mathbb{A}_K)=GL_n^{der}(\mathbb{A}_K)$.
I'm interested in what generality this holds, in other words I'd like to ask:
Question 1:  When is the commutator subgroup $[G(\mathbb{A}_K),G(\mathbb{A}_K)]$ equal to $G^{der}(\mathbb{A}_K)$? 
As I think this question is really a local one, so let me put it this way:
Let $K_v$ be a local field of char 0. $G$ a reductive group over $K_v$, $G^{der}$ its derived subgroup.
Question 2: Is the commutator subgroup $[G(K_v),G(K_v)]$ equal to $G^{der}(K_v)$? 
These questions came up when I wanted to understand 1-dimensional automorphic representations of unitary groups coming from a division algebra with an involution of the second kind and I realized I didn't know what the abelianizations of the adelic points of the groups in question were.

REPLY [7 votes]: I think question 2 has a positive answer when $G^{der}$ is simply connected [EDIT: and without anisotropic factor], but not in general. If $G=PGL_d$ (so $G=G^{der}$) then $G(K)/[G(K),G(K)]=M/M^d$ where $M=K^*$, this is not a trivial group in general.
Relevant refences should be the preliminary chapters in book by Margulis "discrete subgroups of semisimple Lie groups", as well as Platonov-Rapinchuk's book.<|endoftext|>
TITLE: Examples of nice reduced singularities on Hilbert schemes--Edited
QUESTION [5 upvotes]: In his "Murphy's Law" paper, Vakil showed that every "singularity type" (with a precise meaning) occurs on certain Hilbert schemes; for instance, the Hilbert scheme of nonsingular curves in projective space. He also gives a method for constructing such singularities; however, the process to construct even, say, a singularity of nodal type would be extremely involved.  
[As I understand it, one would have to blow up a plane at something like twenty points (conservative estimate), then take a certain eight-fold cover, then find an appropriate line bundle and take six "sufficiently general" sections,...]
For a smooth variety $X$, let $H_X$ denote its Hilbert scheme. A point of $H_X$ corresponds to a subscheme $V$ of $X$.  I am interested in cases in which $V$ is also smooth. [Edit: I am also requiring that $H^1(T_X) = 0$, i.e., that the ambient variety $X$ admit no infinitesimal deformations. To put it another way, the complex structure on the smooth manifold $X$ cannot be deformed. This holds in particular if $X=\mathbb P^n$.] Certainly, explicit examples of such pairs $(V,X)$ corresponding to singular points of $H_X$ have been described; however, in the very few examples I have seen, the technique is to show that $V$ is contained in an irreducible component of $H_X$ that is generically non-reduced.

Can anyone give explicit examples of a smooth projective variety $X$ [such that $H^1(X,T_X) = 0$], together with a smooth subvariety $V$, such that the point $[V]\in H_X$ is both singular and reduced? [A method of constructing explicit examples will not suffice unless you can show, by example, that this method is actually practical to carry out.]

REPLY [9 votes]: Once again, I apologize for my previous comment!  The new formulation of the problem is definitely more fun.  Here is one solution which only uses smooth, rational curves.
Let $n$ be an integer, $n\geq 4$ (in fact, $n=2$ and $n=3$ make sense if you use the Kontsevich spaces instead of Hilbert schemes).
Start with $\mathbb{P}^{n+1}\times \mathbb{P}^n$ with homogeneous coordinates $([x_0,\dots,x_n,x_{n+1}],[y_0,\dots,y_n])$.  Consider the hypersurface $X$ of bidegree $(1,1)$ which is the zero scheme of $x_0y_0+\dots+x_ny_n$.  In fact every smooth $(1,1)$-hypersurface in $\mathbb{P}^{n+1}\times \mathbb{P}^n$ is "conjugate" to this one under an automorphism of $\mathbb{P}^{n+1}\times \mathbb{P}^n$.  Consider the projection $\pi_2:X\to \mathbb{P}^n$; this is a $\mathbb{P}^n$-bundle $\textbf{Proj}_{\mathbb{P}^n}\text{Sym}^\bullet(\mathcal{E})$ where $\mathcal{E}$ is the locally free, rank $n+1$, $\mathcal{O}_{\mathbb{P}^n}$-module $\mathcal{O}_{\mathbb{P}^n} \oplus Q$ and $Q$ is the cokernel of the tautological morphism $\mathcal{O}_{\mathbb{P}^n}(-1)\to \mathcal{O}_{\mathbb{P}^n}^{\oplus (n+1)}$.  From this you can compute that $H^1(X,T_X)$ is zero; it is the same as proving that $H^1(\mathbb{P}^n,\textit{Hom}_{\mathcal{O}_{\mathbb{P}^n}}(\mathcal{E},\mathcal{E}))$ is zero.   
Now consider the Hilbert scheme of (flat families of) curves $C$ in $X$ which have constant Hilbert polynomial $1$ with respect to the invertible sheaf $\pi_1^*\mathcal{O}_{\mathbb{P}^{n+1}}(1)$ and which have Hilbert polynomial $t\mapsto nt+1$ with respect to the invertible sheaf $\pi_2^*\mathcal{O}_{\mathbb{P}^n}(1)$.  In other words, $C$ is contained in a fiber of $\pi_1$, and under $\pi_2$ the curve $C$ projects isomorphically to a curve of degree $n$ and arithmetic genus $0$.  Consider the smooth parameterized curve $V$ which is the image of the following closed immersion.
$$v:\mathbb{P}^1 \to \mathbb{P}^{n+1}\times \mathbb{P}^n, \ \ [t_0,t_1] \mapsto ([0,0,\dots,0,1],[0,t_0^n,t_0^{n-1}t_1,\dots,\widehat{t_0^kt_1^l},\dots,t_0t_1^{n-1},t_1^n])$$
where $(k,l)$ is any pair of integers with $2\leq k,l\leq n-2$ and $k+l=n$.  So the image of $V$ in $\mathbb{P}^n$ is a smooth, rational, degree $n$ curve spanning a $\mathbb{P}^{n-1}$ (in particular, it is not a rational normal curve since it is "linearly degenerate").  Just to remark, since the morphism above is $\mathbb{G}_m$-equivariant for the standard action of $\mathbb{G}_m$ on $\mathbb{P}^1$, it suffices to check that $v$ is a closed immersion at $[1,0]$ and $[0,1]$, and this follows since we included the coordinates $t_0^n$, $t_0^{n-1}t_1$, $t_0t_1^{n-1}$ and $t_1^n$.   
The deformation theory of smooth integral curves in smooth varieties is described, e.g., in Theorem II.1.7, pp. 95-96, of Rational Curves on Algebraic Varieties by János Kollár.  In particular, the Hilbert scheme is a local complete intersection at $[C]$ if its (local) dimension equals the "expected dimension", i.e., $\text{deg}_C(c_1(T_X))+(\text{dim}(X)-3)(1-g(C))$.  In this case, that works out to be $(n+1)^2-4$.  Inside the Hilbert scheme there is an open subscheme $U$ which parameterizes smooth, integral curves; in particular $[V]$ is contained in $U$.  
The claim is that $U$ has precisely two irreducible components each having dimension $n^2+2n-3$, and the Hilbert scheme is reduced at the generic point of each irreducible component.  Moreover, $[V]$ is contained in the intersection of the two irreducible components.  Assuming the claim, it follows that $U$ is a local complete intersection.  Since $U$ is a local complete intersection that is generically reduced, $U$ is everywhere reduced.  Since $[V]$ is contained in the intersection of the two irreducible components, $[V]$ is a singular point of $U$.
The first irreducible component is the closure ${U}_1$ in $U$ of the open subset $W_1$ parameterizing smooth curves $C$ such that $\pi_1([C])$ is not $[0,\dots,0,1]$.  The point is that $\pi_1:X\to \mathbb{P}^{n+1}$ is a $\mathbb{P}^{n-1}$-bundle away from $[0,\dots,0,1]$.  Thus $W_1$ is Zariski locally over $\mathbb{P}^{n+1}\setminus \{[0,\dots,0,1]\}$ equal to a product with fiber isomorphic to the Hilbert scheme of smooth, rational, degree $n$ curves in $\mathbb{P}^{n-1}$.  That fiber space is well-described in many places: it is a smooth, rational variety of dimension $n^2+n-4$.  Hence $W_1$ is a smooth, rational variety of dimension $(n^2+n-4)+(n+1)=n^2+2n-3$.    
The second irreducible component $U_2$ is the Hilbert scheme of smooth, rational, degree $n$ curves in the fiber $D:=\pi_1^{-1}([0,\dots,0,1])\cap X$. Note that $\pi_2:D \to \mathbb{P}^n$ is an isomorphism.  Again the Hilbert scheme of smooth, rational, degree $n$ curves in $\mathbb{P}^n$ is a smooth, rational variety of dimension $n^2+2n-3$.  The scheme $U$ equals the union of its closed subsets $U_1\cup U_2$, and even $W_1\cup U_2$, from which it follows that $\text{dim}(U)$ equals $\text{dim}(U_1)=\text{dim}(U_2)=n^2+2n-3$.  Thus $U$ is indeed a local complete intersection.  Moreover, $U_1$ contains the dense open subscheme $W_1$, which is smooth, hence reduced.  Thus $U$ is reduced at the generic point of $U_1$.  
The last issue is whether or not $U$ is reduced at the generic point of $U_2$.  This is the same as the question of whether or not $H^1(C,N_{D/X})$ is zero for a rational normal curve (of degree $n$) in $D \cong \mathbb{P}^n$.  In fact $N_{D/X}\cong \pi_2^*Q^\vee$, with $Q$ as above.  By an explicit computation, for a rational normal curve $f:\mathbb{P}^1\to \mathbb{P}^n$, the pullback $f^*Q$ is isomorphic to $\mathcal{O}_{\mathbb{P}^1}(1)^{\oplus n}$, and hence $f^*N_{D/X}$ is isomorphic to $\mathcal{O}_{\mathbb{P}^1}(-1)^{\oplus n}$.  In particular, $H^1(\mathbb{P}^1,f^*N_{D/X})$ is zero.  From this it follows that $U$ is smooth at the generic point of $U_2$.
Finally, why is $[V]$ in the intersection $U_1\cap U_2$?  Clearly, $[V]$ is contained in $U_2$.  To see that $[V]$ is contained in $U_1=\overline{W}_1$, consider the family, parameterized by the coordinate $s$, of image curves $V_s$ of the following closed immersions.$$v_s:\mathbb{P}^1 \to \mathbb{P}^{n+1}\times \mathbb{P}^n, \ \ [t_0,t_1] \mapsto ([s,0,\dots,0,1],[0,t_0^n,t_0^{n-1}t_1,\dots,\widehat{t_0^kt_1^l},\dots,t_0t_1^{n-1},t_1^n]).$$ For $s\neq 0$, the images are curves in $W_1$.  For $s=0$, the image equals the curve $V$.  Thus $[V]$ is the image of $0$ under the corresponding morphism $$\tilde{v}:\mathbb{A}^1\to U, \ s\mapsto [V_s],$$ and thus is in the closure of $\tilde{v}(\mathbb{G}_m) \subset W_1$.<|endoftext|>
TITLE: Free left = free right ? 
QUESTION [5 upvotes]: Let  $R \subseteq S$ be an extension of rings with unit. Suppose that $S$ is free as left $R$-module. I wonder what can said about the freeness of $S$ as right $R$-module. To be a little more precise let's consider the following questions: 

Does someone know an example such that $S$ is free as left $R$-module, but isn't free as right $R$-module.
What are conditions that imply that $S$ is free as left module iff it's free as right module ?   
What are categorial examples such that $S$ is free as left as well as right $R$-module ? 

An examples for 2. is  

$R$ is a subring of the center of $S$ 

and examples for 3. are: 

$R \subseteq S$ is a Frobenius extension
$S$ is a Hopf algebra over a field and $R$ a sub-Hopf algebra (freeness holds by the Nichols-Zoeller theorem)

REPLY [8 votes]: Let $M$ be an $(R, R)$-bimodule and let $S$ be the square-zero extension $R \oplus M$ of $R$ with multiplication
$$(r_1 \oplus m_1)(r_2 \oplus m_2) = r_1 r_2 \oplus (m_1 r_2 + r_1 m_2).$$
If $M$ is free as a left $R$-module then so is $S$, and if $S$ is free as a right $R$-module then $M$ is projective. So it suffices to find $M$ which is free as a left $R$-module but not projective as a right $R$-module. 
So let $R = T \times T$ where $T$ is some nonzero ring and let $M = R$ with $R$ acting on the left by left multiplication but acting on the right as follows: the second factor of $T$ acts diagonally by right multiplication and the first factor of $T$ acts trivially. Then $M$ is free as a left $R$-module but torsion as a right $R$-module. Taking $T = \mathbb{F}_2$ gives a minimal counterexample in a fairly strong sense; we have $|R| = 4$ and if $|R| < 4$ then $R$ is a field.<|endoftext|>
TITLE: Homology versus cohomology of Lie groups
QUESTION [8 upvotes]: A central advantage of cohomology theory over homology -- at least in terms of richness of structure and strength as an invariant -- is the additional ring structure from the cup product. Recall that this arises from applying the cohomology functor to the following inclusion map of topological spaces $$X \hookrightarrow X \times X$$ where each $x \in X$ is mapped to $(x,x)$ in the product. The "key" insight here is that homology theory lacks an analogous structure precisely because there is no natural candidate for a continuous map $X \times X \to X$. Fair enough.
But Lie groups provide examples of spaces where there is a great candidate for such a map: the group multiplication. I expected that this would make homology of Lie groups interesting by imposing some nice multiplicative structure on homology generators inherited from the group multiplication. On the other hand, the cohomology ring would reveal nothing that you couldn't already learn from the cohomology of the underlying manifold independent of group structure.

Is this wrong? Why is the literature full of material on Lie group cohomology whereas Lie group homology is relatively sparse?

I suspect that maybe this product structure is not even well defined on the level of homology, but I'm not sure how one would prove that.

REPLY [4 votes]: Dear Pierre,
I apologize if I seem harsh again, but it is not true that the possible diagonal maps (in cohomology)
are irrelevant for the application you are describing.  One of the very nicest ways to compute $H^*(BG;k)$ is to use the Rothenberg-Steenrod spectral sequence which converges from 
$Ext_{H_*(G;k)}(k,k)$ to $H^*(BG;k)$.  It is precisely the algebra structure on $H_*(G;k)$ that is relevant, hence the coalgebra structure on $H^*(G;k)$, and there is no reason to restrict to $k=\mathbf{Q}$. One answer to the original question is to point to this spectral sequence, which has been in common use since the early 1960's. From the perspective of Lie groups in particular, what is calculationally easy but perhaps conceptually surprising is how often the algebra $H_*(G;k)$ is (graded) commutative even though the product on $G$ is almost never homotopy commutative.  Commutative and cocommutative Hopf algebras over $\mathbf{Q}$ are classified, and in this case the relevant $Ext$ groups are trivial to compute.<|endoftext|>
TITLE: Computational number theory
QUESTION [6 upvotes]: I am interested in learning computational number theory and doing some computer experiments. 
Which sort of number theory problems can be solved by using computers?  For example, is it possible to determine the ring of integers in number field extensions or find the discriminant of the extensions? 
To what extent one can use computers to solve number theory problems?
What is the best software for this purpose?  Maple, Matlab, Pari/gp or Mathematica?

REPLY [3 votes]: You might look at Bach and de Shallit's book, for one. 
In general, I'd recommend Python (a real computer language, with built-in long ints) for numerical experimentation.
Sage is an extension of Python, with many built-in number-theoretic bits of data and also algorithms.
It is true that Mathematica and other commercial/proprietary mathematics packages have widespread commercial use, but the non-public-ness of their actual algorithms, etc., is rather disappointing if one cares about what is actually happening. Bad enough that we are not able to personally witness the correctness of a large-integer computation, but even worse if we cannot even view the algorithms involved. Also, as computer languages, they have considerable failings, I think.<|endoftext|>
TITLE: A Lagrangian problem with a countable family of local extrema ?
QUESTION [7 upvotes]: Dear MO contributors,
let $r > 0, L > 0$. I am interested in maximizing the integral:
$$
\int_0^{2\pi} \frac{f(\alpha)^2 f'(\alpha)^2}{\sqrt{f(\alpha)^2 + f'(\alpha)^2}} \ \mathrm{d} \alpha
$$
over continuous and piecewise smooth functions $f : [0,2\pi] \to \mathbf{R}$, subject to three the constraints:
$$
f(0) = f(2\pi), \quad f \ge r,  \quad \int_0^{2\pi} \sqrt{f(\alpha)^2 + f'(\alpha)^2} \ \mathrm{d} \alpha = L.
$$
Although not directly relevant to the question, let me mention that the function $f$ represents a curve in the plane, where $f(\alpha)$ is the distance from the origin at angle $\alpha$ from one axis, and the aim is to maximize a certain torque for a given length of curve.
When I try to write down the Euler-Lagrange equation, I end up with a not-so-inspiring differential equation that is a polynomial of degree 7 in $f$, $f'$ and $f''$. 
Heuristically, I expect an extremal function to develop a number of peaks, and I conjecture that the family of extremal functions can be indexed by the number of peaks (by peaks, I mean local maxima, probably with a discontinuous derivative). More precisely, is it true that there exists at most one extremal function with a given number of peaks ? How does the set
$$
\{n \in \mathbf{N} : \text{there exists an extremal function with } n \text{ peaks} \}
$$
look like ? How many peaks does the global maximizer have ? What is its rough shape ? How much do all this depend on $r$ and $L$ ? (of course, if $L < 2\pi r$, then there are no solutions, and there is a unique solution for $L = 2 \pi r$). Are there good ways to investigate these questions numerically ? Any suggestion about the relevant litterature would also be very welcome.

REPLY [3 votes]: Your problem does not have a maximizing solution.  Here is why:  
Assuming that $f$ is piecewise smooth and not identically constant, we can solve the Euler-Lagrange equations in an interval where $f$ is smooth and $f'$ is nonvanishing as follows (taking advantage of the fact that $\alpha$ does not explicitly appear in the Lagrangian or the constraint):  Suppose that $I=[\alpha_0,\alpha_1]$ is an interval in which $f$ is smooth and such that $f'(\alpha)\not=0$ for $\alpha\in I$.  For simplicity, I'll assume that $f'>0$ on $I$, but the argument is essentially the same for $f'<0$.  Then we have $\mathrm{d}\alpha = \mathrm{d}f/f'$ so that the integral to be maximized over this interval becomes 
(after setting $f_i = f(\alpha_i)$)
$$
I = \int_{f_0}^{f_1} \frac{f^2(f')^2}{\sqrt{f^2+(f')^2}}\,\frac{\mathrm{d}f}{f'}
= \int_{f_0}^{f_1} \frac{f^2\,\mathrm{d}f}{g}\,,
$$
where I have set $g = \sqrt{(f/f')^2+1}$.  Using the same notation, the constraint integral becomes
$$
C = \int_{f_0}^{f_1} g\,\mathrm{d}f\,.
$$
Thus, we are trying to choose $g$ as a function of $f$ so as to maximize $I$ subject to the condition that $C$ is to be held constant.  Once we do this, we can recover $\alpha$ as a function of $f$ by using the fact that
$$
\mathrm{d}\alpha = \frac{\mathrm{d}f}{f'} = \frac{\sqrt{g^2{-}1}}{f}\,\mathrm{d}f\,.
$$
By the method of Lagrange multipliers, we need to choose $g = g(f)>1$ on the interval $[f_0,f_1]$ so as to render the integral $I_\lambda$ stationary, where $\lambda$ is a constant and
$$
I_\lambda = \int_{f_0}^{f_1} \left(\frac{f^2}{g}+\lambda g\right)\,\mathrm{d}f.
$$
Since $g'$ does not appear in the integrand, the Euler-Lagrange equation is particularly simple:  We must have $(f/g)^2 = \lambda$.  Setting $\lambda = c^2>0$, this gives $g = f/c$, so $c < f_0$ (since $g>1$) and we have 
$$
\mathrm{d}\alpha = \frac{\sqrt{f^2{-}c^2}}{cf}\,\mathrm{d}f\,
$$
Setting $f = c\cosh\tau$ where $\tau\in[\tau_0,\tau_1]$ with $0<\tau_0<\tau_1$,
we see that 
$$
\mathrm{d}\alpha = c \frac{\sinh^2\tau}{\cosh\tau}\,\mathrm{d}\tau
= c\,\mathrm{d}\left(\sinh\tau - 2\arctan(e^\tau)\right).
$$
Thus, the parametric curves (for $a$ and $c$ constants)
$$
\bigl(\alpha,f(\alpha)\bigr) = \bigl(a {+} c\sinh\tau {-} 2c\arctan(e^\tau),\ c\cosh\tau\bigr) 
$$
are the solutions to the Euler-Lagrange equations on the intervals where $f'>0$
(with a similar formula when $f'<0$).
Unfortunately, it turns out that these extrema are always local minima, not local maxima, as is easily verified:  Let $g = f/c + \epsilon h(f)$ where the integral of $h(f)$ over $[f_0,f_1]$ is zero and $h(f_0)=h(f_1)=0$.  This is the general perturbation that does not change the constraint integral $C$.  However, the integral $I$ then has the expansion 
$$
I = \tfrac12 c(f_1{-}f_0)^2 + \epsilon^2 c^3\int_{f_0}^{f_1} \frac{h(f)^2}{f}\,\mathrm{d}f + O(\epsilon^3).
$$
Thus, $\epsilon=0$ is always a strict local minimum.
Thus, one concludes that your original maximization problem has no solution.  (Meanwhile, the minimum, namely $I=0$, is clearly achieved by taking $f$ to be an appropriate constant.)<|endoftext|>
TITLE: Are there any natural recursively but not primitive-recursively axiomatized theories?
QUESTION [12 upvotes]: In principle, we could have a recursively axiomatized theory for which the property numbers-an-axiom (even relative to some routine Gödel numbering scheme) is recursive but not primitive recursive. But are there any natural examples?
Of course, any such theory can be primitive-recursively reaxiomatized using Craig's trick. So we know that there can't be theories which are recursively axiomatizABLE but not primitive-recursively aziomatizABLE. But that's not the issue. The question is whether there is a theory T which when presented in a natural way requires open-ended searches to check whether a purported T-proof is indeed a proof according to that specification.  
[I couldn't think of one when I wrote the first edition of my Gödel book, and I still can't as I work on the second edition. But maybe I'm just being dim/ignorant!]

REPLY [12 votes]: Here is another proposal. In this edition, the PS has been completely changed
Let $T_{ZFC}$ = arithmetical truths that $ZFC$ "knows about", i.e., the set of arithmetic sentences $\phi$ such that $ZFC$ proves $\phi^\omega$, where $\phi^\omega$ is the set-theoretical statement that expresses "$\phi$ holds in the von Neumann interpretation". 
$T_{ZFC}$ is an example of a natural r.e. theory whose overt axiomatization is not primitive recursive. Note that $T_{ZFC}$ is much stronger than $PA$, since includes all kinds of statements that are left undecided by $PA$, such as $Con(PA)$, $Con(PA + Con(PA)$, etc.
Moreover, one could argue that this theory is in some sense more natural than $PA$ since it is, implicitly, what mathematicians are really interested in.
Three notes:

There are axiomatizations of $T_{ZFC}$ that do not use the Craig trick; see this FOM posting for more information.
As observed by Kreisel, $T_{ZF}$ = $T_{ZFC + GCH}$ (the argument uses Gödel's constructible universe and absoluteness considerations).
In light of Harvey Friedman's programme of unearthing the deep role of large cardinals in our knowledge of the finite realm, one can also consider natural variants $T_{ZFC^{+}}$, where $ZFC^{+}$ is the result of augmenting $ZFC$ with appropriate large cardinals.
P.S. Since the original question stipulated that the theory, when naturally presented,  be recursive, I offer the following (thanks to Emil Jeřábek and Joel Hamkins for helping to improve this suggestion).

Let $f_G$ be the Goodstein function, i.e., $f_G(m)=n$ iff the length of the Goodstein sequence starting at $m$ is $n$. Note that $f_G$ is recursive, but not primitive recursive. Then perhaps the equational theory of $(\Bbb{N}, 0, 1, +, f_G)$  is recursive but not primitive recursive.
Perhaps even the full theory of $(\Bbb{N}, 0, 1, +, f_G)$ is decidable (i.e., maybe Presburger arithmetic augmented with the Goodstein sequence is a decidable theory).<|endoftext|>
TITLE: Examples of Eigensheaves outside of langlands
QUESTION [11 upvotes]: In geometric Langlands, one looks at correspondences of the form 
$$ Bun_n(X) \leftarrow Hecke \rightarrow X\times Bun_n(X)$$
and calls a sheaf on the lefthand space Hecke eigensheaf, if pulling back and pushing forward this sheaf is leads to the same result as to tensoring it with some sheaf (of a specific form) on $X$. 
Are there other examples of eigensheaves considered in mathematics? 
More precisely are there interesting examples of correspondences of the form
$$Y\leftarrow Z \rightarrow X\times Y$$
and sheaves on $Y$ for which pulling back and pushing forward this sheaf is leads to the same result as to tensoring it with some sheaf on $X$.

REPLY [11 votes]: Another example "on the other side" like Tony's (in fact really a stacky case of Tony's example) is de-equivariantization: given a $G$-space $X$ you can recover quasicoherent sheaves on $X$ from sheaves on $X/G$ (ie equivariant sheaves) as eigenobjects for the natural action of $Rep(G)=QC(BG)$ on $QC(X/G)$. 
This appears in various more disguised forms in math -- for example look at the paper of Arkhipov-Gaitsgory on representations of the small quantum group where exactly this structure appears (similarly in work of Frenkel-Gaitsgory on representations at the critical level, don't remember precisely where right now). To quote the abstract: "We show that the category of $u_\ell$-modules is naturally equivalent to the category of $U_\ell$-modules, which have a {\it Hecke eigen-property} with respect to representations lifted by means of the quantum Frobenius map $U_\ell\ti U(\check g)$, where $g$ is the Langlands dual Lie algebra. "
(Of course these examples are not unrelated to geometric Langlands, but maybe slightly different contexts for your question.)
Another kind of example is given by monodromic sheaves --- namely, you might ask to weaken the condition of equivariance of a sheaf under a group action by twisting, so rather than getting an invariant (aka equivariant) object (eigenobject with trivial eigenvalue) you get a nontrivial eigenvalue. The most familiar case of this is writing twisted D-modules on flag varieties (in the context of Beilinson-Bernstein) -- these are D-modules on $X=G/N$ which are not strongly equivariant for the torus, so the pullback by action to the torus times $X$ is not product with a trivial local system, but rather eigenobject with egienvalue a local system on the torus.
In any case I would claim the "interesting examples" won't come from considering a single correspondence like you suggest, but rather some monoidal category acting (eg via correspondences) on sheaves on your space. That monoidal category can be symmetric as in the Hecke examples, or a "group algebra" as in the monodromic example, or something else, and now you can ask for a kind of twisted invariance (objects that transform according to a given representation of your monoidal category). So I would rather ask, what are interesting examples of monoidal categories acting 
on categories of sheaves, besides the (wealth of) examples given by group actions on varieties?<|endoftext|>
TITLE: Lie algebras over non-algebraically closed fields
QUESTION [13 upvotes]: I am independently studying Lie algebras (in preparation for grad school) from James Humphrey's text "Introduction to Lie Algebras and Representation Theory. Fairly early on in the development (Chapter 2), the assumption that the underlying field is algebraically closed sneaks in and remains in place all the way through to the classification of semi-simple Lie algebras in Chapter 3. I understand where we need the closure in Lie's theorem etc., but I couldn't help wondering

How far can you get in the development of Lie algebras without assuming algebraic closure of the underlying field?

Are there classifications, or partial results etc.? Where can I find them in a format suitable for a beginner? Thanks for your help!

REPLY [12 votes]: Hmmm...   I guess I should try to say something in self-defense, beyond the reference I made to Jacobson's 1962 book Lie Algebras (since republished in paperback by Dover).   It's conventional in many algebraic classifications to start with a splitting field, then descend to smaller fields by something like Galois cohomology.   In the case of semisimple Lie algebras, which emerged from Elie Cartan's work a century ago on Lie groups and differential geometry, most people simplify by working first over an algebraically closed field of characteristic 0 (often specified arbitrarily as $\mathbb{C}$).  Classically the objective was to understand simple Lie algebras (and Lie groups) over $\mathbb{R}$, but this gets quite complicated even though the field extension involved is quadratic.   
The notion of splitting field here is fairly straightforward: you need all eigenvalues of ad-operators to be in your field in order to get a clean structure and classification theory.    This mostly comes down to looking at Cartan subalegras (or "maximal toral subalgebras") in semisimple Lie algebras.    But it's easier just to think about a familiar large field.    Here the classification is relatively elementary, thanks to Killing and Cartan and their successors.   But working over an arbitrary ground field gets quite intricate.   Here some motivation comes from the parallel but more difficult study of semisimple algebraic groups.   
Books and lecture notes abound, but usually at a fairly high level, and some books like those in the defunct Dekker series are hard to find as well as being photocopied from typescript.   Reviews I wrote for the AMS Bulletin long ago of a couple of these are free online from AMS at emath.org, written by Goto-Grosshans and by my graduate advisor George Seligman from a different viewpoint.
I didn't mention in my various reference lists the standard old book by Helgason now republished by AMS: Differential Geometry, Lie Groups, and Symmetric Spaces.   These and more recent books are pretty formidable, as are the much older research papers on structure and classification.    It's all there, but are there readable and accessible modern references?    Good luck.

REPLY [9 votes]: Well, certainly things get more complicated when the field is not algebraically closed, as you can see from the classification of finite-dimensional simple Lie algebras over $\mathbb{R}$.  But there are many cases where one just needs to be more careful with hypotheses.

In the proof of Lie's Theorem: for a solvable Lie algebra $\mathfrak{g}$ and a finite-dimensional representation $\pi : \mathfrak{g} \to \mathfrak{gl}(V)$, you can get the theorem as long as you assume that all of the eigenvalues of all of the endomorphisms $\pi(X)$ lie in the field you are working over.  I don't know how often this happens in practice when the field is not algebraically closed, but the proof does go through.
Engel's theorem goes through without a problem over any field, I believe.
The Poincare-Birkhoff-Witt Theorem works over any field (and in fact, for Lie algebras over any commutative ring $R$ where the underlying $R$-module of the Lie algebra is free).
Cartan's Criterion for Semisimplicity works over any subfield of $\mathbb{C}$, essentially because $\mathfrak{g}$ is semisimple if and only if $\mathfrak{g} \otimes \mathbb{K}$ is semisimple for any extension field $\mathbb{K}$ (note that this is not true for simplicity; if $\mathfrak{g}$ is a simple real Lie algebra which is not obtained from a simple complex Lie algebra by restriction of scalars, then the complexification $\mathfrak{g} \otimes \mathbb{C}$ is not simple.)

Hmmm... I'm sure there's a lot more to say.  Let's just wait until Professor Humphreys shows up.<|endoftext|>
TITLE: Ultrainfinitism, or a step beyond the transfinite
QUESTION [6 upvotes]: Cantor has, in the immortal words of D. Hilbert, given all of  us a paradise (or perhaps, I would  rather say, a great vacation spot), the TRANSFINITE.
$\aleph_0, \aleph_1,\aleph_2\dots$
the lists goes on forever, into higher and higher ethereal realms. In his theological mind,  Cantor thought of these dots as an eternal  ladder, which approaches (without ever  reaching it) the Absolute Infinite, later re-christened as $V$, the Universe of Sets, by Set Theory adepts.
Those same adepts have enriched Cantor's paradise with a great bestiary of enormous cardinals, inaccessibles, Mahlo, Vopenka, Woodin cardinals, etc. Big fellows, no doubt. Yet... In comparison with the size of $V$ they are puny, nil in fact, no more no less as Graham number, or Friedman's TREE(3) stand in comparison to (for finitists) almighty $\omega_0$.  
Now, let us be brave and say: what about breaking through into the trans-transfinite?
What about , for instance, starting from $V$ itself and state that its size is some hyperinfinte number, say $\aleph_{0,1}$ ?
(SIDE NOTE ON NOTATION: The standard aleph series would  now be $\aleph_{0,0}$ , $\aleph_{1,0}$, .... The second subindex controls the degree of hyperfiniteness, much like degrees of unsolvability. I could have put it on top, but then it would cause troubles with cardinal exponentiations ).
Wait, I hear you say loud and clear. Are you crazy?
Don't you know that there is NO SET $X$ such that $X=V$?
Don't you know that there is no max ordinal? 
Yes, ladies and gentlemen, I do know it. But I do reply: and so what? The objection is exactly the same as the one of the finitists vis-a'-vis $\omega$. Someone has broken through the finite, so why not the transfinite? There is no set, but who said that it must be a set? 
In fact, start with a pairs of transitive countable  models of ZFC, $M_0$ and $M_1$, with $M_0\leq M_1$, of different tallness (the ordinal height of the first being strictly smaller than the height of the second). From the point of view of $M_0$, IT is the full universe of sets, and the ideal ordinals of $M_1$ some unimaginable higher level of infinity. Of course, say you, $M_0$ does not see $M_1$. 
True, but we do. And -I think- nothing prevents us from formalizing their reciprocal relation as some new theory of sets (the elements of $M_0$) and classes (the elements of $M_1$). Note that here all sets are classes, but not viceversa. 
Also, being  more reckless, we could generalize the above by stipulating an entire chain of ascending hyper-infinities, and perhaps enrich ZFC with an axiom that says that for each model there is a cofinal (in V) ascending chain of taller models, the Cofinal Tallness Axiom....
OK, now the question(s): 

(set-theory) has anything like the above be attempted? 
(algebra) can we create a system of "numbers" which strictly contains cardinals plus other numbers strictly greater than them? And if yes, what is their arithmetics? 

NOTE: by 2 I mean: axiomatize directly the class CARDINALS. Then find a new class of numbers, say HYPERCARDINALS, which contains CARDINALS as an initial segment, and moreover such that the numbers in HYPERCARDINALS - CARDINALS has some arithmetical property that ordinary cardinals, no matter how large, have not (this will rule out simply having copies of cardinals appended after one another). 

(philosophy) is there any speculation as to a radically NEW notion of infinity, which makes all large cardinals small? 

NOTE: this is of course connected to 2 above, but would interpret the new arithmetical/algebraic  characteristics of the hyper-cardinals as speaking of new properties of hyper-infinite classes. Essentially this interpretation would unravel new conceptualizations of the informal notion "being infinite" . Of course, the challenge here is to steer away from blatant inconsistencies, such as the ones discovered in the early history of Set Theory, and which were eliminated in the formalized ZF approach.  
Any reference, thought, criticism, and what not is most welcome.

REPLY [4 votes]: I would go so far as to say one has to not think of things larger than the collection of classes (in so far as such a thing is even defined) not as just 'bigger sets' but something else. I posit that that something else is ... categories! In particular, not just categories in the fairly vanilla sense as a class of objects (which may be a set) and a set of arrows (or perhaps a class) between any two pairs of objects, but using the first order definition of a category without an equality predicate on objects, and a dependent-type version of equality on objects (can only compare arrows if they are in the same hom-collection). 
A small category with an equality predicate on its objects admits a(n essentially) surjective functor from a discrete category if we assume enough choice. In ordinary foundations (such as ZF(C), NBG or variants), Vopěnka's principle is a large cardinal axiom equivalent to the assertion that there are no subcategories of a locally presentable category (e.g. $Set$) which are simultaneously large (have a class of objects) and discrete. The principle can be seen a shadow in ordinary foundations of the idea that there should be categories which are really just too big to have a collection of objects that behaves like a set. Notice that one can  form the posetal coreflection $Pos(C)$ of a category $C$ (it has the same objects and there is a unique arrow in $Pos(C)$ between any two objects if and only if there is any arrow between the analogous objects in $C$), and even take the core (the largest subgroupoid) of this. But we cannot get a category with an equality predicate on its objects unless we are happy to form some sort of quotient of $Core(Pos(C))$ to get a discrete category, and then it requires serious use of global choice on these super-large 'collections' to turn the canonical functor $Core(C)\to Core(Pos(C))/\sim$ into a functor $Core(Pos(C))/\sim \to Core(C) \to C$ to get an essentially surjective functor from a discrete category.
As a sort of half-way between this notion of category which is too large to be a class, we have the first-order characterisation of the category of classes, otherwise known as algebraic set theory. One could apply the more philosophical ideas from the above paragraphs to the very concrete definitions of algebraic set theory (see for instance section 3.1 of this introduction). One would then have a category of classes which is itself genuinely not a meta-class, nor some sort of collection which behaves like a class, nor just a 'class' corresponding to a large cardinal in model/universe of set theory containing the current model/universe. This would probably require playing around with the axiom (US) (section 3.1 here).
Mike Shulman has some good comments on similar (though less extreme) ideas in this answer.
If one complains that this is just a first step, and really we want a whole hierarchy  of notions of 'bigger than anything we can come up with so far', then Michael Makkai has considered foundations (a sort of type theory, called by him FOLDS) in which it is impossible to consider equality (as above), isomorphism (as might be considered natural in 2-categories, for example), equivalence,... so that we can really only talk about each of these notions if we are working in an $n$-category for some finite $n$, and in general the only available generalised notion of equivalence is full-blown $\omega$-equivalence of $\omega$-categories. But this sort of approach has not been thought of in the sense of making larger and larger hierarchies of objects. (It has come up in Voevodsky's univalent foundations, but only from a homotopy point of view.)<|endoftext|>
TITLE: Homology of homotopy fixed point spectra
QUESTION [8 upvotes]: Let $E$ be a spectrum acted upon a finite group $G$. Is there a general way of computing the homology of the homotopy fixed point spectrum $E^{hG}$ in terms of that of $E$? (I'm aware that there is a spectral sequence for computing $\pi_* E^{hG}$ in terms of $\pi_* E$, but smashing with some other spectrum probably doesn't preserve homotopy fixed points.)
Here's a specific example I have in mind. Take connective $K$-theory $bu$. This has an action of $\mathbb{Z}/2$, which comes from the $\mathbb{Z}/2$-action on $K$-theory (given on the level of cohomology theories by complex conjugation of vector bundles). Then $bu^{\mathbb{Z}/2} = bo$. Actually,  this is only true before taking connective covers. 
Let's say I know how to compute the mod 2  homology of $bu$ (it's $\mathbb{Z}/2[\zeta_1^2, \zeta_2^2, \zeta_3, \zeta_4, \dots]$ as a comodule over the dual Steenrod algebra). Does that give any information about $H_*(bo; \mathbb{Z}/2)$? Is there a good reference for this material?

REPLY [4 votes]: Maybe to expand on the relationship between naive and genuine G-spectra, Tom's answer is stated
in terms of naive G-spectra, but the actual mathematics, the Segal conjecture, is all about genuine G-spectra.  The genuine sphere $G$-spectrum $S_G$ has non-trivial $G$-action and nontrivial $G$-fixed point spectrum.  Consider a finite $G$-CW complex. The $G$-map $EG\times X \to X$ induces a map of genuine function $G$-spectra 
\[  F(X_+, S_G) \to F((EG\times X)_+, S_G). \]
If $G$ is a $p$-group, the Segal conjecture says that this map is an equivalence after $p$-adic
completion.  This remains true after passing to $G$-fixed point spectra.  By definition, that passage
means that you first take underlying naive $G$-spectra and then take actual fixed points.  Tom is 
taking $X$ to be a point, and then the map is
\[ (S_G)^G \to  F(EG_+, S_G)^G, \]
going from the fixed point spectrum to the homotopy fixed point spectrum of the
genuine $G$-spectrum $S_G$.  The target is equivalent to the naive homotopy fixed point spectrum 
$F(EG_+, S)^G$ of the naive sphere $G$-spectrum with trivial action by $G$, and that is
equivalent to $F(BG_+,S)$, giving the cohomotopy groups of $BG$ on passage to homotopy groups.
But all of the real mathematics in validating Tom's answer takes place on the level of genuine $G$-spectra.<|endoftext|>
TITLE: Fundamental problems whose solution seems completely out of reach
QUESTION [53 upvotes]: In many areas of mathematics there are fundamental problems that are embarrasingly natural or simple to state, but whose solution seem so out of reach that they are barely mentioned in the literature even though most practitioners know about them. I'm specifically looking for open problems of the sort that when one first hears of them, the first reaction is to say: that's not known ??!! As examples, I'll mention three problems in geometry that I 
think fall in this category and I hope that people will pitch in either more problems of this type, or direct me to the literature where these problems are studied. 
The first two problems are "holy grails" of systolic geometry---the study of inequalities involving the volume of a Riemannian manifold and the length of its shortest periodic geodesic---, the third problem is one of the Busemann-Petty problems and, to my mind, one of the prettiest open problems in affine convex geometry. 
Systolic geometry of simply-connected manifolds. Does there exist a constant $C > 0$ so that for every Riemannian metric $g$ on the three-sphere, the volume of $(S^3,g)$ is bounded below by the cube of the length of its shortest periodic geodesic times the constant $C$?
Comments. 

For the two-sphere this is a theorem of Croke. 
Another basic test for studying this problem is $S^1 \times S^2$. In this case the
fundamental group is non-trivial, but in some sense it is small (i.e., the manifold is not essential in the sense of Gromov). 
There is a very timid hint to this problem in Gromov's Filling Riemannian manifods.

Sharp systolic inequality for real projective space. If a Riemannian metric in projective three-space has the same volume as the canonical metric, but is not isometric to it, does it carry a (non-contractible) periodic geodesic of length smaller than $\pi$? 
Comments.

For the real projective plane this is Pu's theorem.
In his Panoramic view of Riemannian geometry, Berger hesitates in conjecturing that this is the case (he says it is not clear that this is the right way to bet). 
In a recent preprint with Florent Balacheff, I studied a parametric version of this problem. The results suggest that the formulation above is the right way to bet.

Isoperimetry of metric balls. For what three-dimensional normed spaces are metric balls solutions of the isoperimetric inequality?
Comments.

In two dimensions this problem was studied by Radon. There are plenty of norms on the plane for which metric discs are solutions of the isoperimetric problem. For example, the normed plane for which the disc is a regular hexagon.
This is one of the Busemann-Petty problems.
The volume and area are defined using the Hausdorff $2$ and $3$-dimensional measure.
I have not seen any partial solution, even of the most modest kind, to this problem.
Busemann and Petty gave a beautiful elementary interpretation of this problem: 

Take a convex body symmetric about the origin and a plane supporting it at some point $x$. Translate the plane to the origin, intersect it with the body, and consider the solid cone formed by this central section and the point $x$. The conjecture is that if the 
volume of all cones formed in this way is always the same, then the body is an ellipsoid. 
Additional problem: I had forgotten another beautiful problem from the paper of Busemann and Petty: Problems on convex bodies, Mathematica Scandinavica 4: 88–94. 
Minimality of flats in normed spaces. Given a closed $k$-dimensional polyhedron in an $n$-dimensional normed space with $n > k$, is it true that the area (taken as $k$-dimensional Hausdorff measure) of any facet does not exceed the sum of the areas of the remaining facets?
Comments.

When $n = k + 1$ this is a celebrated theorem of Busemann, which convex geometers are more likely to recognize in the following form: the intersection body of a centrally symmetric convex body is convex. A nice proof and a deep extension of this theorem was given by G. Berck in Convexity of Lp-intersection bodies, Adv. Math. 222 (2009), 920-936.
When $k = 2$ this has "just" been proved by D. Burago and S. Ivanov: http://front.math.ucdavis.edu/1204.1543
It is not true that totally geodesic submanifolds of a Finsler space (or a length metric space) are minimal for the Hausdorff measure. Berck and I gave a counter-example in What is wrong with the Hausdorff measure in Finsler spaces,  Advances in Mathematics, vol. 204, no. 2, pp. 647-663, 2006.

REPLY [4 votes]: Here is a variation of Georges Elencwajg's question, due to Gennady Lyubeznik. Is every closed point (of arbitrary degree over $\mathbb{Q}$) in $\mathbb{P}^2_{\mathbb{Q}}$ set-theoretically the intersection of two curves?<|endoftext|>
TITLE: Geometric Interpretations of Homotopy Theoretical Constructions
QUESTION [10 upvotes]: In homotopy theory there are lots of nice constructions that seem designed to have some effect on the homotopy of a space, i.e. completing, localizing, and taking various homotopy (co)limits. It seems like some of these (for instance Quillen's + construction) have moderately visual interpretations by attaching cells and things like that. In general, does anyone out there have interest in constructing such things geometrically, i.e. drawing a picture of what a sphere looks like after you, I don't know, complete it at a prime, or something like that? Are there sensible ways to visually represent mod p Moore spaces? Do such constructions usually cause things to have to leap into higher, non-visual, dimensions or something? Or is it simply that this kind of thing doesn't really serve any purpose, and we can just prove that it works and that it exists? In general it also seems very interesting to me to know what a sphere localized at a given homology theory might look like.

REPLY [17 votes]: Though it is often satisfying and useful to have an appropriate way to visualize a construction, in fact the real innovation in homotopy theory that lead to the constructions you mention, among many others, was the realization that you need not limit your attention to geometric, visualizable operations and spaces but can be content knowing only certain relevant properties.  
This realization led to what Mike Hopkins (and perhaps others before him) called designer homotopy theory.  The simplest example of a designer homotopy type is the Eilenberg-MacLane spaces---once you convince yourself they exist, you don't concern yourself with what they look like (a horrendous infinite CW complex, perhaps), but use only their characteristic relationship to homotopy groups.  Needless to say this particular example led to spectacular advances in the hands of Serre and others.

REPLY [6 votes]: Many operations on the homotopy groups of a CW complex can be realized by a "geometric" (rather, topological) construction on the space itself. A popular example is the rationalization, studied in rational homotopy theory.
The general idea is to write down a CW model for your space, with explicit attaching maps, and doing the construction on spheres, then re-glue by the attaching maps.
An easy example to look at is the following construction on the 1-sphere:
take the covering $z \mapsto z^n$ from the 1-sphere to itself. A path on the "lower" circle that generates the fundamental group $\pi_1(S^1,1)=\mathbb{Z}$ can be multiplied with $n$ (going around $n$ times) and then lifted to the covering space and there it corresponds to the generator. Attaching a 2-cell along this covering map kills this $n$-times going around map on the homotopy level (since it is liftable, so contractible in the newly attached 2-cell). The result on the fundamental group is that you get $\pi_1(S^1 \cup_{z^n} e^2) = \mathbb{Z}/n\mathbb{Z}$.
You can iterate this with a mapping telescope construction to get an inverse limit over these groups, thus the fundamental group becomes the profinite completion $\hat{\mathbb{Z}}$ of the integers.
Since you asked for how to visualize this, I tend to draw pictures on the blackboard and imagine the paths and the contractions quite visually.
To address the question whether this is useful or just good to know it exists:
Sometimes it can be crucial to know how many cells in which dimensions you need to attach to change the homotopy groups, and knowledge about the attaching maps can be important too. For example, from all attaching maps you can compute CW (co)homology, so you can look at how the (co)homology changes if you, say localize at a prime. Maybe the Quillen + construction is a good example for this, since one really has to attach only a 2-cell and then a 3-cell, to get it right.<|endoftext|>
TITLE: ad (A^n) is a polynomial in ad A ?
QUESTION [8 upvotes]: Let $k$ be a field and $n$ a nonnegative integer. For any matrix $U\in\mathrm{M}_n\left(k\right)$, let $\mathrm{ad} U$ denote the map $\mathrm{M}_n\left(k\right)\to \mathrm{M}_n\left(k\right),\ V\mapsto UV-VU$. Thus, $\mathrm{ad} U$ is an element of the $k$-algebra $\mathrm{End}_k\left(\mathrm{M}_n\left(k\right)\right)$.
Is it true that for every $n\times n$-matrix $A$ over $k$, and for every $m\in\mathbb N$, the endomorphism $\mathrm{ad}\left(A^m\right)$ can be written in the form $P\left(\mathrm{ad}A\right)$ for some polynomial $P\in k\left[X\right]$ satisfying $P\left(0\right)=0$ ?
I know that this holds when $A$ is diagonalizable, and in that case it is used in the proof of Cartan's Lemma from Lie algebra theory. If it holds generally and can be proven neatly, it could be used to tidy up the proof of Cartan's Lemma (which, in the form I know it, is rather ugly, requiring an algebraic extension of the ground field and the use of Jordan's normal form).

REPLY [7 votes]: I don't think it's true even when $A$ is diagonalisable.
Suppose that $A$ is a diagonal matrix, with $(i,i)$th entry $\lambda_i$; thus $A = \rm{diag}$$(\lambda_i)$. Then we can write $\rm{ad}(A) = \rm{diag}$$(\lambda_i - \lambda_j)$, so 
$P(\rm{ad}$$(A)) = \rm{diag}$$ (P (\lambda_i - \lambda_j) )$
for any polynomial $P(t)$.
On the other hand, $A^m = \rm{diag}$$(\lambda_i^m)$ so
$\rm{ad}$$(A^m) = \rm{diag}$$(\lambda_i^m - \lambda_j^m)$.
So if $P(\rm{ad}$$(A)) = $$\rm{ad}$$(A^m)$ then $P(\lambda_i - \lambda_j) = \lambda_i^m -\lambda_j^m$ for all $i,j$ such that $1\leq i,j\leq n$. 
But it can happen that $\lambda_i - \lambda_j = \lambda_a - \lambda_b$ for two pairs of indices $(i,j)$ and $(a,b)$ with  $\lambda_i^m - \lambda_j^m \neq \lambda_a^m - \lambda_b^m$.
For a concrete example we can take $A = \rm{diag}$$(2,1,4,3)$.<|endoftext|>
TITLE: Kervaire invariant: Why dimension 126 especially difficult?
QUESTION [35 upvotes]: Is there any resource that might help non-experts gains some understanding of why
the Kervaire invariant problem remains open now only in dimension $126$? ($126 =2^7-2=2^{j+1}-2$;
whether $\theta_j=\theta_6$ exists in the "$128$-stem"), i.e., why the celebrated Hill-Hopkins-Ravenel 
proof technique fails in this last remaining case?
I read a delightful exposition by Erica Klarreich ("Mathematicians solve 45-year-old Kevaire Invariant Puzzle,"
The Best Writing on Mathematics, 2010, 373ff, Simons Foundation link), which piqued my interest.
But Michael Hopkins' online presentation slides announcing the result
in 2009 ("Applications of algebra to a problem in topology") are beyond my ken.
Perhaps this an area too abstruse for all but the experts?
I'd appreciate pointers to expositions.  Thanks!

REPLY [17 votes]: More detail on Jones's 30 manifold and analogous constructions can be found in my  Extended powers of manifolds and the Adams spectral sequence , Cont. Math. 271, 41--51.  The basic idea in it dates from 1979 or so (immediately upon seeing Jones's construction) but didn't make it into print for another 20 years.  There are also comments about the homotopy theoretic approach to the problem in   the talk I gave in Edinburgh  in 2011 at the workshop on the Kervaire invariant.
The main point is that there is a small set of homology classes in low dimension (dimension 2 for Jones' construction) which one wants to realize as the fundamental class of a manifold with tangent bundle realized by a permutation representation of $\pi_1$.  To get the 30 dimensional Kervaire manifold from $S^7$ one only needs a two manifold and an appropriate representation in $S_4$.  ($D_8$ will do:  only the Sylow 2-subgroup matters.)  To get from $S^7$ to the 62 dimensional Kervaire manifold requires a 6-manifold with a representation of $\pi_1$ in $S_8$.  Unfortunately, no such manifold exists.  In dimension 126, one would need a 14 manifold, and presumably such also fails to exist. So, one needs to find another approach that relaxes the input data needed.<|endoftext|>
TITLE: Same betti numbers as $\Bbb{CP}^n$
QUESTION [5 upvotes]: I am sure that there is an answer out there for the following question. If one is given an n dimensional Kahler manifold $X$ with Betti numbers that are the same as in the case of $\Bbb{CP}^n$, then is $X$ holomorphically diffeomorphic to $\Bbb{CP}^n$? This is of course true in the one dimensional case, but other then that I am clueless. If this is textbook stuff that I might have missed, references will be appreciated :).

REPLY [5 votes]: Dear Common Crane,
let me list several results and question going in the direction of your intuition
(all the information below I learned from Sergey Galkin)
1) Theorem. Hirzebruch-Kodaira, Yau.  A Kahler manifold homeomorphic to $\mathbb CP^n$ is biholomorphic to $\mathbb CP^n$. There is a nice exposition of this result by Valentino Tossati: http://www.math.northwestern.edu/~tosatti/cpn.pdf
2) There is a question of Wilson: 
If $V$ is a complex projective manifold of even dimension $n>k(V)$ and which has the same rational cohomology as $P^n$, is $V$ isomorphic to $P^n$? 
3) In dimension 3 there are "four" complex projective manifolds with same Betti numbers as $\mathbb CP^3$ (these manifolds are $P^3$, the quadric, the manifold called $V_5$ and finally $V_{22}$. The first three manifolds are rigid, but the last one has six-dimensional moduli space of deformations).
4) Finally you can find some interesting recent development related to the question 
in the preprint  http://member.ipmu.jp/sergey.galkin/papers/ipmu-11-0100.pdf
where you will as well find references for articles in 1), 2).<|endoftext|>
TITLE: Operations via Morse Theory
QUESTION [59 upvotes]: I am interested in seeing if and how Morse Theory can "do everything". Some core things are handle decomposition, Bott periodicity, and Euler characteristic. But what do the normal (co)homology operations look like from Morse Homology?
Poincare duality $H_*(M)\cong H^{n-\ast}(M)$ is the symmetry $f\to -f$, i.e. the reversal of flowlines.
Cup product $H^i(M)\otimes H^j(M)\to H^{i+j}(M)$ is given by counting Y-shaped flowlines,  using Morse functions along each of the three edges. The cap product is connected to the above two.
Kunneth isomorphism $H_\ast(M\times N)\cong H_\ast(M)\otimes H_\ast(N)$ is combining flowlines from $f_1:M\to\mathbb{R}$ and $f_2:N\to \mathbb{R}$ to get flowlines for $f_1+f_2:M\times N\to\mathbb{R}$.
Leray-Serre spectral sequence: pull back a Morse function on the base (flowlines of total space project onto flowlines of base space) and use a filtration by ordering the critical-point indices.
Does someone know what goes on for the following?

Slant product
Alexander duality
Steenrod operations (in particular, the Cartan formula)
Massey triple product

My guess for (4) is counting X-shaped flowlines, and then I get suspicious about its relation to $A_\infty$-structures from Lagrangian-Intersection Floer homology.
[[Edit]] There was a MathOverflow post for (2), here. Alexander duality $H_\ast(S^n-M)\cong H^{n-1-\ast}(M)$ arises by taking a height function on $S^n$ and perturbing it to become Morse on the subspace $M\subset S^n$, and then separating the critical points according to its tubular neighborhood and its complement.
[[Edit]] Cohen and Schwarz' paper "A Morse Theoretic Description of String Topology" provides the relative cohomology and the Thom isomorphism, as well as homomorphisms arising from proper embeddings of submanifolds.

REPLY [9 votes]: Steenrod operations in Floer homology are constructed in the thesis of Matthias Schwarz: http://www.math.uni-leipzig.de/~schwarz/diss.pdf
Perhaps a similar construction is feasible for Morse homology?<|endoftext|>
TITLE: Compactness of the set of densities of equivalent martingale measures
QUESTION [12 upvotes]: Consider an incomplete market $(\Omega,\mathcal F,\mathbb P)$ driven by a semimartingale $S=(S_t)_{t\in[0,T]}$. Under the no free lunch under vanishing risk (NFLVR) assumption, the set $\mathcal P^\ast$ of equivalent martingale measures under which $S$ is a sigma-martingale is non-empty:
$$
\mathcal P^\ast\neq\emptyset,\quad\mathbb P^\ast\sim\mathbb P,\enspace \forall \mathbb P^\ast \in \mathcal P^\ast.
$$
Denote densities 
$$
Z_{\mathbb{P^*}}=\frac{d\mathbb P^\ast}{d\mathbb P},\quad \text{for }\mathbb P^\ast\in \mathcal P^\ast,
$$
and the set of densities
$$Z_{\mathcal{P^\ast}}=\{Z_{\mathbb{P^\ast}}\ :\ \mathbb P^\ast\in \mathcal P^\ast\}$$
Now, a certain theorem (a generalized version of the Neyman-Pearson lemma for incomplete markets, see Theorem 4.9 p.55 in this thesis) requires that $Z_{\mathcal P^\ast}$ is compact in $L^1(\Omega,\mathcal F,\mathbb P)$. I would like to see how restricting this assumption is by trying to provide an incomplete market example (i.e., a model where $\mathcal P^\ast$ is not a singleton) where $Z_{\mathcal P^\ast}$ is compact. However, I haven't been able to do so.
I first tried the discrete-time one-period trinomial model, in which the set of martingale measures and the set of associated densities represent convex polyhedrons. However, due to the measure equivalence requirement, the polyhedron is open (at the extreme points, one or more nodes will have zero probability, hence the extreme points are excluded along with the entire boundary). Therefore, it is not compact?
Next thing, I tried the continuous-time jump-diffusion model. The set of equivalent martingale measures can be parametrized with two real numbers, and the Radon-Nikodym derivative can be written down explicitly, however I have no idea how to approach the problem of verifying the compactness property.

Could someone provide an example of an incomplete market (where the set of equivalent martingale measures is not a singleton) where the set of densities is indeed compact?

Alternatively, can it be proved that $Z_{\mathcal P^\ast}$ is compact if and only if $\mathcal P^\ast$ is a singleton?

Update
As pointed below by @weakstar, if there exists a martingale measure $\mathbb Q$ that is absolutely continuous with respect to $\mathbb P$ but not equivalent to $\mathbb P$, a convex combination of $\mathbb Q$ and an equivalent martingale measure $\mathbb P^\ast\in\mathcal P^\ast$ would also be an equivalent martingale measure. In this case, it is possible to construct a sequence of densities $Z_{\mathbb Q^\alpha}$ of equivalent martingale measures converging to density  $Z_{\mathbb Q}$ of a non-equivalent martingale measure which violates the closedness. This example naturally eliminates models like the multinomial model.

REPLY [4 votes]: I guess this answer comes way too late. 
In case you are still trying to see if the set of densities of absolutely continuous martigale measures may be compact in the incomplete case ... the answer is "almost  never".
In " Representing Martingale measures when asset prices are continuous and bounded ", F. Delbaen roughly speaking proves that as soon as your filtration is well-behaved, the aforementioned set of densities is not weakly compact in the complete case. This rules out norm compactness of course.
Regards<|endoftext|>
TITLE: Proof by `universal receiver'
QUESTION [13 upvotes]: Anyone following the news knows about the major breakthoughs that have taken place recently in $3$-manifold topology. These have come via a route whose big-picture I find to be conceptually interesting. To prove a property $P$ (largeness, linear over $\mathbb{Z}$, virtual fibering, LERF, virtually biorderable etc.) for a class of objects $X$ (e.g. fundamental groups of finite volume hyperbolic $3$-manifolds), embed each object of $X$ nicely in an object of a `universal receiver' class of objects $R$ (Right-Angled Artin Groups, or RAAGs), each of which are simple and has good properties. The existence of such an embedding in itself implies property $P$ which you are interested in, maybe with some additional effort (Agol's fibering theorem, tameness, etc.). 
Proving a mathematical statement in this way makes a lot of sense, but I don't recall having seen this proof pattern before anywhere else in mathematics. Well, that's not entirely true- Cayley's Theorem that a group embeds in a permutation group has some corollaries (e.g.  Given a group $G$ and subgroup $H$ with $[G:H]=n$, there exists a exists a normal subgroup $N$ of $G$, with $N\subseteq H$ such that $[G:N]|n!$).

Question: Which other conjectures, that objects in a class $X$ have a property $P$, have been proven by embedding objects in $X$ nicely as subobjects of objects in a universal receiver $R$ whose good properties imply $P$ for objects in $X$?

For a compelling example, it would have to be difficult to prove $P$ for objects in $X$ in any other way. For an even more compelling example, the universal receiver $R$ would be surprising (RAAGs are a surprising universal receiver, I think).

REPLY [2 votes]: Part of the proof of the Green-Tao theorem involves embedding the primes into the "semiprimes", which are sufficiently pseudorandom for results like Szemerédi's theorem to apply. In fact, a lot of recent progress has been made with regards to the relative Szemerédi theorem, and maybe there is some hope that Erdős' conjecture on arithmetic progressions could be proven by embedding any sufficiently dense set of numbers into some pseudorandom set.<|endoftext|>
TITLE: Are all irreducible supercuspidal representation induced from compact-mod-center subgroups?
QUESTION [8 upvotes]: Let $G$ be a reductive group over a local non-archimedean field $F$.
Can every irreducible supercuspidal representation of $G(F)$ be realized as the induction from an open subgroup, which is compact modulo the center?

REPLY [3 votes]: It is perhaps worth noting that all the proofs Broussous mentions are along the lines of: Construct some types that induce irr scs and then show that all irr scs are thus obtained. In other words they are explict constructions. But that is stronger than the original question asked. One might hope that there would be a general proof that all irr scs are induced for reductive p-adic groups that doesn't produce the types.<|endoftext|>
TITLE: Topology of theta nulls
QUESTION [5 upvotes]: Siegel upper half-space, $\mathfrak{h}_g$, consists of symmetric $g\times g$ complex matrices with positive-definite imaginary part. From an element $Z\in \mathfrak{h}_g$ we can construct a theta function $\theta(Z, \cdot): \mathbb{C}^g\rightarrow \mathbb{C}$ which may be thought of as a section of a polarizing line bundle on a certain principally polarized abelian variety whose period matrix is $\Omega = (I\ \ Z)$ (see, for example, Griffiths & Harris). If now we let the first argument vary as well, we obtain a holomorphic function $\mathfrak{h}_g\times \mathbb{C}^g\rightarrow \mathbb{C}$ which induces a function on $\mathfrak{h}_g$ given by $Z\rightarrow \theta(Z,0)$. Such a function is called a theta null. For a little more detail and a bit of context, check out this question on moduli spaces of curves
I am interested in the topology of the vanishing loci of theta nulls in dimensions $g\geq 2$. Specifically, I'd like to know more about their connectivity. Is much known in this area, besides whatever is known for analytic hypersurfaces in general? Any information would be greatly appreciated and, of course, references are welcome.

REPLY [4 votes]: For $g>2$ the locus $M_g^1$ of "vanishing theta nulls" = set of $X$ in $M_g$ with a spin bundle such that $h^0(L) > 1$ is shown to be a connected divisor in:
MONTSERRAT TEIXIDOR I BIGAS
The divisor of curves with a vanishing theta-null
Compositio Mathematica, tome 66, no 1 (1988), p. 15-22.<|endoftext|>
TITLE: Some help in digesting a paragraph in the introduction of Deligne/Rapoport's "Les Schemas de Modules de Courbes Elliptique"
QUESTION [5 upvotes]: http://www.springerlink.com/content/04x54gr171v556m4/fulltext.pdf
On page 149 (DeRa-7), in the middle of the page, I can translate the middle paragraph that starts "3. La surface de Riemann ..." as follows:
3.The Riemann surface $X/\Gamma$ is not compact. Geometrically, this fact is reflected as follows: If $E_\eta$ is an elliptic curve equipped with a level $n$ structure over $\mathbb{C}((T))$, it follows that the minimal model of $E_\eta$ over $\mathbb{C}[[T]]$ has bad reduction. In this case, the special fiber $E_0'$ of the Neron model $E'$ of $E_\eta$ over $\mathbb{C}[[T]]$ is isomorphic to $\mathbb{C}^*\times\mathbb{Z}/kn\mathbb{Z}$ for some suitable $k$. Let $E_0$ be the subgroup of $E_0'$ consisting of the components of $E_0'$ which have order dividing $n$ in $\pi_0(E_0')$. This subgroup is isomorphic to $\mathbb{C}^*\times\mathbb{Z}/n\mathbb{Z}$...
Assuming I've translated it correctly, I've got three questions:

The first sentence seems to be saying that any elliptic curve over $\mathbb{C}((T))$ has bad reduction, which (unless the minimal model isn't what I think it is) is obviously not true, since any elliptic curve over $\mathbb{C}$ is also an elliptic curve over $\mathbb{C}((T))$. Where am I going wrong here?
Earlier the author defined a level $n$ structure as just an isomorphism from the $n$-torsion of the elliptic curve to $(\mathbb{Z}/n\mathbb{Z})^2$. Thus, any such elliptic curve $E_\eta$ has level $n$ structure for every $n$, which would seem to imply that the special fiber of the Neron model is isomorphic to $\mathbb{C}^*\times\mathbb{Z}/kn\mathbb{Z}$, for every $n$, which obviously makes no sense. (I thought the Neron model depends only on the base ring and the elliptic curve?)
I'd appreciate a quick description of what $\pi_0$ of a scheme is (or a good reference for learning about it), and specifically how it's a group (I understand it should intuitively represent the connected components). I've just this week learned about the etale fundamental group $\pi_1$, though I'm not entirely sure how the definition generalizes to higher/lower fundamental groups.

(Also, is there a good translation of this somewhere?)
thanks

will

REPLY [13 votes]: "Il arrive que..." means "sometimes".  So the paragraph says that sometimes the minimal model of $E_\eta$ over $\mathbf{C}[[t]]$ has bad reduction, which is true.
You're starting with an elliptic curve over $\mathbf{C}((t))$, not over $\mathbf{C}$.  There's no reason that $E_\eta$ admits a level $n$ structure over $\mathbf{C}((t))$ (as opposed to over some finite extension).  Geometrically what's going on (morally, in complex analytic geometry) is that you have a family of elliptic curves over the punctured complex unit disk, and you want to choose a level $n$ structure on the elliptic curve at each point so that the level structures vary nicely.  You can do this locally, but not globally (in general).  The algebro-geometric analogue of this fact is that for any fixed $n$, $E_\eta$ admits a level $n$ structure after some finite extension of the base (= passing to some cover of the punctured unit disk) but not necessarily over $\mathbf{C}((t))$ itself.  You can see this concretely and algebraically by thinking aboutWeierstrass equations for an elliptic curves over $\mathbf{C}((t))$, thinking about the equations defining the coordinates of the $n$-torsion points, and seeing that there's no reason for those equations to be solvable if the ground field is not algebraically closed.
$\pi_0(E_0')$ refers to the group of connected components of $E_0'$.  Normally $\pi_0$ is just a set (literally the set of connected components, not just intuitively), but since $E_0'$ has a group structure, you can add two connected components by picking a point on each one, adding the points, and taking the component this sum lands on (you should check that this is well-defined).

Finally, you should note that this paragraph doesn't really make sense, except as motivation: $X/\Gamma$ is a complex analytic object, whereas the rest of the paragraph takes place in the category of schemes.<|endoftext|>
TITLE: Heuristics for the Hodge Conjecture
QUESTION [8 upvotes]: W. V. D. Hodge is famous for his Hodge conjecture, one of the Millennium prize problems. Hodge might have had some rough heuristics or ideas that led him to the formulation of the conjecture. 
I am looking for the history and background behind the formulation of Hodge Conjecture. How did Hodge arrive at his conjecture?
Hodge Conjecture ( What I understood after reading Dan Freed's article ) : 

On a complex projective manifold $\mathbb{X}$, a topological cycle $\mathfrak{C}$ ( on $\mathbb{X}$ ) is homologous to a rational combination of algebraic cycles iff $\mathfrak{C}$ has rotation number $0$. ( Rotation number $e^{i\theta}$ on differential form, volumes are invariant under rotation). 

Hodge Conjecture (Deligne's description): 

On a projective non-singular algebraic variety over $\mathbb{C}$, any Hodge class is a rational combination of classes $\rm{cl(\mathbb{Z})}$ of algebraic cycles. 

The things that interest me:

How are Freed's version and Deligne's version versions equivalent?
How did Hodge arrive at that conclusion? Were there heuristic reasons or intuitive arguments that gives him some hope for a conjecture in that direction? .
How can one state an analogue of the Hodge conjecture in number theory?  Are there any attempts to formulate an analogue in that case? 

I am curious to hear answers, even if highly technical in nature.

REPLY [20 votes]: The best answer I can imagine for a question like this is to quote the man himself:
"The second result of Lefschetz tells us that a necessary and sufficient condition
that a 2-cycle $\Gamma_2$ in $V_2$ be algebraic... This result has many geometrical applications...
It is clearly a matter of great importance to extend Lefschetz's condition for a
2-cycle to be algebraic. The general problem is as follows...."
See  page 184 of
the Proceedings of the ICM 1950
for the full statement:
Hodge, W. V. D., The topological invariants of algebraic varieties, Proc. Intern. Congr. Math. (Cambridge, Mass., Aug. 30-Sept. 6, 1950) 1, 182-192 (1952). ZBL0048.41701.<|endoftext|>
TITLE: For which surfaces is Penner's conjecture known to be true?
QUESTION [14 upvotes]: Robert Penner has proven that, if $A=\{a_1,\dots, a_n\}$ and $B=\{b_1,\dots, b_m\}$ are multicurves in a surface $S$ that together fill $S$, then any product of positive powers of Dehn twists along the curves $a_i$s, and negative powers of Dehn twists along curves the $b_j$s, such that each curve in $A\cup B$ appears at least once, is a pseudo-Anosov homeomorphism of the surface $S$ (see http://www.jstor.org/stable/2001116).
He also conjectures that any pseudo-Anosov mapping class has a power which is representable as a composition of Dehn twists as above.
This conjecture is true for the surfaces $S_{1,1}$ and $S_{0,4}$, essentially as a consequence of the fact that the curve complex of these two surfaces embeds in their Teichmuller space ($\mathbb{H}^2$) through the Farey tessellation, and translations of the triangles of the Farey tessellation are representable as Dehn twists.
I was wondering if there are other surfaces for which the conjecture is known to be true, and where to find a proof.

REPLY [21 votes]: Shin and Strenner have shown that the conjecture is false when 3g + n > 4. 
See http://arxiv.org/abs/1410.6974<|endoftext|>
TITLE: Request for the proof of a result from Ramanujan's letter to Hardy.
QUESTION [5 upvotes]: Srinivasa Ramanujan in his first letter to G.H. Hardy stated many results for which he didn't give proofs. Among them the result taken from this link seems interesting :

If $$\int\limits_{0}^{\infty} \frac{\cos{nx}}{e^{2\pi\sqrt{x}}-1} \ dx = \phi(n)$$ then $\displaystyle\int\limits_{0}^{\infty} \frac{\sin{nx}}{e^{2\pi\sqrt{x}}-1} \ dx = \phi(n)-\frac{1}{2n} + \phi\biggl(\frac{\pi^2}{n}\biggr)\sqrt{\frac{2\pi^3}{n^3}}$. 

The link also mentions that $\phi(n)$ is a complicated function. The following are certain special values and shows some values.
Questions which I would like to ask here are:

Where can I find the proof of the above result?
"The following are certain special values": Whats so special about the values?

REPLY [10 votes]: Thanks "@Charles Matthews". I did email Prof. Berndt (after seeing your comment) and he suggested me to look at this paper:

Some definite integrals connected with Gauss's sums, Mess. Math. 44 (1915), 75-85.

The result appears here in Page 60. with proof.<|endoftext|>
TITLE: Axiom to exclude nonstandard natural numbers
QUESTION [7 upvotes]: In Peano Arithmetic, the induction axiom states that there is no proper subset of the natural numbers that contains 0 and is closed under the successor function. This is intended to rule out the possibility of extra natural numbers beyond the familiar ones. It doesn't accomplish that goal; there remains the possibility that other natural numbers exist and the familiar ones do not form a set. In Internal Set Theory (IST), which is an extension of ZFC that is consistent relative to ZFC, there is a distinction between standard and nonstandard sets, and it can be shown that
(1) 0 is standard;
(2) if $n \in \mathbb{N}$ is standard, then so is its successor;
(3) $\mathbb{N}$ has nonstandard elements.
The induction axiom is not violated because the standard natural numbers do not form a set.
Is there a way to axiomatize set theory so that no such nonstandard natural numbers can exist?
(Note: this question is not about nonstandard models of arithmetic. In IST, $\mathbb{N}$ is a standard set, and within a given model of IST, all models of second-order Peano arithmetic are isomorphic to $\mathbb{N}$.)

REPLY [2 votes]: By Gödel's incompleteness theorem, there can't be any such axiom in a first-order, recursively enumerable theory.
You can axiomatize $\mathbb N$ by adding an infinitary rule of inference, the Hilbert $\omega$-rule:
$$P(0)\wedge P(1) \wedge P(2) \wedge \cdots \over \forall n P(n)$$
to PA for each arithmetic predicate P.  This says, if predicate P holds for each of the natural numbers (0,1,2...) then you can deduce the formula $\forall n P(n)$.  The resulting system is called  ω-logic.  
Obviously this is something of a "cheat" since you no longer have an effective theory.  As one example (maybe there are better ones) of how it can be used, Michael Rathjen's article "The Art of Ordinal Analysis" describes using the $\omega$-rule to analyze stronger and stronger arithmetic theories, and is pretty interesting.<|endoftext|>
TITLE: Not especially famous, long-open problems which higher mathematics beginners can understand
QUESTION [16 upvotes]: This is a pair to
Not especially famous, long-open problems which anyone can understand
So this time I'm asking for open questions so easy to state for students of subjects such as undergraduate abstract algebra, linear algebra, real analysis, topology (etc.?) that a teacher could mention the questions almost immediately after stating very basic definitions.  
If they weren't either too famous or already solved, appropriate answers might include the Koethe Nil Conjecture, the existence of odd order finite simple groups or the continuum hypothesis.
Thanks

REPLY [3 votes]: Given a unit square, can you find any point in the same plane, either inside or outside the square, that is a rational distance from all four corners? 
Or, put another way, given a square $ABCD$ of any size, can you find a point $P$ in the same plane such that the distances $AB$, $PA$, $PB$, $PC$, and $PD$ are all integers?
 
References:

Guy, Richard K. Unsolved Problems in Number Theory, Vol. 1, Springer-Verlag,        2nd ed. 1991, 181-185.
Barbara, Roy. "The rational distance problem", Mathematical Gazette 95, March              2011, 59-61.<|endoftext|>
TITLE: What is the geometric object corresponding to a subalgebra in a polynomial ring
QUESTION [19 upvotes]: Many introductory texts on algebraic geometry set up some sort of algebra-geometry dictionary in which radical ideals correspond to varieties, and so on. I am wondering if there is a geometric way to think about a subalgebra in a polynomial ring?  For instance for invariant rings one may develop some intuition, but what about infinitely generated subalgebras?  A simple example would be the infinitely generated monomial algebra
$k[x,xy, xy^2, xy^3, \dots]$ as a subalgebra in $k[x,y]$ for some field $k$.

REPLY [5 votes]: One should note that there is a formal analogue of this for arbitrary affine schemes. If $A,B$ are commutative rings, then a homomorphism $f:A \to B$ induces a continuous map $f^*:\mathrm{Spec}(B) \to \mathrm{Spec}(A)$.
One can show (see e.g. Exercise 1.21(v), Introduction to Commutative Algebra, Atiyah and Macdonald), the map $f^*$ is dominant, meaning that the image is dense (which, in topology, is good enough as being surjective) iff the kernel of $f$ is contained in the nilradical of $A$ (which is simply the radical of the ideal $0$).
Just as radical ideals rather than arbitrary ideals correspond to algebraic subsets, for any ring $A$, the natural map $\mathrm{Spec}(A/\mathfrak{a}) \to \mathrm{Spec}(A)$ is a homeomorphism whenever $\mathfrak{a}$ is an ideal consisting entirely of nilpotents.
Therefore, the map on $\mathrm{Spec}$'s, i.e. the geometric map associated to a map of rings, is dominant iff the map $f$ realizes $A$ modulo some nilpotents as a subalgebra of $B$. In particular, if $A$ is reduced, the map $f^*$ is dominant iff $f$ is injective.<|endoftext|>
TITLE: Generating functions, Tutte polynomials, and the bivariate series $\sum_n x^n y^{n^2} / n!$.
QUESTION [13 upvotes]: A few years ago I computed the Tutte polynomials of the matroids given by the classical Coxeter groups, and found that their generating functions are all simple variations of the series $\sum_n \frac{x^n y^{n^2}}{n!}$. 
I've wondered if there is a more geometric/algebraic explanation of this. Is this series known? Are there other natural occurrences of it that might be relevant?

REPLY [3 votes]: Here is an attempt at a "soft" answer inspired by a paper that appeared after this question was asked here. The arithmetic Tutte polynomials of the classical root systems by Ardila, Castillo, and Henley. In particular this will not contain anything the OP doesn't already now :-)
There are two methods used for the computation of these (classic or arithmetic) Tutte polynomial generating functions. The finite field method by Ardila mentioned in the question has wide applicability, but in the paper linked above there is a different way to compute this generating function, which I believe give some insight on the presence of $F(x,y)=\sum_{n\geq 0} x^ny^{\binom{n}{2}}/n!$.
Using this generating function we can count one of the most fundamental combinatorial objects: graphs according to the number of connected components, number of edges, and number of vertices (the particular function is $F(x,1+y)^z$). On the other hand, through a series of operations in the sense of combinatorial species, one can obtain the generating function for some combinatorial objects called signed graphs, introduced by Zaslavsky, which serve as a combinatorial model encoding the relevant statistics of the classical root systems. Therefore one should expect the generating functions in question, to be computable from combinations of $F$ together with operations like multiplication, exponentiation, and composition.<|endoftext|>
TITLE: List of Whitehead-like Problems
QUESTION [10 upvotes]: Whitehead problem is a rather well known problem: 

Suppose that $G$ is an abelian group and $\mathrm{Ext}^1(G,\Bbb Z)=0$, is $G$ free?

It wasn't long before it was proved that if $G$ is countable (and thus countably-generated) then the answer is positive. However the question was open for uncountable groups, and it took more than a decade until Shelah came up with the interesting answer: In ZFC the problem is undecidable.

I was wondering what other open problems have somewhat of a Whitehead-like nature, namely they were proved for objects with a countable nature (e.g. separable topological spaces, countably generated), but the question remains open for the general case.
One example which comes to mind is Naimark's problem which was solved for separable $C^\ast$-algebras, and was partially solved in the general case in the sense that it is consistent to have a negative answer.
(As with CW big-lists, please post one problem per answer.)

REPLY [2 votes]: One that - to the best of my knowledge - is still a huge open problem is about non-standard models of $PA$ and Scott systems. A Scott system $S$ is a set of reals such that $(\omega, S)\models WKL_0$; that is, $S$ is closed under Turing reducibility and has paths through all of its trees. Now, given a non-standard model of arithmetic $M\models PA$, we say a real $X$ is coded in $M$ if $$ \exists a\in M\forall b\in M\cap\omega(M\models b\vert a\iff b\in X). $$ (It's a slight abuse of notation to write "$M\cap \omega$" for the standard part of $M$, but this is common.) The set of reals coded in $M$ is called the standard system of $M$ and is denoted by "$SSy(M)$."
A beautiful short argument shows that $SSy(M)$ is always a Scott set: infinite binary trees in $SSy(M)$ come from binary trees of nonstandard height in $M$, by an overspill argument, which must have paths of internally-finite but externally-infinite length; and these long paths corresond to infinite paths in $SSy(M)$. Conversely, it is easy to prove that every countable Scott set is the standard system of some model of $PA$.
The major open question is whether the stronger statement, "Every Scott set is the standard system of a model of $PA$," is true.
(Actually, we can do a bit better than "countable": by a union of chains argument, every Scott set of size $\le\aleph_1$ is a standard system, so in the presence of $CH$ the problem is solved. However, I still think this fits the question, because what's really going on is that countable=easy and $\aleph_1$ is "close enough" to countable to still be tractable.)<|endoftext|>
TITLE: Fundamental group of an  analytic hypersurface
QUESTION [6 upvotes]: Let $M$ denote a complex manifold of dimension $n$ and let $X\subset M$ denote an analytic hypersurface. Then it is a standard fact from several complex variables that around a given point $p\in X$ there are open subsets $V\subset X, W\subset \mathbb{C}^{n-1}$ and and a finite-sheeted covering $\pi: V\rightarrow W$ branched over an analytic set $A$. Frequently, $W$ is taken to be a polydisc (see, for example, Griffiths & Harris). Now set $V' = V\setminus \pi^{-1}(A)$ and $W' = W\setminus A$. If $V'$ is connected, then $\pi$ induces a genuine (holomorphic) covering $\pi':V'\rightarrow W'$. 
For appropriately chosen basepoints $v_0, w_0$ in $V',W'$, respectively, covering space theory says that $\pi'$ induces an injection $\pi_1(V',v_0)\rightarrow \pi_1(W',w_0)$, and in some cases (e.g. involving curves) this might be enough to figure out $\pi_1(V',v_0)$ if $\pi_1(W',w_0)$ is known. If $V$ were smooth, $\pi^{-1}(A)$ would be an analytic subset of complex codimension 1 and the inclusion $V'\hookrightarrow V$ would induce a surjection $\pi_1(V',v_0)\rightarrow \pi_1(V,v_0)$, determining a presentation for $\pi_1(V,v_0)$. However, $V$ may not be smooth, and I am not sure in what generality this map is still a surjection, although clearly I'd like to know.
What I really want to do is determine a presentation for $\pi_1(X,x_0)$. 
It seems that one way to to go about doing this would be to construct an open cover $(V_{\alpha})$ of $X$ consisting of open sets with the same properties possessed by $V$ above AND such that $(V_{\alpha})$ satisfies the hypotheses of the Seifert-van Kampen Theorem (see, for example, Hatcher). In the very best case, one could arrange for the intersections $V_{\alpha}\cap V_{\beta}$ to be simply-connected and for the triple intersections to be path-connected. Then one could read off a presentation of $\pi_1(X,x_0)$ in terms of the presentations of the $\pi_1(V_{\alpha}, v_{0\alpha})$.   
I'd be interested to know when such a cover $(V_{\alpha})$ exists, especially for $n\geq 3$. 
I am also interested in hearing about any known methods of calculating the fundamental group of an analytic hypersurface (not necessarily smooth), in general or in special cases (for example, the Lefschetz Hyperplane Theorem can be used on certain projective hypersurfaces). 
(Note: I'd like to consult Dimca's book "Singularities and Topology of Hypersurfaces", but I'll be away from my library for the next few weeks.)

REPLY [3 votes]: It should be said that van Kampen's  paper  "On the connection between the fundamental groups of some
  related spaces". Amer. J. Math. 55 (1933) 261--267, gives a formula for the case of a union of two spaces with non-connected intersection, and this was needed for his work on algebraic curves: "On the Fundamental Group of an Algebraic Curve". Amer. J. Math. 55 (1933), no. 1-4, 255–267. The non-connected case  seems to me best handled in modern terms using groupoids. (See my web pages.) (An earlier version of the theorem in the connected case and for simplicial complexes was given by Seifert.) 
Since homotopy groups are mentioned in the last comment, I mention the higher order theorems of the Seifert-van Kampen type of which the 2-d version was  given in my paper with Philip Higgins , ``On the connection between the second
relative homotopy groups of some related spaces'', Proc.
London Math.  Soc. (3) 36 (1978) 193-212.  This uses extensively the notion of crossed module, and has been applied  to give explicit calculations of homotopy 2-types, and second homotopy groups. (Again, see references on my web pages). 
In view of the provenance of van Kampen's paper, it would be very interesting to know if such higher theorems are applicable to the situation of the question. 
@Kevin Kordek: September 2013: I should add that Grothendieck was interested in these possibilities. See the last problem stated in the set of problems and "Future directions?" given in our new book on "Nonabelian algebraic topology"; more details and pdf's available from here. Part of his comment is as follows: 
"It seems to me, in any case, that this
$\underset{\to}{\lim}$-operation ["higher order van Kampen
theorem"] in the context of homotopy types is of a very fundamental
character, with wide range of theoretical applications. To give just
one example, relying on the existence of such a formalism, it is
possible to give a very simple explicit algebraic description of the
full homotopy types of the Mumford-Deligne compactifications of the
modular topoi for complex curves of given genus $g$, say, with $\nu$
"marked" points, in terms essential1y of such a (finite) direct
limit of $K(\pi, 1)$-spaces, where $\pi$ ranges over certain
"elementary" Teichm\"uller groups (those, roughly, corresponding to
modular dimension $\leq 2$), and to give analogous descriptions,
too, of all those subtopoi of the previous one, deducible from its
canonical "stratification" at infinity by taking unions of strata.
In fact, such descriptions should apply to any kind of ``stratified"
space or topos, as it can be expressed (in an essentially canonical
way, which apparently was never made explicit yet in this
literature) as a (usually finite) direct limit of simpler spaces,
namely the "strata", and "tubes" around strata, and "junctions"
of tubes, etc."
So   I probably missed out on not pursuing this, partly because of pursuing work with J.-L. Loday on even more powerful Higher van Kampen type theorems.<|endoftext|>
TITLE: Is there a Seiberg-Witten version of Donaldson-Thomas theory?
QUESTION [15 upvotes]: Donaldson invariants are a count of instantons (the solutions to a particular elliptic PDE) on 4-manifolds. One thing which makes the theory difficult is a lack of compactness for the moduli spaces of instantons: sequences of instantons can end up singular in the limit, near points where they form ''bubbles''. Many of the applications of Donaldson theory to 4-manifold topology can be recovered in the analytically simpler Seiberg-Witten framework, where a Weitzenböck identity allows you to prove that the moduli spaces of solutions to the Seiberg-Witten equations are actually a priori compact.
(One aspect of) Donaldson-Thomas theory is a complexification of Donaldson theory, counting certain generalised instantons on Calabi-Yau 4-folds (see Donaldson & Thomas, Gauge theory in higher dimensions, Equation 9: available here). A major problem which people seem to be working hard on solving (for example this recent preprint of Walpuski) is a lack of compactness for these equations. My question  is:

Does there exist a Seiberg-Witten analogue of Donaldson-Thomas theory which circumvents these compactness issues?

If so the trick can't be as simple as the Weitzenböck identity, which allows you to bound the $L^{\infty}$-norm of the (spinor part of the) solution in terms of the scalar curvature, because the scalar curvature vanishes on a Calabi-Yau 4-fold.

REPLY [3 votes]: In 8 dimensional case we do not have direct analogous theory like 4 dimension case by strong weak duality. One can naively consturct the SW theory for CY4(line bundles with sections), but the theory turns out to be quite trivial since we need the virtual dimention of the moduli space is topological and we do not have much choice.
 In 4 dimensions, it is Kroheimer and Mrowka first showed that Donaldson polynomials have recurrence relations for simple type 4 mfds. Then Seiberg and Witten wanted to understand this from Physical perspective and finally got to SW theory. All Gauge theories in 4 dim are expected to be recovered by SW theory. But this is far from clear even for CY3(people seems only consider DT invs for curves(=GW by MNOP) and pts(computed by several groups) so far).<|endoftext|>
TITLE: A generalization of Liouville formula for the determinant of a system of ODE?
QUESTION [6 upvotes]: Let $\Phi(t)$ be an $n\times n$ complex matrix whose columns are (independent) solutions of the  system of ordinary differential equations (ODE):
$\frac{d}{dt}y= A(t) y$, where $A(t)$ is a $t$-dependent $n\times n$ complex matrix.
Then, the Liouville formula provides a very simple relation between the determinant of $\Phi(t)$ at different $t$:
$\det(\Phi(t)) = e^{\int_0^t Tr A(s) ds} \det(\Phi(0))$.
Now, consider the alternative system of ODE:
$\frac{d}{dt}y= A(t) y + B(t) \bar{y}$, 
where $\bar{x}$ represents the complex conjugate of $x$. (Actually, I am interested only in the case where $A(t)$ is diagonal and $B(t)$ is skew-symmetric, if that helps.)
Does it exist any generalization of the Liouville formula applicable to this case? or any other expression for $\det(\Phi(t))$, which is simpler (to compute numerically) than finding the explicit solutions for all the starting conditions in $\Phi(0)$ and then computing the determinant explicitly?
The kind of manipulations used by the classical proof of Liouville formula do not seem very helpful to me, in this slightly modified case, unfortunately.
The solution of this problem would be very beneficial to the method described in arXiv:1205.3996, and to its very relevant physical applications.

REPLY [2 votes]: NB:  I'll slightly rearrange this for clarity: 
As I should have remarked at the beginning, 
writing the fundamental solution to your system in the form
$$
y(t) = \Phi(t)\,y(0),
$$
where $\Phi(0)=I$, is only possible when you allow $\Phi(t)$ to take values in $$\mathrm{Aut}_\mathbb{R}(\mathbb{C}^n,\mathbb{C}^n) \simeq \mathrm{GL}(2n,\mathbb{R}),
$$
i.e., you can't assume that $\Phi(t):\mathbb{C}^n\to\mathbb{C}^n$ is $\mathbb{C}$-linear, but only $\mathbb{R}$-linear.  Consequently, the very meaning of 'determinant' is probably not what you had in mind, since the homomorphism
$$
{\det}_\mathbb{R}:\mathrm{Aut}_\mathbb{R}(\mathbb{C}^n,\mathbb{C}^n)=\mathrm{GL}(2n,\mathbb{R})\to\mathbb{R}\setminus\{0\}
$$
takes (nonzero) values in $\mathbb{R}$, and there is no meaningful way to `lift' this to a 'determinant' that assumes all (nonzero) values in the complex numbers.
One should also note that, if $A:\mathbb{C}^n\to\mathbb{C}^n$ does happen to be $\mathbb{C}$-linear, then one has
$$
{\det}_\mathbb{R}(A) = \bigl|{\det}_\mathbb{C}(A)\bigr|^2.
$$
With that understood, there is a formula for $\det\bigl(\Phi(t)\bigr)$, it's just
$$
\det(\Phi(t)) 
= \exp\left(\int_0^t \mathrm{tr}\bigl(A(\tau)+\overline{A(\tau)}\bigr)\,d\tau\right).
$$
Here's how you can see this:  Since you know that $A(t)$ is diagonal, you can easily write $A(t) = -a(t)^{-1}a'(t)$ where $a(t)$ is a diagonal matrix satisfying $a(0)=I$ (just integrate and exponentiate each diagonal term separately).   
Hence you can regard $a(t)$ as known and write the equation in the form
$$
y'(t) = - a(t)^{-1}a'(t)y(t) + B(t)\,\overline{y(t)},
$$
which gives
$$
(a(t)y(t))' = a(t)B(t)\overline{(a(t)^{-1})}\,\,\overline{a(t)y(t)}.
$$
Set $z(t) = a(t)y(t)$ and write $L(t) = a(t)B(t)\overline{(a(t)^{-1})} C$, where $C:\mathbb{C}^n\to\mathbb{C}^n$ is the $\mathbb{R}$-linear map of conjugation.  Then the equation becomes the $\mathbb{R}$-linear system of ODE
$$
z'(t) = L(t) z(t).
$$
It's important to note that $\mathrm{tr}(L(t))$ vanishes identically.
Of course, $L(t)$, which can be regarded as known, can be written in the form $L(t) = -b(t)^{-1}b'(t)$ where $b(t)$ takes values in the matrix Lie group $\mathrm{Aut}_\mathbb{R}(\mathbb{C}^n,\mathbb{C}^n)$, and satisfies $b(0)=I$.  The general solution of this equation is  then
$$
z(t) = b(t)^{-1}z(0) \qquad \text{so}\qquad y(t) = a(t)^{-1}b(t)^{-1}y(0).
$$
Now, the Liouville formula yields 
$$
{\det}_\mathbb{C}(a(t)) = \exp\left(-\int_0^t \mathrm{tr}(A(\tau))\,d\tau\right)
$$
$$
{\det}_\mathbb{R}(b(t)) = \exp\left(-\int_0^t \mathrm{tr}(L(\tau)+\overline{L(\tau)})\,d\tau\right). = 1
$$
(After all the trace of $L(t)$ as a real endomorphism is identically zero.)
Thus, $y(t) = \Phi(t)y(0)$ where 
$$
{\det}_\mathbb{R}(\Phi(t)) 
= \exp\left(\int_0^t \mathrm{tr}\bigl(A(\tau)+\overline{A(\tau)}\bigr)\,d\tau\right).
$$<|endoftext|>
TITLE: Probability in the Primes
QUESTION [8 upvotes]: Given two randomly chosen positive rational integers, the probability that the two numbers are coprime is $\frac{6}{\pi^2}$. This is also the probability that a positive integer is squarefree. Are there generalizations of these results for Gaussian integers? Or more generally for the ring of integers in an algebraic number field?

REPLY [11 votes]: There are generalizations, see this mathworld article for some results and references. A detailed exposition for arbitrary number fields is given in this paper by G. Collins and J. Johnson<|endoftext|>
TITLE: Functions consistent with two partitions of a finite set
QUESTION [5 upvotes]: Suppose that $P$ and $C$ are two unordered partitions of $[n]$, the set of positive integers from 1 to $n$. Let $c(C,P,x)$ be the number of functions $f$ from $[n]$ to $[x]$ for which 
(1) $f(a)=f(b)$ if $a$ and $b$ belong to the same part of $C$,
(2) $f(a)≠f(b)$ if $a$ and $b$ belong to the same part of $P$. 
Conjecture: The sum $\sum (-1)^{p-1} (p-1)! c(C,P,x)$ is independent of $C$. Here the sum is over all partitions $P$, and $p$ denotes the number of parts of $P$.
Consider the case where $C$ has a single part. Then $c(C,P,x)$ is zero unless $P$ is the trivial partition into $n$ parts, and the sum is $(-1)^{n-1} (n-1)! x$. Thus the conjecture is that this is always the value of the alternating sum, for arbitrary $C$.
This conjecture arises from the same context as questions "Stirling number identity via homology" and "Polynomials akin to Bell polynomials," asked by the same user.

REPLY [4 votes]: Fix a function $f$ such that $(1)$ holds. The contribution of that function to the sum is the sum over all partitions $P$ such that $(2)$ holds of $(-1)^{p-1}(p-1)!$. We will prove that this contribution is $0$ for all nonconstant $f$. Since every constant $f$ is satisfies $(1)$ for every partition, the identity is true.
Fix a nonconstant $f$. Take a partition $P_{-}$ of $[n-1]$ for which $(2)$ holds. How many ways can we extend this partition to $[n]$? We can create a new part, or add $n$ to any part which does not contain any of the $k$ such that $f(k)=f(n)$. Let $a$ be the number of such $k$, then those $a$ must be contained in $a$ different parts of $P_{-}$ since $P_{-}$ satisfies $(2)$. So there are $p-a$ ways to extend $P_{-}$ and keep the number of parts fixed, and one way to extend it and increase the number of parts.
The contribution to the sum of extensions of $P_{-}$ is $(p-a)(-1)^{p-1}(p-1)! + (-1)^p p! = -a (-1)^{p-1} (p-1)!$.
Since each partition $P$ of $[n]$ satisfying $(2)$ is an extension of a unique partition $P_{-}$ of $[n-1]$ satisfying $2$, the total contribution of all partitions of $[n]$ satisfying $(2)$ is equal to $-a$ times the total contribution of all partitions of $[n-1]$ satisfying $(2)$. By induction, this is $0$ unless $f$ is constant on $[n-1]$. But if $f$ is nonconstant on $[n]$ and constant on $[n-1]$ then $f(n)$ must be a new value, so $a=0$, so the product is $0$ anyways. The base case for the induction is $n=1$ where all partitions are constant so there is nothing to prove. This completes the argument.<|endoftext|>
TITLE: Surjectivity of operators on $\ell^\infty$
QUESTION [28 upvotes]: Can anyone give me an example of an bounded and linear operator $T:\ell^\infty\to \ell^\infty$ (the space of bounded sequences with the usual sup-norm), such that T has dense range, but is not surjective?

REPLY [10 votes]: In this paper by Amir Bahman Nasseri, Gideon Schechtman, Tomasz Tkocz, and me it is shown that there is an injective, dense range, non surjective operator on $\ell_\infty$.  The proof is not technically difficult but has some interest.  From the theory of Tauberian operators one deduces that such an operator exists if and only if there is a dense range, injective, non surjective operator Tauberian operator (``Tauberian" in this context just means that the second adjoint of the operator is injective) on $L_1(0,1)$. Although  $L_1(0,1)^* = L_\infty(0,1)$ is isomorphic to $\ell_\infty$, the a priori equivalence is not obvious because of the lack of reflexivity, but we need only the obvious implication.  The main tool for constructing the  $L_1$ operator is a finite dimensional lemma proved by computer scientists. So, in some sense, the original question about operators on a non separable Banach space is connected to computer science!<|endoftext|>
TITLE: Lattices in $SL(n,\mathbb R)$
QUESTION [9 upvotes]: If $\Gamma\subseteq SL(n,\mathbb{R})$ is a lattice (i.e. discrete and finite covolume), does $\Gamma$ necessarily contain some $\mathbb{R}$-diagonalizable copy of $\mathbb{Z}^{n-1}$?
I know that the answer is yes if the lattice is cocompact, and that the answer is also yes in the case $\Gamma=SL(n,\mathbb Z)$.  So I wonder if every lattice satisfies this property.

REPLY [12 votes]: The answer is yes. It is theorem [2.13] of the following paper of Prasad and Raghunathan:
Prasad, Gopal; Raghunathan, M. S. Cartan subgroups and lattices in semi-simple groups. Ann. of Math. (2) 96 (1972), 296–317.
There is also a lot of information in this paper:
http://www.math.bgu.ac.il/~barakw/papers/clorbit.pdf
Note that diagonalizable copies of $\mathbb{Z}^{n-1}$ in $\Gamma$ correspond to closed orbits for the action of the full diagonal subgroup of $SL(n,\mathbb{R})$ on $SL(n,\mathbb{R})/\Gamma$. 
This is related to the Margulis conjecture which (with some caveats) states that the closure of any orbit of the full diagonal on $SL(n,\mathbb{R})/\Gamma$ is algebraic, i.e. is itself the closed orbit of some subgroup. This conjecture is the biggest open problem in homogeneous dynamics (and in particular implies the Littlewood conjecture in number theory).

REPLY [8 votes]: This is a theorem of G. Prasad and M.S. Raghunathan. See Theorem 7.2 in this paper of Steve Hurder's (rigidity of Anosov actions) -- the original reference is a bit less friendly.<|endoftext|>
TITLE: Are coefficients of Maass forms of eigenvalue 1/4 known to be algebraic?
QUESTION [7 upvotes]: I would really like to know whether the following famous conjecture has been solved.  I've read in a few places that it has been solved, but I have been unable to find a reference.  I do know that there was once a very credible proof which was believed for a while and then turned out to be false.  The statement is:

If $f:\mathbb H/\Gamma(N)\to\mathbb R$ is of moderate growth in the cusps and satisfies $\Delta f+\frac 14f=0$, then the Hecke eigenvalues $\lambda_p(f)$ are algebraic.

There is of course a much stronger conjecture (still unsolved) that there is a Galois representation $\rho:\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q)\to GL(2,\mathbb C)$ such that $\lambda_p(f)=\operatorname{tr}\rho(\operatorname{Fr}_p)$.  This would imply the weaker conjecture above.
It was once believed that this stronger conjecture was solved by Blasius and Ramakrishnan (and see a second paper), but their proof relied on a statement (which they could not prove, but which was widely believed to be true) which turned out to be false.  I thought at some point, though, that, perhaps because of work of Richard Taylor, that the weaker boxed statement above had been proven later using different methods.  I have been unable to locate a reference, though, and many papers actually refer to the claimed proof of B&R.
Is the boxed statement known to be true (and does someone know a good reference for the proof)?

REPLY [9 votes]: Your boxed statement is an open problem.  Blasius and Ramakrishnan did not rely on a widely believed statement which turned out to be false.  Their argument accidentally conflated two L-packets for $GSp_4(\mathbb{R})$ which are in fact distinct, due to a miscalculation of the central character of one of the L-packets in question.  This mistake was discovered several years later by Henniart, I believe - see MR1157812.<|endoftext|>
TITLE: Number of neigbour Voronoi cells for a random set of points on S^k or cube [-1, 1]^k? 
QUESTION [6 upvotes]: Consider $S^k \subset  R^{k+1} $.  Sample $N$ points by say uniform distribution. (Example k=120, N=2^24, i.e. N>>k ).
Consider Voronoi cell around each point. 
How many neighbours would a cell have ? I mean neigbours are the cells which have non empty intersection. "How many" means average over distribution. (Clearly it is less than N,
but what is it behaviour ? N/C, sqrt(N) or what ?)
Actually I more interested not about the sphere but about the cube: take unit cube $[-1, 1]^k$.
And take randomly some number $N$ of its vertexes.
The same questions.

Motivatation:
As I tried to explain in this MO quest these problems are related to decoding noise signal. This question can be translated in this language as follows - if there chance to do some "preprocessing" such that it would significantly reduce decoding complexity.
I mean in the answer is much smaller than N, then yes, otherwise, not.

REPLY [4 votes]: This was studied by Miles in the seventies. A good reference is Schneider and Weil's book. The relevant results are in section 10.2<|endoftext|>
TITLE: equidistribution on the unit circle of particular sequences of finite subsets
QUESTION [6 upvotes]: Given a strictly convex function $g : [0, 1] \to \mathbb{R}$, I'm curious about the asymptotic distribution of the points $\exp{(2 \pi i N g(n / N))}$ for $n = 1, 2, \dots, N$, counted with multiplicity, as $N$ gets large.  To be precise, I'm curious what sorts of bounds one can expect for the "discrepancy" 
$$N^{-1}\#\{n : a < Ng(n/N) \pmod{1} < b\} - (b - a),$$ 
and in particular if these points become equidistributed as $N$ goes to infinity.
For what it's worth, I believe the answer is "yes."  I have a guess about how to prove it, but I am not an analytic number theorist, and, even if my proof is right, I'm very curious if people know $1$) better ways to prove it, or $2$) preexisting results that imply it.  
As for my humble stab at a sketch of a proof: First, I believe I have a proof in the case $g(x) = x^2$ using the Erd\"os-Tur\'an Inequality (here).  (In this inequality, one has a bound on the above discrepancy which can shown to go to zero using what I understand are well-known formulas for quadratic Gauss sums.)  Perhaps this is also true for arbitrary quadratic polynomials, and perhaps one can get good bounds on the discrepancy.  For a general strictly convex function, $g$, I think it might be possible to approximate $g$ on small intervals (w.r.t. $N$) by a quadratic polynomial, and use the discrepancies computed in the polynomial cases to bound the discrepancy for $g$.
My proof for $g(x) = x^2$ will be provided only upon request: I am ashamed of it's ugliness.  Thanks in advance.

REPLY [7 votes]: Gerry's reference turned out to be quite useful. Theorem 2.7 of Uniform Distribution of Sequences by Kuipers and Niederreiter states that if $a$ and $b$ are integers with $a < b$, and if $f$ is twice differentiable on $[a,b]$ with $|f''(x)| \geq \rho > 0$ on $[a,b]$, then
$$\left|\sum_{n = a}^{b}{e^{2\pi i f(n)}}\right| \leq \left(\left|f'(b) - f'(a)\right| + 2\right)\left(\frac{4}{\sqrt{\rho}} + 3\right).$$
So if we assume that $g : [0,1] \to \mathbb{R}$ is a continuous twice-differentiable function with $\lambda = \inf_{x \in [0,1]} g''(x) > 0$, then by taking $a = 1$, $b = N$, $f(x) = m N g(x/N)$, we find that
$$\left|\sum_{n = 1}^{N}{e^{2\pi i m N g(n/N)}}\right| \leq \left(m \left|g'(1) - g'(1/N)\right| + 2\right)\left(\frac{4}{\sqrt{m N \lambda}} + 3\right).$$
So now let $\mu_N$ be the probability measure on $[0,1]$ given by
$$\mu_N(B) = \frac{1}{N} \# \left\{1 \leq n \leq N : N g(n/N) \in B \pmod{1}\right\}$$
for each Borel set $B \subset [0,1]$, and let $\mu$ denote the Lebesgue measure on $[0,1]$. Then the Erdős–Turán inequality states that for any positive integer $M$, the discrepency
$$D(N) = \sup_{B \in [0,1]} \left|\mu_N(B) - \mu(B)\right|$$
satisfies
$$D(N) \leq C \left(\frac{1}{M} + \frac{1}{N} \sum_{m = 1}^{M}{\left| \sum_{n = 1}^{N}{e^{2\pi i m N g(n/N)}} \right|}\right)$$
for some absolute constant $C > 0$ (independent of $N$ and $M$). Taking $M = \lfloor N^{1/3}\rfloor$ and using the earlier bound on $\displaystyle\sum_{n = 1}^{N}{e^{2\pi i m N g(n/N)}}$ shows that
$$D(N) = O\left(N^{-1/3}\right)$$
and hence that $\mu_N$ converges weakly to $\mu$ as $N$ tends to infinity.
It may be possible to relax some of these conditions on $g$ by modifying the proof of this theorem in the book of Kuipers and Niederreiter, but I haven't checked too closely yet.<|endoftext|>
TITLE: Character of parity-twisted supersymmetric VOA module -- question inspired by the Stolz-Teichner program
QUESTION [17 upvotes]: I'll begin with some background that is unnecessary for the actual question, but that might be interesting to the reader:
Topological modular forms ($TMF$) is a generalized cohomology theory whose coefficient ring $TMF^*(pt)$ is closely related to the ring $MF_*:=\mathbb Z[c_4,c_6,\Delta]/c_4^3-c_6^2-1728\Delta$ of classical modular forms.
The main goal of the Stolz-Teichner program is to construct $TMF$ via methods of functorial quantum field theory.
More precisely, Stolz-Teichner want to realize cocycles for $TMF^*(X)$ as extended supersymmetric field theories over $X$, that is, functors from the 2-category of 0-, 1-, and 2-dimensional supermanifolds over $X$ (i.e., equipped with a map to $X$) to some algebraic target 2-category (e.g. algebras and bimodules). 
In particular, an element in $TMF^*(pt)$ should be represented by a functor $Z$ from the 2-category of 0-, 1-, and 2-dimensional supermanifolds (no map to $X$) to the 2-category of algebras.
There is a natural map from $TMF^*(pt)$ to $MF_*$ (a $\mathbb Q$-isomorphism),
and, correspondingly, there is a natural way of extracting a modular form from a supersymmetric field theory $Z$. The construction goes roughly as follows.
Given a supersymmetric field theory, consider the value $V:=Z(S^1_{per})$ of $Z$ on the manifold $S^1$, equipped with the periodic supermanifold structure (Ramond sector). The vector space $V$ comes equipped with an action of the semigroup of annuli, which, at the infinitesimal level, means that there are two operators $L_0$ and $\bar L_0$ acting on $V$. Moreover,
the theory being supersymmetric, there is also an odd square root of $\bar L_0$, called $\bar G_0$ (the theories considered here are only half-supersymmetric: no square root of $L_0$).
The modular form associated to $Z$ is given by evaluating the field theory on elliptic curves (with their Ramond-Ramond supermanifold structure). Letting $E_q:= \mathbb C^\times/\mathbb Z^q$, the value of the modular form at the point $q$ is the supertrace of the operator $q^{L_0}+\bar q^{\bar L_0}$ on $V$. A priori, this doesn't look holomorphic... this is where supersymmetry comes to help: the existence of an odd square root of $\bar L_0$ implies that the coefficient of $\bar q^n$ is zero whenever $n\not =0$, and so $Z(E_q)=str(q^{L_0}+\bar q^{\bar L_0})$ is indeed holomorphic as a function of $q$.
My question is about the existence of situations (ignoring $q$) where $str(\bar q^{\bar L_0})$ is a non-zero constant.

(Notational warning: $q$, $L_0$, $M$ below correspond to $\bar q$, $\bar L_0$, and $V$ above)
The question:

Let $V$ be an $N=1$ super vertex algebra that is holomorphic, in the sense that $V$ has a unique irreducible module (namely $V$ itself). Let $M$ be its unique irreducible parity-twisted module (the Ramond sector of $V$). The Ramond algebra (spanned by $L_n$ and $G_n$, $n\in\mathbb Z$) acts on $M$. In particular, we get an even operator $L_0$ and an odd operator $G_0$ acting on $M$, subject to the relation $G_0^2 = L_0 - c/24$. 
That relation implies that the supertrace $str_M(q^{L_0-c/24})$ is a constant (as opposed to a power series in $q$).

Is there an example of a vertex algebra $V$ as above such that $str_M(q^{L_0-c/24})\not =0$?

REPLY [11 votes]: Such an object is described in Dixon, Ginsparg, Harvey, Beauty and the Beast: superconformal symmetry in a monster module Comm. Math. Phys. Volume 119, Number 2 (1988), 221-241.  A reasonably explicit construction is given in Huang's paper A nonmeromorphic extension of the moonshine module vertex operator algebra.
In short: the Leech lattice vertex algebra $V_L$ has a canonical involution $\theta$, and the $\theta$-fixed point subalgebra has a semisimple category of modules with 4 simple objects (I think this is a result of Dong):

$V_L^\theta$ - the fixed point algebra, with character $\frac{J(\tau)+\Delta(\tau)/\Delta(2\tau)}{2} + 12 = q^{-1} + 98580q + 10745856q^2 + \cdots$
$V_L^{\theta = -1}$ - the $-1$ eigenspace, with character $\frac{J(\tau)-\Delta(\tau)/\Delta(2\tau)}{2} + 12 = 24 + 98304q + 10747904 q^2 + \cdots$
$V_L(\theta)^\theta$ - part of the monster vertex algebra, with character $\frac{J(\tau)-\Delta(\tau)/\Delta(2\tau)}{2} - 12 = 98304q + 10747904 q^2 + \cdots$
$V_L(\theta)^{\theta = -1}$ - the odd bit, with character $\frac{\Delta(2\tau)/\Delta (\tau) - \Delta(2\tau+1)/\Delta(\tau+1/2)}{2} = 4096q^{1/2} + 1228800q^{3/2} + 74244096 q^{5/2} + \cdots$

The sum of the first and fourth modules has a canonical product structure that gives you the holomorphic vertex superalgebra you want.  Its parity-twisted module is the sum of the second and third modules, whose characters differ by the nonzero constant 24.<|endoftext|>
TITLE: Tetris-like falling sticky disks
QUESTION [51 upvotes]: Suppose unit-radius disks fall vertically from $y=+\infty$,
one by one, and create a random jumble of disks above the $x$-axis.
When a falling disk hits another, it stops and sticks there.
Otherwise, if the disk center reaches $y=0$, the disk stops
with its center resting on the $x$-axis.
Here is an example of 1000 disks falling at uniformly random
$x$-locations within $[-50,50]$:

         

There are many questions one could ask about this (to me)
beautiful and intriguing structure (e.g., about its contact graph),
but to be specific,
let me concentrate on one quantity: the maximum height
$h_{\max}$ as a function of the number of disks $n$ and
the $x$-range $R$. (In the above example, $R=100$ and $h_{\max}=94.9$.)
It appears that $h_{\max}$ grows linearly, with $h_{\max} \approx n \frac{10}{R}$.
Here is plot, where each point is an average of ten trials:

                   

Two questions:

Q1. Is there a simple explanation of the growth of $h_{\max}$?
Q2. Has this process, or something close to it, been studied before?


Ultimately I am interested in determining packing densities of
randomly jostled shapes, as explored in the earlier MO question
"Average degree of contact graph for balls in a box."
Sticky disks are a very simple model along these lines.

Update (3Mar16).
An article by Ivan Corwin on KPZ universality has just appeared
(AMS Notices PDF),
including this figure to illustrate the "random ballistic" model:

     


Users ansobol and Nechaev and Jeremy Voltz 
previously pointed to the relevance of KPZ universality.

REPLY [8 votes]: That the growth is asymptotically linear is clear. But if you keep the width $2R$ of the strip (over which the centers are chosen uniformly) fixed, then the growth speed $c_R$ is not strictly proportional to $R^{-1}$. This is clear when thinking to the small width case: if $R<1$, then there is only one branch in the tree because two consecutive disks always touch. If $d$ is the distance between their centers, then they will arrange with a height gap equal to $\sqrt{4-d^2}$. Because the distribution of $d$ has density 
$$\frac{2R-d}{2R^2},$$
we find that the growth average speed
$$\frac1n\sum_1^nd_j$$
tends to the expectation of  $\sqrt{4-d^2}$:
$$c_R=\int_0^{2R}\sqrt{4-x^2}\frac{2R-x}{2R^2}dx.$$
Calculus gives
$$c_R=\frac2R\sin^{-1}R+2\sqrt{1-R^2}+\frac4{3R^2}\left((1-R^2)^{3/2}-1\right).$$
As expected, $c_R\rightarrow2$ as $R\rightarrow0$. On the other hand, $c_1=\pi-\frac43$.
Presumably, the OP is interested with $\gamma=\lim_{R\rightarrow+\infty}Rc_R$.

REPLY [6 votes]: Regarding the question

Has this process, or something close to it, been studied before?

I was recently made aware of an intriguing approach of Bob Macpherson and his post-doc Ben Schweinhart at IAS for investigating Brownian trees via computational topology in this pre-print. Since the investigation is topological, it will not yield answers to your question about maximum height, but it does capture other interesting global statistics.
The central idea is to let a diffusion-limited aggregation process run for a while and generate a configuration similar to that of your picture, but lacking any loops whatsoever. Essentially, each time a disc falling from infinity creates a cycle, it is discarded. However, this discarding step is not necessary for subsequent analysis and so would also apply to your (considerably loopier) situation. 
In any case, once the process has run its course and generated a space consisting of the union of balls, you start growing the radii of these balls (allowing overlap of course). As you increase these radii, the homology of the space changes: some loops form, others get filled in and so forth. It is possible to unambiguously associate a (birth, death) interval to each such loop via the theory of persistent homology.<|endoftext|>
TITLE: Are Donaldson-Thomas invariants "A-model" or "B-model" ?
QUESTION [26 upvotes]: Donaldson-Thomas invariants are the (virtual) Euler characteristics of moduli spaces of elements of the derived category of coherent sheaves (with some fixed Chern class, satisfying some stability condition, etc.) which bear some relation to "holomorphic Chern Simons theory", whatever that is. 


Should I think of these as "A-model" or "B-model" invariants?


On the one hand, DT invariants come from the bounded derived category of coherent sheaves, which is what features in the B-model.  On the other, there is the MNOP conjecture which tells me that the DT invariants of a CY 3-fold are ``the same'' as the Gromov-Witten invariants of the SAME 3-fold, which are A-model things.
As I understand it, according to Costello, if I take the cyclic A-infinity category built out of d-b-coh and run it through his machinery, I should get the topological string corresponding to the B-model.  But (modulo my confusion on the matter) according to Kontsevich and Soibelman, if I take the cyclic A-infinity category built out of d-b-coh and run it through their machinery, I should get DT invariants.  So what is going on?

REPLY [20 votes]: Ha, I think I'm going to disagree.
DT invariants are (more-or-less) independent of the complex structure on the CY: they are invariant under deformations of the CY.
However they depend on the (stringy/complexified) Kähler structure, or stability condition, or whatever. Hence wall crossing etc.
So even though they appear to be defined using D(coh), they're rather insensitive to that. What they are sensitive to is the small piece of data that's often forgotten -- a stability condition. So really one should think of them as defined in terms of a point in the space of stability conditions, or the stringy Kähler moduli space.
Ideally they'd be invariants of only the symplectic structure, just like GW invariants, but until MNOP is proved in full generality that's not known even in the projective case.
This makes them sound like "A-model invariants" to me, but then I'm not very sure about the physics language.<|endoftext|>
TITLE: Cayley Transform for all reductive groups a.k.a an algebraic logarithm
QUESTION [13 upvotes]: Is it true that for every reductive algebraic $G$ over ${\mathbb C}$ with a Lie algebra $\mathfrak g$ there is an open neighborhood $U$ of the identity in $G$ and an algebraic function (in a sense of algebraic geometry) $L: U\to \mathfrak g$ which satisfies the following properties of logarithm:
(1) $L$ is $G$-equivariant with respect to the $G$-action on $G$ by conjugation and the Adjoint $G$-action on $\mathfrak g,$
(2) $L(e)=0,$
(3) $dL$ is an isomorphism at $e$,
(4) For some maximal torus $T$ in $G$, $L(T\cap U)$ lies in the Lie algebra of $T.$
For $G=GL(n,\mathbb C)$, the embedding $L:GL(n,\mathbb C)\to gl(n,\mathbb C)$ works.
For $G=SO(n,\mathbb C)$, the Cayley Transform works: $L(A)= (I-A)(I+A)^{-1}$.
Cayley transform has a version for symplectic matrices as well.
Is there a construction which works for all $G$? If not, are there known ad hoc constructions for exceptional groups?

REPLY [7 votes]: Bardsley and Richardson (Etale slices for algebraic transformation groups in characteristic p.
Proc. London Math. Soc. (3) 51 (1985), no. 2, 295–317) give a construction which seems to do what you want. It gives less than a "Cayley map" as in Borovoi's answer.
Let $G$ be connected and semisimple over a field of char. 0.
Bardsley and Richardson construct a mapping $G \to \operatorname{Lie}(G)$ with 
nice properties (which I'll indicate below).
Note that we may as well suppose that $G$ is of adjoint type -- indeed,
if the problem is solved already for the adjoint group $G_{\operatorname{ad}}$,
just take the composite
$$G \to G_{\operatorname{ad}} \to \operatorname{Lie}(G).$$
Now, since the characteristic of $k$ is zero and $G$ is of adjoint type, the adjoint representation
$V = \operatorname{Lie}(G)$ is a faithful representation of $G$ for which the
trace form
defined by $\kappa(X,Y) = \operatorname{tr}(X \circ Y)$ -- a non-degenerate
form on $\mathfrak{gl}(V)$-- remains non-degenerate on the image
$\operatorname{ad}(\operatorname{Lie}(G)) \simeq \operatorname{Lie}(G) \subset \mathfrak{gl}(V)$.
Writing $M$ for the orthogonal complement $M=\operatorname{ad}(\operatorname{Lie}(G))^\perp$ with respect to the form $\kappa$,
we have
$$\mathfrak{gl}(V) = M \oplus \operatorname{ad}(\operatorname{Lie}(G))$$
as $G$-representations.
Write $\pi:\mathfrak{gl}(V) \to \operatorname{ad}(\operatorname{Lie}(G))$ for
the projection on the second factor.
Since $G$ is semisimple, $\operatorname{ad}(\operatorname{Lie}(G)) \subset
\mathfrak{sl}(V)$ so that the identity mapping $I$ satisfies
$\kappa(I,\operatorname{ad}X) = \operatorname{tr}(\operatorname{ad}X)  = 0$ 
for each $X \in \operatorname{Lie}(G)$. Thus $I \in M$.
Write $\lambda$ for the composite mapping
$$G \to \operatorname{GL}(V) \subset \mathfrak{gl}(V) \xrightarrow{\pi} \operatorname{ad}(
\operatorname{Lie}(G))$$.
Since $I \in M$, evidentally $\lambda(1) = 0$. Since $\pi$ is a $G$-module
homomorphism, $\lambda$ is $G$-equivariant.
Moreover, by construction $d\lambda_1$ is the identity mapping. Finally, by $G$-equivariance
the image under $\lambda$ of a maximal torus $T$ is contained in the $T$-fixed
points of $\operatorname{Lie}(G)$, i.e. in $\operatorname{Lie}(T)$.
This verifies the stipulated conditions (1),(2),(3) and (4).
Note that Barsdsley and Richardson go on to show that the restriction of $\lambda$
to the unipotent variety $\mathcal{U} \subset G$ defines a $G$-equivariant isomorphism
$\mathcal{U} \xrightarrow{\sim} \mathcal{N}$ where $\mathcal{N} \subset \operatorname{Lie}(G)$ is the nilpotent variety.
Moreover, by Luna's theorem (a proof valid in positive characteristic is given in Bardsley and Richardson's paper) there are $G$-invariant open subset $U \subset G$ and $U' \subset \operatorname{Lie}(G)$ with $1 \in U$ and $0 \in U'$ such that $\lambda_{\mid U}$ defines a surjective
etale mapping $U \to U'$. So $\lambda$ need not be birational, but it is fairly nice.
Note that Barsdley and Richardson actually formulate the above construction more generally
using representations $V$ of $G$ which are "nice enough" (among other things, the restriction of the traceform on $\mathfrak{gl}(V)$ to the image of $\operatorname{Lie}(G)$ must be non-degenerate), and under suitable assumptions (very roughly: the characteristic should be good for $G$) their construction gives "explicit" Springer isomorphisms $\mathcal{U} \xrightarrow{\sim} \mathcal{N}$ in characteristic $p>0$.<|endoftext|>
TITLE: Repeated Second Eigenvalue of the Adjacency Matrix of a Graph
QUESTION [6 upvotes]: This question is motivated by a talk I went to earlier today.
Suppose we have a $d$-regular graph $G$ with $n$ vertices, with adjacency matrix $A$.  
Let $$\lambda_1\geq \lambda_2 \geq\dots \geq \lambda_n$$ be the eigenvalues of $A$, so in particular $\lambda_1=d$.  If the first two eigenvalues are the same,  that is $\lambda_2=\lambda_1$, then it tells us a lot about the structure of the graph.  In particular, the graph must be disconnected.  (This is an if and only if condition) 
What if the second and third eigenvalues are equal?  That is, suppose that $\lambda_1>\lambda_2=\lambda_3$.  What does that tell us (if anything) about the structure of the graph?  
Additional questions:  If $\lambda_1=\lambda_2=\cdots=\lambda_k<\lambda_{k+1}$, then the graph will have exactly $k$ connected components.  What can we say about $G$ if $\lambda_1<\lambda_2=\cdots=\lambda_{k+1}<\lambda_{k+2}$?  That is, the second eigenvalue has multiplicity $k$.
What if the $n^{th}$ eigenvalue has multiplicity $k$?

REPLY [5 votes]: It you allow weighted adjacency matrices and if you insist (among there things) that the  eigenspace associated to $\lambda_2$ satisfies the "strong Arnold condition", then you are dealing the Colin de Verdiere invariant. For this, the best I can do now is to refer you to the wikipedia article on this invariant.
But you question is what can be said about connected graphs where $\lambda_1$ has multiplicity greater than one. The short answer is: very little. One reason for this that if our graph was associated with some physical of chemical system, than having $\lambda_2$ very close to $\lambda_3$ will have much the same effect as having $\lambda_2=\lambda_3$. There are certainly no results on graph spectra that relate properties of a graph to whether or not $\lambda_2$ is simple.
It is true that the existence of non-trivial automorphisms can prevent all eigenvalues from being simple. The classic result of this type is that if the automorphism group of a graph is vertex transitive, then its eigenvalues can not all be simple. Stronger assumptions may give more, since each eigenspace provides a representation of the automorphism group, and if one of these representations is faithful, then the multiplicity is at least the minimum degree of an irreducible representation of the group.
On the other hand though, having eigenvalues of large multiplicity tells us nothing about the automorphism group. Strongly regular graphs have large multiplicities, but in general no non-trivial automorphisms.<|endoftext|>
TITLE: What is the current state of the mathematics of Higgs fields?
QUESTION [9 upvotes]: Topical. I know there are good mathematical theories in which "Higgs" is used, in a geometrical sense. Would someone care to explain?
To clarify, I'd like to know about Higgs bundles on Riemann surfaces, rather than the standard model of high-energy physics.

REPLY [20 votes]: Since you are  asking about Higgs bundles, I can say a few words here. 
These  were introduced by Hitchin in the mid 1980's, although I'm not sure he used this term.  One can look at the introduction to his paper "The self-duality equations on Riemann surfaces" for some of the motivation and background. In brief outline, he starts from a gauge theoretic perspective: the initial data is a pair consisting of a vector bundle with connection on a compact Riemann surface $X$ and additional vector valued $1$-form (the Higgs field) subject to the equations referred to above. One of his main results is that such a pair is equivalent to a pair $(E,\theta)$ -- now called a Higgs bundle -- consisting of a holomorphic vector bundle and holomorphic section $\theta\in \Gamma(\Omega_X^1\otimes End(E))$
satisfying a suitable stability condition.  One consequence is: 

Theorem. There is a correspondence between irreducible representations of  $\pi_1(X)$ and stable Higgs bundles of degree $0$. 

Simpson rightly christened (a generalization of) this the nonabelian
Hodge theorem. It is indeed a  beautiful and fundamental theorem. It  generalizes an earlier theorem of Narasimhan and Seshadri that corresponds to the case where the Higgs field is zero.  These objects also make their appearance in Geometric Langlands -- but I'm hardly the right person to discuss these aspects. I could go on, but I think I'll stop here.
Perhaps I may add that the underlying geometry is very rich. Both the representation and Higgs sides admit natural moduli problems. Under this correspondence,
the two moduli spaces are homemorphic but are analytically distinct. In fact, the complex structures behave like $i$ and $j$ of the quaternions, and together with a natural symplectic structure, constitute a hyper Kähler manifold (away from the singularities).
Such things are quite rare. On the Higgs side, there is some additional geometry:
A $\mathbb{C}^*$-action obtained by dilating the Higgs field, and the Hitchin fibration
obtained by sending a Higgs bundle $(E,\theta)$ to the characteristic polynomial of $\theta$. The latter turns out to form a completely integrable system, that is the smooth fibres are Langrangian with respect to the symplectic structure.  Arguments from symplectic geometry show that these are tori, although in fact, these fibres turn out to be abelian varieties. What is even more remarkable is that part  of this story carries over into positive characteristic and had played an important role in Ngo's work. I understand very little of this, so instead I'll direct interested  readers to  the recent Bulletin article by Nadler.<|endoftext|>
TITLE: Minimal selector for a family of finite sets
QUESTION [8 upvotes]: A colleague is refereeing a paper in which the following
lemma appears implicitly: 
For any family $\mathcal G$ of nonempty sets let us call 
 a set $B$ a "selector" if $B$ meets all $F\in\mathcal  G$. 
Lemma: For every family $\mathcal G$ of nonempty finite sets there
 is a minimal selector $B$.  (That is, for all $x$ in $B$ 
 there is at least one $F$ in $G$ such that $x$ is the unique
 element of $B\cap F$.) 
A proof is quite easy: The family of selectors is closed 
under intersections of chains, so there must be a minimal
element (using Zorn's lemma in a version that is dual to 
the common one). 
I would like to know 

if the lemma is well known, and/or has a name; 
if the concepts used (selector, minimal selector) 
 have some other (established) name.

(This is a request for references. I will post a mathematical 
question separately, once I know the terminology.) 
I could not find this property in Howard-Rubin's "Consequences of the axiom of choice" but I admit I am not very experienced in using this book, so I may have overlooked something.

REPLY [7 votes]: I think "selector" usually refers to choosing just one element from each set in a family.  For the concept you described here, I've seen names like "blocker" or "blocking set", but I haven't seen them so often that I'd call them standard.  
The blockers of a family of finite sets (all included in some big set $X$) obviously constitute an upward-closed family of subsets of $X$, and this family is closed in the usual product topology of the power set of $X$.  Conversely (and not quite so obviously), every upward-closed, topologically-closed family of subsets of $X$ is the family of blockers of some family of finite subsets of $X$.  Unfortunately, I don't recall ever having seen these facts explicitly written down in the literature.
Note that the existence of a minimal blocker for a family of finite sets is a consequence of the fact that any topologically-closed subfamily of the power set of $X$ has a minimal element.<|endoftext|>
TITLE: Minimal blocks for a family of finite sets
QUESTION [7 upvotes]: In this question I asked for a reference for the following lemma: 
Lemma X: For every family $\mathcal G$ of nonempty finite sets there
is a minimal "blocking set" $B$.   By a "blocking set"  $B$ I mean a set which meets each $F\in\mathcal G$, and "minimal" refers to the subset relation. 
Questions: 

What is the exact strength (or approximate strength) of this lemma in the hierarchy of consequences of AC?   Clearly it implies that for every family of finite sets there is a choice function (as the finite sets can easily be made disjoint, and for a disjoint family a minimal blocking set is the same as a 
choice function.)

In particular, is lemma X equivalent to AC for families of finite sets? Or does it at least follow from the (stronger)  ordering principle?  Or at least from the (even stronger) Prime Ideal Theorem? 
Is there a natural condition on $\mathcal G$ which is weaker than "all elements of $\mathcal G$ are finite", but still implies the conclusion of the lemma above?

REPLY [3 votes]: Dear Martin: I think you can easily extend the result for systems of infinite sets, where there is a set $A$ intersecting all sets in finitely many but at least 1 point. The latter property was investigated by Erdos and Hajnal in On a property of families of sets, Acta Math. Acad. Sci. Hungar. 12 (1961), 87--123, see http://www.renyi.hu/~p_erdos/1961-11.pdf 
Perhaps I also had some papers on this. Does this make any sense?<|endoftext|>
TITLE: Comparison map between de Rham cohomology of analytic and formal neighborhoods of singularities
QUESTION [25 upvotes]: Suppose that $X$ is a complex algebraic (or complex analytic) variety, and $x \in X$ is a singular point.  I am interested in two types of local differential forms at $x$: analytic and formal.
First, let $\mathcal{O}_{X,x}^{\text{an}}$ be the ring of analytic germs of functions at $x$.  I am interested in the complex $\Omega_{X,x}^{\text{an}}$ of analytic germs of differential forms, i.e., linear combinations of elements $h_0 dh_1 \wedge \cdots \wedge dh_k$ for $h_0, \ldots, h_k \in \mathcal{O}_{X,x}^{\text{an}}$, and the corresponding de Rham cohomology $H^\bullet(\Omega_{X,x}^{\text{an}})$.  Explicitly, if $X \subseteq
\mathbf{A}^n$ is a subvariety of affine space cut out by equations $f_1, \ldots, f_m$, then this complex is defined as $\Omega_{\mathbf{A}^n,x}^{\text{an}} / (f_1, \ldots, f_m, df_1,
\ldots, df_m)$, where we quotient by the ideal in the de Rham differential graded algebra generated by the $f_i$ and $df_i$.
Next, let $\hat {\mathcal{O}}_{X,x}$ be the completion of the local ring of algebraic functions at $x$, i.e., the ring of (not-necessarily convergent) formal power series of functions at $x$.  Let $\hat {\Omega}_{X,x}$ be the complex of formal differential forms, i.e., linear combinations of elements $h_0 dh_1 \wedge \cdots \wedge dh_k$ for $h_0, \ldots, h_k \in \hat {\mathcal{O}}_{X,x}$.  For $X \subseteq \mathbf{A}^n$, this is defined as a quotient of $\hat \Omega_{\mathbf{A}^n,x}$ in the same manner as above.
Since one has a canonical inclusion $\mathcal{O}_{X,x}^{\text{an}} \hookrightarrow \hat {\mathcal{O}}_{X,x}$, one obtains a canonical comparison map
$H^\bullet(\Omega_{X,x}^{\text{an}}) \to H^\bullet(\hat \Omega_{X,x}).$
My question is: When is this map an isomorphism?

I am particularly interested in the case that the LHS is
finite-dimensional, e.g., when $X$ has an isolated singularity at $x$
(finite-dimensionality of the LHS then follows from the Theorem of
Section 3.17 of Bloom and Herrera's paper ``De Rham Cohomology of an
Analytic Space,'' (Invent. Math. 7, 275--296 (1969)).

More details and reformulations:

Under the finite-dimensionality hypothesis, the comparison map is definitely surjective: the RHS is the inverse limit of $H^\bullet(\Omega_{X,x} / \mathfrak{m}_{X,x}^N \cdot \Omega_{X,x})$, where $\mathfrak{m}_{X,x} \subseteq \mathcal{O}_{X,x}$ is the maximal ideal, and the LHS surjects to each of these (by lifting closed or exact forms modulo $\mathfrak{m}_{X,x}^N$ to closed or exact analytic forms). So both sides are finite-dimensional and the comparison map is surjective.
Thus, under this hypothesis, the question reduces to: When it is true that, if a closed analytic form $\alpha \in \Omega_{X,x}^{\text{an}}$ is the differential of a formal form in
$\hat \Omega_{X,x}$, then it is also the differential of an analytic form in $\Omega_{X,x}^{\text{an}}$? (Perhaps, an analytic approximation theorem could be applied to answer this.)
Next, I will restrict this question to the special case that interests me: isolated singularities which are locally complete intersections. In this case, by results of Sections 4 and 5 of Greuel's paper ``Der Gauss-Manin-Zusammenhang isolierter Singularitaeten von vollstaendigen Durchschnitten,'' (Math. Ann. 214, 235--266 (1975)), one has the
formula 
$H^\bullet(\Omega_{X,x}^{\text{an}}) \cong \mathbf{C}^{\mu_x-\tau_x}[-\operatorname{dim} X],$
where $\mu_x$ is the Milnor number of the singularity at $x$, and the notation above indicates that the de Rham cohomology of the analytic neighborhood of $x$ is concentrated in degree equal to the dimension of $X$. Also, $\tau_x$ is the Tjurina number, which is the dimension of the singularity ring at $x$: explicitly, if $X$ is locally a complete intersection of dimension $n-m$ cut out at $x \in \mathbf{A}^n$ by functions $f_1, \ldots, f_m$, then the singularity ring is the quotient of $\mathcal{O}_{X,x}^{\text{an}}$ by the ideal generated by the $f_i$ together with the determinants of the $(n-m) \times (n-m)$ minors of the Jacobian matrix $(\frac{\partial f_i}{\partial x_j})$.   In other words, the Tjurina number here is the dimension of the torsion of the germs of differential forms $\Omega_{X,x}^{\operatorname{dim}(X),\text{an}}$ of degree $\operatorname{dim}(X)$.
In this case, I would only want to know whether the same formula holds for the de Rham cohomology of the formal neighborhood, i.e., that the dimension of $H^\bullet(\hat{\Omega}_{X,x})$ is equal to the Milnor number, and not less.

[Readers who are tired of reading can stop here---I will give one more alternative formulation:]

Alternatively, one can work with the de Rham complex modulo torsion, $\tilde{\Omega}_{X,x}^{\text{an}}$, obtained from $\Omega^{\text{an}}_{X,x}$ by modding by the torsion submodule over $\mathcal{O}_{X,x}^{\text{an}}$.  This is equivalent to working with germs of forms modulo those forms that become zero when restricted to the smooth locus, i.e., whose representatives on open neighborhoods of $x$ have zero restriction to smooth open subsets.  In this case, Greuel's formula (still for an isolated singularity at $x$ which is locally a complete intersection) remains the same,
$H^\bullet(\tilde{\Omega}_{X,x}^{\text{an}}) \cong \mathbf{C}^{\mu_x-\tau_x}[-\operatorname{dim} X].$
In the alternative formulation, I would like to know again if the same formula holds replacing analytic germs of forms mod torsion, $\tilde{\Omega}_{X,x}^{\text{an}}$, by formal forms mod torsion.  It follows from Greuel's paper that, still assuming $x$ is an isolated singularity which is locally a complete intersection,  the two questions are equivalent.

REPLY [3 votes]: For the case where x is an isolated singularity (which is probably a necessary hypothesis), we positively answered this question in arXiv:1211.1883, Theorem 4.45.<|endoftext|>
TITLE: Who introduced the concept of Primitive recursive functions?
QUESTION [7 upvotes]: I have thought that Gödel introduced the concept of Primitive recursive functions in his seminal paper "Über formal unentscheidbare Sätze der Principia Mathematica und verwandter Systeme" (I hope I got the title right...). This paper, to best of my knowledge, was published in 1931.
Ackermann published the famous Ackermann function in his paper "Zum Hilbertschen Aufbau der reellen Zahlen" in (again, to best of my knowledge) 1928. His publication seems to enclose primitive recursive functions and the main results is that there exists a total computable function outside PR. This suggests that PR has been known prior 1928, thus prior Gödel's work.
Who introduced primitive recursive functions first?

REPLY [17 votes]: I believe the explicit use of definitions by primitive recursion goes back to Grassman, 1861.  
Dedekind in 1888 not only highlighted such definitions but had a proof that they work as intended, i.e. define unique functions. 
But it is probably Skolem who first clearly recognised the primitive recursive functions as together forming a class of functions of particular interest. Indeed, his 1923 paper is entitled "The foundations of elementary arithmetic established by means of the recursive mode of thought ...", and, as with Gödel in 1931, by "recursive" Skolem here means "primitive recursive". This paper is usually credited with isolating Primitive Recursive Arithmetic as of particular interest for finitism (and hence for the Hilbert Programme).<|endoftext|>
TITLE: Which differential equations allow for a variational formulation?
QUESTION [33 upvotes]: Many ODE's and PDE's arising in nature have a variational formulation. An example of what I mean is the following. Classical motions are solutions $q(t)$ to Lagrange's equation
$$
\frac{d}{dt}\frac{\partial L(q,\dot q)}{\partial\dot q}=\frac{\partial L(q,\dot q)}{\partial q},
$$
and these are critical points of the functional
$$
I(q)=\int L(q,\dot q)dt.
$$
Of course one needs to be precise with what considers a solution to both equations. This amounts to specifying regularity and a domain of the functional. This example is an ODE, but many PDE examples are possible as well (for example electromagnetism, or more exotic physical theories). Once one knows a variational description of the problem, many more methods are available to solve the problem. 
Now I do not expect that any PDE or ODE can be viewed (even formally) as a critical point of a suitable action functional. This is because this whole set up reminds me of De Rham cohomology: "which one-forms (the differential equations) are exact (that is, the $d$ of a functional)?". The last sentence is not correct, but the analogy maybe is? Anyway, my question is:

Are there any criteria to determine if a given differential equation admits a variational formulation?

REPLY [2 votes]: Well this is a classical problem known as "The inverse problem in the calculus of variations". There is a huge amount of references on the problem you can google at.  The problem as if the system $y_i''=F(x,y_j,y_j')$, $i=1,2,..,n$ can be identified with extremals of the equation $\int \phi(x,y_j,y_j') \rightarrow min$ amounts to solving the system of pdes for the partial derivatives of $\phi$, say $\phi_{ij}$. Davis [1928] restated the problem as that of finding an integrating factor $P_{ij}$ such that the system $P_{ij}(F_j-y_j'') = E(\phi)$,where $E$ denotes the Euler-Lagrange operato. There appears some condition on self-adjointness The case $n=2 $ was solved by the first Field medalist Jesee Douglas (1941). He used Riquier-Janet theory. For $n>2$ it remain possible except for cumbersome cases. Spencer and Quillen introduced the Spence cohomology to give suficient conditions for the overdetermined system to become integrable. Some references:
1) The inverse problem on the calculus of variations\ldots W. Sarlet, G. thompson, G.E.  Prince. TAMS 354, Num.7, 2897-2919, 2002.
2) Overdetermined systems of linear PDEs. D.C. Spencer., 1969 (sorry Idon't have the complete reference at hand).
3)J. Douglas. Solution to the inverse problem of the calculus of variations. TAMS 50 (1941), 71-128.
Professor Peter Olver (University of Minnesota) is probably one of the major authorities on the topic.<|endoftext|>
TITLE: Uniform Embedding into Euclidean Space
QUESTION [9 upvotes]: Given a locally compact, separable, metric space $X$.

When does $X$ uniformly embed into some Euclidean space? 

This means, when does there exist some integer $n$ and a closed subset $Y\subset\mathbb{R}^n$ such that $X$ and $Y$ are uniformly equivalent, i.e., there exist a one-to-one map $f:X\to Y$ such that $f$ and the inverse $f^{-1}$ are uniformly continuous?
Background/Motivation
If we just ask for a topological embedding (i.e. $f$ and $f^{-1}$ are continuous), then such an embedding exists if and only if the covering dimension of $X$ is finite. Neccessity is clear, and sufficiency is shown in Corollary 2.6 of Luukkainen: "Embeddings of n-dimensional locally compact metric spaces to 2n-manifolds".
A neccessary condition for uniform embedding of $X$ is that the uniform covering dimension (as defined by Isbell) of $X$ is finite. 
Recall that an open cover of $X$ is called uniform if there exists some $\epsilon>0$ such that for every $x\in X$ the open ball of radius $\epsilon$ around $x$ is contained in some element of the cover.
Recall also that the order of a cover is at most $k$ if the intersection of any $k+1$ different elements of the cover is empty.
The uniform covering dimension of $X$ is at most $k$ if every uniform open cover of $X$ can be refined by a uniform open cover that has order at most $k+1$.
Is this condition also sufficient, i.e., does $X$ uniformly embed into some Euclidean space if and only if it has finite uniform covering dimension?

REPLY [2 votes]: Assume $X$ is uniformly embedded in $\mathbb E^d$ 
and $\varepsilon>\delta>0$.
Then there is $N$ such that any $\varepsilon$-ball in $X$ 
can be covered by $N$ balls of radius $\delta$.
It is easy to construct a rotationally symmetric Riemannian metric on $\mathbb R^2$
which does not have this property for some fixed $\varepsilon$ and $\delta$. Its  uniform covering dimension is still  2...<|endoftext|>
TITLE: Which information can be obtained from Poincaré series ? 
QUESTION [8 upvotes]: If $A= \bigoplus_{i\ge 0}A_i$ is a graded commutative Noetherian algebra over a field, its Poincaré series is given by $P(t) = \sum_{i\ge 0} \dim(A_i)t^i$.  Although the definition of $P(t)$ only depends on the graded vector space underlying $A$, the Krull dimension of the ring $A$ can be obtained from the Poincaré series as the order of the pole of $P(t)$ at $t=1$.
Question: Are there other information about the ring structure of $A$ that can be obtained from $P(t)$ ? 
Since I'm particularly interested in cases where $A$ is the cohomology ring of a finite group, I'm also looking for an example of finite groups whose mod-p cohomology rings are not isomorphic but have identical Poincaré series. 
Thanks in advance.

REPLY [8 votes]: I don't know if it's possible to obtain further information on the ring structure in 
general. However, if $A$ is the mod-p cohomology ring of a finite group, a result 
of Benson and Carlson states that if $A$ is Cohen-Macaulay, then $P(t)$ satisfies 
the functional equation 
$$P(1/t) = (-1)^d P(t)\hspace{90pt}(\ast)$$
where $d$ is the Krull dimension of $A$. Conversely, if a given $P(t)$ doesn't satisfy 
this equation, you know that $A$ isn't Cohen-Macaulay, i.e. the depth of $A$ is less 
than its Krull dimension. 
In case $A$ has Krull dimension 2, than $A$ is Cohen-Macaulay 
iff $P(t)$ satisfies $(\ast)$. These results can be found in the paper Benson, Carlson: Functional equations for Poincaré series in group cohomology. Bull. London Math. Society 26(1994), 438-448.
As an example for groups with identical Poincaré series you can take $E := \mathbb{Z}/2 \times \mathbb{Z}/2$ and $D_8$, the dihedral group of order 8. Their cohomology rings 
$$H^\ast(E;\mathbb{F}_2)=\mathbb{F}_2[x,y],\;\; |x|=|y|=1$$
$$H^\ast(D_8;\mathbb{F}_2)=\mathbb{F}_2[x,y,z]/(xy),\;\; |x|=|y|=1, |z|=2$$
aren't isomorphic (since only one is a domain) but both have Poincaré series 
$P(t)=\frac{1}{(1-t)^2}$. 

Added: There is a paper of R. Stanley (who is also active on MO) that contains some properties of the Poincaré series that may be of interest. I just quote a few: 

If $A$ is Gorenstein of Krull dimension $d$, then $P(1/t)=(-1)^dt^aP(t)$ for some integer $a$. This generalizes $(\ast)$ (with a=0) because the mod-p cohomology ring of a finite group is Cohen-Macaulay iff its Gorenstein.
$A$ is a complete intersection with generators of degree 1 iff $P(t)$ has the form 
$$P(t)=\frac{\prod_{i=1}^l(1+t+\cdots+t^{m_i})}{(1-t)^d}$$
A sequence of homogeneous elements $x_1,...,x_k \in A$ of positive degrees $n_i$ is regular iff $P(A,t) = P(B,t)/\prod_i (1-t^{n_i})$ where $B=A/(x_1,...,x_n)$.<|endoftext|>
TITLE: What do loop groups and von Neumann algebras have to do with elliptic cohomology?
QUESTION [34 upvotes]: Recall that complex $K$-theory is a cohomology theory on topological spaces, which can be described in several equivalent ways: 

Given a finite complex $X$, $K^0(X)$ is the Grothendieck group of vector bundles on $X$. $K^*$ is even-periodic, and this determines the entire cohomology theory. Using the tensor product of vector bundles, $K$ becomes a multiplicative cohomology theory. There is a corresponding ring spectrum. 
The classifying space $BU \times \mathbb{Z}$ for $K^0$ is, by a theorem of Atiyah, the space of Fredholm operators on a countably-dimensional Hilbert space. So we can think of classes in $K^0(X)$ as "families of Fredholm operators" parametrized by $X$: the "index" of such a family should be a virtual vector bundle, which connects to the previous definition. 
$K$-theory is an even-periodic theory, so it is complex-orientable and  corresponds to a formal group on $K^0(\ast) = \mathbb{Z}$. This formal group is the multiplicative one, which turns out to be Landweber-exact. Consequently, one can construct $K$-theory directly from the formal multiplicative group (once one has the spectrum $MU$) via $K_\bullet(X) = MU_\bullet(X) \otimes_{MU_\bullet} K_\bullet$. 
The spectrum for $K$-theory can be obtained by taking the ring spectrum $\Sigma^\infty \mathbb{CP}^\infty_+$ (which is a ring spectrum as $\mathbb{CP}^\infty$ is a topological abelian monoid) and inverting the natural element in $\pi_2$. (This is a theorem of Snaith.)

It's sort of remarkable that $K$-theory can be described both geometrically (via vector bundles or operators) or algebraically (via formal groups or Snaith's theorem). The only  explanation that I can think of for this is that the correspondence between (complex-orientable) ring spectra and formal groups is given more or less in terms of Chern classes of vector bundles, so a cohomology theory built directly from vector bundles would have a good chance of furnishing a fairly simple formal group law. (One can use this sort of argument to prove Snaith's theorem, for instance.)
A much less formal example of a formal group is that associated to an elliptic curve. If $E/\mathrm{Spec} R$ is an elliptic curve, then under appropriate hypothesis (Landweber exactness, or flatness of the map $\mathrm{Spec} R \to M_{1,1} \to M_{FG}$, or  more concretely that $R$ is torsion-free and for each $p$, the Hasse invariant $v_1$ is a nonzerodivisor in $R/pR$) we can construct an "elliptic cohomology" theory $\mathrm{Ell}^*$ which is even-periodic and whose formal group is that of $E/R$. 
The associated formal group can have height up to $2$, so we get something much more complicated than $K$-theory. 
It has been suggested that there should be a geometric interpretation of elliptic cohomology. This seems a lot more difficult, because the formal group law associated to an elliptic curve is less elementary than $\hat{\mathbb{G}_m}$.  There are various programs (which start with Segal's survey, I believe), all of which I know nothing about, to interpret elliptic cohomology classes in terms of von Neumann algebras, loop group representations, conformal field theories, ...
I can understand why a geometric interpretation of elliptic cohomology would be desirable, but it's mystifying to me why researchers in this area are concentrating on these specific objects. Is there a "high-concept" explanation for this, and motivation (to someone without a background in geometry) for how one might "believe" in these visions? Is there a reason loop groups should be "height two" where the unitary group is "height one"?

REPLY [9 votes]: There are various programs (which start with Segal's survey, I believe), all of which I know nothing about, to interpret elliptic cohomology classes in terms of von Neumann algebras, loop group representations, conformal field theories, ...
I can understand why a geometric interpretation of elliptic cohomology would be desirable, but it's mystifying to me why researchers in this area are concentrating on these specific objects.

You probably know most of this by now, but here are some thoughts. As usual I will be working at a very heuristic level throughout this answer.
Von Neumann Algebras. One place to start is the observation is that for $H$ an infinite-dimensional Hilbert space, the von Neumann algebra $B(H)$ has automorphism group the projective unitary group $PU(H)$. $PU(H)$ fits into a short exact sequence
$$1 \to U(1) \to U(H) \to PU(H) \to 1$$
and $U(H)$ is contractible by Kuiper's theorem; thus $PU(H)$ is a $B^2 \mathbb{Z}$, and $BPU(H)$ is a $B^3 \mathbb{Z}$. Hence $H^3(X, \mathbb{Z})$ is, in a suitable sense, a "Brauer group" of $X$ describing bundles of von Neumann algebras (isomorphic to $B(H)$) over $X$.
The significance of this observation is that $H^3(X, \mathbb{Z})$ is a natural cohomology group parameterizing twists of K-theory over $X$; equivalently, there is a natural map from $B^3 \mathbb{Z}$ to $BGL_1(KU)$. Given a bundle of von Neumann algebras over $X$, the corresponding twisted K-theory groups are something like the K-theory of module bundles of Hilbert modules over the bundle of von Neumann algebras, but don't trust me to have the specifics right here.
Now it's also known (see e.g. Ando-Blumberg-Gepner) that $H^4(X, \mathbb{Z})$ is a natural cohomology group parameterizing twists of tmf over $X$; equivalently, there is a natural map from $B^4 \mathbb{Z}$ to $BGL_1(tmf)$. If you could build a (higher) category which deloops von Neumann algebras in a suitable sense, you might hope to realize $BPU(H) \cong B^3 \mathbb{Z}$ as the automorphisms of an object in this category, and then families of those objects over $X$ could be a geometric avatar of these twists of tmf in the same way as above. I believe that conformal nets is explicitly intended to be such a delooping.
Loop group representations. This is the analogue of $G$-equivariant K-theory having something to do with the representation theory of $G$. One picture of tmf whose accuracy I can't comment on is that it should look heuristically like $K(ku)$, the cohomology theory presented by (Baas-Dundas-Rognes) 2-vector bundles. So $G$-equivariant tmf should look heuristically like $G$-equivariant 2-vector bundles, which over a point should look heuristically like representations of $G$ on (suitably dualizable) 2-vector spaces.
Whatever that means, such a thing ought to have a "character" which, rather than being a class function on $G$, or equivalently a function on the adjoint quotient $G/G$, is instead an equivariant vector bundle on $G/G$. Now $G/G$ looks heuristically like the classifying stack $BLG$ of the loop group $LG$, and hence an equivariant vector bundle on $G/G$ looks heuristically like a representation of $LG$. Freed-Hopkins-Teleman makes this precise. The non-equivariant version of this story is Witten's story about tmf having something to do with ($S^1$-equivariant?) K-theory of the free loop space $LX$.
Conformal field theories. This is the analogue of K-theory being presentable by vector bundles with connection. One way to think about a vector bundle with connection on a manifold $X$ is that it is a "$1$-dimensional topological field theory over $X$": that is, it assigns a vector space to a finite set of points equipped with signs in $X$ (the tensor product of either the fibers of the vector bundle or their dual depending on the orientation), and it assigns a map of vector spaces to every oriented cobordism between such points in $X$ (the tensor product of either the holonomies or the evaluation or coevaluation maps).
The historical motivation for generalizing this to $2$-dimensional conformal rather than topological field theories is, I think, to explain the modularity properties of the Witten genus. But again, don't trust me to have the specifics right here. (I guess it's $2$-dimensional topological rather than conformal field theories over $X$ that look more like $2$-vector bundles.)<|endoftext|>
TITLE: Funktorialität in der Theorie der automorphen Formen
QUESTION [15 upvotes]: In 2010 Langlands wrote an article with the title Funktorialität in der Theorie der automorphen Formen:  Ihre Entdeckung und ihre Ziele.  On the IAS website, he says that 

This note ... was written as commentary to accompany the original letter in a collection of documents on reciprocity laws and algebraic number theory, ... to appear.

By the original letter he means of course his famous Letter to André Weil in which the Langlands conjectures were first formulated.
Questions. Has this collection of documents already appeared?  If not, when and where is it likely to appear?  What other documents does it contain?

REPLY [6 votes]: Emil Artin and Beyond – Class Field Theory and $L$-Functions
Heritage of European Mathematics
Della Dumbaugh ( University of Richmond, USA )
Joachim Schwermer (University of Vienna, Austria)
Emil Artin and Beyond – Class Field Theory and $L$-Functions
ISBN 978-3-03719-146-0
DOI 10.4171/146
March 2015, 245 pages, hardcover, 16.5 x 23.5 cm.
68.00 Euro
Table of Contents
I. Class field theory: From Artin's course in Hamburg to Chevalley's "Éléments idéaux
Claude Chevalley's thesis on class field theory and his notion of "Éléments idéaux
Introduction
Letter from Claude Chevalley to Helmut Hasse, June 20, 1935
Letter from Helmut Hasse to Claude Chevalley, June 28, 1935
II. Creating a life: Emil Artin in America
Emigration, immigration and pre-remigration
Introduction
Letter from Solomon Lefschetz to Father John O'Hara, January 12, 1937
III. The collaboration of Emil Artin and George Whaples
The work of Artin and Whaples--A conceptual breakthrough in algebraic number theory
Introduction
George Whaples' application to the Institute for Advanced Study, School of Mathematics, Princeton, NJ, February 10, 1941
IV. Margaret Matchett: Artin's student at Indiana and her thesis
Margaret Matchett and her thesis "On the zeta function for ideles"
Introduction
Margaret Mathcett's doctoral dissertation "On the zeta function for ideles"
V. L-functions by James W. Cogdell
L-functions and non-abelian class field theory, from Artin to Langlands
VI. Automorphic L-functions by Robert P. Langlands
Letter from Robert Langlands to André Weil, January 1967
Funktorialität in der Theorie der automorphen Formen: Ihre Entdeckung und ihre Ziele
Einführung
Bibliography
Index<|endoftext|>
TITLE: Packing and isoperimetrics
QUESTION [10 upvotes]: Suppose a manufacturer bottles small units of liquid and ships them via very large trucks.
If the transportation cost nothing, spherical bottles would minimize the packaging cost (isoperimetric inequality); if packaging cost nothing, cubic bottles (say) would minimize the transportation cost, because they would pack on the trucks with no wasted space.
How would the ideal container vary with the relative cost of packaging (measured as the surface area of one bottle) and transportation (measured by the packing density on an infinitely capacious truck)?
[My "applied" formulation notwithstanding, I mean this as a pure mathematics question, so please idealize and ignore any distracting side issues.]
A reasonable conjecture might seem that one gets the correct family of shapes by starting with a tight lattice packing of spheres and then over-inflating and rescaling the spheres until asymptotically they assume the form of Voronoi cells of a lattice packing.  So one would get constant mean curvature surfaces up to the circular disks at the interfaces between adjacent spheres.  Optimal or not, this would give lower bounds.  But how to compute them?

REPLY [8 votes]: This is the problem of wet foams, soap bubble clusters separated by water. As the liquid fraction goes to zero, you get the dry foam or partitioning problem, optimally the hexagonal honeycomb in 2D (Hales) and conjecturally the Weaire-Phelan structure in 3D or the Kelvin structure if all cells must be congruent. As the liquid fraction increases you pass through the optimal sphere packing (proved by Thue in 2D and Hales in 3D) to spherical bubbles floating in a sea of liquid. The transition in 2D from the hexagonal honeycomb to the circle packing is easy to conjecture but would be very hard to prove. In 3D a similar smooth transition beginning with the Kelvin partition would yield the inefficient BCC lattice sphere packing instead of the optimal FCC lattice sphere packing. A smooth transition beginning with the less efficient rhombic dodecahedra would yield the optimal FCC lattice sphere packing. So the optimal wet foam probably jumps discontinuously from one branch to the other at some point.<|endoftext|>
TITLE: Properties of natural numbers such that there is a "very large largest" number with that property
QUESTION [12 upvotes]: I'm looking for properties (P) such that you would assume that there are infinitely many natural numbers with property (P) (for example because there are very large numbers with that property) but where it turns out that there are only finitely many.
EDIT: Note that the eventual counterexamples question asks for (P) such that the smallest $n$ with property (P) is large; the current question asks for (P) such that the largest $n$ with property (P) is large.

REPLY [5 votes]: A good start might be finite sequences in OEIS. The search keyword:  fini returns 
4660 results including Gerry's narcissistic number and left-truncatable prime.
Another example is A080601 Number of positions that the 3 X 3 X 3 Rubik cube puzzle can be in after exactly n moves with largest term $91365146187124313$. 
The search can be automated filtering only large numbers.
Probably you are not interested in this, but artificial solutions can be constructed easily:
Solutions $x$ of $x^2-y^2=\text{big}$ or trivially roots of $p=\prod_{i \in S} (x-i)$.<|endoftext|>
TITLE: The number of irreducible polynomials over ${\mathbb F}_p$
QUESTION [9 upvotes]: Let $p$ be a prime number. The number of monic irreducible polynomial $P\in{\mathbb F}_p[X]$, in terms of the degree $d$, begins with
$${\rm irr}(1)=p,\qquad{\rm irr}(2)=\frac{p(p-1)}2,\qquad{\rm irr}(3)=\frac{p(p^2-1)}3,\qquad{\rm irr}(4)=\frac{5p^2(p^2-1)}{12}.$$
This seems to be the beginning of a nice sequence of polynomials in $p$. Does someone know the general formula?
Motivation. As F. Brunault pointed out to me, there is a polynomial $\Pi_{n,p}$ that vanishes over ${\bf M}_n({\mathbb F}_p)$. Just take the lcm of all the polynomials of degree $n$ over $F_p$. 
In closed form, it is the product of all the irreducible polynomials of degree $d\le n$, to the power $[\frac{n}{d}]$ (integral part). The degree of $\Pi_{n,p}$ is 
$$\delta(n)=\sum_{d\le n}d[\frac{n}{d}]{\rm irr}(d).$$
In terms of $n$, this degree is
$$\delta(1)=p,\quad\delta(2)=p(p+1),\quad\delta(3)=p(p^2+p+1),\quad\delta(4)=\frac13p(p+1)(5p^2-2p+3).$$
Edit. The values given above for $n=4$ are erroneous.

REPLY [13 votes]: The product of all monic irreducible polynomials of degree dividing $d$ in $\mathbb F_q[x]$ is $$x^{q^d}-x,$$
so from Mobius inversion we get the number of irreducible polynomials of degree $d$ is
$$M _d (q) = \frac{1}{d} \sum _{k | d} \mu(k) q ^{d/k} $$
This is known as the necklace polynomial.

REPLY [10 votes]: Gjergji Zaimi already said it all, but I want to point out a tiny bit longer but equally cute way to derive the same formula. Every monic polynomial over $\mathbb{F}\_q$ decomposes into a product of irreducibles uniquely, hence we have a formal power series equality
 $$
\frac{1}{1-qt}=\prod_{k\ge 1}\frac{1}{(1-t^k)^{M_k(q)}}
 $$
(indeed, the coefficient of $t^n$ on the left is the number of monic polynomials of degree $n$, and the coefficient of $t^n$ on the right counts the decompositions into irreducibles with multiplicities). Taking logarithmic derivatives, one can do the M\"obius inversion and arrive at the formula pointed out by Gjergji.<|endoftext|>
TITLE: Retrieval of algebra structure from spectral sequence
QUESTION [9 upvotes]: Suppose we have a spectral sequence of algebras and know that it degenerates at some $E_r$, take for example the cohomology Leray Serre spectral sequence associated to some fibration $F\hookrightarrow E\rightarrow B$. Suppose we are working over a field so there is no extension problem and so $H^n(E)\cong\oplus_{r+s=n} E_{\infty}^{r,s}$. Under what conditions can we read off algebra structure of $H^*(E)$ from the $E_{\infty}$ page? (e.g. one very special case is when the $E_{\infty}$ page is free as an algebra.)
I think it is not possible in general, since we can have two different algebras with some filtrations such that the associated graded objects are isomorphic, but I would like to know if there are certain conditions under which one can compute the algebra structure.

REPLY [4 votes]: I want to highlight a case that is similar to the free case but is used in several cohomology computations in literature: Let $C=(C^i)_{i\ge 0}$ be a cochain complex with filtration 
$$C^i=F^0C^i \supseteq \cdots \supseteq F^iC^i \supseteq F^{i+1}C^i=0.$$
Assume that $E_\infty^{\ast,\ast}=E_\infty^{\ast,0}\otimes_k E_\infty^{0,\ast}$ and $E_\infty^{0,\ast}=k[x_1,...,x_n]$ is a polynomial algebra with homogeneous $x_i$. Then 

$\qquad\qquad  H^\ast(C) = E_\infty^{\ast,0}\otimes_k k[X_1,...,X_n] \cong E_\infty$ 

The proof is straightforward. As an application let $$B\mathbb{Z}/2 \to E \to B$$ be a fibration with path-connected base such that $\Pi_1(B)$ acts trivially on $H^\ast(B\mathbb{Z}/2;\mathbb{F}_2)$. Write ${H^\ast(B\mathbb{Z}/2;\mathbb{F}_2) =\mathbb{F}_2[z]}$ and let $h$ be minimal with
$d_{2^{h}+1}(z^{2^h})=0$. Assume moreover that $y_i := Sq^{2^i}\cdots Sq^1(d_2z)$ $(i=0,...,h-1)$ is a regular sequence in $H^\ast(B;\mathbb{F}_2)$. Then 

$\qquad H^\ast(E;\mathbb{F}_2)=\mathbb{F}_2[x] \otimes H^\ast(B;\mathbb{F}_2)/(y_1,...,y_{h-1}),\quad \deg(x)=2^h$. 

This was used by Quillen in his computation of the cohomology of the extra-special 2-groups and the Spinor groups.<|endoftext|>
TITLE: Integration of the product of pdf & cdf of normal distribution 
QUESTION [7 upvotes]: Denote the pdf of normal distribution as $\phi(x)$ and cdf as $\Phi(x)$. Does anyone know how to calculate $\int \phi(x) \Phi(\frac{x -b}{a}) dx$? Notice that when $a = 1$ and $b = 0$ the answer is $1/2\Phi(x)^2$. Thank you!

REPLY [13 votes]: This might be a setting where relying on the probabilistic meaning of the functions $\phi$ and $\Phi$ saves ink and tedious computations. 
Let $X$ and $Y$ denote standard normal random variables. Then $\int\limits_{-\infty}^\infty u(x)\phi(x)\mathrm dx=E(u(X))$ for every suitable function $u$  and $\Phi(x)=P(Y\leqslant x)$ for every real number $x$. 
Using this for the function $u:x\mapsto\Phi((x-b)/a)$ and assuming furthermore that $X$ and $Y$ are independent, one sees that the integral to be computed is
$$
(\ast)=E(\Phi((X-a)/b))=P(Y\leqslant(X-b)/a).
$$
Thus,
$$
(\ast)=P(Z\geqslant b),
$$
where $Z=X-aY$ (this step uses the fact that $a\gt0$). Now, the random variable $Z$ is normal as a linear combination of independent gaussian random variables, with mean $0$ and variance $1+a^2$, hence $Z=\sqrt{a^2+1}\cdot T$, where $T$ is standard normal. Thus,
$$
(\ast)=P(T\geqslant b/\sqrt{a^2+1})=1-\Phi\left(b/\sqrt{a^2+1}\right).
$$
Likewise, if $a\lt0$, then $(\ast)=\Phi\left(b/\sqrt{a^2+1}\right).$
In particular, if $b=0$ then, for every $a\ne0$, $(\ast)=\frac12$.<|endoftext|>
TITLE: Groups of matrices that preserve several quadratic forms
QUESTION [9 upvotes]: Given two (or more) quadratic forms (on the same vector space) consider the group of matrices that preserve these forms, i.e. $Q_i=U Q_i U^T$, $i=1,2..,k$ What is known about such groups? (at least for k=2 and the forms are symmetric and one is of full rank) 
The keywords? Where to read?

REPLY [2 votes]: A starting point is that such a $U$ commuttes with $Q_2Q_1^{-1}$. This reduces the analysis to the restriction of $U$ to generalized eigenspaces of $Q_2Q_1^{-1}$.
In particular, suppose that $Q_1$ (or some linear combination of the $Q_j$'s) is positive definite. Then one may choose a basis in which $Q_1=I_n$ and $Q_2$ is diagonal. Then $U$ is block diagonal, with each block being associated with one eigenvalue of $Q_2$. In general, the group is very small.<|endoftext|>
TITLE: A combinatorial formula involving the necklace polynomial
QUESTION [5 upvotes]: This question is motivated by the answers given to my previous one. In combinatorics, the necklace polynomials are given by
$$M(X,n)=\frac1n\sum_{d|n}\mu\left(\frac{n}{d}\right)X^d,$$
where $\mu$ is the Möbius function.
It seems that the following formula holds true (at least, I checked it up to $n=6$):
$$\sum_{k\le n}k\left[\frac{n}{k}\right]M(X,k)=X^n+X^{n-1}+\cdots+X^2+X.$$
Is this classical. Is there any reference. Are there known applications of it ?
To me, here is an application: the lcm of all the monic polynomials of degree $n$ over ${\mathbb F}_p$, which is the simplest polynomial vanishing identically over ${\bf M}_n({\mathbb F}_p)$, has degree $p^n+p^{n-1}+\cdots+p^2+p$.

REPLY [4 votes]: By plugging in the definition of $M$ and rearranging the terms, your polynomial is
$$\sum_dx^d\sum_{m\le n/d}\left\lfloor\frac n{md}\right\rfloor\mu(m).$$
A basic property of Möbius inversion is that
$$\sum_{m\le x}\left\lfloor\frac xm\right\rfloor\mu(m)=\begin{cases}1,&x\ge1,\\\\0,&x<1.\end{cases}$$

REPLY [2 votes]: The LHS of your formula can be rewritten as
 $$
\sum_{k\ge 1, d\mid k}\left[\frac{n}{k}\right]\mu(k/d)X^d,
 $$
which after rearranging terms becomes
 $$
\sum_{d\ge 1}X^d\sum_{s\ge 1}\mu(s)\left[\frac{n}{ds}\right].
 $$
Let us denote $l:=\frac{n}{d}$, then the coefficient of $X^d$ in this sum is
 $$
\sum_{s\ge 1}\mu(s)\left[\frac{l}{s}\right].
 $$
Applying the M\"obius inversion in the form (see Concrete Mathematics (4.61))
 $$
g(x)=\sum_{d\ge 1}f(x/d) \quad \text{ iff } \quad f(x)=\sum_{d\ge 1}\mu(d)g(x/d)
 $$
to the functions 
 $$
f(x)=[x\ge 1]
 $$
and 
 $$
g(x)=[x],
 $$ 
we conclude that 
 $$
\sum_{s\ge 1}\mu(s)\left[\frac{n}{ds}\right]
 $$
is equal to $0$ for $1>\frac{n}{d}$ and is equal to $1$ for $1\le\frac{n}{d}$, which is precisely what your conjectured RHS says.<|endoftext|>
TITLE: Explicit description of a quaternion algebra with a prescribed set of ramified places
QUESTION [8 upvotes]: Let $k$ be an algebraic number field. I understand that given a finite set of non-complex places $S\subset V(k)$ of even cardinality, there exists a unique quaternion algebra $Q$ over $k$ such that $Q$ ramifies at all $v\in S$ (i.e. $Q \otimes_k k_v$ is a division algebra) and $Q$ splits at all the other places. Is there a way to get an explicit description of $Q$ of the form $Q(a,b|k)$ for some parameters $a,b, \in k^{\times}$?
The reason for my question is to find an explicit example of a division quaternion algebra over the field $\mathbb{Q}(\sqrt[4]{2})$ that splits at all the archimedean places and is moreover of the form $Q_0 \otimes_{\mathbb{Q}(\sqrt{2})} \mathbb{Q}(\sqrt[4]{2})$ for a quaternion algebra $Q_0$ defined over $\mathbb{Q}(\sqrt{2})$.
Any hint on the general question or on the specific example is greatly appreciated. Thank you!

REPLY [12 votes]: I wrote something that makes this process algorithmic (and tries to simplifying the resulting algebra as much as possible) in Magma.
Copy and paste the following code into Magma (e.g. the calculator at http://magma.maths.usyd.edu.au/calc/)
_<x> := PolynomialRing(Rationals());  
F<s> := NumberField(x^2-2);  // s = sqrt(2)  
_<xF> := PolynomialRing(F);  
K<w> := ext<F | xF^2 - s>;  // w = 2^(1/4) = sqrt(s)  
ZK := Integers(K);  
// find small primes that are not split in K  
pps := [pp : pp in PrimesUpTo(20,F) | #Factorization(ZK!!pp) eq 1];  
// Compute a quaternion algebra (over F) ramified at 2 finite primes 
// and no infinite places  
B := QuaternionAlgebra(&*pps[1..2]);   
B;  
// Verify ramification  
RamifiedPlaces(B);  
RamifiedPlaces(ChangeRing(B, AbsoluteField(K)));  

The algorithm it uses is probabilistic, since there is no clear "best" quaternion algebra with specified ramification set (and anyway may be expensive to compute).  Anyway, on this run it tells me
Quaternion Algebra with base ring F, defined by i^2 = -s - 1, j^2 = 8*s + 11

So you can take $i^2 = a = -\sqrt{2}-1$ and $j^2 = b = 8\sqrt{2}+11$; these are elements of smaller norm than you could get from $\mathbb{Q}$.<|endoftext|>
TITLE: A request for suggestions of advanced topics in representation theory
QUESTION [6 upvotes]: Please Note: The main points of the question below are in bold in order to minimize the time required to read the question.
Let me begin by stating that I understand representation theory is a vast and deep area with many different subfields. Of course, any learning roadmap request for representation theory would necessarily have many different answers or at least one answer with many different suggestions. I would be more interested in "mainstream topics in representation theory"; one could define this as "the set of topics which every serious representation theorist should know" (although even this is subjective and varies from subfield to subfield). Of course, I am happy for people to suggest topics which they feel are not necessarily "mainstream representation theory"; I would be interested in as many suggestions as possible.
I am interested in representation theory both as a branch of mathematics in its own right and as a set of tools and ideas which one may use to study different (either related or a priori unrelated) areas of mathematics (please feel free to interpret this in a broad sense). My background in representation theory is almost all of (and will soon be exactly) the contents of the book entitled Lie Groups by Daniel Bump. The interdisciplinary nature of representation theory dictates that I have reasonable background in other branches of mathematics; I think that I have such a background but feel free to assume as prerequisites any branch of mathematics when giving suggestions.
I am interested in studying representation theory beyond that which is covered in Daniel Bump's Lie Groups. In other words, I am happy for suggestions for topics that a potential representation theorist should know after reading Bump's book (this is the key point). Of course, I am also interested in hearing suggestions for topics that a potential representation theorist should know even if they are virtually disjoint from Bump's book. I am certainly happy for suggestions to take either the form of a textbook, research monograph, research paper, or some other form that I have not thought about. 
I am not really interested in suggestions for topics that are already subsumed in Bump's book; I certainly do not object to such suggestions but they would not really be in response to this request. (You can view/download free and legally the table of contents of Bump's book at the following website: http://www.springer.com/mathematics/algebra/book/978-0-387-21154-1.)
Thank you very much for all suggestions!

REPLY [2 votes]: Meta-answer:  There are short introductions to a variety of interesting topics in Representation Theory of Lie Groups, a conference proceedings containing lecture notes by Atiyah, Bott, & other luminaries.<|endoftext|>
TITLE: Does it make sense to talk about smooth bundles of Hilbert spaces?
QUESTION [22 upvotes]: Is there a notion of "smooth bundle of Hilbert spaces" (the base is a smooth finite dimensional manifold, and the fibers are Hilbert spaces) such that:

1• A smooth bundle of Hilbert spaces over a point is the same thing as a Hilbert space.
2• If $E\to M$ is a smooth fiber bundle of orientable manifolds (say with compact fibers) equipped with a vertical volume form, then taking fiberwise $L^2$-functions produces a smooth bundle of Hilbert spaces over $M$.
3• If the Hilbert space is finite dimensional, then this specializes to the usual notion of smooth vector bundle (with fiberwise inner product).

I suspect that the answer is "no", because I couldn't figure out how it might work...
If the answer is indeed no, then what is/are the best notion/s of smooth bundle of Hilbert spaces?

REPLY [11 votes]: This is not an answer but rather a comment to Peter Michor's answer. Anyway, I post it as an answer to get more flexibility in text formulation and to get more visibility. Namely, I think there is a crucial error which completely breaks down the argument so that generally it is not possible to perform the construction (2• of OP) of associating a fiberwise $L^2$ bundle over a given fiber bundle. The error lies in the following passage:
 That it is smooth $U\times L^2(F, vol(g)) \to U\times L^2(F,vol(g))$ is seen as follows:
It suffices to show that $(x,f)\mapsto \langle f\circ \rho_x, \lambda\rangle_{L^2}$ is smooth for all $\lambda$ in a subset $\subset L^2$ of linear functionals which together recognize bounded sets.
We may take $C^\infty(F)\subset L^2(F,vol(g))$ as this set. By one of the two smooth uniform boundedness theorems from the book below it suffices to show that for each fixed $f\in L^2$ the function $F\to \mathbb R$ given by
$$x\mapsto \langle f\circ \rho_x, \lambda\rangle_{L^2} = \int_F f(\rho_x(u))\lambda(u)\,vol(g)(u)= \int f(v) \lambda(\rho_x^{-1}(v) ((\rho_x^{-1})^*vol(g))(v)$$ 
is smooth.
But this now obvious since $\lambda$ and $vol(g)$ are smooth. 
  
Specifically, it is not sufficient to check "scalar smoothness" against the set of smooth (not even continuous) functions since it (generally) does not "recognize" bounded sets in $L^2$ . To give an explicit counterexample, understanding $\mathbb S^1$ as $\mathbb R$ mod $1$ , consider the map $f:\mathbb R\times L^2(\mathbb S^1)\to L^2(\mathbb S^1)$ defined by $(t,[\,x\,])\mapsto[\,\langle\,x(t+s):s\in\mathbb R\,\rangle\,]$ . If the argument in Peter Michor's answer were correct, then for any fixed $x$ in $L^2$ the map $c:t\mapsto f(t,[\,x\,])$ should be smooth $\mathbb R\to L^2(\mathbb S^1)$ . However, it is easily seen that it is not even once differentiable if for example one takes $x$ defined by $x(s)=1$ for $|\,s-n\,|\le\frac 14$ and $n\in\mathbb Z$ , and $x(s)=0$ otherwise, since then $\lim_{\,t\to 0\,}\|\,t^{-1}(c(t)-c(0))\,\|_{L^2}=+\infty$ .<|endoftext|>
TITLE: How fast can we *really* multiply matrices?
QUESTION [45 upvotes]: Background: The Strassen Algorithm, described here, has a computational complexity of $\text{O}(n^{2.807})$ for the multiplication of two $n \times n$ matrices (the exponent is $\frac{\log7}{\log2}$). However, the constant is so large that this algorithm is in fact slower in practice than naive matrix multiplication for small $n$. Similarly, the Coppersmith-Winograd algorithm, which has the lowest asymptotic complexity of all known matrix multiplication algorithms, has an exponent of $2.376$ and was discussed here previously.
Question: Recently, I made a claim in a submitted paper that the Smith normal form algorithm has super-cubical complexity and a reviewer countered by saying that actually, the complexity has been reduced to matrix multiplication time = $n^{2.37\ldots}$. I am not an expert on matrix algorithms and would happily change the offending line, but the experience has forced me to wonder, what are the practical implications of saying "X can be done in matrix multiplication time"? More precisely,

Does there exist an actual software implementation of Coppersmith Winograd? If not, is there a theoretical obstacle to its existence?

By a theoretical obstacle I don't mean something like "Well, it would only be better than existing techniques for $n$ larger than the number of atoms in the universe so what's the point?", but rather something like "the algorithm uses the axiom of choice, or the classification of finite simple groups" etc.
PS: Okay, so there is also this paper which apparently reduces the complexity of the Coppersmith-Winograd approach to $2.3737$ from $2.376$, so I stand corrected about CW being the fastest. The question still stands if we replace CW by the method of V. V. Williams.

REPLY [31 votes]: Recently there is a PhD thesis about the practical fast matrix multiplication algorithms like Strassen:
Matrix multiplication is a core building block for numerous scientific computing and, more recently, machine learning applications. Strassen's algorithm, the original Fast Matrix Multiplication (FMM) algorithm, has long fascinated computer scientists due to its startling property of reducing the number of computations required for multiplying $n \times n$ matrices from $\mathcal{O}(n^3)$ to $\mathcal{O}(n^{2.807})$. Over the last half century, this has fueled many theoretical improvements such as other variations of Strassen-like FMM algorithms. Previous implementations of these FMM algorithms led to the "street wisdom" that they are only practical for large, relatively square matrices, that they require considerable workspace, and that they are difficult to achieve thread-level parallelism. The thesis of this work dispels these notions by demonstrating significant benefits for small and non-square matrices, requiring no workspace beyond what is already incorporated in high-performance implementations of matrix multiplication, and achieving performance benefits on multi-core, many-core, and distributed memory architectures.
This work includes several publications:
Strassen's Algorithm Reloaded. In The International Conference for High Performance Computing, Networking, Storage and Analysis (SC16), Salt Lake City, UT, November 2016.
Generating Families of Practical Fast Matrix Multiplication Algorithms. In 31st IEEE International Parallel and Distributed Processing Symposium (IPDPS17), Orlando, FL, May 29-June 2, 2017.
Strassen's Algorithm for Tensor Contraction. In SIAM Journal on Scientific Computing (SISC), 40(3):C305-C326, 2018.
Implementing Strassen’s Algorithm with CUTLASS on NVIDIA Volta GPUs. FLAME Working Note #88, The University of Texas at Austin, Department of Computer Science. Technical Report TR-18-08. August 23, 2018.
The open source code repositories are here:
https://github.com/flame/fmm-gen
https://github.com/flame/tblis-strassen
Other than that, you might be also interested in this paper:
A framework for practical parallel fast matrix multiplication. In Proceedings of the 20th ACM SIGPLAN Symposium on Principles and Practice of Parallel Programming (PPoPP 2015).<|endoftext|>
TITLE: cyclic polygons & trigonometry
QUESTION [6 upvotes]: I posted this question to stackexchange, where it's generated some comments but no progress toward answering it.  I'm going to say somewhat more here than I did there.
At one vertex of a pentagon inscribed in a circle of unit diameter (unit diameter, not unit radius) let the angles between adjacent diagonals be $α,β,γ$.  It follows that at one of the two vertices adjacent to that one, two of the angles, since they are subtended by the same arcs, must be $\beta,\gamma$; call the next one $\delta$, so we have $β,γ,δ$.  At the next vertex we have $γ,δ,ε$, then $δ,ε,α$, and finally $ε,α,β$.  It necessarily follows that
$$α+β+γ+δ+ε=π.\tag{constraint}$$
It's not hard to show that the area of the pentagon is
$$
\frac{\sin(2\alpha)+\sin(2\beta)+\sin(2\gamma)+\sin(2\delta)+\sin(2\varepsilon)}{8}.\tag{1}
$$
It's somewhat more work than that to show that if the "constraint" above holds, then $(1)$ is equal to
$$
\frac 1 2 \left(\overbrace{\sin\alpha\sin\beta\sin\gamma}^\text{sines}\  \overbrace{\cos\delta\cos\varepsilon}^\text{cosines} +\  \cdots\text{nine more terms }\cdots\  - \overbrace{2\sin\alpha\sin\beta\sin\gamma\sin\delta\sin\varepsilon}^\text{all sines} \right).
$$
(For the nine more terms: choose three factors in each term to be sines and then the other two are cosines.)
(As far as I know, this is my own.  I've mentioned it on stackexchange at least once before.)
QUESTION: Can the eleven terms be interpreted as areas, or do they otherwise have a geometric meaning?
One temptation is to think that the eleven terms correspond to the eleven regions into which the diagonals divide the interior of the pentagon.  But the numbers don't correspond to their areas, even if we try to make sense of using "signed" areas in some cases.
With a cyclic quadrilateral we have $\alpha+\beta+\gamma+\delta=\pi$ and
$$
\begin{align}
& \frac{\sin(2\alpha)+\sin(2\beta)+\sin(2\gamma)+\sin(2\delta)}{8} \\
= {} & \frac 1 2 \left(\overbrace{\sin\alpha\sin\beta\sin\gamma}^\text{sines}\  \overbrace{\cos\delta}^\text{cosine}+ \cdots\text{ three more terms }\cdots\right).
\end{align}
$$
Again, the sum of the four terms equals the sum of the areas of the four smaller polygons into which the diagonals divide the polygon, but again, the areas don't coincide at all.
With hexagons we have $\alpha+\beta+\gamma+\delta+\varepsilon+\zeta=\pi$ and
$$
\begin{align}
& \frac{\sin(2\alpha)+\sin(2\beta)+\sin(2\gamma)+\sin(2\delta)+\sin(2\varepsilon)+\sin(2\zeta)}{8} \\
= {} & \frac 1 2 \left(\overbrace{\sin\alpha\sin\beta\sin\gamma}^\text{sines}\  \overbrace{\cos\delta\cos\varepsilon\cos\zeta}^\text{cosines}+ \cdots\text{ 19 more terms }\cdots\right) \\
& {} - \frac 1 2 \left(2\overbrace{\sin\alpha\sin\beta\sin\gamma\sin\delta\sin\varepsilon}\  \overbrace{\cos\zeta}+ \cdots\text{ five more terms }\cdots\right)
\end{align}
$$
This time there are $26$ terms and only $25$ regions into which the polygon gets divided, so the tempting false conjecture is no longer there.
With heptagons, we have $\alpha+\beta+\gamma+\delta+\varepsilon+\zeta+\eta=\pi$ and
$$
\begin{align}
& \frac{\sin(2\alpha)+\sin(2\beta)+\sin(2\gamma)+\sin(2\delta)+\sin(2\varepsilon)+\sin(2\zeta)+\sin(2\eta)}{8} \\
= {} & \frac 1 2 \left(\overbrace{\sin\alpha\sin\beta\sin\gamma}^\text{sines}\  \overbrace{\cos\delta\cos\varepsilon\cos\zeta\cos\eta}^\text{cosines}+ \cdots\text{ 34 more terms }\cdots\right) \\
& {} - \frac 1 2 \left(2\overbrace{\sin\alpha\sin\beta\sin\gamma\sin\delta\sin\varepsilon}\  \overbrace{\cos\zeta\cos\eta} + \cdots\text{ 20 more terms }\cdots\right) \\
& {} + \frac 1 2 \left( 3 \,\overbrace{ \sin\alpha \sin\beta \sin\gamma \sin\delta \sin\varepsilon \sin\zeta \sin\eta}^\text{all sines} \right)
\end{align}
$$
There are $57$ terms.  The number of sub-polygons is $50$.
The pattern goes on, with the alternation of signs and the arithmetically growing coefficients $1,2,3,\ldots$.
QUESTION: Do the terms on the right sides of these identities have a geometric meaning?

REPLY [2 votes]: This won't answer the question, but Joseph O'Rourke got four up-votes for doing some nice graphics to clarify the question.
In the case of the quadrilateral, we have
$$
\sin\alpha\sin\beta\sin\gamma\cos\delta + \cdots\text{ three more terms } \cdots
$$
And then for the first of the four terms:
$$
\begin{align}
& \sin\alpha\sin\beta\sin\gamma\cos\delta \\[10pt]
= {} & -\sin\alpha\sin\beta\sin\gamma\cos(\alpha+\beta+\gamma)\qquad\text{ (since }\alpha+\beta+\gamma+\delta=\pi) \\[10pt]
= {} & \frac 1 8 {\Big(} \sin(2\alpha+2\beta+2\gamma) - \sin(2\alpha+2\beta) - \sin(2\alpha+2\gamma) - \sin(2\beta+2\gamma) \\[6pt]
& {} + \sin(2\alpha)+\sin(2\beta) + \sin(2\gamma) - 0{\Big)}
\end{align}
$$
The last term is the sine of the empty sum (so if these were cosines, we'd see a $1$ there).  Clearly this is reminiscent of usual inclusion-exclusion principle.<|endoftext|>
TITLE: is there a solution to this linear Diophantine system?
QUESTION [6 upvotes]: I have a matrix $A \in \mathbb Z^{n \times m}$, where $m > n$, and a vector $b \in \mathbb Z^n$. Under what conditions does
$$Ax = b$$
have an integer solution? Is there a way to bound the norm of the solution vector $x$ in terms of the norms of $A$ and $b$?
Essentially, I want something like Siegel's lemma, but for the non-homogeneous case.
I am not an expert on this and will appreciate any help. Thanks!

REPLY [7 votes]: This kind of questions arise very often in integer linear optimization. It is well-known that the bitsize of a solution will be polynomially bounded by the sizes of of $A$ and $b$, i.e. by the maximum bitsize of entries of $A$ and $b$, and by $\max(n,m)$. See e.g. Corollary 5.2a in the A.Schrijver's book. 
There are many sufficient conditions known for the existence of an integer solution, e.g. $A$ being totally unimodular (i.e. each square submatrix has determinant 0,1,or -1) is one.
Your problem is in fact easier (in optimization one often assumes $x\geq 0$), and Will gives a good suggestion in his comment above.
The book in Sect. 5.3 also gives more details on the algorithm Igor described in his answer.<|endoftext|>
TITLE: Top cohomology detecting compactness
QUESTION [8 upvotes]: I am looking for a reference for the fact that the top cohomology $H^n(X;A)$ of an $n$-dimensional manifold $X$ is non-trivial precisely when $X$ is compact.
I tried to ask this question on Math.Stackexchange, but there was some issue regarding orientability.
If $X$ is orientable, this should follow from standard Poincaré duality, but in my understanding the assumption of orientability can be removed by using "twisted" Poincaré duality. In this case, the coefficients must be local for $H^n$ to be non-trivial. (Edit: They don't need to be local, see the answer by Johannes Ebert below.) But if the coefficients respect the orientation (i.e. the coefficients are a constant sheaf, which is twisted by the orientation character), then $H^n\neq0$ if and only if $X$ is compact.
I vaguely remember hearing about an isomorphism $H^n\cong H_c^0$, where $H_c^\ast$ denotes compactly supported cohomology. The argument then goes that $H_c^0$ is non-trivial precisely when $X$ is non-compact, which I think of by using covers in Čech cohomology. $0$th cohomology with values in the sheaf of locally constant functions is just constant functions on all open sets of the cover $\{U_i\}$ and these functions must agree on single intersections $U_i\cap U_j$. The cohomology classes are thus given by globally constant functions, which, for compactly supported cohomology can only be the zero function when $X$ is non-compact, and can be any constant function when $X$ is compact.
It seems this argument makes no use of orientability, provided, there exists a twisted version of $H^n\cong H_c^0$...

REPLY [4 votes]: For algebraic varieties, there is an analogue in terms of cohomology of coherent sheaves. It reads as follows: if $X$ is an irreducible quasi-projective algebraic variety over a field $k$, of dimension $n$, then $X$ is projective if and only if $H^n(X,F)\ne 0$ for some coherent sheaf $F$ on $X$. This was conjectured by Lichtenbaum, a first proof was given by Grothendieck (see thm 6.9 in Hartshorne, Local cohomology, Springer LNM), another one by Kleiman (On the Vanishing of $H^n(X,F)$ for an $n$-dimensional variety, Proc. AMS 1967).<|endoftext|>
TITLE: Is there an obvious reason why p-localization of spectra is a finite localization?
QUESTION [8 upvotes]: Is there an obvious reason why $p$-localization of spectra is a "finite" localization in the sense of Haynes Miller? In other words, is there an obvious reason why the localizing subcategory (of the stable homotopy category) consisting of the $p$-acyclic spectra is generated (as a localizing subcategory) by the finite $p$-acyclic spectra? Is there a reference for this?
To be clear, I'm talking about Bousfield localization with respect to $\pi_*(-) \otimes {\mathbb Z}_{(p)}$. It is definitely a smashing localization. 
Recall that a spectrum is $p$-acyclic if $\pi_*(X) \otimes {\mathbb Z}_{(p)} = 0$. Thanks for your attention.

REPLY [8 votes]: The fiber of $S\to S_{(p)}$ is equivalent to $\mathrm{hocolim} \Sigma^{-1}M(k)$, where the limit is taken over the poset of natural numbers prime to $p$ (using the relation of divisiblilty, so $M(k)\to M(kk')$).  Thus, if $X$ is a spectrum such that $X_{(p)}=0$, then $X\approx \mathrm{hocolim} \Sigma^{-1}M(k)\wedge X$.
Thus, if $X$ is $p$-localization acyclic, this proves that $X$ is in the localizing subcategory generated by the $M(k)$, as Tyler suggests, while the converse is obvious.<|endoftext|>
TITLE: Invariant measures and recurrent sets.
QUESTION [11 upvotes]: Suppose $T:X \to X$ is a homeomorphism of a compact metric space. The recurrent set is the set of all points $x \in X$ such that for every $\epsilon>0$ there exists an $n\in \mathbb{Z}$, $n \ne 0$, such that $d(T^nx,x)<\epsilon$. Does there exists a $T$-invariant probability measure $\mu$ on $X$ with support equal to the recurrent set?
A few notes:
1) By the Poincare recurrence theorem, the support of any invariant measure is always contained in the recurrent set but it might be properly contained. 
2) It's easy to prove  that  this is true for symbolic systems. That is, if $A$ is a finite set, $T$ is the shift map on $A^\mathbb{Z}$ and $X$ is a closed shift-invariant subset of $A^\mathbb{Z}$.

REPLY [6 votes]: Just to clarify, the set of recurrent points may not be closed. There exists some transitive homeomorphism, whose uniquely ergodic measure is a Dirac measure. For example start with an irrational vector field $X$ on $\mathbb{T}^2$ and put a stop at $o\in\mathbb{T}^2$. That is, $Y=f\cdot X$ with $f(o)=0$. Then let $\phi_1$ be the time-1 map of the flow induced by $Y$. We can choose $f$ such that $\delta_o$ is the only invariant measure of $\phi_1$, and $\phi_1$ is transitive. In particular the set of recurrent points are dense on $\mathbb{T}^2$. 
Edit: See the following paper for some flow-version examples. In particular see Proposition 1 and 2 there.<|endoftext|>
TITLE: The chromatic number of a Hamming-related graph
QUESTION [5 upvotes]: For integer $1\le k\le n$, let ${\overline H}_n^k$ denote the complement of
the $k$-th power of the Hamming graph on the vertex set ${\mathbb
F}_2^n$; that is, two vectors from ${\mathbb F}_2^n$ are adjacent in
${\overline H}_n^k$ whenever they differ in $k+1$ coordinates at least.
What is the chromatic number of this graph?
Assuming for simplicity that $k$ is even, the Kneser graph
$G_{n,k/2+1}$ is a subgraph of ${\overline H}_n^k$. As a result,
  $$ \chi({\overline H}_n^k) \ge \chi(G_{n,k/2+1})=n-k. $$
Improving upon this estimate (for $k$ close to $n$) would yield an
improved bound for the number of Hamming spheres, needed to cover the
whole space ${\mathbb F}_2^n$ (see this MO post).

REPLY [5 votes]: This is pretty late, but there is a result by Frankl and Rodl (Theorem 1.11 from http://www.renyi.hu/~pfrankl/1987-3.pdf) that shows that the graph on $\mathbb{F}_2^n$ (for $n$ a multiple of four) with vertices being adjacent if they differ in exactly $n/2$ positions has independence number at most $(2-\epsilon)^n$ for some fixed $\epsilon > 0$. Note that they describe the graph with $\pm 1$ vectors with adjacency being orthogonality, but this is equivalent. This gives an exponential lower bound for the chromatic number of this graph, and thus for the graph where adjacency is given by being different in at leat $n/2$ positions. Unfortunately this is for $k$ very far from $n$, but there may be some other useful results in that paper.
There was also a recent paper by Briet and Zuiddam (http://arxiv.org/pdf/1608.06113v1.pdf) that proves that the orthogonal rank (smallest dimension in which vectors can be assigned such that adjacent vertices are orthogonal) of the graph on $\mathbb{F}_2^n$ with adjacency corresponding to differing in at least $n/2$ positions is exponential in $n$. The orthogonal rank lower bounds the chromatic number and their techniques could possibly be adapted for $k$ closer to $n$.
It might also be worth noting that for $k + 1 = n-1$, the chromatic number is known to be 4.
Edit:
I just looked a little more closely at the paper of Briet and Zuiddam mentioned above, and they say the best lower bound on the orthogonal rank of your $\overline{H}_n^k$ is
$$2^{[1-h((k+1)/n)]n - o(n)},$$
where $h(p) = -p\log_2(p) - (1-p)\log_2(1-p)$ is the entropy of the probability distribution $(p,1-p)$. This gives an exponential lower bound on chromatic number for a fixed value of $(k+1)/n$. I think that they say that this lower bound is actually given by an unpublished paper of  Samorodnitsky (http://www.cs.huji.ac.il/~salex/papers/old_sq_measure.ps).<|endoftext|>
TITLE: Untyped Higher Category Theory
QUESTION [6 upvotes]: I am currently trying to wade through the vast lake of higher category theory, a formidable task,or so it seems.
In the process, it has occurred to me that there is a basic analogy in place with various forms of type theories, typed logic, typed set theory, typed lambda calculus, etc.
In higher cats, one has 1-morphisms, 2-morphisms, and so on.
A fairly hierarchical structure, a ladder to infinity of sorts.
Now, whenever there are types, there is (almost) invariably an un-typed variant of the theory, which "forgets" the types. So I wonder if there is something along these lines already somewhere in the categorical endeavor.

I try to be a bit more precise:
imagine you are staring at a
N-category (let us stick to a strict
one, just for sake of simplicity),
from the top, and you forget all the
type labels. You see a fairly
complicated diagram of maps whose
endpoints are other maps, and so on
and so forth. Now try to axiomatize
such a structure. That would be an
untyped higher category (UHC).
Is there a reference for this structure? Now get rid of the strictness, and re-do the experiment. What kind of untyped higher categories are the result of stripping types from general higher cats?
In the example I mentioned, the UHC is
well-founded, in the sense that there
are some fellows (the ground objects) who
only point to themselves (I identify here the objects with their identity maps). Now,
eliminate this distinguished role of
objects and you will have a not well
founded UHC.
Is there a study of
not-well-founded categories, in a
similar spirit as there is a theory of
not well-founded sets?

REPLY [4 votes]: Hi Mirco, I'm putting my comment as an answer in case otherwise you don't find it. I wonder if you need the "n-arrows-only" approach.... where language of partial monoids replace categories. For example where objects and 1-arrows are treated on the same footing as 2-arrows in a double or 2-category in which the identities are provided by the source and target maps. Proposition: A small double category is precisely a set with two commuting partial monoid structures. (And a 2-category similarly but with an extra condition.) Would it be at all useful to you to try to do this for the n-fold or for the n case?<|endoftext|>
TITLE: Sato-Tate measure for GL(3) Automorphic forms
QUESTION [9 upvotes]: As we have known, the Sato-Tate measure for GL(2) turned out to be the half circle measure
$\frac{1}{2\pi} \sqrt{4-x^2}dx$ on [-2,2],
which appears in various versions of equi-distribution problems in GL(2).
My question is what the corresponding measure for GL(3) should be.
Firstly we shall note that Hecke eigenvalues in GL(3) can be imaginary. And we have to replace [-2,2] with {$z\in \mathbb{C}:|z|<=3$} the ball of radius 3 in complex plane. I don't know anything further yet.

REPLY [6 votes]: I can give a closed formula for the pushforward of $SU(3)$ Haar measure under trace. I don't understand the number theoretic issues well enough to know whether I want $SU(3)$ or $U(3)$, though.  
I'll write $(e^{i \alpha}, e^{i \beta}, e^{i \gamma})$ for the eigenvalues of the unitary matrix, with $\alpha+\beta+\gamma=0$, and write $z=x+iy$ for the trace. Set 
$$\Delta = \left(e^{i \alpha} - e^{i \beta}\right) \left( e^{i \alpha} - e^{i \gamma}\right) \left( e^{i \beta} - e^{i \gamma}\right)$$
so
$$|\Delta|  = 8 \sin \left( \frac{\alpha-\beta}{2} \right) \sin \left( \frac{\alpha-\gamma}{2} \right) \sin \left( \frac{\alpha-\gamma}{2} \right) .$$
By the Weyl integration formula, conjugacy class $(\alpha, \beta, \gamma)$ has volume proportional to $|\Delta|^2$.
We have
$$\begin{array}{rcl} 
z &=& e^{i \alpha} + e^{i \beta} + e^{i \gamma} \\
x &=& \frac{z+\bar{z}}{2} \\
y &=& \frac{z - \bar{z}}{2i} \\
\end{array}$$
so $dx dy$ is $\frac{1}{2i} dz \bar{dz}$. We compute
$$
\begin{pmatrix} 
\frac{\partial z}{\partial \alpha} & \frac{\partial z}{\partial \beta} \\
\frac{\partial \bar{z}}{\partial \alpha} & \frac{\partial \bar{z}}{\partial \beta} \\
\end{pmatrix} = 
\begin{pmatrix} 
e^{i \alpha} - e^{i \gamma} & e^{i \beta} - e^{i \gamma} \\
e^{-i \alpha} - e^{-i \gamma} & e^{- i \beta} - e^{-i \gamma} \\
\end{pmatrix}.$$
So the Jacobian of $(\alpha, \beta) \mapsto (z,\bar{z})$ is
$$
\begin{array}{rcl}
(e^{i \alpha} - e^{i \gamma})(e^{-i \beta} - e^{-i \gamma}) - (e^{-i \alpha} - e^{-i \gamma})(e^{i \beta} - e^{i \gamma}) &=& (e^{i \alpha} - e^{i \gamma})(e^{i \beta} - e^{i \gamma})(-e^{-i (\beta+\gamma)} + e^{-i(\alpha+\gamma)})\\
&=& (e^{i \alpha} - e^{i \gamma})(e^{i \beta} - e^{i \gamma})(- e^{i \alpha} + e^{i \beta}) \\
&=& - \Delta \\
\end{array}$$
I think I might have lost a sign somewhere, but the point is that $\Delta dx dy$ is proportional to $d \alpha d \beta$. So the Weyl measure, $|\Delta^2| d \alpha d \beta$, is proportional to $|\Delta| dx dy$. (If you really want to practice your high school trig, try doing that computation with $x$ and $y$ rather than $z$ and $\bar{z}$.)
Now, $\Delta^2$ is the discriminant of the cubic
$$t^3 - (x+iy) t^2 + (x-iy) t - 1.$$
Mathematica tells me that is
$$-27 + 18 x^2 - 8 x^3 + x^4 + 18 y^2 + 24 x y^2 + 2 x^2 y^2 + y^4$$
So the desired measure on $\mathbb{C}$ is proportional to
$$\sqrt{27 - 18 x^2 + 8 x^3 - x^4 - 18 y^2 - 24 x y^2 - 2 x^2 y^2 - y^4} dx dy.$$
The region where the quantity in the square root is nonnegative is exactly the deltoid John Baez describes.

In principal, one should be able to carry out this computation carefully enough to get the constant, but I'm not going to do it. 
Conceptually, this computation is similar to the classical computation that the map sending $(\alpha_1, \ldots, \alpha_r)$ to the coefficients of the polynomial $\prod (z-\alpha_i)$ has Jacobian $\prod_{i<j} (\alpha_i - \alpha_j)$. However, because of the constraint that the eigenvalues have product $1$, the details don't quite match.<|endoftext|>
TITLE: surprisingly difficult filtration problem
QUESTION [14 upvotes]: I am interested in a proof of the following statement which seems intuitive, but is somehow really tricky: 
Let $X$ be a stochastic process and let $(\mathcal{F}(t) : t \geq 0)$ be the filtration that it generates (unaugmented). Let $T$ be a bounded stopping time. Then we have
$\mathcal{F}(T) = \sigma(X(T \wedge t) : t \geq 0)$
I have a proof at hand (Bain and Crisan, Fundamentals of Stochastic Filtering, page 309), but in my opinion there is a major gap. I will try to explain the idea of proof. 
Let $V$ be the space of functions $[0,\infty) \rightarrow \mathbb{R}$ equipped with the sigma algebra generated by the cylinder sets. Consider the canonical map $X^T:\Omega \rightarrow V$ which maps $\omega$ to the trajectory $t \mapsto X(t \wedge T(\omega),\omega)$. Then we have $\sigma(X(T \wedge t) : t \geq 0) = \sigma(X^T)$. 
The difficult part is $\subseteq$. Let $A \in \mathcal{F}(T)$. We want to find a measurable map $g:V \rightarrow \mathbb{R}$ such that $1_A = g \circ X^T$, then we're done. It is now straightforward to show that $1_A$ is constant on sets where the sample paths of $X^T$ are constant. (To be more precise: for $\rho \in \Omega$ consider the set $\mathcal{M}(\rho) = \lbrace \omega : X(\omega,t) = X(\rho,t), 0 \leq t \leq T(\rho) \rbrace$. Then $T$ and $1_A$ are constant on every set of this form). 
The problem is: this is not sufficient! It suffices to construct a map $g$ such that $1_A = g \circ X^T$, but how we can we know that $g$ is measurable? This is where the proof of Bain and Crisan comes up short IMO. 
I can show this result only under the assumption that the map $X:\Omega \rightarrow V$ be surjective: Since $A \in \mathcal{F}(\infty)$, we have a measurable map $g$ such that $1_A = g \circ X$. Let $\rho \in \Omega$. Then $T$ and $1_A$ are constant on $\mathcal{M}(\rho)$. Therefore, $g$ must be constant on the image of $\mathcal{M}(\rho)$ under $X$. Because $X$ is assumed to be surjective, this image contains the function $X^T(\rho)$. Hence, $g \circ X = g \circ X^T$, and we are done.
I think that this result could be a little bit deeper. I have seen two proofs of this for the special case that $X$ is the coordinate process on $C[0,\infty)$, one is given in the book of Karatzas & Shreve, Lemma 5.4.18. The fact that Karatzas proves this late in the book only in this special case somehow makes me think that the general case is not so easy. 
I would really appreciate any comment or other reference for this result.

REPLY [2 votes]: The result is proven in Peskir&Shiryaev, Section 1.2, Theorem 6 under the following assumption on $\Omega$: for each $t\geq 0$ and each $\omega \in \Omega$ there is $\omega'\in\Omega$ such that $X_s(\omega')=X_{s\wedge t}(\omega)\forall s\geq 0$.<|endoftext|>
TITLE: questions on Néron-Tate canonical height
QUESTION [7 upvotes]: I have three questions regarding height pairings:

In [Serre, Lectures on the Mordell-Weil theorem], p. 85 f., it is stated that the following function is a local height function: 
"Let $V/R$ be a scheme over a discrete valuation ring, $D$ a divisor on $V$ and $x \in V(R)$. Assume $x \not\in \mathrm{supp}(D)$. Define $m(x,D)$ as follows:
a) If $x$ does not intersect $D$, $m(x,D) = 0$.
b) If $x$ intersects $D$ in a point $x_0$ of the special fibre, then $m(x,D)$ is the intersection multiplicity of $x$ and $D$ and $x_0$.
Then $x \mapsto m(x,D)$ is a local height associated to $D$." Does anyone have a reference for this?
Why is for a curve $X/k$ and an Abelian variety $B/k$ the Néron-Tate canonical height of the constant Abelian variety $B \times_k X$ over $X$ and $x: X \to B$ and $\mathcal{L}: X \to B^\vee$ 
$$\hat{h}(x,\mathcal{L}) = \mathrm{deg}_X((x,\mathcal{L})^*\mathcal{P}_B)$$ with the Poincaré bundle $\mathcal{P}_B \in \mathrm{Pic}(B \times_k B^\vee)$?  The degree function $\mathrm{deg}: \mathrm{Pic}(X) \to \mathbf{Z}$ is the usual one for a curve.
We have (functoriality of the height) $\hat{h}_{f^*(\mathcal{M})} = \hat{h}_{\mathcal{M}} \circ f$. Can one prove this also with the other definition of the height in 2.?

REPLY [7 votes]: See Chapter 2 of the book by E. Bombieri and W. Gubler, Heights in Diophantine Geometry.
They begin classically: they first define local heights (§2.2), then global heights (§2.3), and finally compare their global heights with Weil's definition (§2.4). 
Later, §2.7 gives the alternate point of view  of Arakelov geometry on local heights,
which should contain what you want. 
Indeed, they define a local height with respect to a Néron divisor (2.7.9)
while Example 2.7.20 explains how models over DVRs give rise to Néron divisors.<|endoftext|>
TITLE: Homotopy of random simplicial complexes
QUESTION [17 upvotes]: A random graph on $n$ vertices is defined by selectiung the edges according to some probability distribution, the simplest case being the one where the edge between any two vertices exists with probability $p = \frac{1}{2}$. I believe this is the Erdős–Rényi model $G(n,p)$ for generating random graphs. 
Similarly, in higher dimensions we can construct random simplicial complexes on $n$ vertices in many ways. One such method is as follows: fix a top dimension $d$, and now define the random simplicial model $S_d(n,p)$ where each $d$ simplex spanning any $d+1$ vertices exists with probability $p$. Some work has been done investigating the homology of such complexes in limiting cases, see for example this paper.
I want to ask

What is known about the properties of the fundamental group (or higher homotopy groups) of random simplicial complexes?

If there is a good reference, that would be enough. I can not find one on google. Thank you for your time.

REPLY [12 votes]: Babson, Hoffman, and Kahle have written a paper on fundamental groups of random 2-complexes. They worked with the Linial-Meshulam model whereby you begin with a complete graph on $n$ vertices and then add independently uniformly random 2-simplices.
Babson has just written a paper on the fundamental groups of clique complexes of Erdős–Rényi random graphs using similar techniques.<|endoftext|>
TITLE: Is the diagonal morphism a strong immersion?
QUESTION [11 upvotes]: Let's call a morphism of schemes a strong immersion if it is an open immersion followed by a closed immersion. This is no standard terminology. The following facts are well-known (see Stacks project, 19.24.3, 22.2.8, 22.2.9, 22.2.10):

Every strong immersion is an immersion.
Every quasicompact immersion is a strong immersion.
Every immersion with a reduced domain is a strong immersion.
There are immersions which are not strong.

Now my question is the following: Let $X$ be an arbitrary scheme. Is the diagonal morphism $\Delta_X : X \to X \times X$ a strong immersion?
According to the facts above, this is true when $X$ is reduced or when $X$ is quasi-separated. One of the main difficulties with such questions is that we cannot work locally (for example, the scheme-theoretic image of $\Delta_X$ might not be a local construction), so that standard methods don't work. Nevertheless, I think it is a quite interesting question.

REPLY [2 votes]: Here is my partial progress, from a while ago. I don't think I really got anywhere.
Let $I$ be the sheaf of ideals that vanish on the image of the diagonal morphism. Then we are required to find a sheaf $J \subset I$ locally generated by sections such that $J=I$ on some open neighborhood of the diagonal. We can just take $J$ to be the maximal subsheaf in $I$ locally generated by global sections. We need to choose some neighborhood of the diagonal. I cannot see any process to choose a neighborhood other than the union of $U_0 \times U_0$ for all $U_0$ affine open. Given a section, we want to extend it to all affine open neighborhoods of $U_0 \times U_0$. Since it is enough to check on a basis of affine opens, it is enough to check that it extends to all affine opens of the form $A \times B$ for $A,B \subset X$ affine open with $U_0 \subset A \cap B$.
Thus we have a scheme $A \cup B$ for $A,B$ affine and an affine open $U_0 \subset A \cap B$. We have a function on $U_0 \times U_0$ that vanishes on the diagonal. Without loss of generality this is $f \otimes 1 - 1 \otimes f$ for $f \in U_0$. We want to extend this to a function on $A \otimes B$. Such a function could take the form $r \otimes u - s \otimes t$ where $r$ and $s$ are functions on $A$, $t$ and $u$ are functions on $B$, and $s$ and $u$ are nonvanishing on $U_0$, $f=r/s=t/u$ on $U_0$ and $ru-st=0$ on all of $U$. It is easy to find pairs satisfying every condition but the last. But there is no obvious way to get them to satisfy the last condition.
One approach would be to first find $r/s$ satisfying the first two properties, then extend $r$ and $s$  from $A \cap B$ to $B$ as fractions, then divide those fractions as $t$ and $u$. Unfortunately, you can only write each function as a fraction on each affine - on different affines they could be different fractions, so $ru-st$ need not be $0$. It is easy to create examples of unextendable functions
But it is hard to produce a counterexample from this, because if you make the function unextendable on $B$, an adversary trying to prove that the morphism is a strong immersion can just try to extend it the same way on $A$. Thus, you need to find an affine scheme with a very bad open subset, and then embed that subset into a very different affine scheme where it is still very bad but in a different way. 
My attempts to forcibly construct a counterexample led to a big confusing mess. You want to set up a "dictionary" where nice well-behaved functions on $A$ become very ugly glued-together messes of fractions of functions $B$, with the functions in those fractions discussed in a second dictionary where they are turned into ugly messes on $A$, so someone who tries to extend a function consistently both ways is left with an infinite regress. But it is not clear that defining this does not create an infinite regress, or if it doesn't what the properties of the resulting object are.<|endoftext|>
TITLE: For which rings R is SL_n(R) a virtual duality group
QUESTION [5 upvotes]: A famous theorem of Borel and Serre says that if $R$ is the ring of integers in an algebraic number field, then $\text{SL}_n(R)$ satisfies virtual Bieri-Eckmann duality.  In other words, there exists a torsion-free finite-index subgroup $\Gamma$ of $\text{SL}_n(R)$ and a $\text{SL}_n(R)$-module $D$ such that
$$H^{\nu - k}(\Gamma;M) \cong H_{k}(\Gamma;M \otimes D)$$
for all $k$ and all $\text{SL}_n(R)$-modules $M$, where $\nu < \infty$ is the cohomological dimension of $\Gamma$.
Question : Are there any other rings $R$ such that $\text{SL}_n(R)$ is known to be a virtual duality group?  Or where it is known not to be one?  I suppose that one necessary condition is that $\text{SL}_n(R)$ has to have finite virtual cohomological dimension, so I'm only interested in counterexamples in which that holds (so, for instance, I'm not interested in $\text{SL}_n(\mathbb{C})$).

REPLY [4 votes]: Actually, Borel and Serre proved the result you mention for S-arithmetic groups: If $K$ is a number field, $S$ a finite set of places of $K$ and $\mathcal{O}_S$ the subring of $K$ of elements that are integral outside $S$ (called $S$-integers) then $SL_n(\mathcal{O}_S)$ is a virtual duality group. This is shown in
Borel, Serre: Cohomologie d'immeubles et de groupes S-arithmétiques. Topology 15(1976), 211-232. Théorème 6.2
To give an example, if $K=\mathbb{Q}$ then $\mathcal{O}_S=\mathbb{Z}[\frac{1}{p_1},...,\frac{1}{p_s}]$ where $p_i$ are rational primes. 
Of interest may also be a result of Behr in the functional field case: Let $K$ be a finite extension of the function field $\mathbb{F}_{p^n}(X)$. Then, with the same notation as above,  $SL_n(\mathcal{O}_S)$ has a duality subgroup $\Gamma_0$ of finite index. However, in contrast to the number field case, $\Gamma_0$   has $p$-torsion. (see Theorem 1 of this paper).<|endoftext|>
TITLE: When are provability predicates provably equivalent?
QUESTION [5 upvotes]: Fix notation
Suppose that $Prf_1(m, n)$ is the numerical relation that holds when $m$ numbers a $T$-proof of the sentence numbered $n$, according to scheme 1 for numbering wffs and sequences of wffs. Likewise $Prf_2(m, n)$ is the relation that holds when $m$ numbers a $T$-proof of the sentence numbered $n$ according to a different numbering scheme 2.
Let $\mathsf{Prf_1}$ represent $Prf_1$ in $T$, and put $\Box_1\varphi =_{def}$ $\exists \mathsf{x}\mathsf{Prf_1(x,\overline{\ulcorner\varphi\urcorner})}$, where $\overline{\ulcorner\varphi\urcorner}$ is $T$'s standard numeral for the number for $\varphi$ under scheme 1. Similarly for $\Box_2\varphi$.
Questions
A) Is it known what are the (most general?) conditions on the relation between coding schemes 1 and 2 for which we have
$T \vdash \Box_1\varphi \leftrightarrow \Box_2\varphi$, for any sentence $\varphi$?
B) What are the nicest/weakest(?) "derivability conditions" on a box $\Box$ in $T$, which if satisfied by both $\Box_1$ and $\Box_2$, mean that $T$ can again prove that equivalence?

REPLY [5 votes]: $\DeclareMathOperator\prf{Prf}\DeclareMathOperator\con{Con}$
As for A, I don’t think there are any useful criteria known that would guarantee the provable equivalence of two proof predicates that would not beg the question.
As for B, no “derivability conditions” in the usual sense the word is used can do this, assuming the conditions hold at least for the standard construction of a proof predicate based on proofs in a common proof system for first-order logic together with a $\Delta^0_1$ list of axioms.
Consider the following construction. Let $\tau(x)$ be any $\Sigma^0_1$-formula defining an axiom set for $T$, and $\prf_\tau$ the associated proof predicate. Pick any $\Pi^0_1$-sentence $\pi=\forall x\,\theta(x)$ with $\theta\in\Delta^0_0$ which is true in $\mathbb N$, but unprovable in $T+\con_\tau$. Define $\sigma(x)=(\tau(x)\lor\exists y\le x\,\neg\theta(y))$, and let $\prf_\sigma$ be the corresponding proof predicate. Then $\prf_\sigma$ is a proof predicate for $T$ (since in $\mathbb N$, $\tau$ and $\sigma$ are equivalent), but $T$ does not prove $\Box_\sigma\bot\to\Box_\tau\bot$: indeed, reasoning in $T$, if $\pi$ fails, then every formula with a sufficiently large Gödel number is a $\sigma$-axiom, and plenty of such formulas are contradictions, hence $\Box_\sigma\bot$. Contrapositively, $T+\con_\sigma$ proves $\pi$, hence using our assumption on $\pi$, $T+\con_\tau$ cannot prove $\con_\sigma$.<|endoftext|>
TITLE: Large cardinals without the ambient set theory? 
QUESTION [5 upvotes]: In an attempt to understand a bit better large cardinals, I have been thinking along the following lines, which could be summarized under the slogan

Talk about  cardinals without the
  (ambient) set theory

the class ON is first-order axiomatizable, and thus it looks like I can carve out of ON the subclass  CARD (for instance, one could add to the theory an equivalence relation,  $\alpha \equiv \beta$ formalizing  equinomerosity, and then define a cardinal in the usual way as the min ordinal in the equivalence class). 
Once I have my definable predicate $CARD(\alpha)$, I can proceed to introduce cardinal arithmetics. For instance, I can define successor as the minimal cardinal greater than the given cardinal.  
Obviously, I need to make some assumptions as to the basic cardinal arithmetics, so that it looks like the standard one in $ZFC$ + (possibly) generalized continuum hypothesis. 
Now, assuming one has done all of the above, it appears  that the "small" large cardinals, such as weak inaccessible, Mahlo, etc are definable in this theory (even in standard presentations, such as Drake,  their definition is arithmetic ). 
But what about the others, the heavy-weight ones? Do I necessarily have to  resort to the ambient set theory ( stationary points, elementary embeddings, etc ) to talk about very large cardinals, or there  is always a direct (algebraic/arithmetical/topological) way to provide their definition? 
Prima facie, it looks like the answer is no, but maybe there is a clever path to answer in the affirmative. Or perhaps, there is some kind of intrinsic boundary, beyond which you need to think of cardinals within the context of set theory
Any thought, refs, or known fact?

REPLY [4 votes]: But anyway, my point is this: just look at ON and totally forget that is the spine of V, and see it as a system of ordered numbers. Now axiomatize what you see, much in the same way as you axiomatize its initial segment N. – Mirco Mannucci Jul 8 at 22:21
for instance, one could add to the theory an equivalence relation, α≡β formalizing equinomerosity, and
talk about cardinals without the (ambient) set theory
Well, this construction [Gavrilovich and Hasson’s Exercices de Style, a homotopy theory of set theory] attempts 
to talk about cardinal invariants and 
use as little set theory as possible: instead it uses axioms of a model category to do that. 
and indeed, as you suggest, there equinomerosity (up to some fixed $\kappa$) is introduced, under the name of cofibrant. The construction does not work with the skeleton, though; but 
perhaps it would be closer to using the skeleton if you modify  the defitinions and replace everywhere 'inclusion' aka 'subset' by 'injective map'; then you'd lose limits in your model category. 
How powerful the language is, is unclear. But you can indeed say $\kappa$ is measurable: 
that's when the corresponding homotopy caetgory is not dense as a partial order.<|endoftext|>
TITLE: Why is the Tangent Groupoid useful in non-commutative geometry?
QUESTION [8 upvotes]: Let $M$ be a smooth manifold. The classical construction is the tangent bundle $TM$. What does the tangent groupoid $GM$ give me that this construction doesn't, and why is it useful in non-commutative geometry?
Heuristically, the tangent groupoid, which actually is a bundle too, thickens the tangent bundle with approximations to it; instead of being over $M$, it is over $M\mathbb{R}$, and the fibre over zero is the standard tangent bundle.
To construct it, we first note two simpler constructions:
a. Any bundle $E \rightarrow M$ can be considered as a groupoid $G$ with object space $M$ and morphisms $G(a,a)=E_a$ with the obvious composition and all other hom-spaces empty. 
b. The pair groupoid $M \times M$ on $M$ has object space $M$ and morphism space $M^2$, with composition $(a,b)(b,c):=(a,c)$ (with all others empty).
Then given a manifold $M$, its tangent groupoid $GM$ is a disjoint union of groupoids $G_t M$ for $t$ in the real line; and where $G_0 M$=$TM$ considered as a groupoid and $G_t M=M \times M$
Now the object space of $GM$ is the disjoint union of the object space of each fibre which is $M$, for each $t$ in the real line. This means we can identify the object space with $M \mathbb{R}$.
We give it a weak topology so that fibres spaces away from the fibre over zero are seen as spaces of approximate tangent vectors, that is:
for $ f \in C^{\infty}M$ we take the weakest topology (the one with the fewest open sets) such that the following are continuous:
a. $(X,m,0) \rightarrow Xf$
b. $(m,n,t) \rightarrow \frac{(fn-fm)}{t}$

REPLY [6 votes]: The tangent groupoid can be used in constructing the index map and proving the Atiyah-Singer index theorem. This may illustrate its importance. Higson and Roe will write a book on it.<|endoftext|>
TITLE: What is the "ray" in ray class group?
QUESTION [17 upvotes]: I have never seen any algebraic number theory book discuss the origin of the term "ray class group." Does anyone know where the word "ray" comes from in this context? I always thought it might be a person, but I never see it capitalized.
For quick background: the ray class group for a modulus $\mathfrak{m}$ of a number field $K$ is the group of fractional ideals of $K$ prime to $\mathfrak{m}$ modulo the principal ideals generated by elements of $K$ congruent to $1$ modulo $\mathfrak{m}$, where "congruent to $1$" for a real place means "positive."

REPLY [25 votes]: There are many introductions to number theory, but few are as original (my term for what others would call weird) as Fueter's "Synthetische Zahlentheorie" published in 1925.
It starts with elementary number theory, and discusses the arithmetic of cyclotomic
fields up to the Dedekind zeta function and applications to quadratic and cubic reciprocity.
In § 5, Fueter defines rays in the field of rational numbers:
 Definition. If a set of numbers has the property that it contains the product and the quotient of any two of its elements, then this set is called a ray.
Actually Fueter does not use the word "set" [Menge] but rather talks about a "domain of numbers" [Bereich von Zahlen]. At the end of this paragraph he makes the following historical remarks:
The necessity of considering sets with the property of rays was first realized by Weber. He
called these sets "number groups". Independently, these groups were introduced by R. Fueter
(Der Klassenkörper der quadratischen Körper und die komplexe Multiplikation, Diss. Univ. Göttingen, 1903; see also Crelle 130 (1905), p. 208), who called them rays. We keep this name here since it is similar in nature to the geometric names of fields [in German: Körper, i.e., solid] and ring, and since the word "group" does
not imply commutativity, which is always satisfied by rays. 
I would have been surprised had there been connections with infinite primes; back in 1903, 
Hilbert had already defined infinite primes, but their prominent role in class field theory only became apparent through the work of Furtwängler and Takagi, which took place after the term "ray" had been coined.<|endoftext|>
TITLE: What kind of algebra has geometric realization as in "Geometric Methods in Representation Theory of Hecke Algebras and Quantum Groups"
QUESTION [5 upvotes]: In Chriss and Ginzburg's book "Representation Theory and Complex Geometry" as well as the paper "Geometric Methods in Representation Theory of Hecke Algebras and Quantum Groups", the group algebra $\mathbb{C}[W]$ and the universal enveloping algebra $U(sl_n)$ have been realized as the cohomology ring $H^\bullet(Z)$ for some certain variety $Z$. Nevertheless, in the paper the author says: "Of course, given an algebra $A$, there is no a-priori recipe helping to find a relevant geometric data $(M,Z)$; ...It is fair to say, however, that it is still quite a mystery why a geometric realization of the algebras $A$ that we are interested in is possible at all."
In the recent years, is there any progress in this area? For example is there any theory that tells us when this $Z$ exists and how to find it? For the universal enveloping algebra $U(g)$ for $g$ other than $sl_n$, does the geometric realization exist?

REPLY [6 votes]: To answer your last question, a geometric realization of $U(\mathfrak{g})$ and its (irreducible highest weight) representations is given for any symmetric Kac-Moody algebra by the work of Nakajima (see his 1998 paper in Duke Math Journal).
I doubt that a complete answer to your general question exists.  However, in recent years, people have been looking at connections between geometric realizations and categorifications.  One thing that both of these tend to have in common is that they give rise to "nice" bases with integrality and positivity properties.  For the case of $U(\mathfrak{g})$, this is the canonical (or global) basis.  For the Hecke algebra, this is the Kazhdan-Lusztig basis.  Thus, something that could be viewed as a condition on an algebra in order for one to expect nice geometric realizations and categorifications would be the existence of such a basis.<|endoftext|>
TITLE: Aproximating dynamical systems by intrinsically ergodic systems
QUESTION [6 upvotes]: Let $X$ be a compact metric space and $f:X \to X$ a continuous map. We say that $(X,f)$ is approximated from below by a sequence of compact metric spaces $(X_i)_{i \geq 1}$ and a  sequence of continuous transformations $(f_i)_{i \geq 1}$ on $X_i$ if we have:

$X_i \subset X_{i+1}$ for every $i \geq 1$;
$X = \overline{\mathop{\bigcup}\limits_{i=1}^{\infty}X_i}$;
$f_{i+1}|_{X_i} = f_i$, $f|_{X_i} = f_i$ for every $i \geq 1$.

And we say that $(X,f)$ is approximated from above by a sequence of compact metric spaces $(Y_i)_{i \geq 1}$ and a sequence of continuous transformations $(g_i)_{i \geq 1}$ on $Y_i$ if we have:

$Y_{i+1} \subset Y_i$ for every $i \geq 1$;
$X = \mathop{\bigcap}\limits_{i=1}^{\infty}Y_i$;
$g_{i}\mid_{Y_{i+1}} = g_{i+1}$, $f\mid_{Y_i} = g_i$ for every $i \geq 1$.

Assume now that $X$ is approximated from above by a sequence $(X_i,f_i)_{i \geq 1}$ and from below by $(Y_i, g_i)_{i \geq 1}$ and $(X_i,f_i)$ and $(Y_i, g_i)$ are intrinsically ergodic systems for all $i \geq 1$, i.e., each of them has a unique measure of maximal entropy.
QUESTIONS:

Is $(X,f)$ intrinsically ergodic?
Is there any example of a dynamical system approximated from above by intrinsically ergodic systems that is not intrinsically ergodic?  In February 2012 I asked a question about limits on intrinsically ergodic systems (see link attached)  Limits of intrinsically ergodic systems, the answer gave me a counterexample for spaces approximated from below.

(If it helps, each $(X_i,f_i)$ and each $(Y_i, g_i)$ in my set-up is an intrinsically ergodic subshift of finite type for every $i$).

REPLY [7 votes]: Every topologically transitive shift space, whether intrinsically ergodic or not, can be approximated from above by intrinsically ergodic systems.
Indeed, given a finite alphabet $A=\{1,2,\dots,p\}$ and a  closed $\sigma$-invariant set $X\subset A^\mathbb{Z}$ (everything works just the same for one-sided shifts), let $\mathcal{L}=\mathcal{L}(X) \subset A^* = \bigcup_{n\geq 1} A^n$ be the collection of all finite words that appear in some sequence $x\in X$.  Thus $X$ determines $\mathcal{L}$ and vice versa.  Let $\mathcal{F} = A^* \setminus \mathcal{L}$ be the set of forbidden words.  Now let $Y_n\subset A^\mathbb{Z}$ be the set of all sequences that do not contain any words in $\mathcal{F}$ of length $\leq n$.  Then $Y_n$ is a shift of finite type and $X= \bigcap_{n\geq 1} Y_n$.
Furthermore, $Y_n$ is topologically transitive and hence intrinsically ergodic by virtue of being an SFT.  To see this, choose any $v,w\in \mathcal{L}(Y_n)$ and write $v=v_1 v_2$, $w=w_1 w_2$ where $v_2$ and $w_1$ both have length exactly $n$.  Then by transitivity of $X$ there exists a word $u$ such that $v_2 u w_1 \in \mathcal{L}(X)$, and by the definition of $Y_n$ we have $v u w\in \mathcal{L}(Y_n)$, which shows that $Y_n$ is transitive.
Thus the counterexample given in the answer to your earlier question works here as well.  Actually, I'll point out that there are quite a broad class of such counterexamples, which can be constructed by looking at coded systems: these are shift spaces defined either in terms of a countable collection of generating words that are allowed to be freely concatenated, or equivalently in terms of a directed graph on countably many vertices with edges labeled from a finite alphabet.  There are plenty of examples of coded systems that are transitive but not intrinsically ergodic (see this question or this answer, for example), and you can approximate coded systems from within by SFTs (or at least sofic shifts) in a very natural way:  just truncate the collection of generators to a finite set, or truncate the graph to a finite subgraph.
In another direction, I believe there is a paper of Gurevich in which certain quantitative conditions are given on the rate of approximation from outside by intrinsically ergodic systems that turn out to be sufficient to guarantee intrinsic ergodicity of $X$.  But I don't have the reference handy at the moment, and I'll have to wait until I'm in my office next week to dig up the paper and see if it's in fact relevant.
Edit: I found the Gurevich paper I was thinking of (actually 2 papers).  References are as follows:

B.M. Gurevic, "Uniqueness of the measure with maximal entropy for symbolic almost-Markov dynamic systems", Soviet Math. Dokl. 13 (1972), No. 3, 569-571.
B.M. Gurevic, "Stationary random sequences of maximal entropy", Chapter 10 (pp. 327-380) of Multicomponent Random Systems, edited by R.L. Dobrushin and Ya.G. Sinai, Advances in Probability and Related Topics, Volume 6, Marcel Dekker Inc (1980).

As you see from the page count, (1) is quite short and just has the statement of the result, no proofs, while (2) is more comprehensive.  Roughly speaking, the main result can be summarised as follows (the result in the paper is more precise because it doesn't assume that various limits exist).
Given a shift space $X$ on a finite alphabet, let $\mathcal{L}_n$ be the set of words of length $n$ that appear in some $x\in X$, and let $Y_n$ be the SFT defined by the condition that $x\in Y_n$ if and only if $x_k \cdots x_{k+n-1} \in \mathcal{L}_n$ for every $k$.  Then $X = \bigcap_n Y_n$, and in particular, $h(Y_n) \to h(X)$, where $h$ is the topological entropy.  Let $\rho_n = h(Y_n) - h(X)$ be the entropy gap; heuristically, $\rho_n$ is the amount of entropy that is destroyed by the restrictions in $X$ of length $>n$.
Furthermore, define $\alpha_n$ by
$$
\alpha_n = \inf \{\tau \mid \forall u,v\in \mathcal{L}_n\ \exists w\in \mathcal{L}_\tau\ s.t. uvw\in \mathcal{L}\}.
$$
That is, in the shift $X$, any two words of length $n$ can be glued together using a word of length $\alpha_n$.  The shift $X$ has specification if and only if $\lim \alpha_n < \infty$, and in this case $X$ is intrinsically ergodic.  (This is due to Bowen.)
Let $R_\alpha = \lim \frac 1n \log \alpha_n$ be the growth rate of $\alpha_n$, and let $R_\rho = -\lim \frac 1n \log \rho_n$ be the decay rate of $\rho_n$.  Thus $R_\alpha$ is the rate at which the gluing time increases (and can be thought of as quantifying how badly $X$ fails to have specification), and $R_\rho$ is the rate at which the entropy gap decays (and can be thought of as quantifying how closely $X$ is approximated by the SFTs $Y_n$ in terms of entropy).
Theorem (Gurevich):  If $h(X)>0$ and $R_\alpha < \frac{R_\rho} {16 h(X)}$, then $X$ is intrinsically ergodic.
Heuristically, "if the failure of specification is slow relative to the approximation by SFTs, then $X$ is intrinsically ergodic".<|endoftext|>
TITLE: What functions can be obtained as a convolution of a Schwartz function and a tempered distribution?
QUESTION [6 upvotes]: Let $\mathcal S (\mathbb R)$ denote the space of Schwartz functions on $\mathbb R$ and $\mathcal S^* (\mathbb R)$ denote the dual space of Schwartz (a.k.a tempered)  distributions.
We consider $\mathcal S (\mathbb R)$ as a Frechet space and $\mathcal S^* (\mathbb R)$ as a direct limit of Banach spaces.
Let $c:\mathcal S (\mathbb R) \otimes \mathcal S^* (\mathbb R) \to  \mathcal S^* (\mathbb R)$ be the convolution map. Let  $\hat c:\mathcal S (\mathbb R) \hat\otimes \mathcal S^* (\mathbb R) \to  \mathcal S^* (\mathbb R)$ be its extention to the completed tensor product. We have an argument that "proves" the following contradictory facts:

$\mathrm{Im} (c)=\mathrm{Im} (\hat c)$ 
$$\mathrm{Im} (c)=(f \in C^\infty(\mathbb R)|\exists \text{ a polinomial }p \text{ s.t. } \forall n\in \mathbb N \text{ the function } \frac{f^{(n)}}{p} \text{ is bounded}  )$$
$$\mathrm{Im} (\hat c)=(f \in C^\infty(\mathbb R)|\forall n\in \mathbb N, \exists \text{ a polinomial }p \text{ s.t. }  \text{ the function } \frac{f^{(n)}}{p} \text{ is bounded}  )$$
$\mathcal T_u(\mathbb R) \subsetneq \mathcal T(\mathbb R)$, were $\mathcal T_u(\mathbb R)$ is the r.h.s of (1) and $\mathcal T(\mathbb R)$ is the r.h.s of (2).

What of those statments are true and what are wrong? Do you have references for any of them?

REPLY [2 votes]: A bit late and far from answering all your questions but perhaps still worth mentioning:
A proof of the inclusion ${\rm Im}(c)\subseteq$ RHS of item 2) can be found in "Topological Vector Spaces and Distributions" by John Horváth. See in particular Proposition 4.11.7 on page 420.
BTW, as Johannes Hahn said the standard notation of for RHS of item 3) is $\mathcal{O}_M$. That for the RHS of item 2) is $\mathcal{O}_C$.<|endoftext|>
TITLE: Altitudes of a triangle
QUESTION [13 upvotes]: The three altitudes of a triangle are concurrent -- this is true in all three constant curvature geometries (Euclidean, hyperbolic, spherical), but, as far as I know, the proofs are different in the three cases. Is there a "uniform" proof?

REPLY [5 votes]: I think this (and other similar facts) can be derived uniformly using elementary analyticity arguments. First you prove it for the round unit sphere. by rescaling this  implies that it's true for the round sphere of any radius. now look at the cosine law in the simply connected space form of constant curvature $k$. since this formula is analytic in $k$, the "size" of the potential failure of the altitudes to intersect at the same point (measured in any reasonable way) will also be analytic in $k$ and since it's constantly zero for $k>0$ it must be constantly zero for all $k$.
It should not be hard to make the above into a rigorous argument.<|endoftext|>
TITLE: Optimal bounds for an alternating sum on a downset
QUESTION [16 upvotes]: Let $n$ be a natural number, and consider the discrete cube $2^{[n]} := \{ A: A \subset \{1,\ldots,n\}\}$ consisting of all subsets of the $n$-element set $[n] := \{1,\ldots,n\}$.  Define a downset in $2^{[n]}$ to be a collection ${\mathcal D}$ of elements $A$ in $2^{[n]}$ with the property that if $A \in {\mathcal D}$ and $B \subset A$, then $B \in {\mathcal D}$.
My question is: what are the largest and smallest possible values for the alternating sum $\sum_{A \in {\mathcal D}} (-1)^{|A|}$, as ${\mathcal D}$ ranges over downsets in $2^{[n]}$, as a function of $n$?  (Here $|A|$ denotes the cardinality of $A$.)
The trivial bounds here are $\pm 2^{n-1}$, by taking only the positive or negative values of ${\mathcal D}$, but of course these values are attained on a "checkerboard" set which is very far from being a downset, and this suggests that significant improvement is possible.
By taking ${\mathcal D}$ to be the set of all subsets $A$ of $2^{[n]}$ of cardinality at most $r$ for some $1 \leq r \leq n-1$, this gives a value of $(-1)^r \binom{n-1}{r}$; setting $r$ close to $(n-1)/2$ then seems to give reasonably good extremals (of size about $2^n/\sqrt{n}$ asymptotically).  In the spirit of Sperner's lemma, one might tentatively conjecture that these are the extremal examples, but I was unable to prove or disprove this.  (I feel like I'm missing some obvious application of downset isoperimetric inequalities or something.)
One motivation for this question is from analytic number theory: partial divisor sums $\sum_{d|a: d \leq x} \mu(d)$ of the Mobius function (which show up from time to time in this subject) can be viewed as an alternating sum over a downset, where $n$ is the number of prime factors $p_1,\ldots,p_n$ of $a$, and ${\mathcal D}$ is the collection of subsets $A$ of $[n]$ for which $\prod_{i \in A} p_i \leq x$.  So any bounds on the general alternating-sum-of-downset problem would imply bounds on partial divisor sums of the Mobius function that depend only on the number of prime factors.
One small observation (using the shifting technology of Frankl) which may or may not be of use: given two natural numbers $1 \leq i < j \leq n$ and a downset ${\mathcal D}$, define the $ij$-shift of ${\mathcal D}$ to be the set formed by replacing any element of ${\mathcal D}$ of the form $A \cup \{j\}$ with $A \cup \{i\}$, if $A$ is disjoint from $\{i,j\}$ and $A \cup \{i\}$ is not already in A.  Note that this is again a downset.  Call a downset ${\mathcal D}$ shift-minimal if it is equal to all of its $ij$-shifts.  Then one can reduce without loss of generality to the shift-minimal case, because shifting does not affect the sum $\sum_{A \in {\mathcal D}} (-1)^{|A|}$.  In other contexts, the reduction to the shift-minimal case can be very powerful, but for some strange reason I was unable to exploit it here.

REPLY [2 votes]: Here is a simple proof which should give exact optimizers for "most" choices of n mod 4 and max/minimizing; and, near-sharp values for the remaining cases.  I'll do it for upsets instead of downsets.
--
Identify the discrete cube with $\{-1,1\}^n$ and let $f : \{-1,1\}^n \to \{0,1\}$ be the indicator of $\mathcal{D}$ which is monotone since $\mathcal{D}$ is an upset. Up to a factor of $2^n$ we are trying to min/maximize $E_x[f(x) \chi(x)]$, where $x$ is uniformly random on the cube and $\chi(x) = \prod_{i=1}^n x_i$.  Let $1 \leq j \leq n$ be uniformly random and let $x^{(j)}$ denote $x$ with its $j$th coordinate negated.  Since $x^{(j)}$ is also uniformly distributed, 
$E_x[f(x) \chi(x)] = E_{x,j}[\frac{f(x) \chi(x) + f(x^{(j)})\chi(x^{(j)})}{2}] = E[\chi(x)\frac{f(x)-f(x^{(j)})}{2}]$
where we used $\chi(x^{(j)}) = -\chi(x)$.  Thus in absolute value, the quantity is at most
$E[\bigl|\frac{f(x)-f(x^{(j)})}{2}\bigr|] = \frac{1}{2n}E_x\left[\sum_{i=1}^n \bigl|f(x)-f(x^{(i)})\bigr|\right] = \frac{1}{2n}E\left[\sum_{i=1}^n f(x)x_i\right]$,
where the last step uses that $f$ is monotone.  But 
$\frac{1}{2n}E\left[\sum_{i=1}^n f(x)x_i\right] = \frac{1}{2n} E\left[f(x)(\sum_{i=1}^n x_i)\right]$
is clearly maximized among $0$-$1$ functions $f$ by the "Hamming ball" which is $1$ when $\sum_{i=1}^n x_i > 0$ and $0$ when $\sum_{i=1}^n x_i < 0$.  (If $n$ is even and the sum is $0$, it doesn't matter what $f$'s values are there.)  
--
For the purposes of checking sharpness, note that the maximizing $f$ (or $f$'s) there happens to be monotone.  The only inequality used is in fact sharp if $n$ is odd and $n$ has the "right" (vis-a-vis min/maxing) remainder mod $4$; if $n$ is even then I think you can get a sharp inequality for any remainder mod $4$ by suitably making $f$ equal to $0$ or $1$ on the middle Hamming layer.  I hope a little trick can handle the case of odd $n$ congruent to the "wrong" value mod 4 but I didn't think about it.
--
For people familiar with "analysis of boolean functions", I think this result could be considered "folklore".  For example, it essentially appears at the end of page 8 here: http://cs.indstate.edu/~jkinne/research/mon.pdf<|endoftext|>
TITLE: Euclidean symmetries of torus links in R^3
QUESTION [5 upvotes]: I have a question about whether Ryan Budney's question:
Torus knots in Euclidean space -- a symmetry argument
can be extended to links. He asks:
Suppose you have a $(p,q)$ torus knot $K$ in $\mathbb{R}^3$ fixed by a subgroup $G$ of $\operatorname{SO}(3)$. Budney asks (more or less) for a nicer proof that $G$ cannot contain $\mathbb{Z}/p\mathbb{Z}$ and $\mathbb{Z}/q\mathbb{Z}$ as subgroups. 
Edit: I think I understand Charlie Frohman's point in the comments to that question now that this follows from knowing the periods of the torus knots AND the structure of the finite subgroups of SO(3), which are very different than the finite subgroups of SO(4).
I'd like to extend this same result to $(p,q)$ torus links: there should be no configuration in $\mathbb{R}^3$ which is both $p$-fold and $q$-fold symmetric. I feel like this sort of thing ``should be known'', but so far my literature search has not turned up anything. 
EDIT: I think you could probably do this in a similar way if you knew the periods of the torus links (as opposed to the torus knots). But is \emph{this} known?

REPLY [2 votes]: A pretty nice way to interpet your question is two consider the action of the symmetry group on $\mathbb{S}^3$. This is done for knots in Boileau, Boyer, Cebanu, and Walsh section 3, but we can get enough of it to work to answer your question. The point is that the (p,q) torus link will also admit a symmetry group with an element $a$ of order $m=lcm(p,q)$. 
The quotients of $\mathbb{S}^3$ under the action of $a$ are dubbed orbi-lens spaces in the language of the above paper because we can describe them similar to the way that Rolfen describes lens spaces in 9B(III) of Knots and Links. Namely, let $w$ be a primitive $m$-root of unity. We consider $\mathbb{S}^3 =$ { $(z_1,z_2)| z_1,z_2\in \mathbb{C} \mbox{ and } |z_1|^2+|z_2|^2=1$ }. The $a$ can be realized as $a(z_1,z_2)=(w^q z_1,w^p z_2)$. 
If $gcd(p,q)=d$, then $a^{p/d}$ will fix the core of a Hopf fibration and $a^{q/d}$ will fix the other core. Also, according the above paper, the quotient has base space a lens space with homology $\mathbb{Z}/d\mathbb{Z}$. We note that the quotient under the action of $a^{m/d}$ will be a lens space and that $p,q \geq 2$ and are not relatively prime so $d\geq 2$ .
Since any finite order element $g$ of $SO(3)$ fixes an axis, the image of $a^{m/d}$ will fix an axis. But the map given by the 1 point compactification of $\mathbb{R}^3$, $f: \mathbb{R}^3 \cup \{\infty\} \rightarrow \mathbb{S}^3$, is 1-1. So we could induce a quotient map on $\mathbb{S}^3$ via the image under $f$ of the points equivalent under $g$ which has base space $\mathbb{S}^3$. That would contradict the above paragraph.<|endoftext|>
TITLE: Why are minimal resolutions of polynomial ideals important?
QUESTION [7 upvotes]: Background: Let $k$ be a field and denote by $P = k[x_1,\ldots,x_n]$ the polynomial ring in $n$ (commuting) variables over $k$. A resolution of an ideal $I \lhd P$ is an exact sequence of $P$-modules $$\ldots \to F_n \to F_{n-1} \to \ldots \to F_0 \to P/I \to 0.$$ This resolution if minimal if the rank of each $F_n$ is minimal.
Here's my question:

What are minimal resolutions of ideals in polynomial rings good for?

Why I'm asking: I've been looking at some papers which apply a purely algebraic version of Robin Forman's discrete Morse theory to various algebraic contexts. One of the most fascinating applications out there involves the construction of minimal (cellular) resolutions of ideals, see for instance this paper. While I understand discrete Morse theory well and know the basic definition of resolutions etc., it is not clear why one would want to construct minimal ones. 
More precisely, does the quest for a minimal resolution help in some practical way, such as with applying Buchberger's algorithm to a generating set of $I$, or is it a platonic search for the simplest algebraic object that resolves $I$? 
Since I am not sure what areas of math are impinged upon by this question, I have only added a minimal number of tags, and I encourage/ask experts to please add other suitable ones.

REPLY [4 votes]: As it has already been pointed out, the key concept is uniqueness.
We know that free (or, if you will, projective) resolutions exist in the case we are interested in. Yet, a module may possess many of them. However, each finitely generated module (over a polynomial algebra over a field) has a minimal free resolution, which is unique up to isomorphism. The degrees of the generators of its free modules not only yield the Hilbert function, but form a finer invariant, bringing us right next to, for example, the notion of Castelnuovo--Mumford regularity (discussed in Eisenbud's book).
For the sake of completeness: while the free (projective) dimension of a graded module in our case is defined to be the minimum length of a resolution by graded free modules (which coincides with the length of the minimal graded free resolution, thus minimality also servers as a notion of dimension), and it specifies in how many steps the module can be resolved, the CM regularity can be viewed as the width of the minimal resolution (in terms of the degrees of the generators at each free component), and it gives an indication of the complexity of the process.
Under similar circumstances, the opposite of a minimal free resolution is a trivial complex. Trivial complexes have no homology at all, thus taking the direct sum of a resolution with a trivial complex yields an other resolution, resulting in lack of uniqueness. It can be shown that every free resolution in our case is the direct sum of the minimal resolution and a trivial complex. Therefore, we may say that minimal resolutions carry most of the information we are looking for.<|endoftext|>
TITLE: Projective spaces as affine varieties
QUESTION [11 upvotes]: This works over the reals but not over the complex field. Consider the set of all $n\times n$ matrices $A$
such that
1. $A^2=A$
2.$A^T=A$
3. $\mathrm{Trace}(A)=1$
The first condition makes $A$ a projection to a subspace of $\mathbb{R}^n$. The second ensures that $A$ is diagonalizable so its eigenvalues are all 0 or 1.The third guarantees that the image of the projection is one-dimensional. Such matrices are in a 1-1 correspondence with one-dimensional subspaces and so constitute $\mathbb{R}P^{n-1}$. 
This seems to represent real projective spaces as affine varieties, with plenty of induced nonzero regular functions. How do we reconcile this with the fact that projective spaces have only constant regular functions?
This does not work over the complexes since $A$ would have to equal its conjugate transpose (to be guaranteed diagonalizable) and conjugation is not algebraic.
This is puzzling me to no end...

REPLY [8 votes]: The fundamental problem seems to be the confusion between the notions of "real algebraic variety" and "algebraic variety over the reals".  Given the similarity in words, it is easy to make this mistake.
You can find a discussion of the distinction in chapter 2 of Coste's lecture notes on real algebraic sets.  Both objects are topological spaces with sheaves of commutative rings, but the former is made by gluing algebraic subsets of $\mathbb{R}^n$, while the latter is made by gluing prime spectra of finite type commutative rings over $\mathbb{R}$.
There is a functor that sends an algebraic variety $X/\mathbb{R}$ to the real algebraic set $X^{ras}$ whose points are the real points of $X$.  Your example demonstrates that this functor does not reflect the property of being affine (i.e., if $X^{ras}$ is affine, $X$ is not necessarily affine).  Neither sheaves of regular functions nor the topological space are preserved by this functor, so you can't reasonably expect global functions to be preserved.
As it happens, there is a more general statement, also mentioned in Coste's lecture notes:

All quasi-projective real algebraic varieties are affine.

In particular, for any quasi-projective variety $Y/\mathbb{R}$, there exists an affine variety $X/\mathbb{R}$, such that $X^{ras} \cong Y^{ras}$.<|endoftext|>
TITLE: Motivic proof of Weil-conjectures?
QUESTION [10 upvotes]: Assuming the standard conjectures (and whatever is needed in addition),
is there a nice proof of the Weil-conjectures written completely in the language of motives?

REPLY [8 votes]: See Theorem 5.6 in Kleiman's article "The Standard Conjectures", in the Motives volume (PSPM 55.1). I'm not quite sure what you mean by "written completely in the language of motives", so it might not be exactly what you are looking for. 
I initially posted this as a comment, but it really should be an answer, so I'm re-posting it as one.<|endoftext|>
TITLE: Denseness of inner automorphisms inside automorphisms of hyperfinite type III_1 factor
QUESTION [15 upvotes]: Let $R$ be the hyperfinite type $III_1$ factor,
and let $Aut(R)$ be its group of automorphisms, equipped with the $u$-topology
(topology of pointwise convergence on the predual).
An automorphism $\alpha\in Aut(R)$ is called inner if it is of the form $\alpha(x)=uxu^*$ for some unitary $u\in R$.
I've heard that inner automorhpisms are dense in $Aut(R)$.
Can someone give me an explicit example of a sequence of inner automorphisms that converges to an automorphism that is not inner?

PS: An answer to the same question for the hyperfinite $II_1$ factor would also be interesting.

REPLY [8 votes]: It is relatively easy to give an explicit sequence of inner automorphisms that converges to the flip automorphism $\sigma$. First of all, realize $R$ as an infinite tensor product of matrix algebras $(R,\varphi)=\bigotimes_n (M_{k_n}(\mathbb{C}),\varphi_n)$. For every $n\in \mathbb{N}$, we find a unitary $u_n\in M_{k_n}(\mathbb{C})\otimes M_{k_n}(\mathbb{C})$ that implements the flip automorphism on $M_{k_n}(\mathbb{C})\otimes M_{k_n}(\mathbb{C})$, i.e. $u_n(x\otimes y)u_n^\ast=y\otimes x$. (This follows from the general observation that every automorphism of a type I factor is inner, but it is a good exercise to find the $u_n$ explicitly).
Now it follows that the sequence of inner automorphisms $\sigma_n=Ad_{u_1\otimes\ldots\otimes u_n\otimes 1\ldots}$ converges to the flip automorphism $\sigma$: Let $\psi$ be any ultraweakly continuous functional on $R\otimes R$. Consider $R\otimes R$ to be represented on the infinite tensor product space $H=\bigotimes_n L^2(M_{k_n}(\mathbb{C})\otimes M_{k_n}(\mathbb{C}), \varphi_n\otimes\varphi_n)$. We know that $\psi$ is of the form $\psi(x)=\sum_k\langle\xi_k,x\eta_k\rangle$ for some $\ell^2$-summable sequences $\xi_k,\eta_k$ in $H$. Since the finite sequences are dense in the $ell^2$-summable ones, we can assume that $\psi(x)=\langle\xi,x\eta\rangle$. Because the finite tensor products are dense in the infinite ones, we can assume that $\xi,\eta\in H_N=\bigotimes_{n=1}^N L^2(M_{k_n}(\mathbb{C})\otimes M_{k_n}(\mathbb{C}), \varphi_n\otimes\varphi_n)$ for some $N\in\mathbb{N}$. Now it follows that $\psi(\sigma_n(x))=\langle \xi,\sigma_n(x)\eta\rangle=\langle\sigma_n^{-1}(\xi),x\sigma_n^{-1}(\eta)\rangle=\langle\sigma^{-1}(\xi),x\sigma^{-1}(\eta)\rangle=\psi(\sigma(x))$.<|endoftext|>
TITLE: Convenient reference for subgroups of a finite semidirect product?
QUESTION [13 upvotes]: Given a finite group $G= H \ltimes N$ (with no particular constraints on $H, N$), it's probably been known for a long time how to describe efficiently the possible subgroups of $G$.   A graduate student was asking me about this, having dug a version out of an obscure research paper in the process of studying an unrelated technical problem.   I'm not sure what exists in group theory books or other such sources.   The answer seems to have the flavor of cocycles and coboundaries but is apparently somewhat complicated to write down. 

Is there a convenient reference giving a recipe for all subgroups of a finite group written as a semidirect product of two known groups relative to a known action of one on the other?

REPLY [6 votes]: Usenko, Subgroups of semidirect products, Ukrainian Mathematical Journal, 1991, Volume 43, Numbers 7-8, Pages 982-988<|endoftext|>
TITLE: Can there be an embedding j:V → L, from the set-theoretic universe V to the constructible universe L, when V ≠ L?
QUESTION [31 upvotes]: Main Question. Can there be an embedding $j:V\to L$ of the
set-theoretic universe $V$ to the constructible universe $L$, if
$V\neq L$?
By embedding here, I mean merely a proper class isomorphism from
$\langle V,{\in}\rangle$ to its range in $\langle L,{\in}\rangle$, or in other words a
quantifier-free-elementary map $j:V\to L$, a class map $j$ for
which $x\in y\iff j(x)\in j(y)$.
This embedding concept is considerably weaker than usually considered in set theory, where one typically has embeddings that are at least $\Delta_0$-elementary if not much more. Of course, we may easily refute the existence of nontrivial fully elementary or even of $\Delta_0$-elementary embeddings $j:V\to L$. Those arguments, however, simply fail with this much weaker embedding concept. One can begin to see this by observing that $$j(x)=\{\
j(y)\mid y\in x\
\}\cup\{\
\{0,x\}\ \}$$ defines an embedding $j:L\to L$ with $j(x)\neq x$ for every $x$. In particular, the existence of a nontrivial embedding $j:L\to L$ in this weak sense is consistent with $V=L$ and carries no large cardinal strength, and does not prove the existence of $0^\sharp$.
The question arises in connection with my paper,

J. D. Hamkins, "Every countable model of set theory embeds into its own
constructible universe", (see also the arxiv entry),

where it appears in the final section with the other questions I
ask here, among others. I have half an expectation, a gnawing
suspicion, however, that this questions may admit an easy answer,
and this is why I am asking it here. But I don't know which way
the answer will go.
The main theorem of the paper shows that every countable model of
set theory $M$ has an embedding $j:M\to L^M$. But the proof
establishes the existence of such embeddings only in an external
way, using the countability of $M$. The main question above
inquires from an internal perspective whether one can ever find
such an embedding as a class inside the model.
The existence of such an embedding as a definable class would of
course imply $V=\text{HOD}$, since one could pull back the canonical
order from $L$ to $V$. More generally, if $j$ is merely a class in
Gödel–Bernays set theory, then the existence of an embedding
$j:V\to L$ implies global choice. So we cannot expect every model
of ZFC or of GB to have such embeddings. Can they be added
generically? Do they have some large cardinal strength? Are they
outright refutable?
There are several more concrete versions of the question.
Question.
Does every set $A$ admit an embedding $j:\langle A,{\in}\rangle \to \langle L,{\in}\rangle$?  If not, which sets do admit such embeddings?
It follows from the main theorem of the paper that every countable
set $A$ embeds into $L$. What about uncountable sets?
Question. Does $\langle V_{\omega+1},{\in}\rangle$ embed into
$\langle L,{\in}\rangle$? How about $\langle P(\omega),{\in}\rangle$ or $\langle \text{HC},{\in}\rangle$?
These latter questions are interesting principally when $V$ has non-constructible reals. I would be very interested in learning the answer.

REPLY [9 votes]: Theorem 1. If $V$ is a non-trivial set generic extension of $W\models\mathrm{ZFC}$ then there is no
$j:V\to W$ as described (i.e. with $x\in y\iff j(x)\in j(y)$ for all $x,y\in V$).
(So in particular, regarding a question at the end of @JoelDavidHamkins' answer, if $V=L[c]$ where $c$ is Cohen generic over $L$, then there is no such $j:V\to L$.)
Proof. Suppose otherwise and let
$\mathbb{P}\subseteq\alpha\in\mathrm{OR}$ with $\mathbb{P}\in W$ and
$G$ be $(W,\mathbb{P})$-generic with $G\notin W$. Let $j:V\to W$ be the embedding. I'll argue somewhat
like in the proof of Theorem 2.3/Lemma 2.2 of Schlutzenberg - Reinhardt cardinals and iterates of $V$ for a contradiction.
Let $\beta$ be a regular cardinal $>\alpha$.
Let $\dot{k}\in W$ be a $\mathbb{P}$-name for $j\upharpoonright\beta$. Working in
$W$, by using the $\mathbb{P}$-forcing relation, we can find some set
$A\subseteq\beta$ of ordertype $\alpha$  and some $p\in G$ such that $p$
decides the value of $\dot{k}(\gamma)$ for each $\gamma\in A$.
Therefore $j\upharpoonright A\in W$. Working in $V$, let $G'\subseteq
A$ be the "translation" of $G$; that is, let $\pi:\alpha\to A$ be the increasing enumeration
of  $A$, and let $G'=\pi``G$. Let $G^*=j(G)\in W$. Then working in $W$, from $G^*$ and $j\upharpoonright A$, we can compute $G'$ (i.e. for $\gamma\in A$, we have $\gamma\in G'$ iff $j(\gamma)\in G^*$; note this uses only the elementarity of $j$ that is assumed). But $A,\pi\in W$ also, and from these and $G'$ we can compute $G$, so $G\in W$, a contradiction.
Remark. By Theorem 2.3 of the paper mentioned above, if $W\models\mathrm{ZF}$ and $V$ is a set-generic extension of $W$ and $j:V\to W$ is elementary, then $W,V$ have the same sets of ordinals. By adapting the foregoing argument with that for Theorem 2.3, the elementarity assumption of Theorem 2.3 can be reduced to the weak elementarity considered here.
(Update) Theorem 2. Suppose ZFC + $j:V\to L$ is an embedding. Then CH holds, and therefore by the earlier answers and comments, GCH holds.
Proof. This is a variant of the proof of GCH $>\aleph_0$ from earlier answers and comments. For $i=0,1$, let $\eta_{\aleph_i}$ be the least ordinal $\eta$ such that there is a map $\pi:\aleph_i\to\eta$ and a map $\pi^+:\mathcal{P}(\aleph_i)\to\mathcal{P}(\eta)^L$ such that for all $\alpha<\aleph_i$ and all $X\subseteq\aleph_i$, we have $\pi(\alpha)\in\pi^+(X)$ iff $\alpha\in X$.
(Correction: I asserted in an earlier version of this that we get such a $\pi,\pi^+$ by restricting $j$, but that was confused, since $j$ need not map ordinals to ordinals. But we can easily modify $j$ to get such a $\pi,\pi^+$. That is, let $\sigma\in L$ be a bijection $\sigma:j(\aleph_i)\to\eta$ where $\eta\in\mathrm{OR}$. Let $\pi=\sigma\circ j\upharpoonright\aleph_i$. For $X\subseteq\aleph_i$ let $\pi^+(X)=\sigma``(j(X)\cap j(\aleph_i))$, noting $\pi^+(X)\in L$. Then $\eta,\pi,\pi^+$ work.)
Claim 1. $\eta_{\aleph_i}$ is an $L$-cardinal, for $i=0,1$.
Proof. Suppose not and let $\eta=\mathrm{card}^L(\eta_{\aleph_i})$. Let $\sigma:\eta\to\eta_{\aleph_i}$ be a bijection with $\sigma\in L$. Let $\pi,\pi^+$ witness the definition of $\eta_{\aleph_i}$. Then using $\pi,\pi^+,\sigma$ we can construct a witness to show that $\eta_{\aleph_i}\leq\eta$, a contradiction.
(That is, define $\pi'(\alpha)=\sigma^{-1}(\pi(\alpha))$,
and define $(\pi^+)'(X)=\sigma^{-1}``\pi(X)$, noting that $(\pi^+)'(X)\in L$.)
Claim 2. $\eta_{\aleph_0}<\eta_{\aleph_1}$.
Proof. Easily $\eta_{\aleph_0}\leq\eta_{\aleph_1}$. So (let and) suppose $\eta=\eta_{\aleph_0}=\eta_{\aleph_1}$. Note that $\mathrm{cof}(\eta)=\aleph_0$ (this is $V$-cofinality), as $\eta=\eta_{\aleph_0}$. Let $\pi,\pi^+$ witness the definition of $\eta_{\aleph_1}=\eta$. Then we can fix $\eta'<\eta$ such that $\mathrm{rg}(\pi)\cap\eta'$ has cardinality $\aleph_1$. Let $A=\pi^{-1}``\eta'$. Let $\sigma:A\to\eta'$ be $\sigma=\pi\upharpoonright A$ and let $\sigma^+:\mathcal{P}(A)\to\mathcal{P}(\eta')^L$ be $\sigma^+(X)=\pi^+(X)\cap\eta'$ (note $\sigma^+(X)\in L$). Now shift $\sigma,\sigma^+$ to have domain $\aleph_1$ instead (but with the same range etc). This shows that $\eta_{\aleph_1}\leq\eta'$, a contradiction.
Now if $\eta_{\aleph_0}<\aleph_1$ then as in the proof of GCH above $\aleph_0$,
we get  $2^{\aleph_0}=\aleph_1$. So suppose $\eta=\eta_{\aleph_0}\geq\aleph_1$. We have $\eta_{\aleph_1}<\aleph_2$, by the proof of GCH above $\aleph_0$ (that is, $j``\aleph_1$ is covered by some set $B\in L$ of ($V$-)cardinality $\aleph_1$, and working in $L$, we can shift $B$ down to its ordertype $\eta'$, and $\eta_{\aleph_1}\leq\eta'<\aleph_2$).
So all together and by the claims,
$$\aleph_1<\eta=\eta_{\aleph_0}<\eta^{+L}\leq\eta_{\aleph_1}<\aleph_2.$$
But then again as in the proof of GCH above $\aleph_0$, we can embed $\mathcal{P}(\omega)$ injectively into $\eta^{+L}$, and therefore $2^{\aleph_0}=\aleph_1$, as desired.
(Update 2) By similar reasoning we can also get a strong form of GCH at uncountable cardinals:
Theorem 3. Assume ZFC + $j:V\to L$ is an embedding where $V\neq L$.
Then for every uncountable cardinal $\kappa$, there is a set $A\subseteq\kappa$ such that $\mathcal{P}(\kappa)\subseteq L[A]$ (and hence $\mathcal{P}(\kappa)\subseteq L_{\kappa^+}[A]$).
Proof. Define $\eta_\kappa$ for $\kappa$ just like $\eta_{\aleph_1}$ was defined above. By covering, $\kappa\leq\eta_\kappa<\kappa^+$, and $\eta_\kappa$ is an $L$-cardinal, like before. Moreover, note that $(\eta_\kappa)^{+L}=\kappa^+$;
in other words, $\eta_\kappa$ is the largest $L$-cardinal which is ${<\kappa^+}$ (and either $\kappa$ is regular and $\mathrm{cof}(\eta_\kappa)=\kappa$, or $\kappa$ is singular and $\eta_\kappa=\kappa$ (by covering)). Let $\pi:\kappa\to\eta_\kappa$ and $\pi^+$ witness the definition of $\eta_\kappa$. Then $\mathcal{P}(\kappa)\subseteq L_{\kappa^+}[\pi]$ (for note that for each $X\subseteq\kappa$, we can compute $X$ from $\pi^+(X)$ and $\pi$, and $\pi^+(X)\in L_{\kappa^+}$; we don't need $\pi^+$ itself to do this). But $\pi$ is coded by some $A\subseteq\kappa$, so we are done.
(Of course if $\eta_{\aleph_0}<\aleph_1$ then we also get $\mathcal{P}(\omega)\subseteq L[x]$ for some real $x$. But I don't see that $\eta_{\aleph_0}<\aleph_1$, and if $\eta_{\aleph_0}>\aleph_1$ then I don't see why there should be such an $x$.)
Hamkins proved already under ZFC + "$j:V\to L$ is an embedding" that $0^\#$ does not exist. This can be refined as follows:
(Update 3) Theorem 4. Assume ZFC + $j:\mathcal{P}(\omega)\to L$ is an embedding. Then $0^\#$ does not exist.
Proof. Suppose otherwise. For each $n<\omega$ let $t_n$ be a term and $\vec{\kappa}_n$ a finite tuple of Silver indiscernibles such that $j(n)=t_n^L(\vec{\kappa}_n)$. Likewise define $t_x$ and $\vec{\kappa}_x$ for $x\subseteq\omega$. Then for $x\neq y$ we have $(t_x,\vec{\kappa}_x)\neq(t_y,\vec{\kappa}_y)$. But because $\mathcal{P}(\omega)$ is uncountable, there will be $x\neq y$ such that $t_x=t_y$ and $\vec{\kappa}_x,\vec{\kappa}_y$  have the same "type" with respect to $\left<\vec{\kappa}_n\right>_{n<\omega}$;
that is, $\vec{\kappa}_x$ sits in $\vec{\kappa}_x\cup\bigcup_{n<\omega}\vec{\kappa}_n$ in terms of ordertype position just as $\vec{\kappa}_y$ sits in $\vec{\kappa}_y\cup\bigcup_{n<\omega}\vec{\kappa}_n$. (That is, consider the function $x\mapsto(t_x,\mathrm{type}(\vec{\kappa}_x))$ in this sense;
there are only countably many values in the range, so we get $x\neq y$ with the same output.) But then by indiscernibility, it follows that $j(x)$ and $j(y)$ agree with each other on membership with regard to $j(n)$ for each $n<\omega$, which contradicts that $j$ is an embedding.<|endoftext|>
TITLE: "You can't push a rope"
QUESTION [11 upvotes]: "You can't push a rope" is a wisdom saying that some engineering teachers pass along to their students.  Since I'm not an engineer, I can only guess at what they mean, but it sounds to me like code for a mathematical principle (but from control theory? numerical analysis? ergodic theory? dynamical systems? statistics?)
1) I would appreciate having any mathematicians who work with engineers hazard a general framework for a rigorous formulation of what the engineers mean by this slogan.
2) I would like to know of any deep theorems that (of the no-go variety) that naturally fall under this rubric (whether engineers know about these theorems or not).

REPLY [10 votes]: The rigorous mathematical context is stability. A straight rope in either tension or compression is a valid solution of the underlying PDE, but in compression this solution is unstable, so it cannot be realized in practice.<|endoftext|>
TITLE: finite complex with non-finitely generated homology with local coefficients
QUESTION [10 upvotes]: I am looking for an explicit example, if one exists, of a (pointed) finite connected CW-complex $X$ such that some homology group with local coefficients $H_n(X,{\mathbb Z}[\pi_1 X])$ is not a finitely generated ${\mathbb Z}[\pi_1 X]$-module.
Such an example would in particular give a finitely presented group $\pi$, and a chain complex of finitely generated free ${\mathbb Z}[\pi]$-modules whose homology groups are not all finitely generated over ${\mathbb Z}[\pi]$. Suggestions on finding an explicit example of such a chain complex (of length 3, without loss of generality) are also welcome.

REPLY [10 votes]: As Ricardo points out in the comments, there's an error in my sketched calculation below.  I also didn't notice the requirement that $X$ should be finite, so the natural $BK$ fails on two counts!  However, it seems possible that a presentation complex for $K$ would do the job.  Stallings shows that $\pi_2$ of any complex with $\pi_1=K$ is infinitely generated as a $K$-module.
==========
I think you want to start with a famous example of Stallings, from the paper 'A finitely presented group whose 3-dimensional integral homology is not finitely generated'.  Stallings constructs a finitely presented group $K$ with the property that `there is no projective resolution of $\mathbf{Z}$ over $\mathbf{Z}[K]$ which is finitely generated in dimension 3' (Corollary 1).
In fact, as observed by Bieri, $K$ can be realizes as an explicit subgroup of the direct product of three free groups, $G=F_2\times F_2\times F_2$: $K$ is the kernel of a map $G\to\mathbf{Z}$ that sends every generator to $1$.  So $K$ defines an `explicit' 3-complex, namely a covering space of the natural $K(G,1)$.  As $K$ is 3-dimensional and finitely presented, it follows from Corollary 1 that $H_3(K,\mathbf{Z}[K])$ is infinitely generated.
Stallings's paper was the starting point for many beautiful constructions.  Highlights include the word of Bieri and Bestvina--Brady .<|endoftext|>
TITLE: Character varieties of finitely generated groups
QUESTION [9 upvotes]: Consider the following situation: $\Gamma_0\leq\Gamma$ are both finitely generated groups and $\Gamma_0$ has finite index in $\Gamma$. The restriction gives a well defined map between the character varieties of these groups:
$$\mathrm{Res}:M(\Gamma,\mathrm{GL}_N(\mathbb{C}))\longrightarrow M(\Gamma_0,\mathrm{GL}_N(\mathbb{C}))$$
where $M(\Gamma,\mathrm{GL}_N(\mathbb{C}))$ is the GIT quotient of the variety $\mathrm{Hom}(\Gamma,\mathrm{GL}_N(\mathbb{C}))$ under the action of $\mathrm{GL}_N(\mathbb{C})$ by conjugation. Is it always true that the map $\mathrm{Res}$ is proper ? If not, can you provide a (simple) counterexample ?
Tanks in advance. Benoît

REPLY [11 votes]: Claim. The restriction map is always proper, where the target group $G$ is the group of $K$-points of a reductive group over a local field $K$, e.g. $G=GL_N({\mathbb C})$.
Proof. First, some generalities, details for which you can find, for instance, here. Let $X$ be the symmetric space or a locally compact Euclidean building corresponding to $G$. For each representation $\rho$ to $G$ of a group $\Lambda$ with generators $\gamma_1,...,\gamma_k$, define the min-max displacement
$$
d_\rho:=\inf_{x\in X} d_\rho(x), d_\rho(x):=\max_j \rho(\gamma_j)(x).
$$ 
Let $x_\rho\in X$ denote a point for which $d_\rho(x)-d_\rho\le 1$. It is not hard to check that the number $d_\rho$ depends only on the projection $[\rho]$ of $\rho$ to $M(\Lambda, G)$. Furthermore, a sequence $[\rho_i]$ is precompact in $M(\Lambda, G)$ iff the sequence $(d_{\rho_i})$ is bounded. Suppose that $[\rho_i]$ is not precompact and the sequence $(d_{\rho_i})$ diverges to infinity. Then you take the asymptotic cone $Cone(X)$ of $X$ centered at the points $x_{\rho_i}$ with scaling factors $(d_{\rho_i})^{-1}$. The result is an isometric action $\rho$ of $\Lambda$ on $Cone(X)$ without a common fixed point. The key fact is that $Cone(X)$ is a Euclidean building, by a theorem of Kleiner and Leeb. By the Cartan-Tits theorem, the action $\rho$ has no bounded orbits. 
Now, we can prove the claim. Suppose to the contrary, that there exists a sequence of representations $\rho_i$ of $\Gamma$ whose projections to $M=M(\Gamma, G)$ diverge, while their restrictions $\rho'_i$ to $\Gamma_0$ project to a relatively compact sequence in $M(\Gamma_0,G)$. Then the above construction yields an action $\rho$ of $\Gamma$ on $Cone(X)$.
Note that the restricted actions $\rho_i'$ have bounded $d_{\rho'_i}$. There are two cases to consider:

The distance between points $x_{\rho_i'}$ and $x_{\rho_i}$ is $O(d_{\rho_i})$. Then 
the sequence  $(x_{\rho_i'})$ represents a point $x$ in $Cone(X)$. This point is necessarily fixed by $\Gamma_0$. Thus, since $|\Gamma:\Gamma_0|<\infty$, the $\Gamma$-orbit of $x$ is bounded, which is a contradiction. 
$d_{\rho_i}=o(d(x_{\rho_i'},x_{\rho_i}))$. Then geodesic segments between $x_{\rho_i'},x_{\rho_i}$ represent a geodesic ray in $Cone(X)$. The point at infinity $\xi$ represented by this ray is fixed by $\Gamma_0$ and, moreover, every element of $\Gamma_0$ acts as a unipotent isometry of $Cone(X)$. Hence, $\Gamma_0$ has a common fixed point in $X$. Now, the argument is the same as in Case 1. QED.<|endoftext|>
TITLE: Shapiro inequality for $n=23$
QUESTION [10 upvotes]: Is the Shapiro inequality for $n=23$ an open problem? The reason why I am asking is I have two contradictory pieces of information from two different articles.
The first article titled "The validity of Shapiro’s cyclic inequality" published by B.A. Troesch in the journal "Mathematics of Computation" in 1989 claims that the inequality does indeed holds for all odd $N  \leq  23$. The link to the article is here.
However, there is another article published in 2002, titled "Shapiro's Cyclic Inequality for Even n" by Bushell, PJ and McLEOD, JB in the journal "Journal of Inequalities and Applications" that claims that the case $n=23$ is still an open problem. Here is a link to the article. Interestingly, this article cites the previous article.
The Shapiro inequality is the claim that if $x_1,x_2,\ldots,x_n \in \mathbb{R}^+$, then
$$\sum_{k=1}^{n} \dfrac{x_k}{x_{k+1} + x_{k+2}} \geq \dfrac{n}2$$
where $x_{n+1} = x_1$ and $x_{n+2} = x_2$ holds true for $n \in \{1,2,3,\ldots,12,13,15,17,19,21,23 \} $.
Here is the wikipedia link for whatever it is worth.

REPLY [7 votes]: There is no incongruity between the two papers that have been cited in the question. 
According to A.M. Fink, what is missing is completely analytical proof of the case $n=23$. So when Bushell wrote that $n=23$ is open, he meant that finding a completely  analytical proof to be an open problem.
In the article linked to above, Fink summarizes Shapiro's inequality, and mentions several milestones of progress on it. Indeed, he calls the inequality "settled", which we may take to assume that the cases where it is true have all been proved, either analytically or by hinging upon computation.<|endoftext|>
TITLE: Applied Problems in Probability which can not be modelled on Polish spaces
QUESTION [9 upvotes]: Probabilist often work on Polish spaces. Does somebody know an ("non-exotic") example, for which it is not possible to work on a Polish space, but instead one has to work on a general measurable space? By non-exotic example I mean something like a stochastic process, which is really used in applications, and cannot be defined on a Polish space...(I posted the this question also here).

REPLY [3 votes]: This answer is identical to the one I gave when the same question was posted at M.SE.:
There are a number of constructions that do not work for Polish spaces, but a certain class of probability spaces, variously known as super-atomless, saturated, nowhere countably generated and a number of other names. A nice overview can be found here. 
A probability space $(\Omega,\Sigma,\mu)$ is saturated if for every two Poilsh spaces $X$ and $Y$, every probability measure 
$\nu$ on $X\times Y$ and every random variable $f:\Omega\to X$ such that its distribution $\mu f^{-1}$ equals the marginal of $\nu$
on $X$, there is a random variable $g:\Omega\to Y$ such that the joint distribution of $(f,g)$ is $\nu$.
The following definition is conceptually different, but can be shown to be equivalent:
A probability space $(\Omega,\Sigma,\mu)$ is super-atomless if there is no $A\in\Sigma$ satisfying $\mu(A)>0$, such that the pseudo-metric space obtained by
endowing the trace $\sigma$-algebra on $A$ with the pseudo-metric $d(A,B)=\mu(A\triangle B)$ is separable.<|endoftext|>
TITLE: Square root for Hamiltonian diffeomorphisms
QUESTION [7 upvotes]: Let $\psi_t: X\to X$, $t \in [0,1]$, be a path Hamiltonian diffeomorphism on a symplectic manifold $X$, given by functions $H_t$. If $H_t \equiv H$ is independent of $t$ then 
$$ \psi_1 = \psi_{\frac{1}{2}}^2 $$
and therefore the Hamiltonian diffeomorphism $\psi_1$ has a Hamiltonian square root.
Is the same thing true for any arbitrary Hamiltonian $\psi_1$, i.e. is there another Hamiltonian $\phi$ such that $\phi^2 = \psi_1$ ?

REPLY [2 votes]: In a short paper posted last week, Peter Albers and Urs Frauenfelder prove that if $(M,\omega)$ is any closed symplectic manifold, then in any $\mathcal{C}^\infty$-neighborhood of the identity in $\text{Ham}(M,\omega)$ there is a Hamiltonian diffeomorphism that does not have a square root in $\text{Ham}(M,\omega)$ (the square root is not required to lie in this neighborhood).
The key is an observation of Milnor's that for any $k > 0$, an obstruction to a self-diffeomorphism of a manifold having a square root is that it has an odd number of $2k$-cycles.<|endoftext|>
TITLE: A Fourier-analytic inequality used by Jean Bourgain
QUESTION [19 upvotes]: I am currently reading Jean Bourgain's 1986 paper A Szemerédi type theorem for sets of positive density in $R^k$ and would appreciate some help in understanding a Fourier-analytic estimate used in that article. I suspect that my question is relatively elementary, but my knowledge of Fourier analysis is not very strong and there are no experts in the topic at my current place of work, so I would very much appreciate a pointer.
In Bourgain's argument, $f \colon \mathbb{R}^d \to [0,1]$ is a nonzero measurable function supported in a fixed bounded measurable set $A$, and the $L^2$ norm of $f$ is fixed. For each $\lambda>0$ we define $P_\lambda \colon \mathbb{R}^d \to \mathbb{R}$ to be the function whose Fourier transform $\hat{P}_\lambda(\xi):=\int_{\mathbb{R}^d} e^{-2\pi i \langle x,\xi\rangle} P_\lambda(x)dx$ is given by $\hat{P}_\lambda(\xi)=e^{-\lambda\|\xi\|}$ for all $\xi \in \mathbb{R}^d$. Parameters $\delta, t>0$ are introduced, and the parameter $\delta$ is subsequently fixed at some small value which depends on $\|f\|_2$ (and possibly on $A$) but not on the precise choice of $f$. It is then claimed that by taking $t$ small enough, the quantity 
$$\|(f * P_{\delta t}) - (f * P_{\delta^{-1}t})\|_2$$
can be made arbitarily small in a manner which is uniform with respect to $f$. It is clear to me that this quantity must converge to zero as $t \to 0$ when $\delta$ and $f$ are fixed, but it is not clear to me why a single value $t$ can be chosen which works simultaneously for all $f$ (where $\|f\|_2$ is fixed and the support of $f$ lies in $A$). Bourgain's paper seems to use a quantitative bound which I infer to resemble
$$\|(f * P_{\delta t}) - (f * P_{\delta^{-1}t})\|_2 \leq C\|f\|_2\frac{\log (1/\delta)}{\log (1/t)}.$$
Certainly it is stated that in order to make the above difference small (relative to $\delta^{1/4}$ and $\|f\|_2$) it is sufficient that $\log (1/t)$ should be a large multiple of $\log (1/\delta)$. Can anyone see more precisely what estimate is being used here, or at least how the above quantity can be bounded uniformly with respect to $f$?
Thanks!

REPLY [33 votes]: I assume you are referring to the argument in page 313 of Jean's paper http://www.springerlink.com/content/a343g53033872345/ .  The point here is that the bound does not hold for all $t$, but for a single $t$ (out of $J$ possible choices $t_1,\dots,t_J$); note that Jean crucially refers in the paper to a "suitable" $t$ rather than an arbitrary $t$.  This is a pigeonholing argument, based on the estimation of
$$ \sum_{j=1}^J \| f * P_{\delta t_j} - f * P_{\delta^{-1} t_j} \|_{L^2}^2$$
which can be done by Plancherel's theorem and routine computations (if the $t_j$ are lacunary, as noted in Jean's paper).  
The use of pigeonholing to turn qualitative results (such as dominated convergence) to quantitative ones (at the cost of losing some control on the parameter for which the bound is attained) is an important trick in the subject; I discuss it at http://terrytao.wordpress.com/2007/05/23/soft-analysis-hard-analysis-and-the-finite-convergence-principle/ .  Another key trick displayed here is to always be aware whether one needs to control the worst-case choice of parameter (i.e. uniform bounds), average-case choice of parameter (e.g. integrated or probabilistic bounds), or best-case choice of parameter (e.g. what comes from the pigeonhole principle).  In this case, because one only needs the bound for a single t, best-case analysis suffices, and one can use many more tricks in this setting than in worst-case or average-case analysis.
Incidentally, I found the reading of Jean's papers as a graduate student to be simultaneously extremely frustrating and extremely rewarding.  Decoding an offhand remark or a mysterious step in his paper was often as instructive (and as time-consuming) as reading several pages of arguments by some other authors.  (But his papers do become much easier to read once one has internalised enough of his "box of tools"...)<|endoftext|>
TITLE: Is there a tropical analogue of a reproducing kernel Hilbert space?
QUESTION [6 upvotes]: In classical functional analysis, one can construct a reproducing kernel Hilbert space by starting with a positive definite kernel, say
$K: [0,1]\times [0,1] \rightarrow \mathbb{R}$.
One then creates linear combinations of the form
$f(x) = \sum^n a_i k(x_i,x)$, together with an inner product
$\langle f,g \rangle = \sum \sum a_i b_j k(x_i,x_j),$
and completes the space as usual.
As far as I can see, the 'essential' properties of a reproducing kernel Hilbert space are
the ability to represent linear operators as integral kernels, and the Riesz representation theorem.
I'm just wondering if there is a similar construction to the one outlined above in the framework of the max-plus algebra. Both of the properties I mentioned above have max-plus analogues (see this introduction). If such a construction exists, how far is it possibe to take it? Is there an idempotent version of Mercer's theorem, for example?

REPLY [8 votes]: See

G.L. Litvinov, V.P. Maslov and G.B. Shpiz. Idempotent functional analysis.
An algebraic approach // Mathematical Notes, v. 69, # 5, 2001, p. 696-729. 
E-print math.FA/0009128  (http://ArXiv.org).
G.L. Litvinov and G.B. Shpiz. Kernel  theorems  and nuclearity in 
idempotent  mathematics.  An  algebraic  approach,  Journal of 
Mathematical Sciences, v. 141, #4, 2007, p. 1417-1428. See also 
E-print math.FA/0609033 (http://arXiv.org), 2006.
G.L. Litvinov. Tropical mathematics, idempotent analysis, classical
mechanics and geometry. - in: Spectral Theory and Geometric Analysis M.Braverman et al., 
Eds.,  AMS Contemporary Mathematics, vol. 535, 2011, p. 159-186. See also E-print 
arXiv: arXiv: 1005.1247 (http://arXiv.org).<|endoftext|>
TITLE: Computing the Zariski closure of a subgroup of SL(n,Z)
QUESTION [20 upvotes]: Suppose $\Gamma$ is a finitely generated subgroup of $SL(n,\mathbb{Z})$, given as a list of generators. We would like to (somewhat efficiently) try to compute the Zariski closure of $\Gamma$, which is a (real) algebraic group. The method should be computer assisted but rigorous. 
In the cases we are considering, $\Gamma$ will usually be Zariski dense in $SL(n,\mathbb{R})$,      so the algorithm we are looking for should be optimized for that case. Also we would like just to know the answer, so having the program run forever if $\Gamma$ is not Zariski dense in $SL(n,\mathbb{R})$ is fine for us: we will just analyze that example further. 
We can probably come up with some ad-hoc method for doing this, but I was wondering if anyone on MO has some interesting ideas or references.

REPLY [3 votes]: In fact, the last word on this is my recent paper; but the punchline is that the subgroup is Zariski-dense if and only if there are two non-commuting elements, such that the Galois group of the characteristic polynomial of one of them is the full symmetric group.<|endoftext|>
TITLE: Adding segments to an annulus - a question regarding the conformal modulus
QUESTION [5 upvotes]: Let $A \subset \mathbb{C}$ be an topological annulus, i.e. a region of $\mathbb{C}$ bounded by two disjoint Jordan curves.
Let $B \subset \mathbb{C}$ be a quadrilateral, i.e. a topological disc with four distinct marked points $(z_1,z_2,z_3,z_4)$ arranged anticlockwise on the boundary.
Both annuli and quadrilaterals, as defined above, have a conformal invariant, in both cases known as the modulus. Let $A_R$ be a geometrical annulus with inner boundary a circle of radius $1$ and outer boundary a circle of radius $R$. If $A$ can be mapped conformally and bijectively onto $A_R$ we say that $A$ has modulus $\ln{R}/2\pi$. Likewise, let $Q_m$ be a geometrical rectangle with vertices $(0, m, i+m, i)$, we say that $B$ has modulus $m$ if there exists a conformal bijection mapping $B$ onto $Q_m$ with $z_1$,$z_2$,$z_3$, and $z_4$ being mapped onto $0$, $m$, $i+m$, and $i$ respectively.
The two concepts are linked by the fact that if we take a geometric annulus centred on the origin with modulus $m$, remove  all the points on the positive real line from the annulus, and then take the preimage under the exponential map, we obtain (infinitely many copies of) a geometric rectangle also with modulus $m$.
My questions relate to the degree in which the choice of branch cut is arbitrary.
More particularly:
Let $C \subset \mathbb{C}$ be another annulus and represent $C$ as the union $R_1 \cup R_2$ of two simply connected regions with the intersection $R_1 \cap R_2$ consisting of two smooth simple curves $\gamma_1$ and $\gamma_2$, disjoint from each other and each of which has one end point, $x_1$ and $x_2$ respectively, on the inner boundary and one end point, $y_1$ and $y_2$ on the outer boundary. Assume we have a continuous function $f_1$ mapping $R_1$ onto $A$, and such that the restriction of $f_1$ to the interior of $R_1$ is a conformal bijection and such that the image of $\gamma_1$ and $\gamma_2$ is a single simple curve from the inner to the outer boundary of $A$, with $f_1(x_1) = f_1(x_2)$ and $f_1(y_1) = f_1(y_2)$. Similarly assume that $f_2$ is a bijection mapping $R_2$ onto $B$, conformal on the interior, with $\gamma_1$ being mapped onto the component of the boundary joining $z_1$ and $z_2$ and $\gamma_2$ being mapped to the component of the boundary joining $z_3$ and $z_4$. See the diagram below.
My first question is whether the modulus of $C$ is determined entirely by the modulus of $A$ and $B$? (This seems unlikely to me, but no harm in asking)
Secondly, if this is not the case can we give any bound for the modulus of $C$ given the modulus of $A$ and $B$.
Thirdly, are there any conditions we can impose to reduce these bounds? For example specifying that the curves $\gamma_1$ and $\gamma_2$ meet the boundary curves of $C$ orthogonally.

(source: qmul.ac.uk)

REPLY [4 votes]: A very similar, but somewhat easier question, is the following: take $R$ a rectangle, say $[0,m]\times[0,1]$ identifying $\mathbb C$ to $\mathbb R^2$ for nicer notation. This has modulus $m$. Now let $\gamma$ be a simple curve connecting the top to the bottom sides of $R$, while staying within the rectangle. It splits it into two simply connected domains $R_1$ and $R_2$ that are topological rectangles. So each one has a modulus, say the moduli are $m_1$ and $m_2$. Do $m_1$ and $m_2$ determine $m$?
The answer is no here, because of the following two examples.
First, looking at $[0,m_1+m_2]\times[0,1]$ split vertically by $\{m_1\} \times [0,1]$, if the answer was yes, then it would have to be $m=m_1+m_2$. 
Second, one can split $[0,m]\times[0,1]$ using a curve that goes back and forth many times between the neighborhood of $\{0\}\times[0,1]$ and the neighborhood ot $\{m\}\times[0,1]$. This will make both $m_1$ and $m_2$ very small, and is incompatible with their sum being equal to $m$.
Something very similar can be done in your case: take $\gamma_1$ straight, and $\gamma_2$ going back and forth between the left and right sides of $\gamma_1$. This will make the moduli of both $A$ and $B$ very small.

One thing which on the other hand is true, in the simply connected case at least, is a sub-additivity relation: with the above notation, one always has $m \geqslant m_1 + m_2$. One way to prove this is via extremal length, as follows. (Not sure the terminology is completely standard.)
Let $\rho : R=[0,m]\times[0,1] \to \mathbb R_+$. The $\rho$-length of a rectifiable curve $\gamma$ is $L_\rho(\gamma) := \int_\gamma \rho ds$ (in terms of the curvilinear coordinate). The $\rho$-width $W_\rho(R)$ of $R$ is the shortest $\rho$-length of a curve connecting the two vertical sides of $R$. The $\rho$-area of $R$ is $A_\rho(R) = \int_R \rho^2 |dz|$ (with respect to the Lebesgue measure on $R$). Then, the extremal width of $R$ is defined as $$W(R) := \sup_\rho \frac {W_\rho(R)^2} {A_\rho(R)}.$$
It is easy to define this for any topological rectangle, and to check that $\rho$ is conformally invariant. In particular it has to be a function of the modulus. In the case of the rectangle, the supremum is reached when $\rho$ is constant, in which case $W_\rho(R)=m$ and $A_\rho(R)=m$, leading to $W(R)=m$. In other words: the extremal width is just the same as the modulus.
But now, given a rectangle $R$ split into two rectangles $R_1$ and $R_2$, one can put maximizing functions $\rho_1$ and $\rho_2$ for the previous variational problem on them. This defines $\rho$ on $R$, which does not do better than the optimal one for $R$: the last sentence is exactly the inequality $m \geqslant m_1 + m_2$.

Maybe things have to be tweaked a little bit in your case, because you are gluing objects of different topologies, but the same philosophy will apply. Given more geometrical information, one might also use extremal length to derive better bounds.
Reference for all that: Ahlfors, Lars V. (1973), Conformal invariants: topics in geometric function theory, New York: McGraw-Hill Book Co., MR 0357743.<|endoftext|>
TITLE: Shortest Casson tower containing a slice disk for the attaching curve
QUESTION [16 upvotes]: A Casson tower is obtained as follows: Start with a properly immersed disk in $\mathbb{B}^4$ - a regular neighborhood of such a disk is called a kinky handle. The boundary of the core disk (necessarily in $\mathbb{S}^3$) is called the attaching circle. At each point of self-intersection of the core immersed disk, we have a simple closed curve by starting at the intersection point, leaving along one sheet and returning along the other. Call the collection of these curves (one for each self-intersection) the set of accessory circles. A Casson tower of height 1 is just a kinky handle. An Casson tower of height $n$ is obtained by attaching kinky handles to each of the accessory circles at the top stage of a Casson tower of height $n-1$. A Casson tower of infinite height is called a Casson handle and Freedman has shown that Casson handles are homeomorphic to $\mathbb{D}^2 \times \mathbb{D}^2$. 
In 1982, Freedman showed that within a Casson tower of height 6 one can embed a Casson tower of height 7 with the same attaching circle, and as a result, Casson towers of height 6 contain Casson handles with the same attaching circle. In 1984, Gompf and Singh improved this by showing how to embed a height 6 tower inside a height 5 tower. I believe that there has been further improvement to this theorem. This paper (on pp 15) says that the improved number might be 4, and this one (on pp 103) appears to claim that it is 3. Unfortunately, neither of these above papers has a reference for these results. 
With this background, I am seeking a reference for the answer to the following question:

What is the least integer $n$ for which it is known that given a Casson tower of height $n$, one can embed within it a Casson handle with the same attaching circle? Equivalently, what is the least integer $n$ such that given a Casson tower of height $n$, the attaching circle bounds an embedded topological disk within the given tower?


References:

M. Freedman, The topology of 4-dimensional manifolds, J. Differential Geom. 17 (1982), 357-453.
R. Gompf and S. Singh, On Freedman's reimbedding theorems, Four-Manifold Theory (C. Gordon and R. Kirby, eds.), Contemp. Math., vol. 35, Amer. Math. Soc., Providence, R.I., 1984, pp. 277-310.

REPLY [3 votes]: A recent paper of Cha-Powell shows that Casson towers (and more generally distorted Casson towers) of height four contain slice disks for the attaching curve. This appears to be all that is known at present. 
The relationship between Casson towers containing slice disks for the attaching curve and topological sliceness of certain ramified iterated Whitehead doubles of the Hopf link (seen through inspecting the Kirby diagrams for Casson towers) indicates that not all Casson towers of height one and two can contain slice disks for the attaching curve.<|endoftext|>
TITLE: College (Euclidean) geometry textbook recommendations
QUESTION [16 upvotes]: I will be teaching a mid-level undergraduate course in Euclidean geometry this fall. Has anyone taught such a course, who can recommend a good textbook?
My students will mostly be future high school math teachers, who have some exposure to proofs and rigor but not extensively so. I hope to cover material such as constructions, semi-advanced theorems (Ceva's theorem, the nine point circle, etc., etc.), and the axiomatic approach. I won't do any projective geometry, or anything similar, as that is covered by a followup course here.
I am hoping to choose a book which covers a variety of approaches (so something short is unlikely to be suitable) and which is suitable for students with uneven preparation (i.e. whose ability to write proofs is shaky at the beginning).
Thank you!
EDIT: Cross posted a related question to Math.SE.

REPLY [3 votes]: There are a lot of good books, most of which have already been recommended in previous answers.
I find the book of Coxeter & Greitzer the most appropriate for a semester course, and I have used it as a main reference in designing an undergraduate class on the subject. If one wants to suggest literature for more challenging problems in Euclidean Geometry (IMO level problem), there is a big list of great books, as for example
T. Andreescu, 106 Geometry Problems From the AwesomeMath Summer Program.

REPLY [2 votes]: I haven't used it myself, but there is Patrick D. Barry's Geometry with Trigonometry (Horwood Publishing, Chichester, England, 2001). From the blurb:

This book addresses a neglected mathematical area where basic geometry underpins undergraduate and graduate courses [...]  This text emphasises a systematic and complete build-up of material, moving from pure geometrical reasoning aided by algebra to a blend of analytic geometry and vector methods with trigonometry, always with a view to efficiency.<|endoftext|>
TITLE: Hsiung on the Complex Structure of $S^6$
QUESTION [22 upvotes]: In 1986 C. C. Hsiung published a paper "Nonexistence of a Complex Structure on the Six-Sphere" and in 1995 he even wrote a monograph "Almost Complex and Complex Structures" to further elaborate on his proof.
Yet answers to the 2009 question on this site all agree that the existence of complex structures on $S^6$ is still an open problem. Some recent preprints answering the question with opposite answers are also cited there.
I would like to know if there are any known mistakes in Hsiung's approach and if so I would appreciate some reference to a paper that points them out.

REPLY [3 votes]: There was a paper published in China (1987?) by Dong (and Guan).
It gave a reason why Hsiung was wrong. That was basically
the note of Guan when he sat in a talk of Hsiung. Dong came late and missed
the talk. Guan gave him his note ant told him that the proof was wrong. After
Guan left to America, Dong published the note and added his own
proof that there is no complex structure on S^6. After
Guan received Dong's note, he warned him about his mistake.
But it was too late and the note was published under the joint
authors Dong and Guan. Later on Guan met Hsiung in Lehigh University.
Hsiung showed Guan his book and said that Yau said that his proof
is correct. Guan again told Hsiung about his mistake without
any conclusion. It is more like politics to Guan.<|endoftext|>
TITLE: When is homotopy orbit space weakly equivalent to orbit space, other than situation of free action?
QUESTION [5 upvotes]: Let $M$ be a closed symmetric monoidal model category. Let $X$ be a cofibrant object (it can also be fibrant if you like) and let $\Sigma_n$ act on $X^{\otimes n}$ by permuting the factors (note that this action is far from being free). There is a natural map from the homotopy colimit of the action (which is the extended power $E\Sigma_n \otimes_{\Sigma_n} X^n$) to the colimit $X^{\otimes n}/\Sigma_n$. In the pointed setting this map goes from $(E\Sigma_n)_+ \otimes_{\Sigma_n} X^{\otimes n}$. Let $(P)$ be the property that this map is a weak equivalence.

Does $(P)$ hold for sSet? [EDIT: Answer is no, see the comments]

Note that $(P)$ is known in at least a couple of instances. For example, Lemma 15.5 of Model Categories of Diagram Spectra by MMSS proves the pointed version for the positive model structures on Orthogonal Spectra and Symmetric Spectra, while EKMM's Theorem III.5.1 proves it for S-alg. I learned these references from Shipley's paper "Monoidal Uniqueness of Stable Homotopy Theory" which is also where I read about the property above. Shipley posits that this property is necessary in order for commutative monoids in $M$ to inherit a model structure. In my current thesis work I've found a more general way, and now I'm trying to see if $(P)$ implies my hypotheses, as well as work out lots of examples.

(1) Are there any other model categories of interest where this property is known to hold or fail?

I imagine this property fails when one knows that commutative monoids don't form a model structure, e.g. in chain complexes over a field of characteristic nonzero (does $(P)$ hold in characteristic zero?), for topological spaces (since a commutative monoid must be a product of Eilenberg-MacLane spaces), or for the usual model structure on symmetric spectra (this was known by Gaunce Lewis, and led to the positive model structure).

(2) Are there any other model categories where you can't have a model structure on commutative monoids? Do these other categories fail to satisfy $(P)$?
(3) Are there any standard hypotheses on a closed symmetric monoidal model category $M$ which will imply $(P)$?

For (3), we can certainly assume $M$ is cofibrantly generated and left proper, even combinatorial, cellular, or simplicial if you like. We can even assume all objects are cofibrant if that will lead to a proof. It's well-known that the map from the homotopy colimit of a diagram to the colimit of the diagram is a weak equivalence if the diagram is Reedy cofibrant, but I feel like this would force the action to be free, which we can't have. Perhaps I'm wrong, though, and there is some way to use Reedy cofibrancy.

REPLY [10 votes]: The equivalence (P) is a deep and subtle property of the
smash product of spectra in modern symmetric monoidal
models for the stable homotopy category.  It is very
unlikely to hold in other contexts.  It was first found
in EKMM [III.5.1] because the use of operads there visibly
builds it into the smash product.  It was later seen to 
hold for orthogonal and symmetric spectra in MMSS [15.5].
I doubt that it holds for the $\mathcal W$-spectra defined in MMSS.
Perhaps this is not the place for a confession,
but the equivariant version of (P) as published by Mandell and
myself for orthogonal $G$-spectra in "Equivariant orthogonal
$G$-spectra and $S$-modules'' and by Mandell for symmetric 
$G$-spectra in "Equivariant symmmetric spectra'' is incorrect.  The correct version replaces $E\Sigma_n$ with the universal principal 
$(G,\Sigma_n)$-bundle $E(G,\Sigma_n)$.  For $E_{\infty}$ $G$-operads 
$\mathcal C$, $\mathcal C(n)$ is just such a universal principal 
$(G,\Sigma_n)$-bundle. 
The essential point is that this is a topological result,
not remotely to be expected in a general model theoretical
context.<|endoftext|>
TITLE: Presentability of the source of a fibration
QUESTION [7 upvotes]: Suppose I have a functor $\mathcal{E} \to \mathcal{B}$ that is both a Grothendieck fibration and an opfibration, $\mathcal{B}$ is presentable, and the fibres $\mathcal{E}_{b}$ for $b \in \mathcal{B}$ are all presentable. Are there any known conditions on this setup under which the category $\mathcal{E}$ will be presentable? (By "presentable" I mean the same thing as "locally presentable".)

REPLY [5 votes]: In section 5.3 of Accessible categories by Makkai and Pare, they prove that if $\Phi : B^{\mathrm{op}} \to \mathrm{Cat}$ is a pseudofunctor such that

each category $\Phi(b)$ is accessible,
each functor $\Phi(\beta) : \Phi(b') \to \Phi(b)$ is accessible,
the category $B$ is $\kappa$-accessible, and
the pseudofunctor $\Phi$ preserves $\kappa$-filtered (weak 2-)colimits,

then the total category of the Grothendieck construction of $\Phi$ is again accessible.  An accessible category is locally presentable iff it is complete iff it is cocomplete, and it is well-known that the total category of a fibration is complete if the base and fibers are complete and the restriction functors are continuous.  Moreover, an accessible functor between locally presentable categories is continuous iff it has a left adjoint, and any adjoint between locally presentable categories is accessible.  Thus, the total category of $\Phi$ is locally presentable if

each category $\Phi(b)$ is locally presentable,
each functor $\Phi(\beta) : \Phi(b') \to \Phi(b)$ has a left adjoint (i.e. the fibration is also an opfibration),
the category $B$ is locally $\kappa$-presentable, and
$\Phi$ preserves $\kappa$-filtered (weak 2-)colimits.

So it sounds like to your hypotheses you need only to add that $\Phi$ preserves $\kappa$-filtered (weak 2-)colimits, for some $\kappa$ such that $B$ is locally $\kappa$-presentable.<|endoftext|>
TITLE: functions from Q to itself with derivative zero
QUESTION [19 upvotes]: Let $f: {\bf Q} \rightarrow {\bf Q}$ be a "${\bf Q}$-differentiable" function whose "${\bf Q}$-derivative" is constantly zero; that is, for all $x \in {\bf Q}$ and all $\epsilon > 0$ in ${\bf Q}$, there exists $\delta > 0$ in ${\bf Q}$ such that for all $y \in {\bf Q}$ with $0 < |x-y| < \delta$, $|(f(y)-f(x))/(y-x)| < \epsilon$.
An example of such a function is the 2-valued function on ${\bf Q}$ that takes the value 0 or 1 according to whether $x<\pi$ or $x>\pi$.
Must $f$ be locally constant, in the sense that for all $x \in {\bf Q}$, there exists $\delta > 0$ in ${\bf Q}$ such that for all $y \in {\bf Q}$ with $|x-y| < \delta$, $f(y)=f(x)$?
I have a feeling that this is not a hard problem (and I am even afraid some of you will think that it is a homework problem!), but it actually arose from my research (see http://jamespropp.org/reverse.pdf), and after an hour of thought I still don't see the answer.  In an ideal world I'd mull it over longer before posting, but since the journal to which I have submitted the paper has given me a deadline for making revisions, and the deadline is approaching, I am swallowing my pride and seeking help.

REPLY [12 votes]: Of course you can't deduce that $f$ is constant. It is worth remarking that  such  $ 
f$ is not necessarily constant even in the quite stronger assumption that it is the restriction to $\mathbb{Q}$ of an  everywhere differentiable function $\mathbb{R}\to\mathbb{R}$.
Indeed, there are everywhere differentiable homeomorphisms $g:\mathbb{R}\to\mathbb{R}$ whose derivative vanishes on a dense set. Moreover, you can put in bijection any pair of countable dense subsets of $\mathbb{R}$ by means of an analytic diffeomorphism (see the linked paper in this answer, or the construction sketched in this other answer). So, composing the above function  $g$ with diffeos $\phi$ and $\psi$ produces an everywhere differentiable map $f:=\phi\circ g\circ\psi$ whose derivative vanishes on $\mathbb{Q}$ and such that $f(\mathbb{Q})=\mathbb{Q}$:  just take $\psi(\mathbb{Q})\subset\{g'=0\}$ and 
$\phi (g(\psi (\mathbb{Q})))=\mathbb{Q}$.<|endoftext|>
TITLE: Irreducible "family" of relative effective divisors of a smooth morphism
QUESTION [7 upvotes]: Let $\pi:X\rightarrow Y$ be a smooth proper (assume projective if needed) morphism of schemes with $Y$ locally noetherian, and let $Z\subset X$ be an irreducible integral closed subscheme  containing no fiber of $\pi$.

Is the locus $Pic_\pi(Z)=\{y\in Y:Z_y \text{ is Cartier in }X_y\}$ closed in $Y$?
If not, what extra hypotheses would make it closed?

As $\pi$ is smooth, Cartier and Weil divisors on fibers are the same, and as it is proper, the dimension of fibers is semicontinuous, so the issue is actually about components of smaller dimension in the fibers. Thus I'd drop the hypothesis on not containing fibers and replace it by the hypotheses that $Z$ dominates $Y$; then the question would be:

Is the locus of $y$ such that $Z_y$ has a (possibly embedded) component of codimension $\ge 2$ in $X_y$ open in $Y$?

I have the feeling that this is related to Zariski's main theorem, although in the first formulation it seems closer to asking whether the subscheme of relative effective divisors is closed in the Hilbert scheme when $\pi$ is smooth. But I can't pin it down.
(In my situation, $Z$ is actually of codimension 2, and everything is over the complex field, but I don't think this is necessary).

REPLY [4 votes]: In fact both sets are constructible in $Y$. 
Suppose for simplicity that $X$ is connected. Then the dimension of the (non-empty) fibers of $X\to Y$ is constant (EGA IV.12.1.1(i), and flatness is enough), denote it by $d$. Let $Y'$ be the (integral) image of $Z$ in $Y$. We can replace $X\to Y$ by $X\times_Y Y'\to Y'$ and suppose that $Z\to Y$ is surjective. Let $e$ be the dimension of the generic fiber of $Z\to Y$. 
Special case: $e=d-1$. By Chevalley's theorem (EGA IV.13.1.1), for any $y\in Y$, the irreducible components of $Z_y$ all have dimension $\ge d-1$. By hypothesis, $Z_y\ne X_y$, hence $\dim Z_y\le d-1$. 
Therefore $ \mathrm{Pic}_\pi(Z) = Y $ (and it is closed in $Y$ if $Z$ doesn't dominate $Y$). 
General case. The set of $x\in Z$ such that $\dim_x Z_{\pi(x)}\le d-2$ is open in $Z$ (EGA IV.13.1.3). The complementary of the image by $\pi$ of this open subset is constructible and is your 
${Pic}_{\pi}(Z)$. 
In case $Z\to Y$ is flat, then $\mathrm{Pic}_\pi(Z)$ is closed by openess of  $\pi|_Z$.
For the second question, if $F$ denotes the set of $x\in Z$ such that all associated components of $Z_{\pi(x)}$ passing through $x$ have dimension $\ge d-1$, then you are considering $\pi(X\setminus F)$. By EGA, IV.9.9.2(iii), $F$ is constructible, so your set is constructible. 
If $Z\to Y$ is moreover flat, then $F$ is open by EGA, IV.12.1.1(i) (I learn recently this reference from an anonymous referee.), hence your set $\pi(X\setminus F)$ is in fact closed.<|endoftext|>
TITLE: Smallest collection of bases for prime testing of 64 bit numbers?
QUESTION [13 upvotes]: I know that no number less than 64 bits will fail the Miller-Rabin tests for all of the first 12 primes.  That is, those 12 tests will provide a fully deterministic primality test for all 64 bit numbers.  (See http://oeis.org/A014233).  I also know that for 32 bit numbers, it suffices to apply the Miller-Rabin tests for the three bases 2, 7 and 61.  (See http://primes.utm.edu/prove/merged.html).  Is there a very small list of possible bases (other than the first 12 primes), which will provide a fully deterministic test for 64 bit primes?

REPLY [12 votes]: According to http://miller-rabin.appspot.com/, the 7-element set {2, 325, 9375, 28178, 450775, 9780504, 1795265022} works for 64-bit integers.<|endoftext|>
TITLE: one-dimensional (sort of) tilings
QUESTION [8 upvotes]: Consider the following one-dimensional tiling problem.  Each "tile" is a sequence of nonnegative integers.  A "region" is also such a sequence.  I can shift the "tiles", or reverse them.  A tiling is a set of shifted and/or reversed tiles that add up to the region.  For instance, if my tiles are these three:
1 2
1 1 1
1 1 3
then I can tile the region
1 5 4 1
by adding up
1 1 1
_ 1 2
_  3 1 1 
Do you think the decision problem of whether a given region can be tiled with a given set of tiles is is NP-complete?  Note that the tempting reduction from Subset Sum doesn't work; I want the inputs given in unary, so the numbers and the size of the region are polynomial.

Cris Moore

REPLY [9 votes]: Yes, the problem is NP-complete. Here is a reduction of 1-in-3 SAT to your problem.
Let $\{C_i:i< n\}$ be the given collection of 3-clauses in variables $\{x_j:j< m\}$. All tiles and the region will have length $n+m+2$, so no shift is possible. For each of the $2m$ literals $a$ (which is $x_j$ or $\neg x_j$), we consider a tile of the form
$$1,\underbrace{0,\dots,0,1,0,\dots,0}\_m,u\_{a,0},\dots,u\_{a,n-1},0,$$
where the $1$ in the left part appears at position $j+1$, and $u_{a,i}$ is $1$ if $a$ occurs in $C_i$, and $0$ otherwise. The region is
$$m,\underbrace{1,1,\dots,1}_{n+m},0.$$
Note that the first and last number prevent reversal from being used. Then the left part forces any possible sum to use exactly one of the two literals $x_j$, $\neg x_j$ for any $j$, and the right part ensures that each clause is satisfied by exactly one of the selected literals.
Another variant of the reduction is that we consider tiles and the region of length $2n$; for each $j< m$, we take the tile
$$u_{x_j,0},u_{x_j,1},\dots,u_{x_j,n-1},u_{\neg x_j,n-1},u_{\neg x_j,n-2},\dots,u_{\neg x_j,0},$$
and the region is
$$\underbrace{1,1,\dots,1}\_{n},\underbrace{2,2,\dots,2}\_{n}.$$
Again, no shift is possible. Moreover, any solution must use all of the tiles, so the only choice is whether we reverse a given tile or not, which selects the variable assignment.<|endoftext|>
TITLE: von Neumann automorphisms: does convergence on a dense algebra imply $u$-convergence?
QUESTION [6 upvotes]: Let $M$ be a separable von Neumann algebra and let $A$ be a (von Neumann-)dense *-subalgebra.
Suppose that $\alpha,\alpha_1,\alpha_2,\dots$ are automorphisms of $M$, such that for every $a \in A$,
$$ \alpha_n(a) = \alpha(a) $$
for all $n$ sufficiently large.
Does it follow that $\alpha_n$ converges to $\alpha$ in the $u$-topology?
(Also: what about if we weaken the hypothesis to just assuming that $\alpha_n(a)$ converges in norm to $\alpha(a)$, for all $a \in A$?)
This question is inspired by a  related MO question, where an explicit example was requested of a sequence of inner automorphisms on the hyperfinite $III_1$- (or $II_1$-) factor which converge to an outer automorphism.
An answer involved a sequence of inner automorphisms satisfying, in particular, the hypotheses of my question, although the proof of convergence uses more information.  Perhaps other examples answering that question would be available if the answer to my question is yes.

REPLY [6 votes]: No. For an example, consider $M=L^\infty([0,1])$ with the Lebesgue measure, take $A$ to be the functions that are piecewise constant on dyadic intervals and $\alpha_n(f)=f\circ \phi_n^{-1}$ where $\phi_n(t)=k/2^n + (2^n t-k)^2/2^n$ if $t \in [k/2^n,(k+1)/2^n[$. In words, $\phi_n$ acts as some fixed transformation (here $t \mapsto t^2$, but it could be anything but the identity) but at a smaler and smaler scale.
Since $\phi_n$ preserves the intervals $[k/2^n,(k+1)/2^n[$, $\alpha_n(a)=a$ for all $a$ $\mathcal F_n$-measurable, where $\mathcal F_n = \sigma([k/2^n,(k+1)/2^n[, 0\leq k<2^n)$. So $\alpha_n$ satisfies your assumption with $\alpha=id$.
However, consider $1 \in L^1([0,1])=L^\infty([0,1])_*$. Then $(\alpha_n)_* 1= \phi_n'$, and $\|1 - \phi_n'\|_{L^1} = \|1 - \phi_0'\|_{L^1} = \int_0^1 |2u-1| dt=1/2$ does not converge to $0$.
But the answers to both your questions become yes if you assume that $M$ is finite and $\alpha_n$ preserves a trace (for example if $M$ is a $II_1$ factor). Or more generally if the $\alpha_n$'s preserve a normal faithful state $\phi$. Indeed, then on the one hand by Hahn-Banach $\{ \phi(a \cdot), a \in A\}$ forms a norm dense subspace of $M_*$, and on the other hand $(\alpha_n)_*(\phi(a \cdot)) = \phi(\alpha_n^{-1}(a) \cdot)$. Hence $(\alpha_n)_*$ converges pointwise in norm on a dense subspace of $M_*$, so it converges pointwise in norm on $M_*$.<|endoftext|>
TITLE: A question about the Mobius Function
QUESTION [5 upvotes]: I have been playing around with the Möbius Function and primorials and I am finding results that I am not yet able to understand which I suspect are very elementary.
Here's the current result which is I am working through.  Any help is greatly appreciated!
Let $p_k$ be any prime.  Let $x$ be any integer.
It seems based on my calculations that $\sum_{i | p_{k}\#} \lfloor{\frac{(x \% i) + (p_k \% i)}{i}}\rfloor\mu(i) \ge -1$ 
where % is the remainder so that $5 \% 3 = 2$ and $7 \% 3 = 1$
But if we let $x,y$ be any integer, we can find that there exists $x,y$ such that:
$\sum_{i | p_{k}\#} \lfloor{\frac{(x \% i) + (y \% i)}{i}}\rfloor\mu(i) < -1$
For example:  
If $x=13$, $y=23$, $p_k = 5$ , then the sum is $-2$
Is there a well known explanation for this?  Thanks.

REPLY [16 votes]: Note that
$$ \lfloor \frac{a \% p + b \% p}{p} \rfloor = \lfloor \frac{a+b}{p} \rfloor - \lfloor \frac{a}{p} \rfloor - \lfloor \frac{b}{p} \rfloor$$
so your expression can be written as $F_{p_k\#}(x+p_k) - F_{p_k\#}(x) - F_{p_k\#}(p_k)$, where
$$ F_n(x) := \sum_{i|n} \lfloor \frac{x}{i} \rfloor \mu(i).$$
Now, the expression $F_n$ can be simplified by elementary number theoretic computations.  Indeed, observe that
$$ \lfloor \frac{x}{i} \rfloor = \sum_{m \leq x: i|m} 1$$
and thus after interchanging summations,
$$ F_n(x) = \sum_{m \leq x} \sum_{i|n, i|m} \mu(i).$$
But by Mobius inversion, $\sum_{i|n, i|m} \mu(i)$ equals 1 when n,m are coprime and 0 otherwise.  Thus $F_n(x)$ is nothing more than the number of natural numbers $m$ less than $x$ that are coprime to $n$.  In particular, this gives
$$ F_{p_k\#}(x+p_k) \geq F_{p_k\#}(x)$$
and
$$ F_{p_k\#}(p_k) = 1$$
which explains your numerically observed phenomenon.
The quantities here are somewhat reminiscent of the quantity $\pi(x+y)-\pi(x)-\pi(y)$ that occurs in the second Hardy-Littlewood conjecture (which, by the way, is widely believed to be false).  Indeed, the quantity $F_{p_k\#}(x)$ (that is, the number of natural numbers up to x that have no prime factor less than or equal to $p_k$) is more commonly denoted $\pi(x,p_k)$ in the analytic number theory literature.
EDIT:  I also encountered some related expressions and inequalities in my paper http://arxiv.org/abs/0908.4323 (and with more advanced versions of these inequalities also in this paper of Granville and Soundararajan: http://arxiv.org/abs/math/0508361 ).<|endoftext|>
TITLE: Graph Minor check
QUESTION [6 upvotes]: Are there any good algebraic/algorithmic tools available to check if a given graph $H$ is a minor of $G$ from the adjacency matrix of $G$?

REPLY [5 votes]: A polynomial time algorithm for this is given in 
Robertson + Seymour, Graph minors. XIII. The disjoint paths problem, Journal of
Combinatorial Theory, Series B, 63 (1995), pp. 65–110.<|endoftext|>
TITLE: Divisibility and factorization in rings that are not integral domains
QUESTION [11 upvotes]: In my course notes for an undergraduate course "Algebra I", I wrote at the point when I'm introducing the notion of divisibility in rings (in a section on unique factorization):

We want to study factorization in arbitrary integral domains$^1$,...
$^1$ (footnote) The study of factorization in arbitrary commutative rings that are not integral domains, is in principle possible, but leads to rather exotic behavior and is therefore less interesting.

I have to confess that I didn't think too seriously about this sentence when I wrote it, but a student asked me whether that doesn't make it more interesting instead of less interesting.
I couldn't really give a sensible answer, hence my question:

Have there been serious attempts of a study of divisibility and/or factorization in arbitrary commutative rings with zero divisors, or is that simply something that doesn't really make sense?

Of course there are immediate problems even with the smallest examples such as $\mathbb{Z}/6$, e.g.
$$ 2 = 2 \cdot 4 = 2 \cdot 4 \cdot 4 = \dots $$
but I could imagine that there might be formal ways to overcome these obvious ambiguities when talking about uniqueness of factorization.

REPLY [16 votes]: The biggest problem complicating factorization when there are zero-divisors present in my experience stems from the fact that there are several ways to define "associate" that are no longer equivalent.
For instance, the reference given by Guntram provides a good introduction to the issues:
$a$ and $b$ are associates, written $a\sim b$ iff $(a)=(b)$
$a$ and $b$ are strong associates, written $a \approx b$ iff $a=\lambda b$ for some $\lambda \in U(R)$
$a$ and $b$ are very strong associates, written $a\cong b$ iff $(a)=(b)$ and if $a=rb$ for some $r\in R$, then $r$ is a unit.
Each gives rise to a different type of irreducible along the lines of if $a=a_1 \cdots a_n$ is a factorization then $a \sim a_i$, $a \approx a_i$, or $a \cong a_i$ for some $1 \leq i \leq n$, respectively $a$ is irreducible, strongly irreducible, or very strongly irreducible.
Examples are given to show how the different types of irreducibles are no longer equivalent.
This means when looking at something like how to define a Unique Factorization Ring (instead of domain) you need to pick what type of atoms you want your factorizations to be broken into and then when rearranging up to associate, what to pick for that.
Of course, this is only part of the story.  Unique Factorization Rings are pretty nice, there are weaker properties that will fail by the existence of zero-divisors, idempotent elements and nilpotent elements.  
Zero-divisors mean, you can factor zero: Say $ab=0$.  Well so does $0=ab=ababab=abxyzw$ or anything after it has already become zero. 
Or nilpotents: $x^n=0$ implies $0=(x^{n-1})^{i}$ for all $i \geq 2$.
So do you allow zero to be factored? Maybe just don't allow zero to appear as a factor? Even that is not enough to deal with it completely.
Suppose $e=e^2$, well then $e=e^i$ for all $i\geq 1$, so does that mean your ring is not even a Bounded Factorization Ring?  This is what happened in your example $\mathbb{Z}/6 \mathbb{Z}$, $4$ is idempotent.
One approach that seems particularly nice was introduced by Fletcher in two papers in Proc. Cambridge Philos. Soc. in 1969 and 1970 called Unique Factorization Rings and The structure of Unique Factorization Rings.  This was U-factorization where you have essential and inessential divisors to help deal with things like nilpotents and idempotents.  M. Axtell has a couple of nice papers on it too in 2002 and 2003
U-factorizations in Commutative Rings with zero-divisors and Properties of U-factorizations.
This makes something like $\mathbb{Z}/6 \mathbb{Z}$ a U-UFR, U-HFR, U-BFR, etc, but not a UFR, HFR, or a BFR.  
Interesting stuff in my opinion.
EDIT: I just wanted to add a few mathematical comments to respond to your mention of Galovich's paper.
This was something that I found very, very confusing when I first began studying factorization in rings with zero-divisors.  Nearly every author uses the term "irreducible" and "associate"; however, the problem is they mean different things in each case.  As far as I can tell D.D. Anderson and Valdez-Leon's method of 3 choices for associate, and several choices of irreducible (irreducible, strongly irreducible, m-irreducible, very strongly irreducible) is the most general and encompasses the others in the following way:
In Galovich's paper he chooses for associate the relation $\approx$.  That is $a$ and $b$ are "associate" in the Galovich sense if $a=\lambda b$ for some unit $\lambda$.  His choice for "irreducible", is what Anderson and Valdez-Leon termed "very strongly irreducible."
In Bouvier's papers we have "associate" is $\sim$ as in Anderson and Valdez-Leon $a$ and $b$ are associate if $(a)=(b)$.  However, the choice for "irreducible" is what is called m-irreducible, which comes from the fact that $a$ is m-irreducible if it is maximal among all principal ideals.  For non-zero elements, the strength of this type of irreducible falls between strong irreducible and very strong irreducible.  For zero, $0$ is m-irreducible if and only if $R$ is a field.  For zero to be irreducible, strongly irreducible, or very strongly irreducible is equivalent to $R$ being a domain.
There is yet another type of atomic referred to as i-atomic which is in terms of factorization into products of principal ideals, and I believe this is equivalent to the U-factorizations studied by Axtell, and Fletcher.
As you can see this is perhaps why most people skip the factorization with zero-divisors present.  Things get complicated rather quickly, but as the student alluded, perhaps this is what makes it more interesting in some ways.
This might be way more information than you want, but I just came across this site and it seems really exciting and useful and I saw your question was about what I study, and I may have gone a little overboard.<|endoftext|>
TITLE: Generating family for the Lebesgue $\sigma$-algebra
QUESTION [9 upvotes]: Let $X$ be a set, and $\cal F$ a family of subsets of $X$, let $\Sigma(\cal F)$ denote the smallest $\sigma$-algebra containing $\cal F$. We can also define $\Sigma(\cal F)$ internally using a transfinite induction up to $\omega_1$. We will say that $\cal F$ is a generating family of $\Sigma(\cal F)$.
For a $\sigma$-algebra $\cal B$, let $\frak g(\cal B)=\min\lbrace|\cal F|:\Sigma(\cal F)= B\rbrace$. Of course this may depend on the space $X$, so let us assume that always $X\in\cal B$, which then can be recovered as $X=\bigcup\cal B$. 
It is not hard to see that the $\cal B(\Bbb R)$, the Borel $\sigma$-algebra of $\mathbb R$, has a countable generating family, namely open intervals of the form $(p,q)$ where $p,q\in\mathbb Q$. That is ${\frak g(\cal B(\Bbb R))}=\aleph_0$.

Question I: (ZFC) Let $\cal L$ denote the Lebesgue measurable $\sigma$-algebra. Can we calculate the exact value of $\frak g(\cal L)$ (as a function of $\frak c$)? Is it $2^\frak c$?

Some observations include the following:

Since $|\Sigma(\cal F)|=|\cal F|^{\aleph_0}$ it is trivial that $\frak g(\cal L)>\frak c$, since ${\frak c}^{\aleph_0}=\frak c$.
By the above we have that under GCH (or at least $2^\frak c=c^+$) we can indeed prove the value is equal to $\frak c^+$.
Plain cardinality arguments need not work, suppose $\aleph_1=\frak c$ and $\aleph_{\omega+1}=2^\frak c$ (with SCH). Take $\aleph_\omega$ many subsets, they generate $\aleph_\omega^{\operatorname{cf}(\aleph_\omega)}=2^\frak c$ many subsets. So less than $2^\frak c$ many sets may generate something of the proper size.
Since the Cantor set is Lebesgue measurable of measure zero, its power set is embedded into $\cal L$, so $\frak g(\cal L)\geq\frak g(\cal P(\Bbb R))$, but on the other hand $\cal L$ is a sub-algebra of $\cal P(\Bbb R)$ so we have to have equality.


Question II: (ZFC) If this value is independent of ZFC, does the assertion $\aleph_2=\frak g(\cal L)$ have interesting consequences (e.g. CH, some combinatorial principle, etc.)? Does the assertion $\frak g(\cal L)=\kappa$ imply $2^\mu=2^\frak c$ for all $\frak c\leq\mu\leq\kappa$?

And a bonus question, can we prove anything on this cardinal in Solovay's model, or in $L(\Bbb R)$ when $L(\Bbb R)\models AD$? Is it even well-defined (i.e. is there a minimal size)? Do note that in both these settings Dependent Choice holds. In a broad ZF context if $\Bbb R$ is a countable union of countable sets then the Borel sets cover $\cal P(\Bbb R)$, but there is also a difficulty establishing measure theory properly.

REPLY [7 votes]: Concerning question II: We have $2^{\mathfrak c} \le \mathfrak g(\mathcal L) ^{\aleph_0}$.    
If CH fails, then $\aleph_2 ^{\aleph_0} = \mathfrak c$, so  $\mathfrak g(\mathcal L) = \aleph_2$ is impossible. 
Hence $\mathfrak g(\mathcal L)  = \aleph_2$ implies CH.  
In fact, it is equivalent to $2^{\mathfrak c} = \aleph_2$ or "GCH for $\aleph_0$ and $\aleph_1$".<|endoftext|>
TITLE: Manifold-Valued Sobolev Spaces
QUESTION [8 upvotes]: I have the following basic question about Sobolev-spaces which take their values in a 
Riemannian manifold $(M,g)$, i.e. 
functions $u:\Omega \to M$, $\Omega \subset \mathbb{R}^n$ bounded, such that for every chart $\varphi : M \to \mathbb{R}^d$, the composition $\varphi\circ u$ is in the usual $H^1$ space.
Question: 1. Is this notion well-defined (this is only easy to show for higher order smoothnes or $n =1$)?
          2. Does the manifold-valued Sobolev space admit the structure of a Hilbert manifold?
          3. Do there exist good references for manifold-valued Sobolev spaces (and their relations with Hilbert manifolds)?
          4. Is there a better way to define manifold-valued Sobolev spaces?
Many thanks for your answers!

REPLY [4 votes]: Not sure if anybody is still interested in this, but one can indeed define Sobolev spaces (of high enough regularity, so that due to the Sobolev embedding theorem the maps your looking at are at least continuous) of mappings from a (compact) manifold with values in a manifold using a chart approach.
While it is true (and has been mentioned above) that if you can check the $H^s$-property in some charts they will not hold in all charts (as I can always blow up things by composing with suitable diffeomorphisms). However, if one restricts to ''nice families of charts'' then the approach works and one obtains a Banach manifold (Hilbert if you are asking for derivatives in $L^2$). This has been worked out in the paper
Inci, Kappeler, Topalov: On the regularity of the composition ofdiffeomorphisms 
Though this was well known for a long time, e.g. see Ebin,Marsden: Groups of Diffeomorphisms and the Motion of an Incompressible Fluid, the new paper is the most complete account on all the technical details (also the point concerning the charts is at least not mentioned in the original Ebin/Marsden paper). Note however, that the results on the manifold structure require your $\Omega$ to be compact (so no chance to choose an open domain in $\mathbb{R}^N$. Otherwise, there is no way this could be a Hilbert or Banach space in any of the usual function space topologies.
I think that similar constructions, at least for the morphisms but not for the manifold structures, may be found in 
J. Eichhorn: Global Analysis on Open Manifolds 2007 
(where compactness of the source space is relaxed by requiring bounded geometries)<|endoftext|>
TITLE: Grouping vectors together
QUESTION [6 upvotes]: Given $n$ unit vectors in $\mathbb{R}^n$ s.t. $0 \leq u\cdot v<1$ for all pair of distinct vectors $u,v$. These vectors span a $d$-dimensional subspace s.t. $d< n$. We conjecture that it is possible to partition the $n$ vectors into $d$ groups such that all the vectors within the same group are pairwise non-orthogonal. It trivially holds when $d\in\{1,2,n-1,n-2\}$. However, we have not been able to show for general $d$. Does the conjecture hold for any $d< n$? If yes, how to prove it? Any thoughts/hints would be appreciated. Thanks in advance.

REPLY [5 votes]: Kahn--Kalai's counterexample to Borsuk's conjecture is a collection of vertices of the unit cube.
Imagine that this cube sits in a coordinate hyperplane $x_1=s$ of $\mathbb R^{n+1}$, so that the origin projects to the center of the cube.
Project this cube centrally to the unit sphere.
For a right choice of $s$, you get a counterexample to your statement.<|endoftext|>
TITLE: Tessellating  $\mathbb{R}^n$ by bricks.
QUESTION [10 upvotes]: Let us call the $\ell_1$-product of intervals $[0,k_1]\times...\times [0,k_n]$ a  brick of size  $k_1+...+k_n$. Consider a tessellation $T$ of $\mathbb{R^n}$ by (shifted) bricks so that every point belongs to at most $n+1$ bricks and the $\ell_1$-distance between any two disjoint bricks is at least 1 (thus every two bricks can share only the boundary points). Let $s(T)$ be the maximal size of a brick in that tessellation. Let $s(n)$ be the minimum of all $s(T)$. For example, $s(1)=1$ (we tessellate $\mathbb{R}$ by intervals $[i,i+1]$), $s(2)\le 3$ (we can tessellate ${\mathbb R}^2$ by $2\times 1$-bricks so that each point belongs to at most 3 bricks). It is easy to have a tessellation with $s(T) \sim 2^n$, so $s(n)$ is at most exponential.
 Question.  What is $s(n)$?  Does $s(n)$ grow exponentially with $n$?
 Update  It looks like the question is completely answered when we assume that all bricks are isometric by Will (upper bound) and Eric (lower bound). What if we tile by different bricks? How about arbitrary convex regions (same properties: every point belongs to at most $n+1$ regions, every two disjoint rejions are at distance at least 1, and, of course, different regions may share only boundary points)? Can we achive livear upper bound on the diameter of a tile? Can there be a constant upper bound?

REPLY [5 votes]: Since Ycor took the liberty to edit this 5 year old question and thus to bring it to the front page again, I'll throw in my one cent (not two yet; that may come later).
If you do a lattice tiling by translates, then, as Eric showed, we have no chance to get below the quadratic rate. However, not all tilings are like that. We cannot gain too much, but I think we still can gain somewhat (though I'm not ready to present an example yet).
First, the foolproof part: why not too much. By our conditions, any shift of the ball $B$ or radius $1/2$ in $\ell_1$ cannot intersect two disjoint bricks simultaneously, so we have all bricks intersecting $B$ also intersecting pairwise, which, by the special feature of bricks, means having a common point (bricks are like intervals on the line in this respect). Thus, the covering number for $(K_j+B)$ is at most $d+2$ but the volume of each brick increased $\sum_{\ell\ge 0}\sigma_\ell(\frac{1}{k_1},\dots,\frac 1{k_d})\frac 1{\ell!}$ times where $\sigma_\ell$ is the usual symmetric sum. By the AM-GM used twice, this is at least (here $q=d/s$)
$$
\sum_{\ell=0}^d \frac{d!}{(\ell!)^2{(d-\ell)!}}q^\ell\ge\max_{\ell\le d/2}\left(\frac{dq}{2\ell^2}\right)^\ell\approx{e^{c\sqrt{dq}}} 
$$
if $q\le d$ and, clearly, increasing beyond that point. Thus $dq\le C\log^2 (d+2)$ and $s\ge cd^2\log^{-2}(d+2)$ regardless of whether we use same size bricks or not and whether we follow any regular pattern or not, or even whether we do a tesselation or just covering, killing all Gerhard's hopes for a linear bound.
To be continued...<|endoftext|>
TITLE: Fullness of pullback functor in algebraic geometry
QUESTION [12 upvotes]: Given $f:X\to Y$ a morphism of schemes (or stacks if it's not harder), I am interested in a geometric reformulation of the condition that the functor $f^*:D^b(Coh(Y))\to D^b(Coh(X))$ is full.  I can only find full and faithful appearing together in the literature, and I need to extricate the two conditions.  Does anyone know a simple formulation, or a good reference?
Intuitively, asking for $f^*$ to be full seems alot like asking that anytime you have a sheaf $F$ on $Y$, and a section of it defined only on $X$ (i.e. a section of $f^*F$), it can be extended to a section on all of $Y$.  And so that would seem to indicate that the image of $f$ should have codimension-two complement.  However, that intuition only really applies to underived $f^*$, and maybe deriving $f^*$ eliminates the codimension 2 requirement?  Also, this intution is assuming that f is mono, so that $f^*$ is just restriction, which I don't think is true a priori.
As a side-note: I'd be interested in the same question (geometric characterization of fullness) for $f_*$ and $f^!$.`

REPLY [11 votes]: In case $X, Y$ are smooth,
$ f^* $
is full if and only if it is full and faithful. This is explained in the introduction of arXiv:1101.5931 (by Canonaco-Orlov-Stellari), which also studies when this implication holds more generally. Thus the pull-back is full and faithful if and only if $Rf_* \mathcal{O}_X = \mathcal{O}_Y$; for example, when $f$ is birational.
Here is a short explanation of their argument for the case of pull-back functors: $f^*$ is full and faithful if and only if $\mathrm{Ext}^i_{Y}(f^* O_{x_1}, f^* O_{x_2})$ is

$\mathbb C$ for $x_1 = x_2$, $i = 0$,
0 for $i \notin [0, \mathrm{dim} Y]$, and
0 for $x_1 \neq x_2$.

(This is due to Bondal-Orlov and Bridgeland.)
Since $f^* $ is full, the 2nd and 3rd condition are automatically satisfied. Thus it will be full and faithful if and only if $f^* O_x \neq 0$ for all $x \in X$. 
Pick $x \in X$ such that $f^* O_x$ is non-zero.
Note that $\mathrm{Hom}^{\bullet}_Y(f^* O_x, f^* O_x) = 
\mathrm{Hom}^{\bullet}_X(O_x, f_* f^* O_x)$ is a quotient of 
$\mathrm{Hom}^{\bullet}_X(O_x, O_x)$, and $f_* f^* O_x$ is supported at $x$. From this one can show that $O_x$ is a sheaf, and in fact isomorphic to $O_x$; all we need is that it's Chern character (in cohomology) is non-zero.
But the Chern character of $f_* f^* O_x$ is independent of $x$, so $f^* O_x \neq 0$ for all $x \in X$.<|endoftext|>
TITLE: Decomposition of Braid Groups
QUESTION [5 upvotes]: I've been trying to google this question, but to no avail. The question sounds elementary but I hope it's suitable for the experts at MO!
Let $B_n$ be the braid group on $n$-strands and $P_n$ be the pure braid group on $n$-strands. We have a well know decomposition of $P_n$ as $F_n \rtimes P_{n-1}$. In this light, my question is simply:
Are there known decompositions of $B_n$ in terms of direct/semidirect products of $B_{k}$ for $1\leq k < n$ and (possibly) other subgroups of $B_n$
I'd even be interested for special cases of $n$
Thanks!

REPLY [3 votes]: The answer is known for $B_3$ and $B_4$:
$B_3=\langle \sigma_1,\sigma_2\mid \sigma_1\sigma_2\sigma_1=\sigma_2\sigma_1\sigma_2\rangle$ decomposes as $B_3=F_2\rtimes \mathbb Z$, where $F_2=B_3'=\langle\sigma_2\sigma_1^{-1},\sigma_1\sigma_2\sigma_1^{-2}\rangle$, and $\mathbb Z=\langle \sigma_1\rangle$.
Similarly,
$B_4=\langle \sigma_1,\sigma_2,\sigma_3\mid \sigma_1\sigma_2\sigma_1=\sigma_2\sigma_1\sigma_2, \sigma_1\sigma_3=\sigma_3\sigma_1,\sigma_2\sigma_3\sigma_2=\sigma_3\sigma_2\sigma_3 \rangle$ decomposes as $B_4=F_2\rtimes B_3=F_2\rtimes (F_2\rtimes \mathbb Z)$, where  the outer $F_2$ is generated by $\sigma_3\sigma_1^{-1}$ and $\sigma_2\sigma_3\sigma_1^{-1}\sigma_2^{-1}$, and $B_3$ has generators as above.
See 
Leonid Bokut, Andrei Vesnin, Grobner–Shirshov bases for some braid groups,
Journal of Symbolic Computation 41 (2006) 357–371
http://math.nsc.ru/~vesnin/papers/bokut-vesnin2006.pdf
or Theorem 2.1 (presentation for $B_n'$ for all $n$) here:
E. A. Gorin, V. Ya. Lin, “Algebraic equations with continuous coefficients and some problems of the algebraic theory of braids”, Mat. Sb. (N.S.), 78(120):4 (1969), 579–610 
http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=sm&paperid=3572&option_lang=eng<|endoftext|>
TITLE: Intersection forms of 4-manifolds with boundary
QUESTION [5 upvotes]: Let $X$ be any simply connected smooth 4-manifold with a fixed Euler characteristic $e$, signature $\sigma$ and boundary $Y$. Assume that the determinant of the intersection form $Q_{X}$ is equal to a fixed integer $k \neq 0, \pm 1$. Is it true that the number of possible intersection forms for such $X$ is finite? Any reference would be appreciated.

REPLY [4 votes]: The classification of integral quadratic forms is discussed in Chapter 15 of Conway-Sloane. In particular, the discussion there implies that there are only finitely many integral quadratic forms of a given determinant and dimension. Section 11 in the chapter discusses methods for computing the number of such forms.<|endoftext|>
TITLE: Kazhdan-Luzstig Polynomials and Lower Intervals in the Bruhat Order
QUESTION [15 upvotes]: I have read in a number of places that the lower Bruhat interval $[e, w]$ is rank-symmetric if and only if the KL-polynomial $P_{e, w}(q) = 1$. All of the proofs I've come across use "rationally smooth Schubert varieties", which I don't really understand.
The KL polynomials can be defined purely in terms of the Iwahori-Hecke algebra of the Coxeter group, and satisfy a number of identities involving sums over Bruhat intervals. I would like to know then if there is a more direct way to prove that $[e, w]$ is rank symmetric iff $P_{e, w}(q) = 1$, using only the Hecke algebra (and Bruhat order).

REPLY [2 votes]: The answer to the question is "yes", allowing for a generous interpretation of "direct way".   This will follow from the recently posted work of Ben Elias and Geordie Williamson on non-negativity of coefficients of Kazhdan-Lusztig polynomials for an arbitrary Coxeter group here.
See the Update to my MO question here, 
which refers to the 1991 conference report by Jim Carrell (with Dale Peterson): in the first section, the equivalence you want is formulated for an arbitrary Coxeter group under the hypothesis that coefficients of relevant K-L polynomials are all non-negative.  (This may be one of the sources you are referring to.) 
At first I had overlooked this type of answer to my own question.  (I'm still looking for other consequences of the non-negativity theorem, of course, but this one is interesting.)   Note that for general Coxeter groups, one needs an approach which doesn't involve the geometry of Schubert varieties.   What Elias and Williamson seem to do is avoid all that algebraic geometry by providing a sophisticated substitute.<|endoftext|>
TITLE: Weak compactness and weak sequential compactness in Banach spaces
QUESTION [5 upvotes]: If $E$ is a Banach space, $A$ is a subset of $E$ and is compact with the weak topology $\sigma(E,E')$, that is the most coarse topology which make every $f\in E'$ continuous, is it true that $A$ is weakly sequentially compact?
As we all know, in $C_1$ spaces, compact concludes sequentially compact. So, we should show that $E$ is a $C_1$ spaces with the topology $\sigma(E,E')$.
Some known conclusions:$\forall x_0\in E$ , the basis of neighborhoods of $x_0$ constitutes of $V$ with the below form 
$$V=\lbrace x\in E;|(f_i,x-x_0)|\lt\epsilon, i\in I\rbrace$$
where I is a finite set.
Can we prove $E$ is a $C_1$ spaces?
Now, for every such $V$, we can find a $\delta>0$, such that $B(x_0,\delta)\subset V$.

REPLY [4 votes]: Yes, it's the  Eberlein–Šmulian theorem<|endoftext|>
TITLE: Riemannian metrics as sections of a vector bundle
QUESTION [10 upvotes]: Let $\pi \colon E \to M$ be a smooth vector bundle. A Riemannian metric on $E$ can be regarded as a global section of the vector bundle $(E\otimes E)^{\ast}$, or more specifically, the subbundle $(S^2E)^{\ast} \subset (E\otimes E)^{\ast}$. However, not every global section $s$ corresponds to a Riemannian metric, as there is no guarantee that $s(x) : S^2E_x \to \mathbb{R}$, when viewed as a symmetric map $E_x \times E_x \to \mathbb{R}$, is positive-definite. So my question is whether we can form a bundle such that positive-definiteness is automatic. More succinctly:

Is there a vector bundle $F$ associated to $E$ such that the global sections of $F$ are precisely the Riemannian metrics on $E$?

If so, can we characterise all Hermitian metrics on a complex vector bundle in a similar way?

REPLY [4 votes]: I agree with the answer given by Peter Dalakov, and with the comments made so far. On the other hand, there is a solution as close as possible to the requirement.
We consider a default Riemannian metric $g_0$ on the vector bundle $E$. This is the special metric mentioned by  Johannes Nordström in his comment. Any other metric can be obtained from this one, by applying a section from the bundle $GL(E)$ (having as fiber at $p\in M$ the general linear group of $E_p$). Any section of $GL(E)$, when applied to $g_0$, will give another Riemannian metric. Also, any section of $GL(E)$ can be written as $\exp(s)$, where $s$ is a section of the vector bundle $SL(E)$.
Hence, we can take as the desired vector bundle the bundle $SL(E)$. Any section $s$ of it will provide a transformation $\exp(s)$ which, when applied to the specially chosen metric $g_0$, gives another metric. Any metric can be obtained this way.

Note 1:
This works similarly for the Hermitian case.
Note 2:
If for the particular problem is necessary to work with generic linear combinations of the metrics themselves, then the solution proposed here will not be of use. Instead, one needs geometric methods which apply to metrics having variable signature, hence also being degenerate. Such methods were developed in arXiv:1105.0201, arXiv:1105.3404, and arXiv:1111.0646. One can define for example covariant derivatives for special tensor fields and differential forms, and also define the Riemann tensor $R_{abcd}$, although these constructions are apparently forbidden by $g_{ab}$ not being always invertible.<|endoftext|>
TITLE: A toolbox for algebraic topology
QUESTION [27 upvotes]: This question has a very general part and a rather concrete part.
General:
When one wants to prove something in algebraic topology (actually in all parts of mathematics) one obviously needs some good ideas, but first one has to have a good set of tools at hand. Introductory books in algebraic topology provide a number of such tools like long exact sequences to name just one. If one proceeds working in that field and reaches research level more and more tools are just treated as "common knowledge". They are used in papers according to the current situation and often left without quotation. 
Over the time one gathers plenty of those tools, but I for my part still take many of them as black boxes. When I use them I always have the feeling of walking on very thin ice. Most advanced books have some of those tools scattered in their body and finding a particular one is often harder than it should be. There they are used to build up a certain theory and often don't reveal themselves as useful tools with applications beyond the topic of the respective book.
Moreover it is one thing to find the reference for a statement one knows to be more or less true, but realising which tool one has to use when one isn't even aware of the precise statement is a different story.
So the first question:
Are there any good books which provide a box of tools used in modern algebraic topology? Maybe something like "AT for the working mathematician". 
They should come with a proof but not necessarily with applications (for the above reason).
Special:
The above is incredibly imprecise and there are so many ways to interpret the question. Hence one example of a statement I actually want to know about, which might also give a hint at what I am looking for.
Second question:
What is the precise statement/where can I find a proof
Given a commutative square of fibrations (cofibrations). Then the fibers (cofibers) in the horizontal direction are homotopy equivalent if and only if the fibers (cofibers) in the vertical direction are homotopy equivalent.
The square is then cartesian, cocartesian, bicartesian?
Edit 1: Now that I think about it it looks like the second question is just an application of the snake lemma. I have to work out the details. Still this statement may stand as an example of what I am looking for.
Edit 2: A book which seems to go in the direction of what I describe might be Goerss/Jardine: simplicial homotopy theory.

REPLY [2 votes]: This is very late, but regarding your second question, in the case of fibrations, see Proposition 7.6.1 in Selick's "Introduction to Homotopy Theory". That section contains a useful discussion of exact sequences in homotopy theory. For the analogous statement for cofibrations to hold, additional assumptions such as simply connectedness are needed.<|endoftext|>
TITLE: Special polynomials over finite fields
QUESTION [5 upvotes]: My field of research is coding theory and I am working on cyclic codes. During my research, I tackled an algebraic problem. After some simple definitions, I asked my question. I will appreciate any helpful answer and comment.
Let ‎$‎‎F_{2^{2m}}‎$ ‎denote ‎the ‎finite ‎field ‎of ‎‎$‎{2^{2m}}‎$‎‏ ‎elements‎, where ‎$‎m‎$‎‏ ‎is a‎ ‎positive ‎integer. ‎Let‎  ‎$‎‎‎‎F_{2^{2m}}[x]‎$ ‎denote ‎the polynomial ring in indeterminate ‎$‎x‎$ ‎with ‎coefficients ‎from‎ ‎$‎‎F_{2^{2m}}‎$.‎
‎
Suppose that ‎$‎f(x)‎$ ‎‎is a polynomial in $‎‎F_{2^{2m}}[x]‎$ ‎and‎ ‎‎$‎f(x)=f_0+f_1x+‎\cdots‎+f_kx^k‎$. We define the conjugate polynomial of ‎$‎f(x)‎$ ‎over‎ ‎$‎‎‎‎F_{2^{2m}}‎$ as follows:‎
‎$‎‎\overline{f(x)}‎={f_0}^{2^m}+f_1^{2^m}x+‎\cdots ‎+f_k^{2^m}x^k.$‎
In particular, if a polynomial is equal to its conjugate polynomial ‎over‎ ‎$‎‎F_{2^{2m}}‎$, then it is called self-conjugate polynomial.‎
‎
Let ‎$‎n‎$ ‎be ‎an ‎odd ‎positive ‎integer. Since ‎‎$‎gcd(n,‎2^{2m})=1‎$‎, the polynomial ‏‎$‎‎‎x^n+1‎$ ‎can ‎be ‎factorized ‎into ‎distinct ‎irre‎ducible polynomials over ‎$‎‎F_{2^{2m}}‎$.‎
‎‎
It ‎is ‎obvious ‎that ‎for ‎any ‎monic ‎irreducible ‎polynomial ‎dividing ‎‎$‎x^n+1‎$ ‎over‎‎ ‎$‎‎‎‎F_{2^{2m}}‎$, its  conjugate polynomial ‎is ‎also a ‎monic ‎irreducible ‎polynomial ‎dividing ‎‎$‎x^n+1‎$ ‎over‎‎ ‎$‎‎‎‎F_{2^{2m}}‎$. ‎‎
‎
For ‎example, let ‎‎$‎‎\omega‎‎$ be a‎ ‎primitive ‎element ‎of‎ $‎‎F_4‎$‎. ‎‎‎‎‎The factorization of ‎$‎‎x^5+1$ over $‎‎F_4‎$ is ‎
‎$‎‎x^5+1=(x+1)(x^2+‎\omega ‎x+1)(x^2+‎\omega‎^2x+1)‎‎$‎‎‎‎‎‎
It ‎is ‎obvious ‎that ‎$‎x+1‎$‎‏ ‎is a‎ ‎‎self-conjugate polynomial ‎and ‎‎$x^2+‎\omega ‎x+1‎$ ‎is ‎the‎ conjugate polynomial of ‎$x^2+‎\omega‎^2x+1‎$ ‎over $‎‎‎‎F_4‎$‎‎.‎
 ‎‎
For another ‎example,  let $‎‎\omega‎‎$ be a‎ ‎primitive ‎element ‎of‎ $‎‎F_{16}‎$‎. ‎The factorization of ‎$‎‎x^{11}+1$ over $‎‎‎‎F_{16}‎‎$ is‎
‎$‎‎x^{11}+1=(x+1)(x^5+‎\omega^5 ‎x^4+x^3+x^2+‎\omega‎^{10}x+1)(x^5+‎\omega^{10} ‎x^4+x^3+x^2+‎\omega‎^{5}x+1)‎‎$
It ‎is ‎obvious ‎that ‎$‎x+1‎$, ‎‎$x^5+‎\omega^5 ‎x^4+x^3+x^2+‎\omega‎^{10}x+1‎$ and ‎$x^5+‎\omega^{10} ‎x^4+x^3+x^2+‎\omega‎^{5}x+1‎$‎ ‎are‎ ‎‎self-conjugate polynomials over $‎‎‎‎F_{16}‎‎$.‎
Because of my researches I think that if $‎f(x)‎$ ‎‎is a self-conjugate monic ‎irreducible ‎polynomial ‎dividing ‎‎$‎x^n+1‎$ ‎over‎‎ ‎$‎‎F_{2^{2m}}‎‎$, then the degree of $‎f(x)‎$ is odd, ‎but I ‎‎could ‎not ‎prove ‎it.‎‎‎
‎
Is this conjecture true in general? If the answer is no, please give me an example?

REPLY [4 votes]: First, a polynomial $f$ is self-conjugate iff its coefficients belong to the fixed field of the Frobenius $x\mapsto x^{2^m}$, which is $F_{2^m}$.
Second, an $f\in F_{2^m}[x]$ which is irreducible over $F_{2^{2m}}$ must have an odd degree $d$: the splitting field of $f$ is the unique degree $d$ extension of $F_{2^m}$, which is $F_{2^{md}}$; if $d$ were even, then this field is a degree $d/2$ extension of $F_{2^{2m}}$, hence $f$ can’t be irreducible over the latter.
So, yes, any irreducible self-conjugate polynomial has an odd degree, $x^n+1$ has nothing to do with it.<|endoftext|>
TITLE: angle between subspaces
QUESTION [11 upvotes]: Let $E$ be a finite dimensional real inner product space.  I want to define the angle between two subspaces $E_1$ and $E_2$.  This has a fairly obvious meaning if $E_1$ is 1-diemsnional: Take the angle between any non-zero vector in $E_1$ and its orthogonal projection onto $E_2$.
There are a number of other cases that can be treated ad-hoc, if one is a hyperplane, or the dihedral angle between planes in $R^3$.
In general, it isn't quite clear what the right definition is.  I see two possibilities:

If $p=\dim E_1\le \dim E_2$, consider the two subspace $\lambda^p(E_1)$ and $\Lambda^p(E_2$   of $\Lambda^p(E)$ (which is also an inner product space, and proceed as above, since $\Lambda^p(E_1)$ is a line.
$Hom(E,E)$ is itself an inner product space with the inner product 
$$
\langle A,B\rangle=trace  A^\top B.
$$
Let $A_i$ be the orthogonal projection onto $E_i$ and take the angle between $A_1$ and $A_2$.

Are either of these definitions standard?  Are they equivalent (I think so)?  Is there another definition, perhaps more immediate?

REPLY [3 votes]: I know this is a few years late, but still it might be useful ...
A nice way to express Dan's answer is to let $A$ and $B$ be matrices whose columns form orthonormal bases for $E_1$ and $E_2$ respectively. Then the cosines of the principal angles are the singular values of the matrix $A^TB$ (or of $B^TA$ of course).<|endoftext|>
TITLE: Proving the existence of good covers
QUESTION [14 upvotes]: Usually one proves the existence of good covers in compact manifolds by Riemannian methods: we pick an arbitrary Riemannian metric, prove that geodesically convex neighborhoods exist, that they are closed under finite intersections, and diffeomorphic to balls; this is, for example, the argument that Bott and Tu sketch in their book.

Is there a non-Riemannian approach to this?

While this is not necessary for most things, it is a nice fact that good covers can be found which realize the covering dimension bound. 

Is there a differential-topological way to find them?

REPLY [3 votes]: I think you can obtain a good cover of $C^2$ manifold (compact or not) from the charts/atlas definition and a little bit of topology (locally finite atlas and a relatively compact "shrinking" of it).
The very simple idea (akin to that in Vitali's answer) is that under a $C^2$ diffeomorphism between open subsets of euclidean $n$-space, the (pre-)image of a sufficiently small ball centered at a point will be convex, as soon as the curvature of its boundary "dominates" the second derivative of the diffeomorphism (or its inverse).
In formulas, if $\phi$ is the diffeomorphism, this boils down to the fact that the $C^2$ function $x\mapsto |\phi(x)-\phi(x_0)|^2$ has a positive definite hessian at $x_0$, hence is convex near $x_0$. 
With a little more care, I think you can still conclude if $\phi$ is only $C^{1+Lip}$.<|endoftext|>
TITLE: Annihilators in Matrix Rings 
QUESTION [9 upvotes]: I think this is not a research question, but in stackExchange remained unanswered.
Let $R$  be a finite commutative ring. For $n>1$  consider the full matrix ring $M_n(R)$ . For a matrix $A\in M_n(R)$  is true that the cardinality of the left annihilator (in $M_n(R)$ ) of $A$  equals the cardinality of the right annhilator?

REPLY [13 votes]: Let $k$ be a finite field, and let $R := k[X,Y] / (X^2,XY,Y^2)$. Then $R = k \oplus kx \oplus ky$ is a finite ring of order $|k|^3$ with maximal ideal $\mathfrak{m} := kx \oplus ky$ of square zero.
Now let $A = \begin{pmatrix} 0 & x \newline 0 & y \end{pmatrix} \in M_2(R)$. Then
$\begin{pmatrix} 0 & x \newline 0 & y \end{pmatrix} \begin{pmatrix} a & b \newline c & d \end{pmatrix} = \begin{pmatrix} cx & dx \newline cy & dy\end{pmatrix}$ whereas $\begin{pmatrix} a & b \newline c & d \end{pmatrix}\begin{pmatrix} 0 & x \newline 0 & y \end{pmatrix}  = \begin{pmatrix} 0 & ax +by \newline 0 & cx + dy\end{pmatrix}$.
The equation $ax + by = 0$ with $a,b \in R$ implies $a, b \in \mathfrak{m}$ as can be seen by writing $a = \lambda + u, b = \mu + v$ for some $\lambda,\mu \in k$ and $u,v \in \mathfrak{m}$. 
Hence rann$(A) = \begin{pmatrix} R & R \newline \mathfrak{m} & \mathfrak{m}\end{pmatrix}$ has size $|k|^{10}$, whereas lann$(A) = \begin{pmatrix} \mathfrak{m} & \mathfrak{m} \newline \mathfrak{m} & \mathfrak{m}\end{pmatrix}$ has size $|k|^8$.<|endoftext|>
TITLE: The space of pure states
QUESTION [5 upvotes]: Let $A$ be a unital $C^{\ast}$-algebra and let $P(A)$ be a space of pure states on $A$ (a state $\omega$ is called pure it is an extreme point in space of states). 
a) Is the space $P(A)$ compact with respect to the weak-$\ast$ topology?
In the commutative case it appers that $P(A)$ is the space of nonzero linear characters and this space is compact, is this also true in the noncommutative case?
b) In concrete situation, $A=B(H)$, the space of bounded linear operators on the Hilbert space
is it true that $P(A)$ is extremely disconnected?
I would be grateful if anyone could help.

REPLY [5 votes]: a) The set of pure sates may not be *-closed in the dual space of $A$, see for example a paper of J.Glimm: http://www.ams.org/journals/tran/1960-095-02/S0002-9947-1960-0112057-5/ So of course it may be not compact.
Maybe it is reasonable to consider the spectrum of $A$ insead, i.e. the space $\hat A$ of all classes of irreducible representations of $A$. Look at Dixmier "C*-algebras and their representations", chapter 3. In particular, 3.1.8 tells you that if $A$ is unital then $\hat A$ and the space of primitive ideals are both compact but maybe not Hausdorff.
b) For $A=B(H)$, he set $P(A)$ is much bigger than simply the set of states of tpe $f_x(a)=(ax,x)$ for $x\in H$, $a\in A$. But the set of these $f_x$ is arc connected, because the unit sphere in $H$ is arc connected. So $P(A)$ cannot be totally disconnected.
[Hope I didn't write anything silly at this late hour]<|endoftext|>
TITLE: How fast can extreme eigenvalues of the average of random matrices converge to their expectation?
QUESTION [18 upvotes]: Suppose that $X_1,X_2,\ldots,X_m$ are independent $d\times d$ random matrices and let $\overline{X} := \frac{1}{m}\sum_{i=1}^m X_i$. One of the questions studied under the theory of random matrices is that if and at what rate $\lambda_{\max}\bigl(\overline{X}\bigr)$ (i.e., the largest eigenvalue of $\overline{X}$) converges to $\mathbb{E}\left[\lambda_{\max}\bigl(\overline{X}\bigr)\right]$ as $m\to +\infty$.
To obtain such guarantees, some assumptions are usually made about the $X_i$'s. For instance, it may be assumed that $X_i$'s are self-adjoint, have independent sub-gaussian or sub-exponential entries, etc. As far as I've read the literature, there's one particular assumption---namely $\lambda_{\max}\bigl(X_i\bigr)\leq R$ almost surely---that is made in almost every studied case. For instance, using this boundedness assumption, the matrix versions of the Chernoff's and Bernstein's inequalities as well as some other inequalities (see e.g., Vershynin'12, Tropp'11, MacKey et al'12, Hsu et al'12) can guarantee concentration inequalities of the form $$\Pr\left\{ \left|\lambda_{\max}\bigl(\overline{X}\bigr)-\mathbb{E}\left[\lambda_{\max}\bigl(\overline{X}\bigr)\right]\right|\geq \epsilon \right\} \leq d\exp\left(-\epsilon^2m/R\right).$$ They may vary in the exponent of $\epsilon$, the coefficient of the exponential, or the denominator inside the $\exp\left(\cdot\right)$, but they all decay at a rate in the order $m/R$.
Now, here's the issue. Suppose that $X_i=x_ix_i^\mathrm{T}$ where the $x_i$'s are iid vectors whose entries are iid Rademacher random variables. Then we get $\lambda_{\max}\left(X_i\right)=d$, so the best $R$ we can choose is $d$. Thus the probability bound expressed above decays at a rate not faster than $m/d$. The same rate holds if instead of Rademacher we use uniformly distributed zero-mean random variables. However, it is known, e.g. from Compressed Sensing literature, for these examples that the above probability bound for $\overline{X}$ decays at least at a rate of $m$ independent of $d$.
My question is why the bounds obtained in random matrix theory literature do not yield the sharpest convergence rates for these simple cases? Is this a by-product of the employed techniques or is there something fundamental that prevents it ?
Furthermore, is there any way to get rid of the assumption $\lambda_{\max}\left(X_i\right)\leq R$ a.s., altogether? This seems to be too restrictive, e.g., it doesn't even permit Gaussian random variables. It might be possible to resolve this issue using the relaxed versions of the concentration inequalities such as e.g. the extension of McDiarmid's inequality proposed by Kutin'02, but I think the first caveat still persists.

REPLY [6 votes]: A possible relevant post What kind of random matrices have rapidly decaying singular values?. In that post I discussed the distribution of maximal eigenvalue of a random matrix based on the result [Johnstone]. However that answer does not fully answer your question since there I emphasized the fact that the maximal eigenvalues of most random matrices in applications(Wishart and Uniform on a Grassmanian) will follow Tracy-Widom distribution asymptotically as long as the size of matrix grows with the $p$ in a reasonable manner. 
$\blacksquare$1. Why the concentration bound does not work for Rademacher example?
This post posed a slightly more complicated problem. $\bar X_{n}$ does not necessarily be Hermitian nor almost bounded in each of its entries. As you observed that assumptions are always put onto the random matrices in order  to make it into a covariance matrix from some random vector.([Mackey] and [Tropp] requires Hermitian in order to use Stein-pair argument;[Vershynin] mostly requires a lot of isotropic conditions to reduce his argument to a lower dimension where the tail is tractable.) Most of the existing results about the concentration bound focus on the case where the random matrix comes from a covariance matrix because of not only applied interest but also its positivity. See a discussion in [Guédon et al.].
This is the first obstacle why the results you pointed out might not be applicable to $\lambda_{max}(\bar{X})$. The example you constructed using Rademacher random variables has the problem that it is not positive definite, in fact its rank is always one regardless of the dimension of $x_i$ you choosen, therefore it will correspond to a degenerate distribution whose support will concentrate on a lower dimension space. 
I do not think your comparison is of any meaning by "applying" results for covariance matrices onto your example or Rademacher r.v.s. In fact in your example, suppose we have $x_i=(r_{i1},r_{i2},r_{i3})$ where $r_{ij}$ are iid Rademacher random variables, then it is readily seen that the eigenvalues of $X_i=x_i x_i^{T}$ will always be $(3,0,0)$, in that case it is meaningless to talk about converging rate. For $x_i=(r_{i1},r_{i2},\cdots, r_{in})$ it is not hard to verify it has eigenvalues of $x_i x_i^{T}$ will be $n$ with multiplicity one and $0$ with multiplicity $(n-1)$. So no matter how larger the Rademacher r.v. is, it is actually a one dimensional degenerated distribution and $X_i=x_i x_i^{T}$ cannot be a covariance matrix.
Therefore to your question 

My question is why the bounds obtained in random matrix theory
  literature do not yield the sharpest convergence rates for these
  simple cases? Is this a by-product of the employed techniques or is
  there something fundamental that prevents it ?

The semi-positivity of "simple cases" that prevents it from getting the sharpest rate control. As for the case where $x_i$ are uniformly distributed random variables with zero mean, we can also prove that its eigenvalues are $O(1)$(big O) with multiplicity 1 and $o(1)$(small O) with multiplicity $(n-1)$ so that positivity is preserved yet only one eigenvalue contribute to the concentration bound in $O(1)$ order.
So I think the failure of attaining sharpest bound is due to the nature of your example instead of the technique applied in the proof.
$\blacksquare$2. What is the concentration bound for Gaussian matrices?
A sharp concentration bound is available in [Bandeira&van Handel].
Back to the second part of your question, you are correct that $\lambda_{max}(X_i)≤R$ will not even allow Gaussian random matrices, but we usually study a more general statistical model called matrix model[Pastur et.al], which has densities in form of $$P_{n,\beta}(d_{\beta}M)=Z_{n,\beta}^{-1}exp\left\{ -\frac{1}{2}n\beta trV(M)\right\} d_{\beta}M$$[Pastur et.al] (4.1.1) where $V(\lambda)\geq(2+\epsilon)log(1+|\lambda|)$ is a continuous function ensuring the integrability of the matrix measure and where $\beta=1$ is equivalent to the real symmetric matrices and $\beta=2$ is equivalent to Hermitians. For the Gaussian model you pointed ou, we can realized it for  $V(\lambda)=\frac{\lambda^2}{2w}$ so if you admit that integrability is essential to such a matrix model then the concentration bound must be of the form shown and cannot even be discrete.[Pastur et.al] pp.105-106
In fact the eigenvalues for Gaussian matrices has following distribution density as derived in Pastur et.al$$\tilde{Q}_{n,\beta}^{-1}exp\left\{ -\frac{n\beta}{2}\sum_{l=1}^{n}V(\lambda_{l})\right\} \left|\prod_{1\leq j<k\leq n}\left(\lambda_{j}-\lambda_{k}\right)\right|^{\beta}\prod_{j=1}^{n}d\lambda_{j}\left[H_{\beta}\right]$$
So the bound must not even be sharper than this distribution law, which is providing an exponential bound. 
If as you said in your post, the $X_i$'s are self-adjoint, then in Gaussian case the $\bar{X}$ will again be distributed as a Gaussian distribution, and the distribution above will also apply to your $\bar{X}$. Generally, [Johnstone] gave a relatively good bound (under a mild constraint on dimension $p$). See the linked post (What kind of random matrices have rapidly decaying singular values?).
My point here is that we  should first figure out the distribution of $\bar{X}$ and then discuss what kind of concentration bound applies. It is also possible that although the assumptions of concentration bound are satisfied by $X_i$ but not satisfied by $\bar{X}$ 
$\blacksquare$3.Extension of [Kutin]'s weakly bounded difference result.
After a careful examination of [Kutin], I found your argument not very convincing. The weakly difference bounded conditions (Def. 2.3 in [Kutin]) also required that $|X(\omega)-X(\omega')|<b_k$ when $\omega$ and $\omega'$ differ only in the k-th coordinate besides the McDiarmid-type condition $$Pr_{(\omega,v)\in\Omega\times\Omega_{k}}(\left|X(\omega)-X(\omega')\right|<c_{k})\geq1-\delta$$, if you use $\Omega_k$ as the probability spaces associated with the Gaussian random variables $X_k$, although the $\delta$ probability part of the weakly bounded difference condition holds, the exact bounded part $|X(\omega)-X(\omega')|<b_k$ of the weakly bounded condition fails for two independent Gaussian random variable for any bounded sequence ${b_k}$.
So I do not think you can apply [Kutin]'s result directly. And by the way if you have a chance to read this, could you add the literature you mentioned in OP? Thanks
Reference
[Vershynin]http://www-personal.umich.edu/~Vershyninv/papers/non-asymptotic-rmt-plain.pdf
[Tropp]Tropp, Joel A. "User-friendly tail bounds for sums of random matrices." Foundations of computational mathematics 12.4 (2012): 389-434.
[Johnstone]Johnstone, Iain M. "Multivariate analysis and Jacobi ensembles: Largest eigenvalue, Tracy–Widom limits and rates of convergence." Annals of statistics 36.6 (2008): 2638.(Slides: http://www.stat.harvard.edu/NESS10/Johnstone-slides.pdf)
[Pastur et.al]Pastur, Leonid Andreevich, Mariya Shcherbina, and Mariya Shcherbina. Eigenvalue distribution of large random matrices. Vol. 171. Providence, RI: American Mathematical Society, 2011.
[Kutin]Kutin, Samuel. "Extensions to McDiarmid’s inequality when differences are bounded with high probability." Department Computer Science, University of Chicago, Chicago, IL. Technical report TR-2002-04 (2002).
[Bandeira&van Handel]Bandeira, Afonso S., and Ramon van Handel. "Sharp nonasymptotic bounds on the norm of random matrices with independent entries." The Annals of Probability 44.4 (2016): 2479-2506.
[Guédon et al.]Guédon, Olivier, et al. "The central limit theorem for linear eigenvalue statistics of the sum of independent random matrices of rank one." Spectral Theory and Differential Equations. Amer. Math. Soc. Transl. Ser 2.233 (2014): 145-164.<|endoftext|>
TITLE: Source for: Geodesics in CAT(0) spaces
QUESTION [9 upvotes]: I am seeking a good introductory reference that could lead to an understanding of
the properties of geodesics in
complete CAT(0) metric spaces.
I am especially interested in learning the differences between geodesics in these spaces
and those in an $n$-dimensional Euclidean space with its usual Euclidean metric,
which is of course CAT(0).
I am ultimately interested in simplicial, cubical, and polyhedral complexes,
but I am willing to start anywhere.  Thanks for educating me!
          
 (source)
          
[Image from "Shortest path problem in rectangular complexes of global nonpositive curvature" (Elsevier link)]

REPLY [2 votes]: A more recent reference for geodesics is the following preprint: 
Koyo Hayashi. A polynomial time algorithm to compute geodesics in CAT(0) cubical complexes. Oct 2017, arXiv:1710.09932.
Even though the title says "cubical complexes", from the abstract:

Our algorithm is applicable to any CAT(0) space in which geodesics between two close points can be computed, not limited to CAT(0) cubical complexes.<|endoftext|>
TITLE: Does list of distances define points uniquely?
QUESTION [7 upvotes]: There are N points on a plane. Is it feasible to reproduce their relative location
having only the list of distances. Assuming that translation, rotation and mirror are allowed 
in the result. The list contains only distances between every pair of points, but not which points these are.
For a simple triangle like A=(0,0) B=(1,0) C=(1,1) the distance are:
|AB| = 1
|AC| = sqrt(2)
|BC| = 1
The list would be: 1, 1, sqrt(2).

REPLY [4 votes]: Complementing the counterexample in Vladimir S Matveev's answer, the recent preprint Determining Generic Point Configurations From Unlabeled Path or Loop Lengths by Gkioulekas, Gortler, Theran, and Zickler, finds a positive answer under certain circumstances to a generalized question.
First, they treat all dimensions $d\geq2$, and second, the list of distances provided (the "measurement set") does not necessarily correspond to the distances between pairs of points, but may also be the lengths of arbitrary paths and loops on the complete graph on the point set.
The main result roughly states that provided: (1) the measurements come from a generic point set (thus ruling out examples like the one in Vladimir S Matveev's answer) and (2) the list of measurements "allows for trilateration" (meaning that there are enough measurements to inductively construct full-dimensional simplices), there is a unique point configuration (up to congruence) consistent with the measurements.<|endoftext|>
TITLE: Almost all loops have a trivial automorphism group; almost all groups have a non-trivial automorphism group.  What goes on in between?
QUESTION [6 upvotes]: NB: For this question, everything is finite.
Recently I've been fascinated by the following two observations:

Almost all loops have a trivial automorphism group (McKay & Wanless, 2005, in the context of Latin squares).
Almost all groups have a non-trivial automorphism group (in fact, all groups of order $n \geq 3$ admit a non-trivial automorphism).

However, the only difference between loops and groups is "cancellation" vs. "associativity".  (Actually, in a loop, an element's left inverse might not equal its right inverse,  but the equality of left and right inverses in a group does not need to be included as a group axiom.)

Question:  How can we strengthen the loop axioms so as to preserve the property that "almost all X have a trivial automorphism group"?
Question:  How can we weaken the group axioms so as to preserve the property that "almost all Y have a non-trivial automorphism group"?
Question:  Is the a set of axioms "between" the loop axioms and the group axioms for which "almost all Z have a trivial automorphism group" and "almost all Z have a non-trivial automorphism group" are both false?

REPLY [2 votes]: I suggest a general algebraic approach for this problem, since to me the major thing  that is changing is the idea of 'many".
Although you haven't said so, you seem to be asking about a demarcation in the lattice of varieties of algebra with one binary operation where one side has algebras with more than one automorphism versus those that have only one.  The latter are called rigid, and since no nontrivial variety has only rigid algebras (think of powers), you will need to have a good technical definition of 'many'. Perhaps pseudovarieties are the classes of interest.
I recommend looking at studies of rigid algebras.  If you are interested in equational formulations which promote rigidity, you could do worse than looking at versions of primal algebras, which are very rigid.  The varieties they generate are called arithmetical, and therein might lie part of the answer you seek.
Gerhard "Ask Me About System Design" Paseman, 2012.07.14<|endoftext|>
TITLE: Bruhat order and the Robinson-Schensted correspondence
QUESTION [20 upvotes]: The Robinson-Schensted correspondence is a bijection between elements of the symmetric group $S_n$ and pairs of standard tableaux of the same shape.  The symmetric group is partially ordered by the Bruhat order, so this bijection induces a partial ordering "$\leq$"on the set of pairs of standard tableaux.  Is there a natural "tableaux-theoretic" description of $\leq$?  That is, can we give a necessary and sufficient condition for $(S,T)\leq (S',T')$ intrinsic to the set of pairs of tableaux?
(this was posted at mse a few weeks ago to no response)

Added later:
Each involution in $S_n$ is mapped to a pair $(P,P)$, so Bruhat order on involutions induces an order on standard tableaux.  Characterising this order is listed as an `unsolved problem' in Björner and Brenti's Combinatorics of Coxeter Groups.

REPLY [9 votes]: The question is natural but looks difficult to approach strictly within the combinatorial definitions.   Maybe it's helpful here to suggest a broader geometric framework, which applies more generally to reductive algebraic groups but involves here a general linear group having the symmetric group $S_n$ as Weyl group.    Recall that the "Bruhat order" was defined by Chevalley in the more general setting as the order on Bruhat cells in the flag variety given by inclusion of one cell in the closure of another.   These cells are naturally labelled by elements of $S_n$.   
On the other hand, the associated geometry of the flag variety and its cotangent bundle (work of Springer, Steinberg, Spaltenstein) leads to Springer fibers with equidimensional components.  For the general linear case, these Springer fibers in the flag variety correspond to partitions (Jordan forms of unipotent elements).   In this set-up, pairs of typical flags in components of a Springer fiber relate naturally to pairs of standard tableaux of the shape determined by an element of $S_n$, correlating with the Robinson-Schensted correspondence.   This is brought out by Steinberg in a short paper here.   The Bruhat order is lurking in all of this machinery, but the connection with Robinson-Schensted isn't made explicit enough.    
A further related 2005 paper by Anna Melnikov appears here, treating the combinatorics of orbital varieties (but only in a very special case).<|endoftext|>
TITLE: Orthogonal Groups over finite fields
QUESTION [13 upvotes]: Hello
Let $\mathbb{F}_p$ be the finite field with $p$ elements. One can show that over finite fields, there are just two non-degenerate quadratic forms. 
So here I want to pick 
any non-degenerate symmetric matrix $B$, and then look at the special orthogonal group defined by
$$
SO_{n}(\mathbb{F}_p,B):=\{ A\in SL_n(\mathbb{F}_p): ABA^T=B \}
$$
Is it true that the commutator subgroup of $SO_{n}$ is the whole group? In the other words I would like to know if $SO_{n}$ over finite fields is a perfect group.

REPLY [4 votes]: I think it's worth adding that there is a very detailed analysis of the orthogonal groups over arbitrary fields (not just finite fields, and including characteristic 2) in Dieudonné's "La Géométrie des Groupes Classiques". (The first edition of this book already goes back to 1954, but I'm referring to the 3rd edition from 1971.)
In particular, Chapter II, Section 8, seems to answer the question you're interested in when the characteristic is not 2, and Section 10 (in particular item (8) on p.67) answers your question when the characteristic is 2.<|endoftext|>
TITLE: What is the status of irrational numbers within finitism/ultrafinitism?
QUESTION [10 upvotes]: According to constructivism, "it is necessary to find (or "construct") a mathematical object to prove that it exists". There are several formulas to calculate $\pi$, such as: 
    
 (image source)
so I take it $\pi$ exists according to constructivism.
According to finitism, which is a form of constructivism, "a mathematical object does not exist unless it can be constructed from natural numbers in a finite number of steps."
Where does that leave irrational numbers, such as $\pi$? Do they simply not exist according to finitism? How does one reason about the ratio between a circle's circumference and its diameter, if one is working within a finitistic/ultrafinitistic framework?

REPLY [5 votes]: Even while considering the universe to be finite, one can do mathematics symbolically as a game with a system of rules. If the game doesn't have enough pieces
we just add new pieces, with new properties or allowed moves as required. 
All that matters is that the enlarged system is compatible with the old system; that 
the smaller game is a subgame of the large one; that the smaller system can be embedded in the larger
one. 
When does something exist ? Well if there isn't something from amongst the objects under consideration that has the properties we want then we just create new symbols and define how they relate to the old ones.
If we were a pythagorean and the only numbers that exist are rational numbers, then we wouldn't call $\sqrt 2$ a number, but if we were also finitist symbolicists then we could embed any collection of numbers into a collection that contains not only numbers but also "splodges" which is what we're going to call $\sqrt 2$. It's important to always keep in mind that $\sqrt 2$ isn't a number - it's a splodge. In this new system of arithmetic we've invented we can add numbers to splodges to get new splodges like $1+\sqrt 2$.
What a fun game. Let's add some more splodges. We're bored with algebraic splodges so
let's add some non-algebraic splodges like the one in your question. 
Of course that expression is a bit cumbersome so we'll give it the shorthand symbol $\pi$ instead.
Given a splodge $x$, it would make calculus easier if there were a splodge $x+o$ that was nearer to $x$ than any other splodge, however that isn't possible so we embed the splodges in a larger system called the hypersplodges that contains not only splodges
but also vapors, and contains not only the concept nearer but also the concept "nearer". Vapors like $x+o$ are "nearer" to $x$ than any splodge could ever be, and when you're finished using them they evaporate leaving just a splodge.
We want a splodge that satisfies $x^2+1=0$ however there isn't one, so we embed the splodges in a larger system called weirdums in which we've added a piece called $i$ with the rule that $i^2=-1$, and under the new system we can "add" splodges to weirdums to get new weirdums like $1+i$.
In solving differential equations we'd like a function which is zero everywhere except at a single point but which has a non-zero area under the curve. There is no function that behaves like this so we'll go to a larger system that contains not only functions but also spikes which do have the desired property because the larger system contains a rule about spikes which says they do. Conveniently certain calculations involving spikes cancel out leaving just functions.
A finitist or ultrafinitist shouldn't recognise the concept of infinite sets therefore the only sets are finite-sets and since all sets are finite the adjective finite is superfluous therefore from this point onwards we just use the term "set". Some people want to consider sets that contain things they haven't put in there themselves - which of course can't be done because a set only contains the items we've put there. So we embed the system of sets in a larger system that includes not only sets but also dafties. In this daft system the rules are that a daftie can have an "affinity" for things whether those things have been previously mentioned or not. Dafties have an affinity for things in the same way that sets contain things. A compatible embedding of a system of sets in the daft system means that a set has an affinity for the items it contains when the set is considered as part of the daft system, therefore by daft reasoning one can say things not only about dafties but also about sets. To each set one can attach a number. You can't do this with dafties so we embed the numbers in a larger system containing sinners and attach a sinner to each daftie. Sometimes there is a need for something that looks like a daftie but has no sinner - such things are called messes. A mess can have an affinity for collections of dafties that no daftie could have an affinity for. This could go on, but you need gobbledegook theory. The set system has zero gobbledegook. The daft system is level-1 gobbledegook. The messy system is level-2 gobbledegook. Finitists like to maintain a zero level of gobbledegook. Analysts are usually happy with one-level of gobbledegook and category-theorists are comfortable with any amount of gobbledegook.
There are two ways to compatibally extend a system: 
1) a conservative extension adds new items but doesn't say anything
new about the old items that couldn't be said before; 
2) a progressive extension does say new things about the old items but only about things that were previously undecidable
P.S. We can combine the vapors, splodges and linedups in a system called the messysplodges but they haven't been studied much because they're a bit messy.<|endoftext|>
TITLE: Sine and Archimedes' derivation of the area of the circle
QUESTION [7 upvotes]: The elementary "opposite over hypotenuse" definition of the sine function defines the sine of an angle, not a real number.  As discussed in the article "A Circular Argument" [Fred Richman, The College Mathematics Journal Vol. 24, No. 2 (Mar., 1993), pp. 160-162.  Free version here.  Thanks to Aaron Meyerowitz's answer to question 72792 for the reference.], angles might be measured either by the area of a sector of unit radius having the angle or by the arc length of such a sector.  If the former convention is adopted then it can be proven using a completely unexceptionable Euclidean argument that $\lim_{x\to 0} \sin(x)/x = 1$.  
Also, whichever convention is adopted (or so it seems to me), using completely unexceptionable Euclidean arguments, it is possible to prove the angle addition formulas for sine and cosine.  Using these two ideas, it is straightforward to find the derivatives of sine and cosine, and from there one can derive an algorithm for computing digits of sine and cosine (and for computing $\pi$) using the relatively sophisticated mean-value version of Taylor's theorem.
The equivalence of the two definitions of sine (or of angle measurement) apparently depends on something like Archimedes' postulate: "If two plane curves C and D with the same endpoints are concave in the same direction, and C is included between D and the straight line joining the endpoints, then the length of C is less than the length D."  (Again, thanks to Aaron Meyerowitz.)  Of course, it is just this postulate that Archimedes needed to prove that the area of a circle is equal to the area of a triangle with base the circumference of the circle and height the radius.  And something like it is surely necessary to derive any algorithm for computing digits of $\pi$.  (Except, and this confuses me a bit, it seems that if we used the area definition of angle, we could derive an algorithm for computing sine without depending on this postulate, and from there we could get an algorithm for computing digits of $\pi$ since $\sin(\pi)=0$.)
I am looking in general for elucidation of the conceptual connections between the ideas I have so far discussed and of their background.  But here are two more specific questions.

First, in what sense is a postulate like Archimedes' needed in the foundations of geometry?  (I wonder, in particular, if in a purely formal development we might get by without it, but we would somehow be left without assurance that what we had axiomatized was really geometry.)  Also, are more intuitive alternatives to Archimedes' postulate?
Second, what is really needed to get an algorithm for computing digits of sine?  Does it really require such complicated technology as Taylor series?  It seems like if one uses the area definition of angle, one might be able to give an algorithm using unexceptionable Euclidean techniques and without so much as invoking the notion of limit.

REPLY [5 votes]: To answer your question 1: it really depends on which set of axioms you consider 'canonical'. If you look at it from the viewpoint of the first-year courses in calculus and vector analysis and the like, you build up the Euclidean plane as $\mathbb{R}^2$, and you define curve length as some kind of integral, prove its independence of parametrization, etc. Within this framework, one will never need to use an axiom such as that of Archimedes, because you simply "define" curve length, and the question whether this definition is actually reasonable, i.e. corresponds to all manner of intuitive ideas about arc-lengths, is usually left to the imagination of the student.
Likewise, in the framework of Euclidean geometry, you need to have some kind of definition of arc-length. That this is put in the form of an axiom, and not simply given as a definition, signals the fact that Archimedes intends to justify his definition. He apparently finds it reasonable, and perhaps somehow consistent with everyday geometric intuition, that, when one has two curves with identical endpoints, which are both convex "in the same direction", and the first of which lies wholly or partly outside the second, and nowhere inside it, the first curve has greater length than the second. Perhaps he thinks of the first curve as a rubber band, that has been stretched out starting from the same position as the second curve. (I don't know whether this makes the axiom more intuitive, but I've tried to imagine his reasoning and this is what it feels like for me.)
Anyway, the point that I am trying to get across here is, you need to do something. Arc-length is not a given once you just start from the usual Euclidean axioms. Think about it: straight line segments can be compared, owing to the isotropy of Euclidean space (which incidentally is an unstated axiom in Euclid's Elements), which is how we can talk about the length of a line segment -- or at least about length relative to some sort of reference line segment. On the other hand, how would you compare a curved line with a straight line? You need to come up with a definition, at least, but an axiom, which is supposed to appeal to intuitive geometrical reasoning, is even better.
[By the way, this story does not hold for surface areas, since areas bounded by straight lines and those bounded by curved lines can be compared! Here Euclid can appeal to his "common notion" that the whole is greater than the part -- which by the way shows he clearly hadn't heard of open intervals, but that is another matter. For example, a circle is smaller than a circumscribed polygon (this is how Euclid phrases it, but he's referring to their areas), and greater than an inscribed polygon. Using this kind of reasoning, he shows that circles are to each other as the squares on their diameters, which shows that, in this case at least, arguments mixing lengths and areas are more straightforward than ones mixing lengths of straight and curved lines.]<|endoftext|>
TITLE: When is the infimum of an arbitrary family of measurable functions also measurable?
QUESTION [11 upvotes]: Let $(X,\Sigma,\mu)$ be a measure space and consider a family of $\mu$-measurable functions $f_i:X \to \mathbb{R}$ for $i$ lying in some index set $I$. Define $$f(x) = \inf_{i \in I} f_i(x)$$
I think every good analysis book mentions or proves that if $I$ is countable, then $f$ is also $\mu$-measurable. What is not so clear is:

If $I$ has cardinality of the continuum, is $f$ still $\mu$-measurable?

Since I strongly suspect that the answer is "no" although a counterexample is not immediately coming to mind, here is the real question:

What are minimal conditions on $f_i$ that will make $f$ $\mu$-measurable even when $I$ is uncountably large?

By minimal conditions I am hoping for some weak properties, such as those that hold $\mu$-almost everywhere rather than those that require something strong from the global structure of $I$, such as a (partial or total) ordering.
If you know of a book or paper that deals with this, please let me know. I understand there is a good chance this type of thing is not considered research material by folks here; in this case I will delete the question.

REPLY [2 votes]: The infimum of any family of measurable functions is measurable if we interpret the infimum as the lattice infimum. For details see: https://mathoverflow.net/a/316658/121665<|endoftext|>
TITLE: Combinatorics of folding digit strings
QUESTION [6 upvotes]: Say that a string of $n$ digits, each from $\lbrace 0,1,2,\ldots,b-1 \rbrace$,
is foldable if, were each
digit on its own stamp in a sequence of connected stamps,
one could fold the stamps so that like digits are on top
of one another, forming a "stack" of $0$s, adjacent to a stack of $1$s,
adjacent to a stack of $2$s, and so on.
For example, for $b=2$ binary strings, the stamps should
form two piles: $0$s and $1$s, like this:
          
For binary strings, it is clear that a foldable string must
have runs of an even number of $0$s and $1$s everywhere except
at the two ends of the string; these end runs can be of even or odd length.
It is not too difficult to work out that of the $2^n$ binary
strings of length $n$, the number foldable is
$$f(n) = 3 \cdot 2^{n/2} -2 \;\;, n \; \mathrm{even}$$
$$f(n) = 2 \cdot 2^{\lceil n/2 \rceil} - 2 \;\;, n \; \mathrm{odd}$$
For example, for $n=10$, $3 \cdot 2^5 -2 = 94$ of the $1024$ strings are
foldable; for $n=11$, $126$ are foldable.
I am having difficulty generalizing the count
to strings of digits from larger sets $\lbrace 0,1,2,\ldots,b-1 \rbrace$,
$b>2$.  For $b=3$, there should be three consecutive "stacks", of $0$s, $1$s, and $2$s.
Although an exact formula would be nice, I am particularly interested in whether the exponential
growth remains $2^{n/2}$.
Perhaps someone has seen this before, perhaps in another guise?
If so, I'd appreciate a reference.  Thanks!
Addendum. In case anyone wants to try to understand the exact count for $b=3$, here are
some tentative computational enumerations, 
with $n$ the number of digits, $f$ the number of foldable strings,
and $u$ the number of unfoldable strings:
$$n=1 \;,\; f=3 \;,\; u=0 \;,\; f+u = 3$$
$$n=2 \;,\; f=7 \;,\; u=2 \;,\; f+u = 9$$
$$n=3 \;,\; f=13 \;,\; u=14 \;,\; f+u = 27$$
$$n=4 \;,\; f=23 \;,\; u=58 \;,\; f+u = 81$$
$$n=5 \;,\; f=39 \;,\; u=204 \;,\; f+u = 243$$
$$n=6 \;,\; f=65 \;,\; u=664 \;,\; f+u = 729$$
$$n=7 \;,\; f=107 \;,\; u=2080 \;,\; f+u = 2187$$
OEIS identifies the $f$-sequence as A154691: 
Expansion of 
$$ \frac {1+x+x^2}{(1-x-x^2)(1-x)} .$$

REPLY [3 votes]: Edit: On second thought, this is pretty much exactly Anthony's answer, only slightly more explicit. Didn't see this when I wrote this up, sorry.
There's an approach to this problem using elementary linear algebra, which gives you an explicit formula for $f(n)$ each $b$ (however no nice formula for an all $b$ at once), and an algebraic integer $a$ such that $f(n)=c\cdot a^n + \textrm{lower order terms}$. For instance, for $b=3$ you obtain $a = \frac{1+\sqrt{5}}{2}$ and for $b=4$ you obtain $a = \sqrt{3}$. The actual formulas get really messy, for instance for $b=3$ you get (thanks to Maple)
$$
f(n) = 8/5\cdot {\frac {-120-60\,\sqrt {5}+10\, \left( -1 \right) ^{1+n} \left( 
\sqrt {5}+1 \right) ^{-n}{2}^{n}+11\, \left( \sqrt {5}+2 \right) ^{n-1
} \left( 3+\sqrt {5} \right) ^{2-n}\sqrt {5}{2}^{n}+5\, \left( -1
 \right) ^{n} \left( \sqrt {5}-1 \right) ^{-n+1} \left( 3+\sqrt {5}
 \right) ^{-n+1}{4}^{n}+ \left( -1 \right) ^{1+n}{2}^{1+n}\sqrt {5}
 \left( \sqrt {5}+1 \right) ^{-n}+25\, \left( \sqrt {5}+2 \right) ^{n-
1} \left( 3+\sqrt {5} \right) ^{2-n}{2}^{n}}{ \left( \sqrt {5}-1
 \right) ^{2} \left( 3+\sqrt {5} \right) ^{2} \left( \sqrt {5}+1
 \right) }}
$$
the first ten values of which are $3, 7, 13, 23, 39, 65, 107, 175, 285, 463, \ldots$
So, the following is a description of how to obtain the growth behavior and explicit formulas (some details are missing since this got a bit longer than I expected):
First notice that the diagrams that you draw in your question are uniquely determined by the foldable sequence in all except in $b$ cases, namely the sequences $[i,i,\ldots,i]$ for which there are two possible diagrams. To see this just notice that whenever a sequence contains a subsequence [i-1,i] or $[i,i-1]$ this subsequence is represented in the diagram by a blue line crossing the $i$-th vertical red line in your digram, and the rest of the diagram is hence determined (See it this way: whenever you "add a digit" to your sequence this corresponds to a unique extension of the blue line, the only ambiguous part was the placement of the first line segment). In conclusion we focus our interest on the number $g(n)$ of possible "folding diagrams", and we have 
$$
   g(n) = f(n) + b
$$
(in fact this is where the summand $-2$ comes from in your formula; it will turn out that $g(n)$ is essentially a linear combination of exponential functions)
Now we decompose 
$$
   g(n) = g_{0,0}(n) + g_{1,0}(n)+ g_{1,1}(n)+g_{2,1}(n)+g_{2,2}(n) + \ldots + g_{b-1,b-2}(n) +  g_{b-1,b-1}(n) + g_{b,b-1}(n)
$$
where $g_{i,j}(n)$ denotes the number of all folding diagrams corresponding to a sequence which starts with the digit $j \in \{ 0,\ldots,b-1 \}$ and is rooted at the $i$-th vertical red line in your diagram (counting from the left starting at $0$; so, $j\in \{0,\ldots, b\}$ and for a given $i$ the value of $j$ has to be $i$ or $i-1$). 
Now the $g_{i,j}$ satisfy a linear recursive formula (this is obvious: once we specify the first line segment of a folding diagram we can concatenate it with any folding diagram of length $n-1$ which starts at the right vertical red line to obtain a folding diagram of length $n$):
$$
  g_{i,i}(n) = g_{i+1,i}(n-1) + g_{i+1,i+1}(n-1) \quad \textrm{ for } i\in\{0,\ldots,b-2\}
$$
$$
  g_{i,i-1}(n) = g_{i-1,i-1}(n-1) + g_{i-1,i-2}(n-1) \quad \textrm{ for } i\in\{2,\ldots,b\}
$$
$$
   g_{1,0}(n) = g_{0,0}(n-1) \quad \textrm{and} \quad g_{b-1,b-1}(n) = g_{b,b-1}(n-1)
$$
Note that by definition  $g_{i,j}(1)=1$ for all $g_{i,j}$'s which occur in these recursion relations.
So we can put this recursion relation into a matrix $A_b$, and we will get a formula
$$
  \left(\begin{array}{c}g_{0,0}(n)\newline g_{1,0}(n) \newline g_{1,1}(n)\newline \vdots \newline g_{b,b-1}(n) \end{array}\right) = A_b \cdot  \left(\begin{array}{c}g_{0,0}(n-1)\newline g_{1,0}(n-1) \newline g_{1,1}(n-1)\newline \vdots \newline g_{b,b-1}(n-1) \end{array}\right) = A_b^{n-1}\cdot 
\left(\begin{array}{c} 1\newline 1 \newline 1 \newline \vdots \newline 1\end{array}\right)
$$
Since $g(n)$ was just the sum of all $g_{i,j}(n)$ and $f(n)=g(n)-2$ we get
$$
f(n) = [1,\ldots,1]\cdot A_b^{n-1} \cdot [1,\ldots,1]^\top-b
$$
It should be clear now that by transforming $A_b$ into Jordan normal form we can get an explicit formula for each $b$ (as a linear combination of $\leq n$-th powers of the eigenvalues of $A_b$).
So when it comes to the growth behavior, we're interested in the eigenvalue of $A_b$ of biggest absolute value. 
Now the matrices $A_b$ have a very simple form:
$$
   A_3 =  \left[ \begin {array}{cccccc} 0&1&1&0&0&0\newline 1&0&0&0
&0&0\newline 0&0&0&1&1&0\newline 0&1&1&0&0&0
\newline 0&0&0&0&0&1\newline 0&0&0&1&1&0
\end {array} \right] \quad A_4 =  \left[ \begin {array}{cccccccc} 0&1&1&0&0&0&0&0\newline 1
&0&0&0&0&0&0&0\newline 0&0&0&1&1&0&0&0
\newline 0&1&1&0&0&0&0&0\newline 0&0&0&0&0&1&1
&0\newline 0&0&0&1&1&0&0&0\newline 0&0&0&0&0&0
&0&1\newline 0&0&0&0&0&1&1&0\end {array} \right] 
$$ 
Notice how $A_{b}$ sits as a submatrix in the lower right corner of $A_{b+1}$, and this should be suffice  to prove the following recurrence relation for the characteristic polynomials $\chi_b(x)$ of $A_b$ (to be honest I just checked it up to $b=10$ but it should be doable):
$$
  \chi_{b+1}(x) = x^2\cdot(\chi_b(x) - \chi_{b-1}(x))
$$
Together with the information that $\chi_2(x)=x^2-2$ and $\chi_3(x)=x^4-2x^2$ this gives you all characteristic polynomials for all the $A_b$ fairly easily.
So we get
$$
   f(n) = c \cdot a^n + \textrm{lower order terms}
$$
where $a$ is the largest absolute value of a root of $\chi_b(x)$.
A few technicalities are required to make sure that $c\neq 0$: one should only consider roots of $\chi_b(x)$ which aren't multiples of $\sqrt{-1}$ (these will cancel each other out) and secondly, one needs to make sure that the vector $[1,\ldots,1]^\top$ doesn't lie in an $A_b$-stable proper subspace of $\mathbb R^{2b}$ (this can probably be shown, but it's too messy for me right now and I already spent more time on this answer than I should've).<|endoftext|>
TITLE: A mysterious Heisenberg algebra identity from Sylvester, 1867
QUESTION [29 upvotes]: I am trying to understand two papers by James Joseph Sylvester:
P92: "Note on the properties of the test operators which occur in the calculus of invariants, their derivatives, analogues, and laws of combination; with an incidental application to the development in a Maclaurinian series of any power of the logarithm of an augmented variable."
and
P95: "On the multiplication of partial differential operators."
[The numbering is from volume 2 of Sylvester's Collected Works. These, incidentally, are spread over four volumes: volume 1, volume 2, volume 3, volume 4 (first two courtesy of anonymous book scanners), with some duplicates on the Internet Archive. All of them are out of copyright.]
On the first page of P95 (aka page 11 of the linked two-paper PDF), Sylvester states that
"If $\phi$ be any such function [i. e., a polynomial or power series in infinitely many commuting variables $x$, $y$, $z$, ..., $\delta_x$, $\delta_y$, $\delta_z$, ... (here, $\delta_x$, $\delta_y$, $\delta_z$, ... are just symbols, not differential operators!) which is multilinear with respect to $\left(\delta_x,\delta_y,\delta_z,...\right)$], [we have]
$e^{\displaystyle t\phi\star} = \left[e^{\displaystyle \left(e^{\displaystyle t\phi\star}-1\right)\phi}\right]\star$."
Here, as far as I understand, the $\star$ operation is defined as follows (see page 1 of P92, aka page 1 of the linked PDF): If $\psi$ is any polynomial or power series in infinitely many variables $x$, $y$, $z$, ..., $\delta_x$, $\delta_y$, $\delta_z$, ..., then $\psi\star$ means the differential operator we obtain if we collect all the $\delta_x$, $\delta_y$, $\delta_z$, ... variables at the right end of every monomial and replace them by the partial derivative operators $\frac{\delta}{\delta x}$, $\frac{\delta}{\delta y}$, $\frac{\delta}{\delta z}$, .... I cannot say that I am sure about this, though, because no matter how I try to obtain a small, verifiable example for the formula, I get some nonsense which is either wrong or I am not able to check.
Sylvester studied these in the context of classical invariant theory, but nowadays quantum field theorists are interested in these differential operators as elements of the Heisenberg algebra. Is there any modern (readable) reformulation of the above identity? Has anyone else tried to comprehend its meaning? Is it related to the identity $\left(\exp a\right)\left(\exp b\right)\left(\exp a\right)^{-1} = \exp\left(\left(\exp\left(\mathrm{ad} a\right)\right)\left(b\right)\right)$ which holds for any two elements $a$ and $b$ of a ring for which these exponentials make sense? (This is speculation based on nothing more than the appearance of nested exponentials in both identities.)

REPLY [7 votes]: This can be interpreted cleanly using the notion of Pre-Lie algebra.
Indeed, vector fields on an affine space form a Pre-Lie algebra. To prove the desired identity, it is enough to consider the free Pre-Lie algebra on one generator.
This identity is known to be related to a pre-Lie version of the Baker-Campbell-Hausdorff formula. This is written in several places in the literature.
Here are a few references: 

Agracev, A. A. and Gamkrelidze, R. V., Chronological algebras and nonstationary vector fields
Dominique Manchon, A short survey on pre-Lie algebras, E. Schrödinger Institut Lectures in Math. Phys., Eur. Math. Soc.
Pierre Cartier, Vinberg algebras, Lie groups and combinatorics. (English summary) Quanta of maths, 107–126, Clay Math. Proc., 11,<|endoftext|>
TITLE: Is there an algebro-geometric description of $\nu$?
QUESTION [16 upvotes]: Motivation: According to the "chromatic" picture of stable homotopy, we should think of the moduli stack $M_{FG}$ of formal groups as a "good approximation" to the stable homotopy category (more precisely, we should think of the category of quasi-coherent sheaves on the stack $M_{FG}$ as the "good approximation"). Every finite complex $X$ defines a module $MU_* X$ over the complex bordism ring $MU_*$, which is the Lazard ring classifying a universal formal group law; this is also a comodule over $MU_\ast  MU$, which corresponds to strict automorphisms of a formal group law. Taking account of the grading lets one say "formal group" instead of "formal group law."
Let's say for instance that we have a two cell complex, with cells far away in dimensions from one another. Then this approach means that we get a two-dimensional vector bundle on $M_{FG}$. I'm wondering if we can understand these algebro-geometrically. For instance, line bundles on $M_{FG}$ can be understood (the Picard group is $\mathbb{Z}$, generated by the Lie algebra $\omega$ of a formal group), and perhaps 2-dimensional vector bundles are not too far off. 
Here's the specific situation I have in mind. Let $\nu: S^3 \to S^0$ be the second (stable) Hopf map, which generates the 3-stem. Then the cofiber of $\nu$ is the desuspension $\Sigma^{-4} \mathbb{HP}^2$, and this (as an even, two cell complex) has free $MU$-homology. The homology of this corresponds to some vector bundle on $M_{FG}$, which is an extension of $\mathcal{O}$ and $\omega^4$. Alternatively, $\nu$ is detected in the 1-line of the ANSS as a class in $\mathrm{Ext}^1_{M_{FG}}(\omega^4, \mathcal{O})$, which has order twelve, I think. Is there a description of this class purely in terms of formal groups? I'd be interested as well in the pull-back of this bundle to $M_{1,1}$ (i.e., under the map $M_{1,1} \to M_{FG}$ sending an elliptic curve to its formal group.) More concretely, this describes cooperations in elliptic homology on $\mathbb{HP}^2$. 
(As a simpler example, we can do this for $\eta$, the first Hopf map. Then we are looking at $\Sigma^{-2} \mathbb{CP}^2$, and the relevant bundle on $M_{FG}$ is the next order version of the Lie algebra. Given an even-periodic homology theory $E$, then $\widetilde{E}_0(\mathbb{CP}^2)$ is dual to $\widetilde{E}^0(\mathbb{CP}^2)$, which corresponds to functions on the formal group of $E$ mod functions which vanish to degree 3 or higher at the origin. $\widetilde{E}_0(\mathbb{CP}^2)$ is dual to this.)

REPLY [4 votes]: The answer seems to be yes: given an even-periodic cohomology theory $E$, then $E^*(\mathbb{HP}^\infty)$ can be described as even functions on the formal group of $E$ (just as $E^*(\mathbb{CP}^\infty)$ corresponds to functions on the formal group of $E$). 
In fact, there is a natural map $\mathbb{CP}^\infty \to \mathbb{HP}^\infty$ (sending a complex line to the associated quaternionic line in some $\mathbb{H}^\infty$, or as the map $BU(1) \to B( \mathrm{Sp}(1))$ from the inclusion $U(1) \to \mathrm{Sp}(1)$) and the pull-back on ordinary   cohomology has the following property: if $(-1): \mathbb{CP}^\infty \to \mathbb{CP}^\infty$ is the natural involution, then the cohomology of $\mathbb{HP}^\infty$ is identified with the invariant elements of $\mathbb{CP}^\infty$. The same holds for an even-periodic cohomology theory with $E^0(\ast)$-torsion-free at least, by the following logic. The map $$\mathbb{CP}^\infty \stackrel{(-1)}{\to} \mathbb{CP}^\infty \to \mathbb{HP}^\infty$$ is homotopic to the plain inclusion (the inclusion $U(1) \to \mathrm{Sp}(1)$ and $U(1) \stackrel{z \mapsto \overline{z}} U(1) \to \mathrm{Sp}(1)$ are homotopic as group-homomorphisms---in fact, they are conjugate by $j$). So $E^\ast(\mathbb{HP}^\infty) \to E^*(\mathbb{CP}^\infty)$ is a monomorphism into the subset of even functions on the formal group, and counting dimensions shows that it is an isomorphism (note that the AHSS degenerates). 
So, it follows that $E^\ast(\mathbb{HP}^2)$ can be identified with even functions on the formal group, modulo functions that vanish to order six and higher. More interestingly, though, the fact that $12$ times the associated extension is trivial (that is, that there is a $\mathbb{Z}/12$ in the ANSS at that point corresponding to $\eta$---there is a $\mathbb{Z}/2$ above it because $\pi_3(S^0) = \mathbb{Z}/24$) can be seen algebro-geometrically, at least when one pulls the extension back to $M_{1,1}$.  See M. Hopkins's ICM address for a discussion of this.<|endoftext|>
TITLE: Higher categories as data structures
QUESTION [11 upvotes]: Still wading through higher category theory. I find the subject a bit intimidating, not so much for technical reasons, but because I lack sufficient intuition as to the motivation(s)/heuristics one should use a particular definition instead of another one.
The plethora of formal  possibilities is so great that I would love to have a road map of sorts (such as: if you want to do this, follow this choice, if you wanna do that, here is the menu) 
I do know that there are motivations, for instance in the context of abstract homotopy theory, abstract quantum field theory, etc. 
But I wonder:

from a DATA MODELING's standpoint,
  is there any research geared toward
  using higher cats as advanced data
  structures?

After all, graphs and ordinary 1-dim cats are extremely useful in this respect, so it seems to me that their higher version should also play a big role. 
Any good refs,  thoughts? 
CODA: The ideal situation I have in mind would be something like a nice handbook,  titled
--- higher categories for the working computer scientist--
or
---higher categories  for the working data modeler---.

REPLY [3 votes]: There is work on higher categories and rewriting theory. This considers higher categories as a way of encoding the resolutions of a theory which may itself be of a class of categories. The relevant names are Burroni, Metayer, Lafont, Malbos, Guiraud, so look them up and checkout their preprints etc.<|endoftext|>
TITLE: generalisations of the Seifert-van Kampen Theorem? 
QUESTION [25 upvotes]: I have been reading Jacob Lurie's book "Higher Algebra", version May 8, 2011. One is grateful to him for covering such a lot of ground and for making it all so readily available. 
My attention was drawn to the Section A.3 on "The Seifert-van Kampen Theorem" p. 845. 
It starts by stating the classical theorem determining the fundamental group of  pointed space which is a union of two open sets with path-connected intersection. (The most general theorem of this type is for the fundamental groupoid on a set $A$ of base points for a space $X$ which is the union of a family $\mathcal U$  of open sets and such that $A$ meets each path-component of all 1-,2-,3-fold intersections of the sets of $\mathcal U$). 
It states: " In this section, we will prove a generalization of the Seifert-van Kampen theorem, which describes the entire weak homotopy type of $X$ in terms of any sufficiently nice covering of X by open sets: Theorem A.3.1." However this Theorem makes no mention of groups or connectivity conditions. 
So my question is: How does one deduce the SvKT  as stated there, or its more general version, from Theorem A.3.1?
Theorem A.3.1 itself seems closely related to classical theorems on excision, showing the singular complex is chain homotopy equivalent to the singular complex of $\mathcal U$-small simplices. (I don't  have the earliest reference for this, but I like the proof by R. Sch\"on from Proc. AMS 59 (1976).) 
A particular point for the deduction of the most general version of the SvKT is: why the number 3? One explanation is that it has to do with the Lebesgue dimension of $\mathbb R^2$.

REPLY [16 votes]: I don't actually see how to deduce the version of the classical SvKT on a set of base points $A$ directly from Lurie's version.  It seems that to apply Lurie's theorem, we would need the stronger hypothesis that $A$ meets every path component of every finite intersection of open sets in $\mathcal{U}$ (which is reasonable, since the conclusion of Lurie's theorem is a stronger statement about fundamental $\infty$-groupoids rather than just fundamental 1-groupoids).  But I think we can adapt his proof to derive the classical version.
Here's the first thing I'm going to try to prove: let $\chi:C\to \mathcal{O}(X)$ be a functor, with $C$ a small category and $\mathcal{O}(X)$ the poset of opens in $X$.  For each $x\in X$, define $C_x$, as Lurie does, to be the full subcategory of $C$ spanned by those objects $c$ with $x\in\chi(c)$.  Assume that for every $x$, the nerve of $C_x$ is simply connected.  Then we have $\Pi_1(X) \cong \mathrm{colim}_{c\in C}\; \Pi_1(\chi(c))$, the colimit being a weak 2-colimit of groupoids.
Mimicking Lurie's argument in the 1-truncated case, we have a (pseudo 2-)functor $F:\mathrm{Gpd}^{\mathcal{O}(X)^{\mathrm{op}}} \to \mathrm{Gpd}$ defined by Kan extension from the functor $\Pi_1 : \mathcal{O}(X) \to \mathrm{Gpd}$.  The (2,1)-topos $\mathrm{Sh}_{(2,1)}(X)$ is the (bicategorical) localization of $\mathrm{Gpd}^{\mathcal{O}(X)^{\mathrm{op}}}$ at the covering sieves, and Lurie's A.3.2 shows that $F$ inverts these covering sieves and hence factors through $\mathrm{Sh}_{(2,1)}(X)$.  In particular, this induced functor $F:\mathrm{Sh}_{(2,1)}(X) \to \mathrm{Gpd}$ preserves (bicategorical) colimits.
Thus, it suffices to show that our functor $\chi:C\to \mathcal{O}(X)$ has colimit $X$ (the terminal object) when composed with the Yoneda embedding into $\mathrm{Sh}_{(2,1)}(X)$.  And since $\mathrm{Sh}_{(2,1)}(X)$ has enough points (being sheaves on a topological space), it suffices to check this on all stalks.  (At a finite categorical dimension, there is no hyper-incompleteness to worry about.)  But at the stalk over $x\in X$, the $C$-diagram is trivial at those $c\in C_x$ and empty at the others, so its colimit is simply the groupoid reflection of $C_x$, which was assumed to be terminal (this is equivalent to the nerve of $C_x$ being simply connected).
This completes the proof of the 1-groupoidal version of Lurie's theorem.  Now let's deduce a more classical statement.  Let $X$ be our space and $\mathcal{U}$ an open cover of it.  Define $C$ to be the category of 1-, 2-, or 3-fold intersections of open sets in $\mathcal{U}$, whose morphisms are the canonical inclusions from an $(n+k)$-fold intersection to an $n$-fold intersection, and let $\chi$ be the obvious functor.
For any $x\in X$, the category $C_x$ is obviously nonempty (because $\mathcal{U}$ covers $X$) and connected (because if $x\in U$ and $x\in V$, then $x\in U\cap V$).  It is not much harder to see that it is simply connected: any two parallel zigzags of inclusions can be made equal by passing through at most triple intersections.  Thus, we have
$$\Pi_1(X) \cong \mathrm{colim}_{c\in C} \; \Pi_1(\chi(c)).$$
Now let $A\subseteq X$ be a subset which meets all path components of all 1-, 2-, and 3-fold intersections of open sets in $\mathcal{U}$.  Then $\Pi_1(\chi(c))$ is equivalent to its full sub-groupoid $\Pi_1(\chi(c),A)$ spanned by objects that are points of $A$, as is $\Pi_1(X)$.  Since 2-dimensional colimits are invariant under equivalence of groupoids, the above statement passes to these groupoids as well.
Now the generalized SvKT of Ronnie and his coauthors amounts to asking that $\Pi_1(X)$ be the strict colimit of the functor $c\mapsto \Pi_1(\chi(c),A)$, in the 1-category of groupoids, so it basically suffices to show that this strict 1-colimit is also a (weak) 2-colimit.  Now $C$ is a direct category, and $\mathrm{Gpd}$ is a model category with the canonical model structure (weak equivalences are equivalences, cofibrations are injective on objects), so $\mathrm{Gpd}^C$ inherits a Reedy model structure for which the adjunction
$$ \mathrm{colim} : \mathrm{Gpd}^C \;\rightleftarrows\; \mathrm{Gpd} : \Delta $$
is Quillen.  It follows that the 1-colimit of a Reedy cofibrant diagram is also a 2-colimit.  Unfortunately, $c\mapsto \Pi_1(\chi(c),A)$ is not Reedy cofibrant, but it is "partly" so.  For instance, for any $U\in \mathcal{U}$, consider the latching object
$$ L_U = \mathrm{coeq}\left( \coprod_{V,W} \Pi_1(U\cap V\cap W,A) \;\rightrightarrows\; \coprod_{V}\Pi_1(U\cap V,A) \right) $$
Then the map $L_U \to \Pi_1(U,A)$ is injective on objects; thus our functor is at least Reedy cofibrant "at the top level".  It will suffice to show that if $G\in \mathrm{Gpd}^C$ is "sufficiently Reedy cofibrant" in senses like this, then $\mathrm{colim}(G)$ can be calculated in a homotopy-invariant way.
Let $G\in \mathrm{Gpd}^C$, and enumerate the elements of $\mathcal{U}$ (perhaps transfinitely) as $(U_\alpha)_{\alpha<\lambda}$.  We will define a transfinite sequence of groupoids
$$ H_0 \to H_1 \to H_2 \to \cdots $$
such that $H_\alpha$ is the colimit of $G$ restricted to the subcategory of $C$ determined by the $U_\beta$ with $\beta<\alpha$, and their pairwise and triple intersections.  Of course we can take $H_0 = 0$.  Now given $H_\alpha$, define $H_{\alpha+1}$ to be the pushout of $H_\alpha$ and $G(U_{\alpha})$ along
$$ K_\alpha = \mathrm{coeq}\left( \coprod_{\beta,\gamma < \alpha} G(U_\alpha \cap U_\beta \cap U_\gamma) \;\rightrightarrows\; \coprod_{\beta<\alpha} G(U_\alpha \cap U_\beta) \right). $$
For limit $\alpha$, we of course define $H_\alpha$ to be the colimit.  It is easy to verify that each $H_\alpha$ is the colimit as asserted, and thus $\mathrm{colim}_{\alpha<\lambda} \; H_\alpha = \mathrm{colim} \; G$.
Now suppose $G$ has the property that the map $K_\alpha \to G(U_\alpha)$ is injective on objects (a cofibration) for every $\alpha$.  Then the pushout defining $H_{\alpha+1}$ is a homotopy pushout and thus homotopy-invariant.  Moreover, the map $H_\alpha \to H_{\alpha+1}$ is again a cofibration, so the colimits at limit stages are also homotopy colimits and thus homotopy-invariant.  Therefore, for $G$ with this property, strict colimits are 2-colimits.  But our functor $c\mapsto \Pi_1(\chi(c),A)$ does have this property (it is a slightly stronger version of being "Reedy cofibrant at the top level").
Thus, its strict colimit is also a homotopy colimit, so $\Pi_1(X,A)$ is equivalent to this strict colimit.  The theorem of Ronnie and coauthors asserts that it is in fact isomorphic to this strict colimit, but it is easy to check that it has the same set of objects as the strict colimit (namely $A$), and an equivalence of groupoids which is bijective on objects is an isomorphism.<|endoftext|>
TITLE: Is there a mathematical explanation for the Aharonov-Casher effect?
QUESTION [9 upvotes]: Recall that the Aharonov-Bohm effect can be interpreted mathematically as follows.
Consider an electromagnetic field A on some smooth manifold M, i.e., A is an element in the first differential cohomology of M, or in other words, an isomorphism class of complex hermitian line bundles with a metric connection over M.
The field strength of A is defined to be its curvature.
It can happen that the field strength is zero, i.e., a charged particle traveling through M has no forces acting on it, yet we can still observe a nontrivial change in phase if the particle travels around a loop with a nontrivial holonomy.
This change of phase can then be observed in a physical experiment.
In other words, a line bundle with connection can be flat without being trivial, in particular, it can have nontrivial holonomy.
Recently I learned that there is a dual (in a certain physical sense) effect to the Aharonov-Bohm effect,
namely, the Aharonov-Casher effect,
in which a charged particle is replaced by a neutral particle with a magnetic moment.
Can we interpret the Aharonov-Casher effect mathematically in a similar way to the Aharonov-Bohm effect?

REPLY [4 votes]: This is the same mathematical effect (from reading their paper on it):  
The action-functional is effectively (ignoring kinetic term) $\bar{v}\cdot \bar{E}\times\bar{\mu}$. But this is equivalent to the precession of the magnetic moment in a magnetic field, $\bar{E}\times\bar{\mu}=e\bar{A}$ (a paper of Kan and Koh, 1992, actually explains this in great detail). From here the Lagrangian for the AC-effect is effectively the Lagrangian for the AB-effect, and is due to the vector potential... this is what the dual aspect is.
Clarification: In fact, their paper came across this effect by simply manipulating the view of the AB-effect in the case of a solenoid (the standard example). A solenoid can be represented as a bunch of magnetic moments lined up, and this is what they do to get the AC-effect. They explicitly attribute this to the vector potential, quote, "Is it possible to generate a situation in which a neutral particle exhibits the A-B effect? We will show that
this is indeed possible and is actually a necessary consequence of the physics described by Eq. (1)." Equation 1 here is the standard Lagrangian for particle motion, and involves the vector potential (that is how you get a potential term in the Lagrangian).
Aside: This is related to how us physics students learn [in Electrodynamics] that the only physical quantities are the E-field and B-field, and the vector potential $A$ and scalar potential $\phi$ are simply mathematical constructs to help computations... this is indeed true, as the electromagnetic field is described by virtual photons. Yet the AB-effect (and hence AC-effect) shows how through the vector potential we realize a topological condition on our fields!<|endoftext|>
TITLE: Geometric Meaning of higher pushforward for open immersions
QUESTION [12 upvotes]: Here by higher pushforward I mean the right derived functor of the pushforward functor.
I'm wondering if higher pushforward of open immersions between schemes have some geometric meaning. For a proper flat map, the higher pushforward is like a sheaf which contains information of the cohomology of the fibers. For the higher pushforward of open immersions, the sheaves are supported on the complement of the open, so there are no such thing like "fibers", what's a better way to understand higher pushforward in this case?

REPLY [11 votes]: I think a common way people understand this is via local cohomology and cohomology with support.
Suppose that $Z \subseteq X$ is a closed subset with $\alpha : U = X \setminus Z \to X$ the open immersion.  Since you are interested in the pushforward, I'm going to assume that $X = \text{Spec}R $ is affine (this is harmless for your purposes I think).  I'm going to assume that $M$ is a coherent $O_X$-module and we are going to pushforward $M \mid_U$ (this is also harmless).  Then as it sounds like you already know, there are isomorphisms for all $i > 0$:
$$
R^i \alpha_* (M\mid_U) = \underline H^{i+1}_Z(M)
$$
where by the underlined $\underline{H}$ I just mean the sheafy local cohomology module.  
And so now you are basically asking, what can we say about local cohomology modules?  
Well, of course it depends on $M$, but people study this from a number of perspectives.  Unfortunately for you, the simple geometric interpretation I typically hear involves as sheaf cohomology of complements of closed sets...  There are notable results in this direction though.  For example, see the paper of Barth in AJM in 1970, Hartshorne and Speiser in the Annals 1977, or the work of Peskine and Szpiro (1973), or just SGA2.   
However, a large percentage of modern commutative algebra is interested in the study of these modules.  
Let me highlight a couple recent important problems/questions/areas of work, especially those with geometric connections.

Depth and Serre's conditions, you probably already know that these are connected if you've read Hartshorne, but there are lots of other deeper connections as well.  You could for instance see the book Cohen-Macaulay Rings by Bruns and Herzog.  You can also look up various forms of local duality.  Of course, depth conditions also appear in the study of Moduli spaces, see for example recent work of Koll\'ar.
$M = O_X$.  This is the most common situation that I see studied.  In characteristic $p > 0$, there are literally dozens of papers that study the action of Frobenius on these modules (see for example the recent paper of Enescu-Hochster in ANT or Lyubeznik's paper on $F$-modules in Crelle).  In any characteristic, people certainly also study these as things which $\mathcal{D}$-modules act upon.
Associated primes.  A lot of attention has been paid to what associated primes can occur for local cohomology modules.  For example, it was a longstanding open conjecture that there are always finitely many such associated primes.  Anurag Singh disproved this, MRL (2000) (although its true in some cases, see the work of Huneke and Sharp and Lyubeznik).  Later Moty Katzman gave an equicharacteristic example.  
Hodge Theory:  You can put mixed Hodge structures on local cohomology modules, and these can have various interpretations.  For singularities, see the work of Steenbrink in the early 1980s.
Singularities and section rings.  If $Z$ is a closed point and $M = O_X$, then the these modules are Matlis dual to canonical modules.  In particular, things like rational singularities can be detected that way.  Even more specifically, if $R$ is standard graded (ie a cone), then these local cohomology modules have graded pieces that correspond to sheaf cohomology on the variety you took the cone over.  Because of this, many questions about projective geometry are studied this way by people with a more algebraic bent.  Especially I'd say stuff about Castelnuovo-Mumford regularity.<|endoftext|>
TITLE: Differentials in the Adams Spectral Sequence for spheres at the prime p=2
QUESTION [12 upvotes]: How does one compute the differentials in the Adams Spectral Sequence for spheres at the prime 2 in the range $13\le t-s\le 20$?  There seem to be 6 nonzero differentials, and at this point I only understand $d_2(h_4)=h^2_3h_0$.
There seem to be two methods that are used or referenced in various texts, but I haven't figured out exactly how to apply either in this context.  The first is the Massey Product/Toda Product (apparently they are the same, but Massey is algebraic and works in $E_2$, and Toda is topological and works in $\pi_*^s$).  The second is by building a cofiber sequences $S^0\to S^0\cup_f e^i\to S^i$ which gives a long exact sequence in both the $\pi^s_*$ and the spectral sequence itself.
If possible, could somebody point me to a resource where they use these methods in this range, or give me a hint on how I can try to do this?
Thanks a bunch
-Joseph

REPLY [17 votes]: With the aid of machine computations, you can readily determine the Adams differentials up to $t-s=30$ using the multiplicative structure, the relation between Steenrod operations in $\text{Ext}_A$ and the $E_\infty$ (or $H_\infty$) ring spectrum structure, a comparison with the mapping cone $C_\sigma = S \cup_{\sigma} e^8$ and a comparison with the image-of-$J$ spectrum $j = \text{hofib}(\psi^3-1 : ko \to bspin)$.
In this limited range this sidesteps the use of Toda brackets and the solution of the Adams conjecture, which were frequently used by Barratt, Mahowald and Tangora for their calculation in the range $t-s \le 45$.  This is not an argument for not learning about those methods and results!  Nonetheless, this is how I recently went through the calculation in class:
In the range $t-s \le 30$ the Adams $E_2$-term is generated as an algebra by the $h_i$ for $0 \le i \le 4$ in Adams filtration $s=1$, $c_0$ and $c_1$ in filtration $s=3$, $d_0$, $e_0$, $f_0$ and $g = g_1$ in filtration $s=4$, $Ph_1$ and $Ph_2$ with $s=5$, $r$ with $s=6$, $Pc_0$, $i$, $j$ and $k$ with $s=7$, $Pd_0$ and $Pe_0$ with $s=8$, $P^2h_1$ and $P^2h_2$ with $s=9$, $P^2c_0$ with $s=11$, $P^2d_0$ with $s=12$, and $P^3h_1$ and $P^3h_2$ with $s=13$.  This can easily be checked with Bruner's ext-program. The notation is derived from Peter May's thesis, as presented in Martin Tangora's 1970 Math. Z. paper.  Only $f_0$ is ambiguously defined modulo decomposables.
Many differentials can now be determined by the relation between the $E_\infty$ structure on $S$ and the Steenrod operations on $\text{Ext}_A$.  This is due to D.S. Kahn, J. Milgram, J. Makinen and R.R. Bruner, in increasing generality. See Bruner's chapter in the $H_\infty$ book of Bruner, May, McClure and Steinberger for further details.
The $E_\infty$ structure applied to $\sigma : S^7 \to S$ gives a map $\Sigma^7 RP_7^\infty = D_2(S^7) \to S$ that extends $\sigma^2$, and gives you the relations $2\sigma^2 = 0$ (which forces the differential $d_2(h_4) = h_0h_3^2$) and $\eta\sigma^2 = 0$.
The $E_\infty$ structure applied to $2\sigma : S^7 \to S$, together with the Steenrod operations on $\text{Ext}_A$, tells you that $h_1 h_4$ is a permanent
cycle.
The $E_\infty$ structure applied to $\eta\sigma : S^8 \to S$ tells you that $h_2 h_4$ is a permanent cycle, if you accept a Cartan formula for power operations.  This deduction can be replaced by the arguments below.
The $E_\infty$ structure applied to $\epsilon : S^8 \to S$ gives the differential $d_2(f_0) = h_0^2 e_0$, and tells you that $c_1$ is a permanent cycle.  This relies on the algebraic facts that $Sq^2(c_0) = h_0 e_0$,
$Sq^1(c_0) = f_0$ and $Sq^0(c_0) = c_1$.  (Bruner's original paper uses a
different indexing system for the $Sq^i$ than I use here.)
Using $h_0$- and $h_1$-linearity of $d_2$, you can deduce $d_2(h_0 f_0) =
h_0^3 e_0$ and $d_2(e_0) = h_1^2 d_0$.
Multiplying by $d_0$, you get $d_2(d_0f_0) = h_0^2 d_0 e_0$, and using
$h_2$-, $h_0$- and $h_1$-linearity of $d_2$ you find $d_2(k) = h_0 d_0^2$, $d_2(j) = h_0 Pe_0$, $d_2(Pe_0) = h_1^2 i$ and $d_2(i) = h_0 Pd_0$.  Many more $d_2$-differentials for higher $t-s$ follow in this way.
Together with the Adams differential $d_2(h_5) = h_0 h_4^2$, these are the only $d_2$'s affecting $t-s \le 30$. This is clear from the multiplicative structure (and the fact that $d_2 \circ d_2 = 0$).
To get at the $d_3$-differentials, you can use naturality along $i : S \to C_\sigma$.  There is a class $\beta$ in bidegree $(t-s,s) = (15,3)$ of the $E_2$-term of the Adams spectral sequence for $C_\sigma$, such that $h_2 \beta
= i_*(f_0)$.  (This notation is compatible with the $tmf$-notation used by A. Henriques and T. Bauer.)  By naturality $d_2(i_*(f_0)) = h_0^2 i_*(e_0)$, and by $h_2$-linearity $d_2(\beta) = h_0 i_*(d_0)$ in the Adams spectral sequence for $C_\sigma$.  Thus $d_2(h_0 \beta) = h_0^2 i_*(d_0)$.  It follows that $\pi_{14}(C_\sigma)$ (implicitly $2$-completed) has order dividing $2^2=4$.
Looking at the long exact sequence
$$
\pi_7(S) \overset{\sigma}\longrightarrow \pi_{14}(S) \to \pi_{14}(C_\sigma) \to \pi_6(S) \to \pi_{13}(S)
$$
you can deduce that $\pi_{14}(S)$ has order exactly $4$, so that $d_0$ and
$h_3^2$ survive to $E_\infty$ and detect $\kappa$ and $\sigma^2$, respectively. This also implies that $h_0 d_0$ must be a boundary, and $d_3(h_0 h_4) = h_0 d_0$ is the only possibility, which implies $d_3(h_0^2 h_4) = h_0^2 d_0$.
You can also deduce that $i_*(h_4)$ is a permanent cycle for $C_\sigma$,
and this implies that $d_3(h_2h_4) = 0$ for $S$.
The final $d_3$-differential in this range, $d_3(r) = h_1 d_0^2$, can be obtained from the $E_\infty$ structure applied to $\kappa : S^{14} \to S$, using $Sq^2(d_0) = r$.  The next possible $d_3$-differentials start in
$t-s=31$.
The only possibly nonzero $d_4$-differential is that on $h_4 c_0$, which might hit $Pd_0$, but $h_1 h_4 c_0$ is a permanent cycle and $h_1 Pd_0$ is nonzero, so this is excluded by $h_1$-linearity of $d_4$.
The only possibly nonzero later differentials are those on $h_2 h_4$ and $g$, which might conceivably hit $h_1 Pc_0$ or an $h_0$-power times $Ph_2$, respectively. To eliminate this possibility, one can use naturality with respect to the map $e : S \to j$.
The cohomology of $j$ is the unique nonsplit extension of the $A$-modules given by the cokernel and desuspended kernel of
$$
(\psi^3-1)^* : \Sigma^4 A/A\{Sq^1, Sq^2Sq^3\} = H^*(bspin) \to H^*(ko) = A/A\{Sq^1, Sq^2\} \,,
$$
sending the generator $\Sigma^4 1$ to $Sq^4$.  Its Adams $E_2$-term can be calculated in this range by Bruner's programs, and in the range $t-s \le 24$ it is easy to see that the only differential pattern that is compatible with the known abutment, i.e., $\pi_*(j)$ completed at $2$, is one with $E_3 = E_\infty$.
The map of $E_3$-terms induced by $e : S \to j$ is surjective for $t-s \le 24$, except for $t-s=15$ (the homomorphism $\pi_{15}(e)$ involves a shift of Adams filtration).  In particular, the imagined targets for differentials on $h_2 h_4$ and $g$ in the Adams spectral sequence for $S$ map to survivors in the Adams spectral sequence for $j$, hence cannot be hit by differentials for $S$, after all.<|endoftext|>
TITLE: cohomological dimension of a group acting on a product
QUESTION [6 upvotes]: I recently came across an interesting result of Kobayashi [Corollary 5.5], a special case of which is the following: Suppose $\Gamma$ is a discrete torsion free subgroup of $SL_n(\mathbb{R})$ which acts properly discontinuously on the homogeneous space $X=SL_n(\mathbb{R})/SL_{n-1}(\mathbb{R})$. Then the cohomological dimension of $\Gamma$ is less than or equal to $n$.
The homogeneous space $X$ above is diffeomorphic to a fiber bundle with base space $S^{n-1}$ and fibers $\mathbb{R}^n$. This motivates my question:
Suppose $G$ is a discrete torsion free group acting properly discontinuously on $M \times \mathbb{R}^n$ where $M$ is a compact manifold. What can be said about the cohomological dimension of $G$? Is it less than or equal to $n$?
Kobayashi's proof uses spectral sequences, a tool which I am not familiar with. So before spending time learning about these objects, I was wondering if there is an obvious obstruction to the generalization of Kobayashi's result mentioned above.

REPLY [6 votes]: I think Kobayashi's approach also works in the OP's situation: 

If $M$ is an oriented compact connected manifold then $cd_\mathbb{R}(G) \le n$. 

Proof: As in Kobayashi the base ring is the field of real numbers. Let $X = M \times \mathbb{R}^n$ and let $A$ be an $\mathbb{R}G$-module. Since $G$ is torsion-free and acts properly discontinuously, the action is actually free. Thus by [Cartan-Eilenberg: Homological Algebra, XVI §9] there is a spectral sequence 
$$E_2^{i,j}=H^i(G;H^j(X;A)) \Rightarrow H^{i+j}(X/G;A)$$
(see also Kobayashi, p. 14). Before diving into technical details let's sketch the basic idea. Suppose $\dim M=m$ and $cd_\mathbb{R}(G) = d$. Since $X \simeq M$, $H^j(X;A)=0$ if $j>m$. Futhermore $H^i(G;-)=0$ if $i > d$. Hence the $E_2$-term looks like 
$$\begin{array}{lcccr}
- & - & - & - & \bullet \newline 
| &   &   &   & |   \newline 
| &   &   &   & |   \newline 
- & - & - & - & - \newline 
\end{array}$$
Outside the rectangle all entries are zero and the bullet has coordinate $(d,m)$. For positional reasons $E_2^{d,m}=E_\infty^{d,m}$. 
Now suppose $E_2^{d,m} \neq 0$. Hence the abutment $H^{d+m}(X/G;A) \neq 0$. But $X/G$ is a manifold of dimension $m+n$. Hence $H^i(X/G;A)=0$ if $i> m+n$ and consequently $d+m \le n+m$, i.e. $d \le n$ as to be shown. 

The details: By $X \simeq M$ and Poincare duality we have isomorphisms $$H^m(X)\cong H^m(M) \cong H_0(M) \cong \mathbb{R}$$ as $\mathbb R$-vector spaces. Under this isomorphism the action of $g \in G$ on $H^m(X)$ corresponds to scalar multiplication on $\mathbb R$ by an $\rho(g) \in \mathbb R^\times$ satisfying $\rho(g)\rho(h)=\rho(gh)$ for $g,h \in G$. Moreover, by universal coefficients 
$$H^m(X;A) \cong H^m(X) \otimes_\mathbb{R} A \cong A \hspace{110pt}(1)$$ 
as vector spaces and hence 
$E_2^{i,m}=H^i(G;A)$. 
Assume for the moment we know that the action $\odot$ of $G$ on $A$ in $(1)$ is given by 
$$g \odot a = \rho(g)(g \cdot a).\hspace{150pt}(2)$$
Let $(B,\ast)$ be an $\mathbb R$-module such that $H^d(G;B) \neq 0$. Define a $G$-action $\cdot$ on $A := B$ by $$g \cdot a := \rho(g^{-1})(g\ast a)$$ (this is indeed an action because multiplication in $\mathbb R$ is commutative). Hence $\odot = \ast$ and $E_2^{d,m}=H^d(G;B) \neq 0$ as desired. 

It remains to show $(2)$. This is a straightforward, but tedious calculation using the definition of the various $G$-actions. So I think it's best  -- to leave it to the reader ? No, but I think it's best to write it down here only if one is really interested in the demonstation.<|endoftext|>
TITLE: Integer cohomology of the Grassman manifold of n planes in $R^\infty$
QUESTION [6 upvotes]: I can't seem to find a reference on the web that gives the $\mathbb{Z}$ cohomology of the Grassmann manifold of real n-planes in infinite dimensional Euclidean space and also the Bockstein maps associated with the coefficient sequence
$$0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z/2Z} \to 0.$$
The real question is which products of Stiefel-Whitney classes are really $\mathbb{Z}$ classes.

REPLY [5 votes]: I don't know if these have everything that you want, but see the following:
Brown, Edgar H., Jr.
The cohomology of BSOn and BOn with integer coefficients.
Proc. Amer. Math. Soc. 85 (1982), no. 2, 283–288.
Feshbach, Mark The integral cohomology rings of the classifying spaces of O(n) and SO(n). Indiana Univ. Math. J. 32 (1983), no. 4, 511–516.<|endoftext|>
TITLE: Sparse ramsey theory
QUESTION [8 upvotes]: It is known that for any graph H and all $k∈N$, there exists a graph $G$ such that any $k$-coloring of the edges of $G$ yields a monochromatic copy of H and ω(G)=ω(H) (the two graphs have the same clique numbers). 
My question is: Given any graph $H$ with finite girth, is there a $G$ with the same girth as $H$ such that any $2$-coloring of the edges of $G$ yields a monochromatic copy of $H$? 
I think this is an open problem but if someone can confirm that and give some references concerning this I would be most obliged.

REPLY [7 votes]: Here there is a set of lecture notes from a course given by Imre Leader in 2003. On page 18 of these notes this is given as an open problem.<|endoftext|>
TITLE: Automorphism group
QUESTION [7 upvotes]: Let $G=PSL(2,q)$ where $q$ is prime power. What is Aut$(G\times G)$ and Aut$(G\times G\times G)$? Also if $G=A_{n}$ where $A_{n}$ is the alternating group of degree $n$, then what is Aut$(G\times G)$? 
Thanks in advance

REPLY [13 votes]: In general, if $S$ is a finite non-Abelian simple group, and $E$ is a direct product of $n$ copies of $S$, then ${\rm Out}(E) = {\rm Aut}(E)/E$ is isomorphic to ${\rm Out}(S) \wr S_{n}.$ This is because every minimal normal subgroup of $E$ is isomorphic to $S$ (in fact, is one of the obvious simple direct factors of $E$) and the automorphism group of $S$ permutes the minimal normal subgroups of $E$.
  To provide more detail in order to make up for the lack of a reference: The $n$ "obvious" simple direct factors of $E$ are called the components of $E.$ The direct product of $n$ copies of ${\rm Aut}(S)$ obviously sits inside ${\rm Aut}(E).$ Furthermore, the assumed isomorphisms between the $n$ components may be included to show that ${\rm Aut}(S) \wr S_{n}$ embeds in ${\rm Aut}(E).$ On the other hand, the permutation action of ${\rm Aut}(E)$ on the components of $E$ gives a homomorphism from ${\rm Aut}(E)$ to $S_{n}.$
The kernel of this homomorphism is the intersection $K$ of the normalizers of the individual components. Since $E$ contains its centralizer in ${\rm Aut}(E)$, the group $K/E$ is isomorphic to a subgroup of a direct product of $n$ copies of ${\rm Out}(S).$
Hence this establishes that $|{\rm Aut}(E)| \leq | {\rm Aut}(S) \wr S_{n}|.$ But we have the inequality the other way round, so ${\rm Aut}(E) \cong {\rm Aut}(S) \wr S_{n}.$

REPLY [3 votes]: See this wikipedia article. The result mentioned there implies that the automorphism group is the wreath product power of the automorphism group of $G.$<|endoftext|>
TITLE: Measures that are not OD
QUESTION [5 upvotes]: Is anything known about the consistency strength of the statement:
"There is a normal measure (on a cardinal) that is not ordinal-definable"?
In particular, is it consistent relative to the existence of a measurable cardinal?
It looks like it's consistent relative to the existence of a supercompact cardinal.
If $\kappa$ is supercompact then we can force to make it Laver indestructible.
So assume that $\kappa$ is still $(\kappa+2)$-strong after we add $(2^{2^\kappa})^+$ many Cohen subsets of $\kappa^+$, more than the number of measures on $\kappa$ in $V$.
  Solovay proved that if $\kappa$ is $(\kappa+2)$-strong then for every set $X \in V_{\kappa+2}$ there is a normal measure on $\kappa$ whose ultrapower contains $X$. So letting $X$ range over the Cohen subsets of $\kappa^+$ that we added, a counting argument shows that we must get some normal measures on $\kappa$ that are not in $V$.  Cohen forcing is homogeneous, so these measures cannot be ordinal-definable.
I don't know how strong this kind of indestructibility is, or whether it's necessary.
I am also interested to know anything about countably complete measures on any set that are not ordinal-definable from that set.

REPLY [6 votes]: It was proved, long ago, that one can force over a model with a measurable cardinal $\kappa$ and get a model with lots ($2^{2^\kappa}$, I think) of normal measures.  I believe that (1) most of those measures won't be OD in that model and (2) the relevant paper is 
Kenneth Kunen and Jeff Paris, Boolean extensions and measurable cardinals, Annals Math. Log., Vol. 2 (1971), pp. 359-377.
Unfortunately, I'm traveling and would find it difficult to check these things right now.

REPLY [6 votes]: This is equiconsistent with a measurable cardinal.  
Start with a measurable cardinal $\kappa$ in $V$, and assume without loss of generality that $2^\kappa=\kappa^+$. Indeed, we might as well assume $V=L[\mu]$ is the canonical inner model of one measurable cardinal. Next, perform the Easton support iteration of forcing that adds a Cohen subset to every inaccessible $\gamma\leq\kappa$, and let $V[G*g]$ be the corresponding forcing extension, where $G$ is the forcing up to $\kappa$ and $g$ is the stage $\kappa$ forcing. The standard lifting arguments show that $\kappa$ remains measurable in $V[G*g]$. 
Specifically, fix any ultrapower $j:V\to M$ by a normal measure on $\kappa$ in $V$. The forcing $j(P)$ is isomorphic to $P*P_{\rm tail}$, and one may find in $V[G*g]$ an $M$-generic filter $j(G*g)\subset j(P)$ satisfying the lifting criterion, and thereby lift the embedding to $j:V[G][g]\to M[j(G)][j(g)]$. The filter $g$ is used at stage $\kappa$ in $j(G)$. There are in fact numerous lifts of $j$ to the forcing extension, and since these are all still ultrapowers by normal measures in $V[G*g]$, this is a model where $\kappa$ carries $2^{2^\kappa}$ many normal measures.
Each of these measures is determined by and determines the filter $j(G*g)\subset j(P)$ that was used in the construction. Since the forcing is almost homogeneous, it follows that the $\text{HOD}^{V[G*g]}\subset \text{HOD}^V$, and moreover even if we add $G$ as a parameter, we have $\text{HOD(G)}^{V[G][g]}\subset\text{HOD(G)}^{V[G]}$, since the stage $\kappa$ forcing is almost homogeneous. Thus, in particular, if one of the measures in $V[G][g]$ is ordinal definable, then so would be the corresponding $j(G)$, and so we would have $j(G)\in V[G]$. But $g$ appears explicitly at stage $\kappa$ of $j(G)$, and so this is impossible.  
This argument therefore shows that $V[G][g]$ is a model where $\kappa$ is measurable, carries $2^{2^\kappa}$ many normal measures, and none of these measures is ordinal definable there.<|endoftext|>
TITLE: Showing that a family of polynomials has positive and real roots.
QUESTION [13 upvotes]: Hi everybody, for my research I am dealing with the following function: 
$$\alpha_n(x):=\left.\frac{\partial^{2n+1}}{\partial z^{2n+1}}\frac{\sinh(z)}{\cosh(z)-1+x}\right|_{z=0},\quad n\in \mathbb{N},$$
It is possible to show that
$$\alpha_n(x)=\frac{P_n(x)}{x^{n+1}},$$ where $P_n(\cdot)$ is a polynomial of order $n$ in $x$, having integer coeffients.
To make few concrete examples
$$\alpha_0(x)=\frac{1}{x}$$ 
$$\alpha_1(x)=\frac{-3+x}{x^2}$$ 
$$\alpha_2(x)=\frac{30-15 x+x^2}{x^3}$$ 
$$\alpha_3(x)=\frac{-630+420 x-63 x^2+x^3}{x^4}$$ 
$$\alpha_4(x)=\frac{22680-18900 x+4410 x^2-255 x^3+x^4}{x^5}$$ 
and so on.
What I would need to show (and it is veryfied for all the special cases I was able to compute, like those above) is that all the roots of $P_n(x)$ (and therefore those of $\alpha_n(x)$) are real and strictly greater than 2. 
An explicit albeit complicated expression for $\alpha_n(x)$ can be obtained, namely:
$$\begin{equation*}
\begin{array}{ll}
\alpha_n(x)=&x^{-n-1}\sum_{j=0}^{n} x^{n-j}\sum_{k=j+1}^{n} (2k)!\genfrac{\lbrace}{\rbrace}{0pt}{}{2n+1}{2k}\sum_{i=j}^{k}\left[ {i+1\choose j+1}\binom{2k}{2 i+1}-{i\choose j+1}\binom{2k}{2 i}\right](-2)^{j+1-2k}+\\
&x^{-n-1}\sum_{j=0}^{n} x^{n-j}\sum_{k=j}^{n}{(2k+1)!\genfrac{\lbrace}{\rbrace}{0pt}{}{2n+1}{2k+1}}\sum_{i=j}^{k} \left[ {i+1\choose j+1}\binom{2k+1}{2 i+1}-{i\choose j+1}\binom{2k+1}{2 i}\right](-2)^{j-2k},
\end{array}
\end{equation*}$$
where the number between  the curly brakets are the Stirling number of the second kind; Moreover, 
$$\alpha_n(2)=-\sum_{k=1}^{2n+1}{k!\genfrac{\lbrace}{\rbrace}{0pt}{}{2n+1}{k}}(-2)^{-k}\neq0.$$
If someone is interested, I can post more on how I got these expressions.
Thanks in advance to everybody that will try to help me.
Best Regards 
Enzo

REPLY [4 votes]: This paper by P. Bränden and this paper by M. Visontai & N. Williams give a somewhat general approach to proving real rootedness of polynomials (especially those coming up combinatorially).<|endoftext|>
TITLE: Is a random subset of the real numbers non-measurable?  Is the set of measurable sets measurable?
QUESTION [29 upvotes]: One might say, "a random subset of $\mathbb{R}$ is not Lebesgue measurable" without really thinking about it.  But if we unpack the standard definitions of all those terms (and work in ZFC), it's not so clear.
Let $\Sigma \subset 2^\mathbb{R}$ be the sigma-algebra of all Lebesgue measurable sets.  Give $2^\mathbb{R}$ the product measure.  (It's a product of continuum many copies of the two-point set.)  We want to say that $\Sigma$ is a null set in $2^\mathbb{R}$...but is $\Sigma$ even measurable?
Laci Babai posed this question casually several years ago, and no one present knew how to go about it, but it might be easy for a set theorist.
Also, a related question: Think of $2^\mathbb{R}$ as a vector space over the field with two elements and $\Sigma$ as a subspace. (Addition is xor, that is, symmetric set difference.) What is $\dim\left(2^\mathbb{R}/\Sigma\right)$?
It's not hard to see that $\dim\left(2^\mathbb{R}/\Sigma\right)$ is at least countable, so if $\Sigma$ were measurable, it would be a null set.  But that's as far as I made it.

REPLY [9 votes]: The problem of choosing subsets at random has been studied in a rather different context in mathematical economics. Suppose we choose a subset of $[0,1]$ by independently throwing a fair coin for each number. Heuristically, such a set should have measure $1/2$. For what we do is randomly choose an indicator function with pointwise expectation $1/2$. By some intuitive apeal to a law of large numbers, the sample realizations should have the same expectation. This kind of reasoning is widely used in economics. A large population is modeled by a continuum and even when each person faces individual uncertainty, there should be no aggregate uncertainty. 
For the reason given by Will Sawin, the naive approach doesn't work quite well. For Lebesgue measure, some intuition comes from Lusin's theorem to the effect that every measurable function is continuous on a "large" subset. Continouity is a condition to the effect that the value at a point is closely related to the value at nearby points. If you choose independently at each value, you wouldn't expect to get a function continuous on a large set.
The general tradeoff between independence and measurable sample realizations is strongly  expressed in the following result of Yeneng Sun: 
Proposition: Let $(I,\mathcal{I},\mu)$ and $(X,\mathcal{X},\nu)$ be probability spaces with (complete) product probability space $(I\times X,\mathcal{I}\otimes\mathcal{X},\mu\otimes\nu)$ and $f$ be a jointly measurable function from $I\times X$ to $\mathbb{R}$ such that for $\mu\otimes\mu$-almost all $(i,j)$ the functions $f(i,\cdot)$ and $f(j,\cdot)$ are independent. Then for $\mu$-almost all $i$, the function $f(i,\cdot)$ is constant.
Note that the independence condition in this result is quite weak. Sun calls it almost sure pairwise independence. But an important discovery by Sun was that if joint measurability and almost sure pairwise independence were compatible, one could obtain an exact law of large numbers for a continuum of random variables by an application of Fubini's theorem. In particular, such a law of large numbers holds for extensions of the product spaces that allow for the conclusion of Fubini's theorem to hold and still allow for nontrivial (a.s. pairwise) independent processes. He called such extensions rich Fubini extensions and gave one example of such a product space: The Loeb product of two hyperfinite Loeb spaces. So one can get natural random sets for some spaces. The reference is: The exact law of large numbers via Fubini extension and characterization of insurable risks (2006) 
A systematic study of rich Fubini extensions was done by Konrad Podczeck in the paper On existence of rich Fubini extensions (2010), in which he has essentially shown that one can choose random subsets of a probability space if and only if the probability space has the following property, which he called super-atomlessnes (and which is known by a lot of other names such as saturation): 
For any subset $A$ with positive measure, the measure algebra of the trace on $A$ does not coincide with the measure algebra of a countably generated space. 
Lebesgue measure on the unit interval does not satisfy this condition, but there exists extensions of Lebesgue measure that are superatomless. 
Conclusion: One cannot obtain random Lebesgue measurable sets in a sensible way by choosing independently elements, but one can choose random sets in an extension of Lebesgue measure this way.<|endoftext|>
TITLE: What is the significance that the Springer resolution is a moment map?
QUESTION [19 upvotes]: Let $\mathcal{B}$ be the flag variety and $\mathcal{N} \subset \mathfrak{g}$ is the nilpotent cone. We know that the Springer resolution
$$
\mu: T^*\mathcal{B}\rightarrow \mathcal{N}
$$
is the moment map, if we identify $\mathfrak{g}$ with $\mathfrak{g}^* $ by the Killing form and consider $\mathcal{N} \subset \mathfrak{g}$ as a subset of $\mathfrak{g}^*$.
As far as I know, the geometric construction of Weyl group and $U(sl_n)$ does not involve moment map or even symplectic geometry, as in the paper "Geometric Methods in Representation Theory of Hecke Algebras and Quantum Groups"
My question is: what is the consequence of the fact that the Springer resolution is a moment map?

REPLY [20 votes]: One reason to emphasize the Springer resolution's role as a moment map is that it is the semiclassical shadow of Beilinson-Bernstein localization. More precisely passing to functions, the moment map description asserts that the Springer map is describing the Hamiltonian functions on the cotangent to the flag variety which generate the action of the Lie algebra. We may now quantize the cotangent bundle $T^* G/B$ to the ring of differential operators on $G/B$, and likewise quantize the dual space $g^*$ to the Lie algebra to the universal enveloping algebra $Ug$, so that the moment map describes the map from $Ug$ to global differential operators on the flag variety. What's truly significant about the Springer map (it's a birational, proper, symplectic [crepant] resolution of [rational] singularities) now translates into the Beilinson-Bernstein equivalence (for generic parameters) between $Ug$-modules and (twisted) D-modules on the flag variety, the cornerstone of geometric representation theory. There's now an entire subject (wonderfully represented in a workshop last week in Luminy) seeking to generalize all the features of this setup to other symplectic resolutions and their quantizations, viewed as the settings for "new representation theories" (the prime examples being Hilbert schemes and other quiver varieties).<|endoftext|>
TITLE: Picard group of a singular projective curve
QUESTION [14 upvotes]: Let $X$ be a singular irreducible projective curve over an algebraically closed field and $\pi : \widetilde{X} \to X$ the normalization morphism. In the book on Neron models by Bosch et al. (I have slightly simplified the setup), the Picard group of $X$ is computed by taking the long exact sequence associated to $$0 \to \mathcal{O}_X^* \to \pi_*\mathcal{O}_{\widetilde{X}}^* \to \pi_*\mathcal{O}_{\widetilde{X}}^*/\mathcal{O}_X^* \to 0.$$ This seems to depend on the assertion that $\text{Pic}(\widetilde{X}) \cong H^1(X,\pi_*\mathcal{O}_{\widetilde{X}}^*)$, which is not obvious to me. If $\mathcal{O}_{\widetilde{X}}^*$ were coherent, then this would be a consequence of the exactness of $\pi_*$ on coherent sheaves ($\pi$ is finite), but of course $\mathcal{O}_{\widetilde{X}}^*$ is not even a sheaf of $\mathcal{O}_{\widetilde{X}}$-modules. How can we resolve this difficulty?

REPLY [3 votes]: The proof of the isomorphism can be found in Proposition 2.8 of this 2013 paper by Hartshorne and Polini: https://arxiv.org/pdf/1301.3222.pdf<|endoftext|>
TITLE: "Must read" papers in numerical analysis
QUESTION [35 upvotes]: In 1993, Prof. L.N. Trefethen published a NA-net posting with a list of thirteen paper he used for teaching the seminar Classic Papers in Numerical Analysis. 
In Trefethen's words, ... this course provided a satisfying vison of the broad scope of numerical analysis and the sense of excitement at what a diversity of beautiful and powerful ideas have been invented in this field.
Prof. Trefethen's list (links):

Cooley & Tukey (1965)              the Fast Fourier Transform
Courant, Friedrichs & Lewy (1928)  finite difference methods for PDE
Householder (1958)                 QR factorization of matrices
Curtiss & Hirschfelder (1952)      stiffness of ODEs; BD formulas
de Boor (1972)                     calculations with B-splines
Courant (1943)                     finite element methods for PDE
Golub & Kahan (1965)               the singular value decomposition
Brandt (1977)                      multigrid algorithms
Hestenes & Stiefel (1952)          the conjugate gradient iteration
Fletcher & Powell (1963)          optimization via quasi-Newton updates
Wanner, Hairer & Norsett (1978)   order stars and applications to ODE
Karmarkar (1984)                  interior pt. methods for linear prog. 
Greengard & Rokhlin (1987)        multipole methods for particles

Most readers of this note, according Prof. Trefethen, will have thought of other classic authors and papers that should have been on the list.


The question is: In your opinion, what are other classic authors and papers that should be in a must read list of papers in numerical analysis?

REPLY [2 votes]: Any mention of de Boor should also include Jos Stam's work: http://www.dgp.toronto.edu/~stam/reality/Research/pub.html
(Especially his siggraph 1999 course notes, which cover a number of useful approaches for dealing with b-splines.)<|endoftext|>
TITLE: Classical Limit of Feynman Path Integral
QUESTION [9 upvotes]: I understand that in the limit that $\hbar$ goes to zero, the Feynman path integral is dominated by the classical path, and then using the stationary phase approximation we can derive an approximation for the propagator which is a function of the classical trajectory (see this document, pg 46).  
I am under the impression that this further implies that the particle follows the classical trajectory but I don't understand how the above mentioned fact implies this.
The propagator describes the time-evolution of the wavefunction, so I would think that this classical limit form of the propagator should give a time-evolution in which the wavefunction follows the classical trajectory, but I have not been able to find such work.  Moreover, even this statement itself is problematic since the wavefunction describes a probability distribution and not a single trajectory.
$\textbf{New Edit:}$ In section 7 of Feynman's paper introducing the path integral (see here) he discusses the classical limit. It appears that the key to understanding why the fact that the classical path dominates the path integral further implies that the particle follows the classical trajectory may be found in Feynman's remark on pg 21: "Now we ask, as $\hbar → 0$ what values of the intermediate coordinates $x_i$ contribute most strongly to the integral? These will be the values most likely to be found by experiment and therefore will determine, in the limit, the classical path." However, I don't understand why "These will be the values most likely to be found by experiment" ?

REPLY [3 votes]: Physics note: for any non-zero value of $\hbar$, and for any chaotic potential $V(x)$, a wavefunction which is localized around a classical configuration $x_a$ (which could be a point particle's position, an entire field configuration, etc.) will still evolve to a state which is not localized around any configuration $x_b$ within a few multiples of the Lyapunov time. (The number of multiplies depending only logarithmically on the smallness of $\hbar$.)  Since most real-life potentials are chaotic, you will find that the wavefunction of isolated macroscopic objects evolves to highly non-classical states in a relatively short time.
My advisor wrote a colorful paper on this a while back, pointing out that even a massively macroscopic variable like the orientation of Saturn's moon Hyperion would find itself in a grossly non-classical state on human time-scales: Why We Don't Need Quantum Planetary Dynamics: Decoherence and the Correspondence Principle for Chaotic Systems.
Whether or not this is a problem is disputed.  For physicists who think more needs to be said to explain classicality, the most popular answer involves decoherence.<|endoftext|>
TITLE: solving linear equations made difficult
QUESTION [31 upvotes]: (Note: This is a what's-in-the-literature question, not a what's-mathematically-true question, but I believe both are considered valid kinds of MathOverflow question.)
I saw this amusing derivation on the blackboard at MSRI a few months ago (I'm paraphrasing and reformatting slightly, though my attempts at formatting may not work as intended):
"Problem: Solve $x = ax + b$ for $x$.
Solution:
$$x = a(ax+b) + b = a^2 x + ab + b = a(a(ax+b)+b)+b= a^3 x + a^2 b + ab + b = \cdots$$
 (assuming $|a| < 1$) 
$$= \lim_{n \rightarrow \infty} a^n x   +  b \sum_{i=0}^{\infty} a^i
   = 0 + b/(1-a).$$
This also holds by analytic continuation for all $a \neq 1$."
Has anyone seen this before?  I took a photograph of the blackboard, and I am inclined to submit it to Mathematics Magazine, but first I want to know the provenance.
Curt McMullen was in residence at MSRI at the time, and he seemed a likely culprit, but when I pointed it out to him he seemed amused, and he denied authorship, so I don't have any suspects at present.
It would be embarrassing to publish this and then receive letters saying "This argument appears almost word-for-word in Littlewood's Miscellany" (or something like that).

REPLY [5 votes]: Actually this calculation has a formal sense in every ring, by working in the ring of formal power series in $a$ (here $1-a$ is invertible with inverse $\sum_{i \geq 0} a^i$). There are many "pseudo-analytic" proofs in ring theory (one was discussed here). I've made this CW because I cannot answer the question whether this has appeared in the literature, but I am pretty sure that it has.<|endoftext|>
TITLE: Are reflective subcategories of complete infinity categories complete?
QUESTION [7 upvotes]: It is well known that reflective subcategories of complete categories are complete, and that limits in the subcategory are computed by taking the limit in the ambient category and applying the reflector (however it will act trivially,  see the comments). Has this been proven yet for $(\infty,1)$-categories? I know that if the ambient  $(\infty,1)$-category is (locally) presentable, and the subcategory is accessible that this is in HTT, however this is a very special case, and the latter condition is often hard to verify even when dealing with the presentable case. Has anything been worked out on this?

REPLY [3 votes]: Here's a proof which is certainly overkill, but it has the merit of using references so you can read the proofs in detail.
We have $i: \mathcal{C} \subset \mathcal{D}$ a fully faithful subcategory with $r$ a reflector.
Step 1. The inclusion $i$ is monadic. Proof: It is clearly conservative, and it preserves and reflects $i$-split simplicial objects since $i$ is fully faithful so we can realize the splitting already in $\mathcal{C}$. By Barr-Beck (HA.6.2.2.5) the functor $i$ is monadic.
Step 2. Monadic functors 'create' limits. Proof: This is the statement of HA.4.2.3.3. where the '$\mathcal{C}$' in that corollary corresponds to $\text{End}(\mathcal{C})$ here, the $\mathcal{M}$ corresponds to our $\mathcal{C}$, and the algebra $A$ corresponds to the monad $i \circ r$.<|endoftext|>
TITLE: Is $M_{1,n}$ affine?
QUESTION [5 upvotes]: A famous conjecture of Looijenga states that the moduli space of curves $M_{g,n}$ is the union of $g- \delta_{0,n}+ \delta_{0,g}$ open affine subsets, where $g,n$ are non-negative integers satisfying $2g-2+n>0$, and $\delta$ is the Kronecker delta.
I know of proofs of this conjecture in the case $(g,0)$ for $2 \leq g \leq 5$ (Fontanari-Looijenga and Fontanari-Pascolutti), and in the case $(0,n)$ for all $n$'s.
Is this conjecture true for $M_{1,n}$ ? Namely, is it known if $M_{1,n}$ is affine?
(Are there other cases when the conjecture is known?)

REPLY [9 votes]: Yes, $M_{1,n}$ is affine. More generally, $M_{g,n+1}\to M_{g,n}$ is an affine morphism for any $n \geq 1$.<|endoftext|>
TITLE: Deformations of smooth projective hypersurfaces and the Jacobian ring 
QUESTION [11 upvotes]: It is a well-known result of Griffiths that the pieces of Hodge filtration of a smooth hypersurface $X:=  (f=0)$ of degree $d$ in $\mathbb{P}^{n}$ are isomorphic to graded pieces of the Jacobian ring associated to $X$ i.e the ring $R= \mathbb{C}[x_{1},...x_{n}]/J$ where $J$ is the ideal generated by the partial derivatives of $f$ . The isomorphism is given by means of the so-called Poincare residue. There is another isomorphism $R^{d} \cong H^{1}(X,T_{X})$ (both are isomorphic to the space of infinitesimal deformaions of $X$) . I don't know how this isomorphism is explicitly given.  In other words, I am wondering that in a concrete example how one can explicitly find the image of an element of $H^{1}(X,T_{X})$  in $R^{d}$under this last isomorphism. More exactly if I have a 1-Cech cocycle of the bundle $T_{X}$, how can I explicitly associate a homogenous polynomial of degree $d$ in $R^{d}$ to it? For simplicity you can consider the family $y^{4}= x(x-1)(x+1)(x-\lambda)$ of plane curves of degree $4$.

REPLY [5 votes]: For $d\geq 5$, for a smooth plane curve of degree $d$, it is simply not true that the natural map $R_d \to H^1(X,T_X)$ is surjective: there are plenty of abstract deformations which are not plane curves.  I am including below the proof that the natural map is surjective when $n\geq 4$, when $n=3$ and $d\leq 3$, and when $n=2$ and $d\leq 4$.  
The basic idea is to look at the short exact sequence of sheaves
$$ 0 \to T_X \to T_{\mathbb{P}^n}|_X \to \mathcal{O}_{\mathbb{P}^n}(d)|_X \to 0.$$
Using the long exact sequence of cohomology, once you prove that $h^1(X,T_{\mathbb{P}^n}|_X)$ equals $0$, then it follows that the connecting map 
$$H^0(X,\mathcal{O}_{\mathbb{P}^n}(d)|_X) \to H^1(X,T_X)$$ 
is surjective.  Of course the image of $H^0(X,\mathcal{O}_{\mathbb{P}^n}(d)|_X)$ is $R_d$.  Finally, using the long exact sequence of cohomology associated to the Euler sequence,
$$ 0 \to \mathcal{O}_{\mathbb{P}^n}|_X \to \mathcal{O}_{\mathbb{P}^n}(1)^{\oplus n+1}|_X \to T_{\mathbb{P}^n}|_X \to 0,$$
to prove that $h^1(X,T_{\mathbb{P}^n}|_X)$ equals $0$, it suffices to prove that $h^1(X,\mathcal{O}_{\mathbb{P}^n}(1)|_X)$ and $h^2(X,\mathcal{O}_X)$ both equal $0$.  These cohomologies can be computed from the usual computation of $H^q(\mathbb{P}^n,\mathcal{O}_{\mathbb{P}^n}(r))$ using the long exact sequence of cohomology associated to the short exact sequence,
$$
0 \to \mathcal{O}_{\mathbb{P}^n}(r-d) \to \mathcal{O}_{\mathbb{P}^n}(r) \to \mathcal{O}_{\mathbb{P}^n}(r)|_X \to 0
$$
For $n\geq 3$, always $h^1(X,\mathcal{O}_{\mathbb{P}^n}(1)|_X)$ equals $0$, and the second vanishing holds so long as $\text{dim}(X) \geq 3$, i.e., $n\geq 4$.  It is not hard to prove that the second vanishing also holds if $n=3$ and $d\leq 3$.  It definitely fails if $n=3$ and $d=4$, i.e., it fails for quartic K3 surface (although, in fact, the natural map is surjective again for $d\geq 5$).  
This brings us to the case that $n=2$, i.e., $X$ is a plane curve.  By adjunction and Serre duality on $X$, $h^1(X,\mathcal{O}_X(d))$ equals $0$.  So the natural map is surjective if and only if $h^1(X,T_{\mathbb{P}^n}|_X)$ equals $0$.  By the Euler sequence again, this holds if and only if the map
$$
H^1(X,\mathcal{O}_X) \to H^1(X,\mathcal{O}_X(1))^{\oplus 3} 
$$
is surjective.  If $d$ equals $1$, $2$ or $3$, the second group is already zero.  Finally, if $d$ equals $4$, your case, then by Serre duality the transpose map is
$$
H^0(X,\mathcal{O}_X)^{\oplus 3} \to H^0(X,\mathcal{O}_X(1))
$$
So you are reduced to the question of whether the restriction map
'$$H^0(\mathbb{P}^2,\mathcal{O}_{\mathbb{P}^2}(1)) \to
H^0(X,\mathcal{O}_X(1))
$$'
is surjective, i.e., whether the next term $H^1(\mathbb{P}^2,\mathcal{O}(-3))$ equals $0$, which it does.
Once again, for $d\geq 5$, the natural map is not surjective.
Regarding your original question, how to "transform" a cocycle into an element of $R_d$: on the basic open affine covers of $\mathbb{P}^2$, write the corresponding Cech 1-cocycle for $T_{\mathbb{P}^2}|_X$ as the coboundary of Cech 0-cochain, which you can do since the cohomology group vanishes. Then take the image 0-cochain in $\mathcal{O}(d)|_X$.  The coboundary of this 0-cochain will be the image of your original 0-cochain in $\mathcal{O}_X(d)$, which is zero since the composite map $T_X \to \mathcal{O}_X(d)$ is zero.  So your 0-cochain is a 0-cocycle, i.e., a global section in $R_d$.<|endoftext|>
TITLE: Is there a sense in which the homotopy theory of simplicial sets is the "paradigmatic" one?
QUESTION [14 upvotes]: I could not come up with a better title for my question. 
What I am asking is this (preemptive excuses to all experts in  homotopy theory for naivetes of all kinds you may find herein): 
the category of simplicial sets has always been to me something distinguished, on multiple counts: 
to begin with,  it is simple to describe, almost childish,  yet it has a seemingly unfathomable richness.
Its homotopy theory, although usually presented via the geometric realization functor (which is historically correct), is in fact entirely self-contained, and purely combinatorial in character.
Also, simplicial sets plays a very special role in category theory (after all, categories are just some almost trivial example of simplicial sets) and, even more important, in higher dimensional cats. 
Somehow, I have the lingering feeling that, in abstract homotopy theory, simplicial sets (or, more properly simplicial objects in some ambient category ) should be, at least for some suitable notion of "regularity" of homotopy theories, paradigmatic. I am thinking of something like:
If an homotopy theory is "regular...." (whatever that may mean, fill the dots, the "regular" would stand for combinatorial in essence), then it is representable in the homotopy theory of simplicial objects for some ambient category.
Am I dreaming or there is something along these lines?

REPLY [6 votes]: This requires some topos theory I'm afraid, but gives, I think, a satisfactory explanation.
The category $\mathrm{Set}^{\Delta^{\mathrm{op}}}$ $\textit{classifies intervals}$: the simplicial set $\Delta[1]$ is linearly ordered, with smallest and largest elements $0,1:\Delta[0]\to\Delta[1]$, and for any topos $\mathscr E$ and any linearly ordered object $I\in\mathscr E$ with distinguished smallest and largest elements $\bot,\top:1\to I$ there is a unique up to isomorphism $f_I:\mathscr E\to\mathrm{Set}^{\Delta^{\mathrm{op}}}$ with $f_I^*(\Delta[1])=I$. In particular, one can form various $\mathscr E$ containing one or other category of spaces, and for the corresponding interval $I$ there, $f_I^*(S)$ is the geometric realization of the simplicial set $S$, while for a space $X$, ${f_I}_*(X)$ is the singular simplicial set of $X$.
Thus simplicial sets are determined (up to equivalence of categories) by the fact that they classify all possible notions of a continuous path, hence of a homotopy.
Reference - e. g. in ncatlab; first occurrence of this I've seen is in Johnstone's first topos theory book, where it is attributed to Joyal.<|endoftext|>
TITLE: Distinguishing two local versions of the axiom of choice
QUESTION [9 upvotes]: Two equivalent formulations of the axiom of choice are:

Every family $(X_i)_{i \in I}$ of pairwise disjoint nonempty subsets of a set $X$ has a choice function.
Every family $(X_i)_{i \in I}$ of nonempty subsets of a set $X$ has a choice function.

However, the usual proof of the implication (1) → (2) replaces the set $X$ with the set $X \times I$ and extracts a choice function for $(X_i)_{i \in I}$ from a choice function for the pairwise disjoint family $(X_i\times\lbrace i \rbrace)_{i \in I}$ of nonempty subsets of $X \times I$. Since $X \times I$ can be much larger than $X$, there is no reason to believe that (1) → (2) for a fixed set $X$.
For a given set $X$, (2) has a maximal instance where the family $(X_i)_{i \in I}$ consists of all nonempty subsets of $X$. We therefore see that (2) is equivalent to:

There is a choice function $\mathcal{P}(X)\setminus\lbrace\varnothing\rbrace\to X$.
The set $X$ is wellorderable.

For a given set $X$, (1) is equivalent to: 

Every surjection $q:X \to Y$ has a right inverse.
Every equivalence relation on $X$ has a transversal.

It appears that (2) is indeed stronger than (1) for a fixed set $X$ and I feel that this should be well known, but I don't recall a model of ZF (or ZFA) where some set $X$ satisfies (1) but not (2). Does anyone know such a model? A model of ZF where $X = \mathbb{R}$ satisfies (1) but not (2) would be most interesting.

REPLY [7 votes]: Suppose that $X$ is a strongly amorphous set, that is an amorphous set that every partition has only finitely many non-singletons parts. It is clear that there is no choice function from every family of non-empty subsets of $X$, as $X$ cannot be split into two disjoint infinite sets.
If $\cal F$ is a pairwise-disjoint family of subsets of $X$ then without loss of generality $\cal F$ is a partition of $X$ (otherwise simply add the complement of $\bigcup\cal F$). If this partition is finite, then of course there is a choice function, however if the partition is infinite then all but finitely many sets in the partition are singletons, and the choice function is trivial (choose from the non-singletons, and the singletons allow only one choice).

Some remarks on the case when $X=\mathbb R$ (which may not be consistent, though):

Since every countable family can be made disjoint this implies countable choice for sets of real numbers.
In fact, this means that every ordinal $\kappa$ which $\mathbb R$ can be mapped onto, can be mapped into $\mathbb R$, so there are $\aleph_1$ many real numbers. In particular $\aleph(\mathbb R)=\Theta(\mathbb R)$ (where $\Theta$ is the least ordinal that there is no surjection onto it from $\mathbb R$).
Every family of at most $\frak c$ many sets must have a choice function. We can consider a bijection of $\mathbb R$ with $\mathbb R^2$, and if $\lbrace A_r\mid r\in\mathbb R\rbrace$ is a family, then the preimage of $\lbrace r\rbrace\times A_r$ form a disjoint family.
To complement the above one, every partition of $\mathbb R$ has size of at most $\frak c$ since there exists a choice function back into $\mathbb R$.

The above rule out the 'usual' models where $\mathbb R$ is not well-orderable, e.g. Feferman-Levy, Solovay, ZF+AD, Cohen's first model. It could be very well that the above already show that there cannot be a model in which $X=\mathbb R$ but I cannot see a reason why (or a model where it is true).<|endoftext|>
TITLE: n-categories enriched in (n+1)-categories
QUESTION [10 upvotes]: Recall the notion of an $n$-cateogry $C$ enriched in a symmetric monoidal category.  Instead of a set of $n$-morphisms $mor(a, b)$ (where $a$ and $b$ are compatible $(n{-}1)$-morphisms), we have an object $mor(a, b)$ in some symmetric monoidal category $S$.  Composition of $n$-morphisms in $C$ takes the form of an $S$-morphism $mor(a, b)\otimes mor(b, c)\to mor(a, c)$.
I've come across some examples which seem to be well-described by the following generalization of enrichment.  We replace the symmetric monoidal category $D$ with an $(n{+}1)$-category $D$.  Each $k$-morphism of $C$ is assigned a $k$-morphism of $D$, for $0\le k < n$.  (These assignments must of course satisfy various compatibility conditions.  I'm only trying to sketch the rough idea here.)  If $a$ and $b$ are compatible $(n{-}1)$-morphisms of $C$, $mor(a, b)$ is an $n$-morphism of $D$ (whose domain and range are determined my the assignments for $a$ and $b$).  Composition of $n$-morphisms of $C$ takes the form of an $(n{+}1)$-morphism in $D$ from $mor(a, b) \bullet mor(b, c)$ to $mor(a, c)$, where $\bullet$ denotes composition of $n$-morphisms in $D$.
(To recover the more familiar symmetric-monoidal-category-enriched case, let $S$ be a symmetric monoidal category and let $D$ be the $(n{+}1)$-category whose $n$-morphisms are objects of $S$, whose $(n{+}1)$-morphisms are morphisms of $S$, and whose $k$-morphisms are trivial for $k < n$.)
My question is:

Does anyone know of references which discuss versions of enrichment similar to this, or which give natural examples of it?

All I've been able to find so far is Tom Leinster's article Generalized Enrichment for Categories and Multicategories, which discusses a very general version of $n$-categories enriched in $(n{+}1)$-categories.  I'm not certain whether what I describe above can been seen as a special case of Leinster's definition.

REPLY [8 votes]: The case $n=1$ is well-known in the category-theoretic literature and is called (of course) "categories enriched in a bicategory".  It was introduced by Betti and Carboni (Cauchy-completion and the associated sheaf) and Walters (Sheaves and Cauchy-complete categories), and studied by Street (Enriched categories and cohomology) and other authors as well.  Important examples include:

If $S$ is a site and $D$ is the bicategory whose objects are those of $S$ and whose morphisms are sieves, then $D$-enriched categories that are both symmetric and Cauchy-complete can be identified with sheaves on $S$. See Walters' paper Sheaves on sites as Cauchy-complete categories.
If $D$ is the bicategory of spans in a category $S$ with pullbacks, then $D$-enriched categories are very closely related to locally small fibrations over $S$.  See Variation through Enrichment by all four authors mentioned above.
If $D$ is the bicategory constructed from a monoidal fibration as in this paper of mine, then $D$-enriched categories are a sort of category that is "both indexed and enriched".

Personally, I think it is better in most cases to enrich in a double category than in a bicategory: in both examples above, $D$ underlies a double category.  Recognizing this doesn't change the notion of "$D$-enriched category" but it does give a wider notion of $D$-enriched functor, which in particular makes the second example into an honest equivalence with locally small fibrations.  There is a similar equivalence for the third example.  (I don't know of anywhere that this is written down yet.)
Leinster's notion of "category enriched in an fc-multicategory" is a bit more general even than this: an fc-multicategory (which I prefer to call a virtual double category) is to a double category in the same way that a multicategory is to a monoidal category.  I do know of some examples that require this full generality, but they are kind of contrived.  For instance, if instead of a monoidal fibration you start with a "multicategorical fibration", then instead of a double category you get a virtual double category.
I'm not quite sure how you want to generalize this to $n>1$.  One possibility would be to observe that when $D$ is a bicategory, a $D$-enriched category is the same as a lax functor into $D$ whose domain is the chaotic 1-category on some set of objects.  But for $n$-categories with $n>2$, there is probably no unique notion of "lax functor"; you would have to choose at which level(s) you want it to be lax.  I guess you might also choose different inputs to make the domain "chaotic on".
One version which is certainly useful is to keep the domain as a chaotic 1-category, and make the functors lax at the bottom level but pseudo everywhere above that.  In some sense this is the most straightforward "categorification", which produces an obvious generalization of the notion of bicategory enriched in a monoidal bicategory.  I'd be interested to hear what other examples you have in mind!<|endoftext|>
TITLE: G-equivariant Whitehead's Theorem
QUESTION [7 upvotes]: Suppose $X$ is a CW complex and $Y$ is a subcomplex.  Let $G$ be a compact Lie group that acts on $X$ and $Y$.  Suppose further that the CW structures on $X$ and $Y$ are $G$-stable.  Moreover assume that $\pi_n(X/G)\cong \pi_n(Y/G)$ for all $n\geq 0$ and are induced by the cellular inclusion $Y/G\hookrightarrow X/G$.
Whitehead's Theorem implies that there is a strong deformation retraction (SDR) from $X/G$ to $Y/G$.
In this setting, does there exist a $G$-equivariant SDR from $X$ to $Y$?
If not, what if one further assumes the existence of a SDR from $X$ to $Y$ (not assumed $G$-equivariant).  Would that then imply the existence of a $G$-equivariant SDR from $X$ to $Y$?
EDIT:  After Tom Goodwillie answered both questions negatively, I have decided to add another assumption; namely, assume that the fixed point set $X^G$ is contained in $Y$ (or perhaps assume that $X^G$ $G$-equivariantly retracts to a subspace of $Y$).

REPLY [10 votes]: The usual statement is that if $X\to Y$ is an equivariant map of $G$-CW complexes and if for every closed subgroup $H$ the induced map of fixed-point spaces $X^H\to Y^H$ is a homotopy equivalence then in fact the map has an inverse up to equivariant homotopy. 
This is part of the following picture: The category of $G$-spaces has a model structure in which a $G$-map $X\to Y$ is called a weak equivalence (resp. fibration) if and only if for every subgroup $H$ the resulting map $X^H\to Y^G$ is a weak equivalence (resp. fibration). As generating cofibrations you can use the inclusions $G/H\times S^{n-1}\subset G/H\times D^n$ for all $H$ and $n$. In particular $G$-CW complexes are cofibrant. And strong homotopy equivalence in the model category sense becomes equivariant homotopy equivalence in the obvious sense. Thus the statement above becomes an instance of the general principle that for cofibrant objects every weak equivalence is a strong equivalence.
By the way, a reasonable question is, is the analogous statement true with orbits instead of fixed points? That's not true, though. Take a CW space $Z$ that is acyclic but not contractible. Let $X$ be the suspension of $Z$, and let $G$ of order $2$ act on it by switching the two cones, with $Z$ as fixed point set. Then both $X$ and the orbit space are contractible, but $X$ is not equivariantly contractible because the fixed point set is not contractible.<|endoftext|>
TITLE: Determining rational functions by their critical points
QUESTION [14 upvotes]: Fix an integer $d > 1$ and $2d-2$ points $P_1, \ldots, P_{2d-2}$ in the Riemann sphere (not necessarily distinct). Thanks to the work of Eisenbud and Harris on limit linear series (Inventiones, 1983), we know that there are only a finite number of rational functions $\phi(z)$ of degree $d$ with complex coefficients that are ramified exactly at the points $P_i$, up to postcomposition by invertible rational functions. (The latter preserve the ramification points, so they must be taken into account.) Suppose that $\phi_1, \ldots, \phi_n$ are a maximal collection of such functions with the property that $\phi_i = \sigma \circ \phi_j$ for some $\sigma \in \mathrm{Aut}(\mathbb{P}^1)$ implies $i = j$. 

Is there an invariant that allows one to distinguish between $\phi_1, \ldots, \phi_n$? 

If $P_1, \ldots, P_{2d-2}$ are all distinct, then Goldberg (Advances in Math, 1991) showed that the number of postcomposition classes of rational functions ramified at exactly these points is positive and bounded above by the Catalan number $\rho(d) = \frac{1}{d}\binom{2d-2}{d-1}$. Moreover, if the $P_i$ are in general position, then the number of classes is exactly $\rho(d)$. 
For example, when $d = 3$, we may normalize the ramified points to be at $0, 1, \infty$, and $c$, and we may further assume that $0, 1$, and $\infty$ are fixed after postcomposing by a suitable automorphism. Then Goldberg's result says there are precisely 2 rational functions of this shape for a general choice of $c \in \mathbb{C}$. How do we distinguish between them? 
One can work out this example explicitly to see that 
$$ \phi(z) = \frac{\alpha z^3 + (1-2\alpha)z^2}{(2-\alpha) z - 1}.$$ 
The fourth critical point is $c = \frac{2\alpha - 1}{\alpha(2-\alpha)}$, so we require that $\alpha \neq 0, \pm 1,2^{\pm 1}$ in order to have four distinct critical points. A given $c$ generically determines two values of $\alpha$. But a priori, what data can I specify in order to nail down one or the other of these two functions?

REPLY [6 votes]: If all of the points $P_i$ lie on a circle $\gamma \subset \mathbb{CP}^1$, there is a beautiful description in
Eremenko and Gabrielov
Rational functions with real critical points and the B. and M. Shapiro conjecture in real enumerative geometry.
Ann. of Math. (2) 155 (2002), no. 1, 105–129. 
Let $\phi: \mathbb{P}^1 \times \mathbb{P}^1$ be one of the rational maps. The circle $\gamma$ divides $\mathbb{P}^1$ into two hemispheres; call them $N$ and $S$. The image $\phi(\gamma)$ is contained in a circle; call it $C$. (This is NOT obvious.)
Consider $\phi^{-1}(C) \cap N$. (An earlier version of this answer wrote $\phi^{-1}(\phi(\gamma))$, but I want the inverse image of the whole circle $C$, which might be larger.) This is a collection of noncrossing arcs whose end points lie at the points $P_i$. Ermenko and Gabrielov show that there is precisely one rational function $\phi$ for each possible connectivity of these arcs. For example, there are $5$ possible ways to draw $3$ noncrossing arcs connecting $6$ points on the boundary of a disc, and there are $5$ degree $4$ maps with $6$ specified critical points.
I would be very interested in knowing a generalization of this result to the case where the points do not lie on a circle; I am fairly certain none is known.<|endoftext|>
TITLE: How can I tell whether a manifold is homogeneous?
QUESTION [9 upvotes]: I have been influenced by this question with many beautiful answers.
Are there any useful practical criteria to say positively that a real connected paracompact smooth manifold $X$ is homogeneous?
I can think of something silly like existence of a finite dimensional Lie algebra of vector fields spanning $T_aX$ at each $a\in X$ but I don't see any practical way of constructing such subalgebra besides giving a homogeneous structure:-)

REPLY [3 votes]: as mentioned in various comments there are many geometric conditions of various type that imply  homogeneity (or local homogeneity).  Constant curvature, parallel curvature tensor, almost flat metric, quarter-pinched curvature, nonnegative bisectional curvature  and so on. But in purely topological terms the best you can hope for is necessary conditions of the kind mentioned in the question you linked. Necessary and sufficient conditions are pretty much impossible because topological recognition problems are hard.
For example,  if you look at compact 2-connected homogeneous spaces (or even biquotients) then there are only finitely many of them in every dimension since nonabelian simple Lie groups have only finitely many irreducible representations in every dimension. so in principle the recognition problem for such manifolds ought to be straightforward: just check that that $\pi_1(M)=\pi_2(M)=0$ and compare $M$ to a finite list. However, that last step is actually quite tricky. Surgery theory sort of tells us how to do it but the method is hardly practical.
To give a specific example from my own experience. When classifying biquotients with singly generated rational cohomology rings I and Wolfgang Ziller had to deal with one specific biquotient $M^{11}=G_2//SU(3)$ given by a representation $\rho:SU(3)\to G_2\times G_2$ where the representation on the left has  index 2 and on the right has index 3. We were able to show that $M^{11}$ is almost diffeomorphic to the unit tangent bundle $T^1S^6$ which is the homogeneous space $SO(7)/SO(5)$. this means that they differ at most by a connected sum with an exotic sphere. but we couldn't decide if they are actually diffeomorphic. to do that one needs to compute the Eells-Kuiper invariant of $M$ which would couldn't do as that requires writing $M$ as a boundary.<|endoftext|>
TITLE: Normalizers in symmetric groups
QUESTION [62 upvotes]: Question: Let $G$ be a finite group. Is it true that there is a subgroup $U$ inside some symmetric group $S_n$, such that $N(U)/U$ is isomorphic to $G$? Here $N(U)$ is the normalizer of $U$ in $S_n$.
Background: If true, this would for instance give a trivial proof of the Fried-Kollar Theorem that every finite group is the full automorphism group of a number field.
Results: If $U\le S_n$ acts regularly with respect to the natural action of $S_n$, then $N(U)/U\cong\text{Aut}(U)$. However, many finite groups are not the automorphism group of another finite group, like most cyclic groups. On the other hand, it is easy to get $N(U)/U\cong G$ for each abelian $G$ by choosing $U$ a direct product of semidirect products $C_{p_i}\rtimes C_{m_i}$ for suitable distinct primes $p_i$ and divisors $m_i$ of $p_i-1$, with the natural intransitive action of $U$ with orbit lengths $p_1, p_2,\dots$.
Added recently (answering Stefan Kohl's question from the comments): $Q_8$ is a normalizer quotient in $S_{81}$. Let $U=\mathbb F_3^4\rtimes H$ be the primitive group of degree $81$ where $H=C_5\rtimes C_8$ with $C_8$ inducing an automorphism group of order $2$ on $C_5$. Then $N_{S_{81}}(U)/U=Q_8$. This can be seen by hand, or using GAP:

gap> u:=PrimitiveGroup(81,27);;
gap> nu:=Normalizer(SymmetricGroup(81),u);;
gap> w:=nu/u;;
gap> Order(w);
8
gap> IsQuaternionGroup(w);
true

Remark 1: $N_{S_{81}}(U)$ is the semiaffine group ${A\Gamma L}_1(\mathbb F_{81})$.
Remark 2: $U$ in Magma is PrimitiveGroup(81,26).

REPLY [2 votes]: I have come across a fairly recent result which pertains to this question. It is in this paper:
Guralnick, Robert M.; Maróti, Attila; Pyber, László, Normalizers of primitive permutation groups, Adv. Math. 310, 1017-1063 (2017). ZBL1414.20002.
An arXiv version is here. The paper examines the situation when $U$ is primitive. They show that, in all but a finite number of situations, $|N(U)/U|<n$. Indeed, they strengthen this bound if you add in a particular infinite family. The main result is this one:
Theorem: Let $U$ be a primitive subgroup of $S_n$, and let $N=N_{S_n}(U)$.  Then $|N/U|<  n$ unless $U$ is an affine primitive permutation group and the pair $(n, N/U)$ is one of:
$$(3^4,O^−_4(2),(5^4,Sp_4(2)),(3^8,O^−_6(2)),(3^8,SO^−_6(2)),(3^8,O^+_6(2)),(3^8,SO^+_6(2)),(5^8,Sp_6(2)),(3^{16},O^−_8(2)),(3^{16},SO^−_8(2)),(3^{16},O^+_8(2)), \textrm{ or }(3^{16},SO^+_8(2)).$$ Moreover if $N/U$ is not a section of $\Gamma L_1(q)$ when $n=q$ is a prime power, then $|N/U|< n^{1/2}\log n$ for $n≥2^{14000}$.
@YCor's comment on the original question suggests that the primitivity assumption is not too onerous. It would be interesting (but probably very hard) to try and understand what might happen when $U$ is transitive and imprimitive.<|endoftext|>
TITLE: Axioms for Riemann $\zeta$ function
QUESTION [9 upvotes]: Are there any set of axioms that completely characterize the Riemann zeta function?
 i.e. like Ressayre axioms for the exponential function in an exponential field or functional equations.

REPLY [5 votes]: There are in fact at least two axiomatic characterizations of $\zeta(s).$  One of them is given by Hecke and one of them was given by Hamburger.
Hamburger's Theorem states:  Suppose $D(s)$ is a dirichlet series, convergent in some half plane and whose coefficients have polynomial growth.  If $$$$
1) There exists a polynomial $P(s)$ so that $P(s)D(s)$ is entire and of finite order.
2) $D(s/2)$ is also dirichlet series. That is, the coefficients of $D(s)$ are supported on squares.
3) If $\xi(s)= \pi^{-s}\Gamma(s)D(s)$ then $\xi(1/2-s)=\xi(s).$
Then $D(s)=C\zeta(2s).$
Hecke's version states: Suppose $D(s)$ is a dirichlet series, convergent in some half plane and whose coefficients have polynomial growth.  If $$$$
1) $(s-1/2)D(s)$ is entire and of finite order. 
2) The coefficients of $D(s)$ have arbetrary support.
3) If $\xi(s)= \pi^{-s}\Gamma(s)D(s)$ then $\xi(1/2-s)=\xi(s).$
Then $D(s)=C\zeta(2s).$
A little extra information: These theorems cannot be combined. The so-called "Big Mac Theorem" where$$$$
1) There exists a polynomial $P(s)$ so that $P(s)D(s)$ is entire and of finite order.
2) The coefficients of $D(s)$ have arbetrary support.
3) If $\xi(s)= \pi^{-s}\Gamma(s)D(s)$ then $\xi(1/2-s)=\xi(s).$
produces infinitely many linearly independent dirichlet series!
Sources:
http://www.rowan.edu/open/depts/math/HASSEN/Papers/paper1.pdf<|endoftext|>
TITLE: Product of Fibonacci numbers
QUESTION [9 upvotes]: Consider the counting function
$$
f(x)=|\{n\le x:n\text{ is a product of Fibonacci numbers}\}|
$$
so for example $f(4)=4=|\{1,2,3,4\}|$ since 1, 2, and 3 are Fibonacci numbers and $4=F_3\cdot F_3.$ (See A065108.)
What is known, asymptotically, about the growth of $f$?
It's clear that for any $k$, $f(x)\gg(\log x)^k$ (this can be made effective without too much work), and it doesn't seem likely that $f(x)\gg x^k$ for any $k>0$.

REPLY [9 votes]: I think the following is related to Pietro Majer's comment:
Upper bound: The first $2$ Fibonacci numbers are $1$, so we can leave them out of products. The $n$th Fibonacci number is greater than $\phi^{n-2}$ for $\phi = \frac{1+\sqrt 5}{2}$. So, for any product of Fibonacci numbers less than $x$, there is a corresponding sum of exponents which is less than $\log_\phi x$. This means the number of integers up to $x$ which are products of Fibonacci numbers is at most the number of partitions of size up to $\log_\phi x$.
The number of partitions of $m$ are known to be asymptotically $\frac{1}{4\sqrt3 ~m} \exp (\pi \sqrt{\frac23m})$. the number of partitions is increasing, so the number of partitions of size up to $m$ is at most $m$ times this, or asymptotically $\frac{1}{4\sqrt3} \exp (\pi \sqrt{\frac23 m})$. So, 
$$f(x) \lt \frac{1+o(1)}{4\sqrt3} \exp (\pi \sqrt{\frac 23 \log_\phi x}).$$
This is not sharp since some Fibonacci numbers are products of others: $8 =2^3, 144 = 2^4 3^2$. I don't know of any other such examples, so I think the growth rate is bounded above and below by functions of the form $\exp(c\sqrt{\log n})$.
Lower bound: $F_k \lt \phi^{k-1} \lt \phi^k$. We would like to construct distinct products of Fibonacci numbers out of some restricted partitions of $n \lt \log_\phi x$. 
Note that $(F_m, F_n ) = F_{(m,n)}$. So, the Fibonacci numbers of prime index are relatively prime. We can't use the second Fibonacci number because that's $1$. However, partitions of $n \lt \log_\phi x$ into odd prime parts correspond to distinct products of Fibonacci numbers up to $x$. A crude lower bound is that there are at least $(1+o(1))\frac{\sqrt {2n}}{\log \sqrt{2n}}$ odd primes up to $\sqrt{2n}$ by the prime number theorem, and the subsets correspond to at least $\exp (c \frac {\sqrt{n}}{\log n})$ distinct partitions of numbers up to $n$ into odd primes hence to distinct products of Fibonacci numbers. So, for some $c \gt 0$, there is a lower bound for $x$ large enough for the formula to make sense of the form
$$\exp(c \frac{\sqrt{\log_\phi x}}{\log_e\log_\phi x}) \lt f(x).$$
See also this MO question on the number of partitions into odd primes.

REPLY [4 votes]: A follow-up to the comment of Anonymous which addresses your question exactly: see slides 9-13 here for an investigation of your question. Basically, it can be proved that the decomposition into a product of Fibonacci numbers is more or less unique (if you ignore the "special" Fibonacci numbers $f_1=1$, $f_2=1$, $f_6=8$, and $f_{12}=144$), and it all reduces to the asymptotics of the partition function in the spirit of Hardy--Ramanujan--Rademacher.<|endoftext|>
TITLE: Completion and Tensor Product of Algebras
QUESTION [7 upvotes]: Let $A$ be a commutative ring with 1, $I$ an ideal in $A$, $B$ an $A$-algebra. I am trying to prove the following isomorphism of $A$-algebras:
$$ \big( A^* \otimes _A B \big) ^* \cong B^* $$
"$^*$" denotes the $I$-adic completion: every $A$-algebra $X$ may be endowed with the $I$-adic topology, defined by the ideal $IX$ in $X$, and the $I$-adic completion of the algebra is the completion with respect to this (uniform) topology.
I have so far been able to prove that the image of $B$ in $T :=A^*\otimes_A B$ under the map $1 \otimes id \colon B \to T$ is dense in $A$. At this stage I considered the map $(1 \otimes id)^* \colon B^* \to T^*$, and tried to show that it is an isomorphism, but I'm having troubles both with  the injectivity and the surjectivity of this mapping. 
Are these $A$-algebras always isomorphic? If so, how can this be proven? If not, how can a counterexample be constructed, and what do I have to require (Noetherity? Flatness? Finiteness?) for them to be isomorphic?

REPLY [3 votes]: Thank you for your answer, Konstantin!
In fact, since asking the question, my advisor, Prof. Dan Haran, has found a proof which does not use any Noetherity conditions:
For any $A$-algebra $C$, we denote by $\varphi_C$ the canonical mapping from $C$ to its $I$-adic completion $C^* $, and for an $A$-homomorphism $\psi \colon B \to C$ we denote by $\psi^* $ the induced $A$-homomorphism from $B^* $ to $C^* $.
Using the fact that the image of $A$ under $\varphi_A$ is dense in $ A^* $, one can directly see that the image of $B$ under $1 \otimes id$ is dense in $T : = A^* \otimes_A B$.
Therefore the image of $ B^* $ under the induced mapping $ (1 \otimes id)^* $ must also be dense in $ T^* $.
The algebra mapping $\beta \colon A \to B$ induces the mapping $\beta^* \colon A^* \to B^* $, and using the universal property of the tensor product (as a co-product in the category of commutative $A$-algebras), we obtain a unique $A$-homomorphism $\psi \colon T \to B^* $ such that $\psi \circ (1 \otimes id) = \varphi_B $ and $\psi \circ (id \otimes 1) = \beta^* $.
Now $ \varphi_B^* = \varphi_{B^* } \colon B^* \to (B^* )^* $ is an isomorphism, and $ \psi^* \circ (id \otimes 1 )^* = \varphi_B^* $ shows that $ (id \otimes 1 )^* $ is an injection. Therefore we can view $ B^* $ as a complete dense $A$-subalgebra of $ T^* $, which means that:
$$ T^* = \overline{B^* } \cong (B^* )^* \cong B^* $$
(where $\overline{B^* }$ is the closure of $B^* $ in $T^* $ ).<|endoftext|>
TITLE: What is the largest tensor rank of $n \times n \times n$ tensor?
QUESTION [16 upvotes]: The tensor rank of a three dimensional array $M[i,j,k], i,j,k\in [1,\ldots,n]$ is the minimal number of vectors $x_i,y_i,z_i$, such that $M=\sum_{i=1}^d x_i\otimes y_i\otimes z_i$. 
From dimension argument it easily follows that there exists a tensor of tensor rank at least $\frac{1}{3}n^2$. One can also easily show that every tensor is of tensor rank at most $n^2$. 
So I know that the maximal tensor rank is between $\frac{1}{3}n^2$ and $n^2$. Does any one know what is the maximal tensor rank?
p.s. As far as I understand maximal border rank is $\frac{1}{3}n^2$.

REPLY [11 votes]: For tensors in $\mathbb{R}^3 \otimes \mathbb{R}^3 \otimes \mathbb{R}^3$ or in $\mathbb{C}^3 \otimes \mathbb{C}^3 \otimes \mathbb{C}^3$, the maximum rank is $5$. See Bremner, Hu, On Kruskal's theorem that every $3 \times 3 \times 3$ array has rank at most $5$, 2013 (MR3089693). (Their proof in the complex case also works for any algebraically closed field of characteristic not equal to $2$. Interestingly, over the field with two elements there are $3 \times 3 \times 3$ tensors of rank $6$.)
For tensors in $\mathbb{R}^n \otimes \mathbb{R}^n \otimes \mathbb{R}^n$ or in $\mathbb{C}^n \otimes \mathbb{C}^n \otimes \mathbb{C}^n$, for $n \geq 4$, the maximum rank is at most $\binom{n+1}{2}$. See Atkinson, Lloyd Bounds on the ranks of some 3-tensors, 1980 (MR0570374). (They state the result only over $\mathbb{C}$, but say that "however, many of our techniques remain valid for arbitrary fields of characteristic zero". I have not tried to figure out if their proof works over $\mathbb{R}$.) Alternatively, see Theorem 1 and Theorem 3 of Sumi, Miyazaki, Sakata, About the maximal rank of $3$-tensors over the real and the complex number field, 2010 (MR2652318). (Theorem 4 of this paper also gives the upper bound $5$ in the $n=3$ case.)
I think it is still open what the actual maximal rank is for $n \geq 4$, but this narrows down the range to be between $n^3/(3n-2)$ and $\binom{n+1}{2}$.<|endoftext|>
TITLE: Reference request for the list of maximal subgroups of SU(3,1)
QUESTION [5 upvotes]: Is there a reference with the list of maximal subgroups of SU(p,q) for "small" values of p and q? (such as SU(3,1) as suggested in the title of the question)

REPLY [4 votes]: Mohamed Selim Taufik, On maximal subalgebras in classical real Lie algebras: "This paper is concerned with the classification of irreducible maximal subalgebras of the classical real Lie algebras su(p,q), sv(p,q) and si(p,q). We use the results of E. B. Dynkin, who classified the maximal subalgebras of the classical Lie algebras in the complex case. (...)"
Boris P. Komrakov Maximal subalgebras of real Lie algebras and a problem of Sophus Lie: "A classification of the maximal proper subalgebras of the simple real finite-dimensional Lie algebras is presented without proof. Contributions by A. A. Morozov, E. B. Dynkin, M. Berger and M. S. Taufik are mentioned."<|endoftext|>
TITLE: K-theory, monoidal vs. exact
QUESTION [7 upvotes]: My question is somewhat related to this one. However I think it adds something new to the table so I decided to post it sperately.
There is a construction of K-theory for symmetric monoidal categories (smc) and one for exact categories. From what I understand they are (a priori) two different pair of shoes. 
For a symmetric monoidal category $S$ there is the classifying space $BS$. We define the K-theory $K(S)$ to be the group completion of this space. If we consider $BS$ to be a one-object topological category it is up to an equivalence of categories a strict monoidal category and a group completion is given by $BS\to\Omega BBS$.
If we now take our favorite category $P(R)$ of projective modules over a ring then the above construction isn't very interesting ($BS$ is contractible since $0\in S$ is an initial object). But we can take $iP(R)$ the isomorphism category and then $K(iP(R))$ magically coincides with the K-theory of $P(R)$ as an exact category (I write magically since I can read the proof but still don't really know why taking the isomorphism category is the sensible thing to do).
Something I like about the monoidal approach is Thomasons mapping cone construction. Given a functor of smc $A\to B$, such that $A=iA$ one can construct a smc $C$ such that there is a long exact sequence:
$$\cdots \to K_{i+1}(C)\to K_{i}(A)\to K_{i}(B)\to K_{i}(C)\to\cdots$$
Now I am in the situation that $A$ and $B$ are in fact exact categories. $A=P(R)$ and $B=P(R,\mathbb G_m)$ the category of projective modules equipped with an automorphism. 
So to apply Thomason's construction I may just take $A=iP(R)$, fair enough. May I also take $B=iP(R,\mathbb G_m)$ and get the K-theory of $P(R,\mathbb G_m)$ as an exact category (I have the feeling that we pass to direct sum K-theory here, or something like that)? If I now construct $C$ then I notice that $iC=iP(R,\mathbb G_m)$, so I definitely may not just take the isomorphism category here. Somehow I got the feeling that I am just jumping between the definitions and never really now whether it's justified.
So the question: For an exact category $D$, how are $K(D)$ (K-theory of $D$ as an exact category) $K(D)$ (as a symmetric monoidal category) and $K(iD)$ (again monoidal) related?

REPLY [7 votes]: As you note, the classifying space of any exact category is contractible because of the presence of an initial object, so your second K(D) (which I take to be the group completion of the classifying space of D under direct sum) is trivial.  But the other two (namely K(D), meaning the Q-construction K-theory of D, and (iD)^{gp}, the group completion of iD under direct sum) are closely related.  Indeed, there is always a map (iD)^{gp} --> K(D), and this map is an equivalence provided that every short exact sequence in D splits.
This is a non-trivial result.  As far as I'm aware, it's due to Quillen, but the first proof of it was given in Grayson's paper Higher Algebraic K-theory: II [after Daniel Quillen].  Actually the result is not explicitly stated in that paper, but it follows immediately from the first two theorems on its page 11 (well, I suppose one also needs to compare the S^{-1}S construction of that paper with the usual group completion -- e.g. using the group completion theorem -- or else redo the arguments with the usual group completion).  The proof is not that long, but it's sort of tricky, the trick being the theorem at the top of page 10.
Hope this helps!<|endoftext|>
TITLE: Is there a simple relation between the entropy of a matrix and its characteristic polynomial?
QUESTION [5 upvotes]: Assume $M$ is an invertible positive matrix of rank $N$.  The Von Neumann entropy $H$ of a matrix $M$ with eigenvalues $\{ \lambda_n \}$ is
$H[M] = -\sum_{n=1}^N \lambda_n \ln \lambda_n$.
In principle, the eigenvalues are encoded in the characteristic polynomial 
$\phi_t (M) = \mathrm{det}(tI-M) = \prod_n(t-\lambda_n) = t^{N} + a_{N-1} t^{N-1} \cdots + a_1 t +a_0 $.
The trace $\mathrm{Tr}\, M$ is given by the coefficient $a_{N-1}$ in the characteristic polynomial: 
$\lim\limits_{t \to \infty} \dfrac{\phi_t (M)-t^N}{t^{N-1}} = a_{N-1} = (-1)^{N-1} \, \mathrm{Tr}\, M = (-1)^{N-1} \sum_n \lambda_n$.
Is there a similar relationship between the entropy and the characteristic polynomial?

REPLY [6 votes]: I'm not sure if this qualifies as simple (or if this is helpful at all), but we have
$$
\frac{\phi'_M(t)}{\phi_M(t)}=\sum_n\frac{1}{t-\lambda_n}
$$
Using the residue theorem, we can write
$$
H[M]=\frac{-1}{2\pi i}\oint\frac{\phi'_M(z)}{\phi_M(z)}z\log(z)\,dz
$$
where the integral is taken over a closed contour containing all of the eigenvalues of $M$ (I guess we're either working on the Riemann surface of $\log(z)$, or we chose a branch of $\log(z)$).<|endoftext|>
TITLE: How many simplicial complexes on n vertices up to homotopy equivalence?
QUESTION [24 upvotes]: Fix a number $n$, and define $\gamma(n)$ to be the number of simplicial complexes on $n$ unlabeled vertices up to homotopy equivalence. It is unlikely that an explicit formula exists, but what is known about the growth of $\gamma(n)$ as $n$ increases?
This seems to be a fairly basic generalization of "how many non-isomorphic graphs on $n$ unlabeled vertices?" but while this problem even has an OEIS entry, I can't find any decent references or calculations for $\gamma$.
Note: I do not mean to ask about the Dedekind number which simply counts all possible simplicial complexes on $n$ vertices without regard to homotopy equivalence.

REPLY [13 votes]: Andrew Newman just posted a preprint to the arXiv showing that the answer is doubly exponential in $n$.
In particular, he showed that the number of homotopy types of simplicial complexes on $n$ vertices is at least
$$\exp \left( \exp \left( 0.004n \right) \right),$$
for all large enough $n$.
This matches the upper bound from Dedekind numbers, up to the constant $0.004$.
Newman's result depends on showing that this many different torsion subgroups are possible for homology in dimension $d$, where $d \approx \delta n$ for some small constant $\delta > 0$. The existence proof is partly constructive and partly depends on the probabilistic method.
https://arxiv.org/abs/1804.06787<|endoftext|>
TITLE: Serre's Open Image Theorem Without Shafarevich's Theorem
QUESTION [11 upvotes]: In Abelian l-adic Representations and Elliptic Curves (1968), J. P. Serre showed that the adelic representation $$\rho_{E}\colon G_K \to \mathrm{GL}(\hat{\mathbb{Z}}^2)$$ associated to an elliptic curve $E/K$ over a number field $K$ has open image. To do it, he uses Shafarevich's Theorem on the finiteness of isomorphism classes of elliptic curves in a given isogeny class to show that the $\ell$-adic representation $$\rho_{E,\ell}\colon G_K \to \mathrm{GL}(T_\ell(E))$$ is irreducible for all $\ell$ and that the mod $\ell$ representation $$\bar{\rho}_{E,\ell}\colon G_K \to \mathrm{GL}(E[\ell])$$ is irreducible for almost all $\ell$.
My question is, do we now have a method of proving this theorem without using Shafarevich's Theorem? The latter depends on Siegel's Theorem, which depends on Roth's Theorem in Diophantine Geometry.

REPLY [4 votes]: First, you forgot to assume that $E$ does not have CM. However, this actually suggests a difficulty in a Shafarevich-free proof.
Let $K = \mathbf{Q}(\sqrt{-1})$, and let $C/K$ be an elliptic curve with CM by $\mathbf{Z}[\sqrt{-1}]$. Now consider the following thought experiment. Can you rule out the existence of an elliptic curve $E/K$ without CM such that
$\rho_{E,\ell} = \rho_{C,\ell}$ for all primes $\ell$?
This is certainly implied by the Tate conjecture (in a case proved by Faltings), but not only is this harder than the original proof, it also really uses/implies a (generalization of) Shafarevich's conjecture. Certainly $E$ admits isogenies $E \rightarrow E'$ of degree $p$ for any prime $p$ which splits in $K$, but ruling this out is exactly Shafarevich again. I'm not sure you can overcome this obstacle.
On the other hand, it is elementary to (essentially) reduce to this case, basically using Serre's original argument. Namely, one reduces to the case that the $\rho_{E,\ell}$ are abelian, and a classification of crystalline characters of the right weight (plus purity) essentially reduces to this CM-like case.<|endoftext|>
TITLE: Multiplicity one conjecture
QUESTION [7 upvotes]: I recently became interested in Maass cusp forms and heared people mentioning a "multiplicity one conjecture". As far as I understood it, it says that the dimension of the space of Maass cusp form for fixed eigenvalue should be at most one. 
Since Maass cusp forms always are defined for a Fuchsian lattice, I wonder 
1) for which lattices this conjecture had been conjectured?
2) what is the motivation for this conjecture?
3) to whom this conjecture is due?
4) is it published somewhere?
5) is it proven in some cases?

REPLY [3 votes]: At the risk of blowing my own trumpet, I feel like I ought to mention a recent preprint of mine that addresses this question.
Marc Palm answered your question 2, 3, and 4 reasonably well. (For question 4, the reference for where it is published is Cartier's paper.) For question 1, I know that Bolte and Johansson certainly expect the conjecture to be true for $\Gamma_0(q) \backslash \mathbb{H}$ with $q$ squarefree, provided one first removes all the Maaß oldforms (those coming from lower level), as they will always give rise to spectral multiplicity at least $2$. So for $q > 1$, the conjecture should be modified to only be about the eigenvalues of Maaß newforms.
Bolte and Johansson (and later Strömbergsson) describe a spectral correspondence between the eigenvalues of Maaß newforms on $\Gamma_0(q) \backslash \mathbb{H}$, $q$ squarefree, and eigenvalues of the automorphic Laplacian for the group of units of norm one in a maximal order in an indefinite quaternion division algebra over $\mathbb{Q}$. Bolte and Johansson conjecture that the spectrum of this automorphic Laplacian is simple (see the  Hypothesis on p.61), and hence that the eigenvalues of Maaß newforms on $\Gamma_0(q) \backslash \mathbb{H}$ are simple when $q$ is squarefree. Here we are looking at the congruence subgroups
\[\Gamma_0(q) = \left\{\gamma \in \mathrm{SL}_2(\mathbb{Z}) : \gamma \equiv \begin{pmatrix} * & * \\\ 0 & * \end{pmatrix} \pmod{q}\right\},\]
\[\Gamma_1(q) = \left\{\gamma \in \mathrm{SL}_2(\mathbb{Z}) : \gamma \equiv \begin{pmatrix} 1 & * \\\ 0 & 1 \end{pmatrix} \pmod{q}\right\},\]
\[\Gamma(q) = \left\{\gamma \in \mathrm{SL}_2(\mathbb{Z}) : \gamma \equiv \begin{pmatrix} 1 & 0 \\\ 0 & 1 \end{pmatrix} \pmod{q}\right\}.\]
On the other hand, I show in my preprint that if $q$ is odd but not squarefree, then the new part of the spectrum of the Laplacian on $\Gamma_0(q) \backslash \mathbb{H}$ is never simple; there is always a positive proportion of eigenvalues $\lambda \leq T$ for which the corresponding eigenspaces of Maaß newforms are at least two-dimensional. For $\Gamma_1(q) \backslash \mathbb{H}$, the situation is even worse: there is spectral multiplicity even if $q$ is squarefree (provided $q \neq \{1,2,3,6\}$), and the dimension of an eigenspace can be proven to grow with $m$ if $q = p^m$ for an odd prime $p$. The proof doesn't use anything about the eigenvalue $1/4$, as in GH from MO's answer, but rather looks at twists of newforms that have the same level $q$ after twisting; this gives rise to spectral multiplicity.
There is one remaining case where I know that spectral multiplicity must occur; that of $\Gamma(p) \backslash \mathbb{H}$, with $p$ an odd prime; Randol shows that there must be infinitely many eigenvalues with multiplicity at least $\frac{1}{2} \left(p + (-1)^{(p - 1)/2}\right)$. I don't think anything is known for noncongruence subgroups.
I still think that for any congruence subgroup $\Gamma$, the multiplicity of an eigenvalue of the Laplacian on $\Gamma \backslash \mathbb{H}$ ought to be uniformly bounded as $\lambda \to \infty$, with the bound depending on $\Gamma$. Unfortunately our only tool for attacking this question (finding an upper bound on the multiplicity) seems to be via getting better error terms in the Weyl law, and we are very far off making that approach be useful.<|endoftext|>
TITLE: Viennot-type geometric description for dual RSK correspondence?
QUESTION [10 upvotes]: Is a geometric construction of the dual RSK correspondence along the lines of Viennot's "light and shadows construction" written up somewhere? This is a bijective correspondence between 0-1 matrices and pairs of SSYT with mutually transpose shapes.

REPLY [3 votes]: It's all written up rather nicely in Heather Dornom's Honours thesis from 2005. She gives a version of the matrix ball construction that works in these cases and also explains growth models for the RSK correspondence and its dual.<|endoftext|>
TITLE: Is the class of $k$-gonal curves dominant
QUESTION [6 upvotes]: Before I start, let me make a note on terminology. Curves are always smooth projective connected curves over an algebraically closed field of characteristic zero.
Let $\mathcal C$ be a class of curves. We say that $\mathcal C$ is dominant if, for all curves $X$, there exists a curve $Y$ in $\mathcal C$ and a finite morphism $Y\to X$.
Bogomolov and Tschinkel proved that the class of hyperelliptic curves and their unramified covers is dominant. Manin proved that the class of modular curves $X(n)$ and their unramified covers is dominant. Both proofs rely on Abhyankar's Lemma.
Let $k\geq 2$ be an integer. Let $\mathcal C_k$ be the class of $k$-gonal morphisms, i.e., the class of curves for which the gonality equals $k$.
Q1. Is $\mathcal{C}_2$ dominant?
Q2. Is $\mathcal{C}_k$ dominant?
Q3. Is $\cup_{2 \leq j \leq k} \mathcal{C}_j$ dominant if $k>>0$?
Let me repeat this in words. Let $X$ be a curve. Does there exist a $k$-gonal curve $Y$ and a finite morphism $Y\to X$?
I'm mainly interested in the case $k=2$. In this case, it suffices to answer the following question.
Q1b. Let $X$ be a curve. Does $\mathbf{P}^1$ admit a closed immersion into the symmetric product $X^{(2)}$?

REPLY [6 votes]: The points of $X^{(2)}$ correspond to divisors of degree 2; two points of $X^{(2)}$ are in the image of a morphism $\mathbb P^1 \to X^{(2)}$ if and only if the two corresponding divisors are linearly equivalent. Hence there is a non-trivial morphism $\mathbb P^1 \to X^{(2)}$ if and only if $X$ is hyperelliptic.
More generally, if $Y \to X$ is non-constant morphism and $Y$ is $k$-gonal, then $X$ is also $k$-gonal.<|endoftext|>
TITLE: Kostant partition function: asymptotics and specifics
QUESTION [9 upvotes]: Let $\Phi$ denote a root system and let $\mathfrak P$ denote the associated Kostant partition function. Thus $\mathfrak P(\lambda)$ is the number of ways of writing $\lambda$ as a sum of elements of $\Phi^+$. For example, if $\Phi=A_3$ then the longest root $\alpha_0$ can be written as $\alpha_0=(\alpha_1+\alpha_2+\alpha_3)$, $(\alpha_1)+(\alpha_2)+(\alpha_3)$, $(\alpha_1+\alpha_2)+\alpha_3$ or $\alpha_1 + (\alpha_2+\alpha_3)$. Thus $\mathfrak P(\alpha_0)=4$. There is a recursion $$\mathfrak P(\mu)=-\sum_{1\neq w\in W} (-1)^{l(w)}\mathfrak P(\mu+w(\rho)-\rho)$$ due to Kostant.
I can't find very many general results on $\mathfrak P$. 
One knows that $\mathfrak P((p-1)\rho)$ is the dimension of the $0$ weight space of the Steinberg module (see Jantzen, RAGS, II.10.12). I conjecture that one also has $\mathfrak P(r\rho)$ is the dimension of the $0$ weight space in $L_{\mathbb C}(r\rho)$, where $L_{\mathbb C}(r\rho)$ is the irreducible representation of high weight $r\rho$ for a complex Lie algebra with root system $\Phi$ (which may be zero if $r$ is odd). A reference even for this basic statement would be good!
[RE-EDIT: This paragraph, including the conjecture, is completely wrong. See Chuck Hague's and Jim Humphreys' remarks below. Of course one must have $\dim H^0(\lambda)_0\leq \mathfrak P(\lambda)$ for all $\lambda$, so this gives a lower bound.]
I'd also like some data on asymptotics. I have used the Maple `Coxeter/Weyl' package to compute $\mathfrak P(2\rho)$ for $A_n$ up to $n=5$. The numbers are $1,3,15,219,7834\dots$ and correspond to sequence A007081 on the OEIS. 
[EDIT: These calculated values are in fact the dimensions of the $0$ weight space of $H^0(2\rho)$. Since the conjecture that $\mathfrak P(2\rho)=\dim H^0(2\rho)_0$ is incorrect these do not coincide with the values of $\mathfrak P(2\rho)$ as claimed.]
I guess $\mathfrak P(2\rho) < n^n$. [EDIT: This looks very optimistic---suspect it's wrong for $n=4$.] Any ideas? What about $\mathfrak P (2r\rho)<(nr)^{nr}$ or something better? [EDIT: Similarly.] Lower bounds?

REPLY [2 votes]: I have some answers.
I couldn't find a computer program giving values of $\mathfrak P$ so I wrote one. Assuming it's giving the correct numbers, $\mathfrak P_{A_n}(2\rho)$ is 
$$1,3,26,898,128826,82462230$$ with natural logs
$$0,1.10,3.26,6.80,11.77,18.22.$$ The latter looks roughly quadratic to me. Note that the sequence does not come up on the OEIS.
As for proving some bounds... 
First of all, [Cline, Parshall, Scott; Reduced standard modules and cohomology] has a statement of a bound on page 5258. They state that it's easy to prove $\mathfrak P(2(h-1)\rho)\leq N!P(N(h-1))$ where $N=|\Phi^+|$ is the number of positive roots and $P$ is the usual partition function. Wikipedia tells me that $$P(n)\sim \frac{1}{4n\sqrt 3}\exp\left(\pi\sqrt{\frac{2n}{3}}\right)$$ so this gives $\log \mathfrak P(2(h-1)\rho)$ is $O(n^{3/2})$.
I think I can improve on this (just for type $A_n$ but it should be representative):
Observation 1: Looking up in the tables in Bourbaki, one gets $2\rho=n\cdot1\alpha_1+(n-1)\cdot2\alpha_2+\dots$. So the lowest coefficient is $n$ and the highest coefficient is roughly $(\frac{n+1}{2})^2$.
Observation 2: Once the coefficients of the non-simple roots in a sum of positive roots has been chosen, one completes the expression by adding in a determined number of simple roots to make up the difference. So $\mathfrak P(2r\rho)$ can be obtained by counting sets of positive non-simple roots whose sum is less than or equal to $2r\rho$ (in the dominance ordering).
Lower bounds: just use roots of height 2. Choose $\lfloor n/2\rfloor$ coefficients $b_i$ so that $$b_1(\alpha_1+\alpha_2)+b_2(\alpha_3+\alpha_4)+\dots\leq 2r\rho.$$ By Observation 2, each $b_i$ can be anything up to $rn$. Get $$\mathfrak P(2r\rho)\geq (rn)^{n/2}.$$ Probably not very good, but at least it shows it's exponential.
Upper bounds: There are $n(n-1)/2$ non-simple roots and by Obs 2 each can occur with any coefficient which at most $r(n+1)^2$. So get $$\mathfrak P(2r\rho)\leq (r(n+1))^{n(n-1)}.$$
Combining these two gives that $\log \mathfrak P(2\rho)$ having growth rate strictly more than linear and strictly less than cubic. The upper bound says that $\log \mathfrak P(2\rho)$ is $O(n^2\log n)$. The lower bound says that $\log \mathfrak P(2\rho)$ grows at least as fast as $n\log n$.
I think one can probably do better, particularly on the lower bound. I may have another look at it at some later stage. But basically I'm content with the statement that $\log \mathfrak P(2\rho)$ is roughly quadratic with the rank.<|endoftext|>
TITLE: Given a Grothendieck topos, what does its localic groupoid look like? 
QUESTION [9 upvotes]: Possible Duplicate:
Toposes (topoi) as classifying toposes of groupoids 

For example, if a topos E is the object classifier, or the preseaf topos on a small category C, is there a way of describing its localic groupoid? More generally, is there a way of describing the localic groupoid of the classifying topos of a geometric theory T in terms of T? 
(By 'the localic groupoid of a Grothendieck topos E' I mean the localic groupoid, G, such that  E = BG, where BG is the category G equivariant sheaves - such a thing is not uniquely  determined, so I'll take any description.)
Has someone written down this dictionary somewhere already? 
Thank you, Christopher

REPLY [2 votes]: DISCLAIMER: This answer does not provide references,  nor is a formally thought-out proof. 
Only some (hopefully) useful heuristics.
You begin with a topos E with associated theory T(E). 
From T one creates a locale L (and a localic groupoid G(L) ), such that there is a surjection Sh(L)----> E, where Sh(L) is the topos of ALL sheaves on L. The topos E is then recovered by isolating the equivariant sheaves, ie G(L) -equivariants.  Now, let T(Sh(L)) the theory of the topos Sh(L). 
The question is basically: how T(sh(L)) relates to the original T(E). 
Conjecture: there is some kind of modality on T(sh(L)) such that T(E) is gotten as the "fixed points" for that modality. 
The intuition behind this conjecture is that the topos E is obtained by considering only well-behaved sheaves in the pool of all sheaves. They are, in a sense, the ones which are invariant with respect to some shuffling of the topos sh(L) by some suitable action. 
NOTE: My (unpublished) dissertation had a  somewhat germane theme: what I was after was some "toposophical" semantics of  general modal logics, and the trick was to consider topoi endowed with an extra lex endofunctor. The endofunctor was then used to isolate a fixed-points subcat (or more generally a cat of coalgebras for the endo) which had the nice property that the subobject functor had a built-in modal operator (induced by the endo). I do not know if the Joyal -Tierney representation produces something along similar lines, but the guess is yes.<|endoftext|>
TITLE: Symplectic formulation of statistical physics
QUESTION [17 upvotes]: Does there exists a symplectic formulation of statistical physics?
I know that thermodynamics can be written in a symplectic language and of course classical mechanics is intrinsically formulated symplectic, but I do not know anything which tries to relate them 'symplectilly'. Partial results are also welcome!

REPLY [3 votes]: I invite you to read the following papers about "Lie group Thermodynamics" of Jean-Marie Souriau. Souriau has discovered that Gibbs equilibrium is not covariant with respect to Dynamical groups, then he has considered Gibbs equilibrium on a Symplectic Manifold with covariant model with respect to a Lie group action. Souriau has introduced a geometric (planck) temperature in the Lie Algebra of the group:
[1]Barbaresco, F. Koszul Information Geometry and Souriau Geometric Temperature/Capacity of Lie Group Thermodynamics. Entropy 2014, 16, 4521-4565.
http://www.mdpi.com/1099-4300/16/8/4521/pdf
[2]Barbaresco F., Koszul information geometry and Souriau Lie group thermodynamics, AIP Conf. Proc. 1641, 74 (2015)
http://djafari.free.fr/MaxEnt2014/papers/Tutorial7_paper.pdf
More information at GSI'15 conference:
www.gsi2015.org
F. Barbaresco
GSI General Chair<|endoftext|>
TITLE: Ample vector bundles on complex tori
QUESTION [6 upvotes]: Let $X$ be a $n$-dimensional complex torus and $\omega$ a Kähler form on $X$. Then, it is well known that a real $(1,1)$-class $[\alpha]\in H^{1,1}(X,\mathbb R)$ is a Kähler class if and only if for all $1\le j\le n$ one has
$$
\int_X\alpha^j\wedge\omega^{n-j}>0.
$$ 
In particular, if $L\to X$ is a holomorphic line bundle, then $L$ is ample (and hence $X$ is an abelian variety) if and only if
$$
c_{1}(L)^j\cdot[\omega]^{n-j}>0,\quad 1\le j\le n.
$$
Question. Is it known any similar numerical criterion for higher rank holomorphic vector bundle on complex tori?
I would be interested also in weaker forms of this question. For instance, adding the hypothesis of semi-stability for the vector bundle, or possibly just requiring the vector bundle to be big instead of ample.
Any hint or comment is very welcome!

REPLY [3 votes]: Dear Simone,
This is just a comment.
The answer to your question, if one exists, is surely that there exist universal polynomials $p_j^r$ ($1 \leq j \leq r = {\rm rank} E$) in the Chern classes of the vector bundle $E \to X$ such that
$$
\int_X p_j{}^r(c_1(E), \dots, c_j(E)) \wedge \omega^{n-j} > 0
$$
for all $j$ implies that the bundle $E$ is ample. Here "universal" means that the polynomials in question only depend on the dimension $n$, but otherwise not on the variety $X$. For line bundles, these polynomials are known and are, as you wrote, $p_j{}^1(x) = x^j$ for all $j$.
The problem is that the condition you wrote for line bundles is a corollary of a more general theorem of father, one that characterizes Kahler classes amongst real $(1,1)$-classes. Our hope of approaching the problem should thus be to find suitable positivity criterion for higher degree classes. The ideal outcome would be a higher rank version of the Kodaira condition; so we'd know that if a "degree vector" $(u_1, \dots,u_r)$ of integral cohomology classes satisfies some conditions, then there exists an ample vector bundle $E$ of rank $r$ such that $c_j(E) = u_j$ (compare with $L$ ample iff $c_1(L)$ Kahler and integral). The trouble is that finding these conditions amounts to finding the universal polynomials $p_j{}^r$, and thus answering a much more general question.
In short, we have no idea what a "positivity condition" for a collection of cohomology classes $(u_1, \ldots, u_r)$ looks like. I agree with your approach of simplifying the problem and starting the search for these on complex tori. However, I think that if we knew the answer on complex tori, we'd know it on general complex manifolds too.<|endoftext|>
TITLE: Why depth, dimension, etc?
QUESTION [7 upvotes]: Recently I came across a paper by Stephen McAdam "GRADE SCHEMES AND GRADE FUNCTIONS" (Transactions of the American Mathematical Society, Vol. 288, No. 2 (Apr., 1985), pp.
563-590). So far I have just skimmed it, but from what I've understood he tries to set up unified axiomatic approach to study codimension, depth, asymptotic grade, etc. All these functions can be considered as monotone functions from the poset of prime ideals $P$ of the ring into the natural numbers satisfying certain properties (see Lemma 1.2 and Theorem 2.4 in the paper).
I wonder is it possible to recover (if not fully, then to some extent) the poset $P$ from knowledge of such functions? i.e. does there exist a theorem (in poset theory?) which states that study of $P$ is equivalent to study of "grade functions" whose abstract characterization is given in the paper.
More interestingly, I would like to know what is the meaning of such functions in terms of category of $R$-modules. There are some answers in this direction, see for example http://arxiv.org/abs/1202.5605 (CLASSIFICATION OF RESOLVING SUBCATEGORIES AND GRADE
CONSISTENT FUNCTIONS HAILONG DAO AND RYO TAKAHASHI). I have not read this paper either.
I have asked a similar question here: Characterizing Posets by Functions Into Natural Numbers
Thanks!

REPLY [4 votes]: A problem with the first question is that the definition of grade function in McAdam involves some ring-theoretic concepts. For example, I do not see how to express condition (iii) of Theorem 2.4 (about conforming pairs) just using the poset. So your first question is not (yet) well-defined. 
About the second question, "categorifying" such functions. I think it is possible. In particular, when $R$ is Cohen-Macaulay, then $\text{grade}(P) =\text{heigh}(P)$ for any prime, thus by Theorem 2.4 any grade function (restricting to $\text{Spec} R$) will be grade-consistent in the sense of the second reference. By the results there, these functions will classify certain subcategories of $\text{PD}(R)$, the category of f.g modules with finite projective dimension.  
As a simplest non-trivial example, let's look at $R= k[x]$, $k$ any field. Note that $\text{PD}(R)$ will just be $\text{mod}(R)$. Then any grade function $f: \text{Spec} R \to \mathbb N$ will have to satisfy: $f(0)=0$, $f(P)=0$ for $P\in V$, a finite subset of $\text{MaxSpec}\ R$, and $f(P)=1$ other wise. Such function is uniquely determined by $V$. 
Then these functions bijectively correspond to the resolving subcategories:
$$X_V = \{M\in \text{mod}(R) | M \ \text{is locally free on V}\} $$  
via the following  rule: a category $X$ defines the function:
$$f_X(P) = \max \{\text{pd}_{R_P} M_P | M \in X \} $$<|endoftext|>
TITLE: Better error bounds for partial sums of reciprocals of primes?
QUESTION [6 upvotes]: One of Mertens' theorems gives that
$$\sum_{ p \text{ prime,} p \leq k } 1/p  - \log{\log{k}} = B + E(k)$$
where $B$ is a constant near $0.26$ in value and $E(k)$ is an error
term whose size is dominated by something close to $4/\log{k}$, when $k$
is large enough to make the sum meaningful.
I want to work with partial sums of the above with $j \lt p \leq k$,
so that I can say the partial sum of the reciprocals of primes
greater than $j$ and at most $k$ is $\log{(\log{k}/\log{j})} + E(j,k)$
where $j$ is not too small (perhaps $j \gt 3$ or $5$), $k$ not too
large, say $j^\alpha \lt k \lt j^\beta$ where often 
$1 \lt \alpha \lt \beta \lt e$, and $E(j,k)$ is
comfortably small.  Unfortunately $4[1/\log{k} + 1/\log{j}]$ looks too big
for me; I am hoping to have (for large enough $j$) $E(j,k)$ bounded by
something that is $O(1/j)$ or better.
I have access to Mark Villarino's treatment of Mertens' theorem.  As
of 2005, it seems $E(k)$ is no better than $O(1/\log{k}^2)$
I also hope to obtain recent work of Pintz and Diamond on oscillations
ia related
formula which is Mertens product formula, but I do not see yet how 
it will me help me with this formula.
As I am still a tyro at number theory, I don't even know how realistic
my hopes are for $E(j,k)$ to be $O(1/j)$.  Can someone who is familiar with
the recent literature give me a guide post?  Either references or
heuristics showing what sort of bounds to expect for $E(j,k)$ or even $E(k)$
would be welcome.
UPDATE 2012.07.24
I want to acknowledge the contributions of joro,
Christian Elsholtz, and Eric Naslund.  joro and
Eric helped me realize that expecting $O(x^{-1/2})$
error even conditionally is expecting a bit much, and Christian
helped me realize that Dusart still has some nice
unconditional refinements.  I will likely accept
Christian's answer, but not before I do some computations
of my own.  In particular, Rosser and Schoenfeld have in
Theorem 20 of their 1962 paper on functions relating to primes a nice
difference of $2/(x^{1/2}\log x)$ which is valid for 
$1 \lt x \lt 10^8$ between lower and upper estimates for
the sum of interest, and I may end up using or refining that 
estimate in combination with Dusart's results for larger $x$ for my
own nefarious purposes.  I am hoping in particular that the
oscillations will be slow enough that my desired partial sums
from $j$ to $j^\alpha$ will have very small error.
END UPDATE 2012.07.24
Gerhard "Ask Me About System Design" Paseman, 2012.07.19

REPLY [4 votes]: Under Riemann hypothesis you can get better bound.
Sharper bounds for the Chebyshev functions $ \theta (x)$ and $ \psi (x)$. II  Lowell Schoenfeld 
http://www.ams.org/journals/mcom/1976-30-134/S0025-5718-1976-0457374-X/
(6.21) p.4
$$ \left|\sum_{p \le x}\frac1p-\log\log{x}-B\right|< \frac{3 \log{x} +4}{8 \pi \sqrt{x}}$$
if $13.5 \le x$.

REPLY [4 votes]: Up to a factor of logarithm, $E(x)$'s oscillation has an amplitude which is of the same magnitude as that of $\frac{1}{x}\left(\psi(x)-x\right)$, that is the error in the prime number theorem. Specifically $$E(x)=O\left( e^{-c(\log x)^{3/5-\epsilon}}\right)$$ unconditionally, and $$E(x)=O\left( x^{-\frac{1}{2}+\epsilon}\right).$$ under RH.
Proof: Let $W(x)=\pi(x)-\text{li}(x)$ be the error term in the prime number theorem. Then 
$$\sum_{p\leq x}\frac{1}{p}=\int_{2}^{x}\frac{1}{t}d\left(\pi\left(t\right)\right)=\int_{2}^{x}\frac{1}{t\log t}dt+\int_{2}^{x}\frac{1}{t}dW(t).$$ 
Integration by parts yields 
$$\int_{2}^{x}\frac{1}{t}dW(t)=\frac{W(t)}{t}\biggr|_{2}^{x}+\int_{2}^{x}\frac{W(t)}{t^{2}}dt.$$
Since 
$$\int_{2}^{x}\frac{W(t)}{t^{2}}dt=\int_{2}^{\infty}\frac{\pi(t)-\text{li}(t)}{t^{2}}dt+O_{\epsilon}\left(e^{-c\left(\log x\right)^{\frac{3}{5}-\epsilon}}\right),$$ 
it then follows that $$\sum_{p\leq x}\frac{1}{p}=\log\log x+b+O_{\epsilon}\left(e^{-c\left(\log x\right)^{\frac{3}{5}-\epsilon}}\right),$$ 
where $b=\int_{2}^{\infty}\frac{\pi(t)-\text{li}(t)}{t^{2}}dt+\frac{W(2)}{2}-\log\log2.$

REPLY [3 votes]: In case you also want computationally explicit bounds look at Theorem 6.10 of:
Pierre Dusart: Estimates of Some Functions Over Primes without R.H.
http://arxiv.org/abs/1002.0442
Probably the proof is for you more interesting than the Theorem itself.
On the one hand side it shows (as Eric did) the connection with the error term of the prime number theorem. On the other hand, with the constants detailed in Theorem 5.2 
you can construct explicit bounds of the type.
If $x>x_k$, then $\vert \sum_{p < x} \frac{1}{p}- B - \log \log x \vert <
\frac{\eta_k /k}{(\log x)^k} +
\frac{\eta_k (1 + \frac{1}{k+1})}{ (\log x)^{k+1}} $.
The asymptotic bounds given by the error term of the prime number theorem are of course stronger. (Explicit bounds of the type that Eric mentioned exist, but are probably not useful for computations).<|endoftext|>
TITLE: A mixing property of linear map over finite fields
QUESTION [9 upvotes]: Let $F$ be a finite field of odd size $q$, and $\phi_0 : F \mapsto F$ be any map from $F$ to itself. For each $a \in F$, set $\phi_a : x \in F \mapsto \phi_0 (x) + ax $. 
When $\phi_0 : x \mapsto x^2 $ , each image $\phi_a (F)$ has size $\frac{q+1}{2}$. It turns out that for any $\phi_0$, there's always some $a \in F$ such that $| \phi_a(F) | \geq \frac{q+1}{2}$.
But the only proof I know is somewhat artificial : it relies on the observation that $ K =\bigcup_{a \in F} \phi_a(F)^n $ is a Kakeya set of dimension $n$ over the finite field $F$. By subsequent improvements of Dvir's work on such sets, it is known that $K$ must have $\geq \left( \frac{q^2}{2q-1} \right)^n $ elements. As $n$ is arbitrary and $\frac{q^2}{2q-1} > \frac{q+1}{2} -1 $, this yields the above claim. 
I've found no direct proof so far. It seems MO might be the right place to ask for such a proof.

REPLY [5 votes]: Lemma 21 from a joint paper of S. Kopparty, S. Saraf, M. Sudan, and myself gives exactly what you are looking for: for any prime power $q$ (including the case of $q$ even), there exists $a\in F$ with $|\phi_a(F)|>q/2$. The proof for $q$ even is very similar to that given above by Peter Mueller, the proof for $q$ odd is slightly different.<|endoftext|>
TITLE: Comparability implies well-orderability?
QUESTION [5 upvotes]: I am trying to prove a small proposition that got me completely stumped, and I cannot find a single counterexample. 

(ZF) Suppose that $E$ is such that for every $A\subseteq\mathcal P(E)$ either $|E|\leq|A|$ or $|A|\leq|E|$, then $E$ can be well-ordered. 

It is not a biconditional statement since we have models of ZF (e.g. Solovay's model) where $\omega$ serves as a counterexample to this, but I still make true or false of the above statement.
Is this result known, or known to be false?
If the above is indeed false, how about a stronger requirement:

(ZF) Suppose $E$ is such that the cardinalities below $|\mathcal P(E)|$ are linearly ordered, then $E$ can be well-ordered.

REPLY [8 votes]: It is open whether the continuum hypothesis for an infinite set $E$ implies the well-orderability of $E$. Of course, if $CH(E)$ holds, then the assumption in your (first) statement holds. 
($CH(E)$ is the statement that any subset $A$ of $\mathcal P(E)$, either $A$ injects into $E$, or else $A$ is in bijection with $\mathcal P(E)$.)
This is a question that dates back to Ernst Specker, "Verallgemeinerte Kontinuumshypothese und Auswahlaxiom", Archiv der Mathematik 5 (1954), 332–337. There is a nice presentation in Akihiro Kanamori, David Pincus, "Does GCH imply AC locally?, in "Paul Erdős and his mathematics, II (Budapest, 1999)", Bolyai Soc. Math. Stud., 11, János Bolyai Math. Soc., Budapest, (2002), 413–426.
I do not know about the stronger statement you ask for. Part of the difficulty comes from the "bad" cardinal arithmetic we should have below $|{\mathcal P}(E)|$. For example, Specker proved that if CH holds for both $X$ and $\mathcal P(X)$, then $\mathcal P(X)$ is well-orderable.<|endoftext|>
TITLE: Curves of constant curvature on an ellipsoid
QUESTION [10 upvotes]: It is not difficult to see that the curves of constant geodesic curvature on a geometric sphere
are all circles: simple, closed curves that are geometric circles lying in a plane:
     
My question is:

Q. What are the curves of constant (positive) geodesic curvature on an ellipsoid?

Earlier Dmitri Panov asked a more general MO question,
"Curves of constant curvature on $S^2$."
My question is more specific.  I would be interested to know how large is the class of simple (non-self-intersecting), closed,
constant-curvature curves on an ellipsoid, whether there are nonsimple closed curves,
whether there are infinitely long curves, etc.
Dmitri's question revealed that many general questions are open, but
perhaps there has been a special study
made of the ellipsoid?  I have so far not found any literature specifically on this.
Thanks for pointers!
Addendum. I've made a few experiments which suggest that simple,
closed curves of constant curvature might not be uncommon. My calculations were incorrect---Sorry to mislead!
Further edit. I've now rewritten the calculations, which (I think!) are now correct within
numerical accuracy.  Here is a portion of a 
(non-simple) constant-curvature curve on the
ellipsoid $\frac{x^2}{2^2}+y^2+z^2=1$:

REPLY [10 votes]: You may want to have a look at the article Foliation by constant mean curvature spheres, by Rugang Ye, Pacific Journal of Mathematics, 147 (1991), 381–396.
In this article, the author shows, given a Riemannian surface $M$, that every nondegenerate critical point $p$ of the Gauss curvature has a punctured neighborhood that is foliated by closed curves of constant geodesic curvature.  (Think of the rings of a bulls-eye centered on $p$.)
In particular, for an ellipsoid with three distinct axes, the six critical points of the Gauss curvature (where the axes meet the surface) are nondegenerate, and hence there exists a foliation by closed, constant geodesic curvature curves centered on each one.  This gives three distinct $1$-parameter families of closed constant geodesic curvature curves on the ellipsoid with three distinct axis lengths.  I don't know whether the classic differential geometers were able to compute these curves explicitly or not.
Of course, there will be many other closed curves with constant geodesic curvature on the ellipsoid that don't fall into these families.  (Just think of the ones with geodesic curvature $0$, i.e., the closed geodesics, for example.)  However, the 'generic' curve of constant geodesic curvature on the ellipsoid does not close.
On the other hand, it was asserted (without explicit proof) by Darboux (see Livre VI, Chapitre VII, Paragraph 622, footnote 1 of his Leçons sur la Théorie Générale des Surfaces) that the only surfaces for which all of the curves of constant geodesic curvature are closed are the surfaces of constant Gauss curvature.  I don't know where a proof might first have appeared in the literature, but I once asked Victor Guillemin about this, and he produced a proof.  In Ye's article quoted above, he shows that, if one has a sequence of closed curves of constant geodesic curvature whose curvatures go to infinity and that converge to a point $p$ in the surface, then $p$ must be a critical point of the Gauss curvature, and this provides an alternative proof of Darboux' claim.<|endoftext|>
TITLE: Diameter of symmetric group
QUESTION [30 upvotes]: Let $\Sigma_n\subset G$ be a set of generators of the symmetric group $S_n$. It is a well-known conjecture that the diameter of the Cayley graph $\Gamma(S_n,\Sigma_n)$ is at most $n^C$ for some absolute constant $C$. (The diameter of the Cayley graph is just the maximum of $\ell(g)$ for $g\in S_n$, where $\ell(g)$ is the length of the shortest word on $A \cup A^{-1}$ equal to $g$.)
For $\Sigma_n$ of bounded size, the diameter cannot be less than a constant times $\log |S_n|$, i.e., a constant times $n\log n$.
It is clear and well-known that, for $\Sigma_n = \{(1 2),(1 2 \dotsb n)\}$, the diameter of $\Gamma(S_n, \Sigma_n)$ is at least a constant times $n^2$. (It is also at most that.)
Are there any examples of generating sets $\Sigma_n$ for which the diameter is larger than $n^{2+\epsilon}$ for every (or infinitely many) $n$? Larger than $n^2 (\log n)^A$ for some $A>0$ and infinitely many $n$?

REPLY [11 votes]: What follows is an incomplete answer.  I am in the middle of Russian woods, so would rather have somebody else trace all the refs, etc. but looking at the bounty expiration date decided that it's worth stating what is known.
The answer is NO to all, but that's a conjecture not a theorem.  I have seen this conjecture stated several times in various forms, here are two I find interesting:

1) the diameter of every connected
  Cayley graph $\Gamma$ on $S_n$ is $O(n^2)$,
2) for every O(1) generators of $S_n$,
  the mixing time of the nearest neighbor r.w. on the corresponding Cayley graph $\Gamma$ is $O(n^3\log n)$.

Since the mixing time is greater than the diameter, the second implies also a bound on the diameter as well.  The second conjecture was stated by Diaconis and Saloff-Coste someplace, and is also sharp for a transposition and long cycle as in the question (see Saloff-Coste's survey).  The first conjecture is a dated folklore and I remember reading it in various places; it appears e.g. in this paper (p. 425) by Gamburd and me.
UPDATE 
See this recent paper by Diaconis ("Some things we've learned..", 2012) where he reiterates conjecture 2) in Question 2 on p.9.<|endoftext|>
TITLE: A mixing property for finite fields of characteristic $2$
QUESTION [17 upvotes]: In connection with this MO post, here is a question somewhat implicitly contained in a joint paper of S. Kopparty, S. Saraf, M. Sudan, and myself.
Let ${\mathbb F}$ be a finite field, and suppose that $\varphi_0\colon{\mathbb F}\mapsto{\mathbb F}$ is a function from ${\mathbb F}$ to itself. For each $a\in{\mathbb F}$, consider the function $\varphi_a\colon{\mathbb F}\mapsto{\mathbb F}$ defined by $\varphi_a(x)=\varphi_0(x)+ax$ ($x\in{\mathbb F}$).

Is it true that if $q:=|{\mathbb F}|$ is even, then there exists $a\in{\mathbb F}$ such that the image of $\varphi_a$ has size larger than $2q/3$?

(A negative answer would yield an improvement on the known bounds for the smallest size of a Kakeya set in the vector spaces ${\mathbb F}^n$.)

It may be worth explaining where the coefficient $2/3$ comes from. In the aforementioned paper, we show that if $q$ is an even power of $2$, then for $\varphi_0(x)=x^3$ one has $\max_a |\varphi_a({\mathbb F})|\le(2q+1)/3$, whereas if $q$ is an odd power of $2$, then for $\varphi_0(x)=x^{q-2}+x^2$ one has $\max_a |\varphi_a({\mathbb F})|\le2(q+\sqrt q+1)/3$. The question is whether one can get better bounds for an appropriate choice of the function $\varphi_0$.

REPLY [10 votes]: The use of Birch/Swinnerton-Dyer in my previous and David Speyer's answer is vaste overkill!
Actually in David's example one can compute exactly the value set sizes. I do only the relevant case $m=1$. (As the size is exact, there is no need for the asymptotic consideration $m\to\infty$, only $r\to\infty$ matters.)
Theorem. Set $r=2^k$, $F=GF(r^2)$ and $\phi_0(x)=x^{r+1}$. Then $\lvert\phi_a(F)\rvert=\frac{r(r+1)}{2}=\frac{r+1}{2r}\lvert F\rvert$ if $a\ne0$, and $\lvert\phi_0(F)\rvert=r$.
Proof. The latter statement is trivial, and in the former statement it suffices to assume $a=1$, for if $b^r=a$, then $\phi_a(bx)=b^{r+1}\phi_1(x)$. Let $T(x)=x^r+x$ be the trace map from $F$ to $GF(r)$. Note that $x^{r+1}\in GF(r)$ for all $x\in F$. Thus if $\phi_1(x)=\phi_1(y)$ for $x,y\in F$, then $\delta=y-x\in GF(r)$. Given $x\in F$, we count the number of $0\ne\delta\in GF(r)$ with $\phi_1(x)=\phi_1(x+\delta)$. A short calculation gives the equivalent relation $\delta=T(x)+1$. Comparing dimensions, we see that $T$ is surjective as $GF(r)$ is the kernel of $T$. Thus there are exactly $r$ elements $x$ with $T(x)+1=0$. So $\phi_1$ assumes $(r^2-r)/2$ values twice, and $r$ values once. From $(r^2-r)/2+r=r(r+1)/2$ the claim follows.<|endoftext|>
TITLE: Coverage, itself considered as a presheaf
QUESTION [5 upvotes]: A coverage $J$ on a category $C$ assigns to an object $U$ of $C$ a set of covering families $J(U)$. The covering families are required to be stable under pullback, which amounts to requiring that for every arrow $f: V \to U$ in $C$, there is a function $J(U) \to J(V)$ which sends a covering family $\{x_i: U_i \to U\}_{i \in I}$ of $U$ to a covering family $\{y_j: V_j \to V\}_{j \in J}$ of $V$ such that each composite $fy_j$ factors through some $x_i$. This apparently defines a presheaf on $C$, as the functor laws seem to hold.
I am interested in this presheaf and its properties. Does it have a name? For example, for every presheaf on a site we may ask whether it is a sheaf. One might want to know what are the requirements for a coverage, when considered as a presheaf as above, to be a sheaf w.r.t. itself.

REPLY [7 votes]: First of all, Andreas' comment is right: a coverage gives no specified way to "pull back" a covering family of $U$ to a covering family of $V$.  However, if you consider what Sketches of an Elephant calls "sifted" coverages, meaning that all covering families are sieves, then there is a canonical choice: the pullback of a sieve $R$ on $U$ along $f:V\to U$ is the sieve $f^*(R)$ consisting of all $h:W\to V$ such that $f h\in R$.
For an arbitrary sifted coverage, this pullback sieve may not be a covering family, but it always contains a covering family and thus lies in the "saturation" of the coverage.  If a sifted coverage $T$ is closed under pullback of sieves, in this sense, then it does yield a presheaf on $C$, which is in fact a sub-presheaf of the subobject classifier in the presheaf topos $[C^{\mathrm{op}},\mathrm{Set}]$ (which is defined by $\Omega(U) = $ the set of all sieves on $U$).  (If $T$ is a Grothendieck topology, then this sub-presheaf $T$ is the classifier of dense sub-presheaves.)  See also C2.1.10 in the Elephant.
I claim that this sub-presheaf $T\subseteq \Omega$ is $T$-separated iff the coverage $T$ contains at most one covering sieve of every object.  (In particular, if $T$ is a Grothendieck topology, then it must be the trivial topology.)  This condition is clearly sufficent; for necessity, suppose $R$ and $S$ are two $T$-covering sieves of an object $U$.  Then for any $f:V\to U$ in $R$, the pullback sieve $f^*(S)$ is covering.  It follows that any $T$-separated presheaf is also separated for the sieve generated by all composites $f h$ with $h\in f^*(S)$ (this sieve lies in the saturation of $T$ to a Grothendieck topology).  But this sieve is precisely $R\cap S$.
Thus, if $T\subseteq \Omega$ is $T$-separated, it is also separated for $R\cap S$ for any $R,S\in T$.  However, for any $f:V\to U$ in $R\cap S$, we have $f^*(R) = f^*(S)$ being the maximal sieve on $V$.  Thus, since $T$ is separated for $R\cap S$, we must have $R=S$; hence $U$ admits at most one $T$-covering sieve.
Now assuming $T$ satisfies this condition so that $T$ is $T$-separated, then $T$ is a $T$-sheaf whenever if $R$ is a (the) covering sieve of $U$ and for each $f:V\to U$ in $R$ we have a (the) covering sieve $S_V$ of $V$, then there is another covering sieve of $U$ (which, of course, must also be $R$) such that $f^* R = S_V$.  But when $f\in R$, then $f^*R$ is the maximal sieve on $V$, so this means that the domain of every morphism in a covering sieve is covered only by its own maximal sieve --- which is already implied by $T$ being a functor.  Such objects are called $T$-irreducible (C2.2.18 in the Elephant).
Thus there are three classes of objects in $C$: the irreducible ones, which are covered by their maximal sieve; those that are covered by some non-maximal sieve whose domains are all irreducible; and those that are not covered by any $T$-sieve.  The irreducible objects are themselves a sieve in $C$, by functoriality, as are the objects that are covered by any $T$-sieve at all.  The objects that are not covered by any $T$-sieve will be covered only by their maximal sieve in the Grothendieck topology generated by $T$, so they will be irreducible there.
In particular, the Grothendieck topology generated by $T$ is rigid in the sense of C2.2.18: every object is covered by the family of morphisms out of irreducible objects.  It follows that the category of $T$-sheaves is equivalent to the category of presheaves on the irreducible objects for this topology (which are those that are $T$-covered by the maximal sieve or that are not covered by any $T$-sieve).
This is perhaps not a complete answer to your question, but it shows that the condition of a sifted coverage being a sheaf for itself is very restrictive.<|endoftext|>
TITLE: the use of parentheses to mean "I won't tell you this again"
QUESTION [15 upvotes]: A reader of one of my drafts found fault with my use of parentheses; I put the word "bounded" in parentheses in a statement of a certain theorem, and he replied "But the statement isn't true if the assumption of boundedness is dropped!"
That reader seemed to be thinking that parentheses mark things that are in some way inessential (as is sometimes the case in non-mathematical prose).  But, as I wrote to him:
Here I am using parentheses to mean "Of course the interval must be bounded!  In case some of you are nodding off, I'll include the stipulation of boundedness, but I might not include it next time."  I wonder if that use of parentheses has a name?
Does this use of parentheses have a name, or any sort of pedigree that might dignify it, within or beyond mathematical writing?
I have no idea how to tag this post; it's a question about the (possibly nonexistent) subfield of modern Rhetoric that is concerned with the ways mathematicians use language to communicate ideas to other mathematicians.  I'll be grateful if someone will suggest appropriate tags and add them (and I'll make a note of what the tag is, in case I need it again).

REPLY [12 votes]: I think it might be beneficial to see the actual context in which the comments were made (by me; not as a referee, but just someone that Jim wrote to and asked for comments on his nice paper, which by the way, has a fair bit of its provenance in various MO threads). 
The work in question is on the arxiv here.  Various properties of an ordered field $R$ are being considered and compared.  The last two are:

(17) The Shrinking Interval Property: suppose $I_1 \supset I_2 \supset \ldots$ are (bounded) closed intervals in $R$ with lengths decreasing to zero.  Then the intersection of the $I_n$'s is nonempty.

and

(18) The Nested Interval Property: Suppose $I_1 \supset I_2 \supset \ldots$ are (bounded) closed intervals in $R$.  Then the intersection of the $I_n$'s is nonempty.  

I was not thrilled with the use of (bounded) in (17), but I let it go.  I objected to the use of (bounded) in (18).
Note that "(bounded)" is playing different roles in the two statements.  In (17), it is a superfluous hypothesis: if the lengths of the intervals are decreasing to zero then necessarily all but finitely many of them are bounded.  In (18) it certainly isn't.  I found this lack of parallelism especially confusing: so confusing that the first time I read it I honestly did arrive at the (ridiculous) conclusion that Jim Propp was unaware that e.g. $\bigcap_{n=1}^{\infty} [n,\infty) = \varnothing$.<|endoftext|>
TITLE: Existence of an elliptic curve and a distortion map 
QUESTION [5 upvotes]: I would like to know if the following is true or wrong or unknown.
For any integer $m$, there is an elliptic curve over the complex numbers such that there is a (distortion) map $\phi$ so that $\{ T, \phi(T) \}$ is a basis of$E[m]$, for some point $T \in E[m]$.
I know if $m\equiv 3 (4)$ then we can take $E: y^2=x^3+x$ and $\phi$ is a complex multiplication by $i$. 
So I'd like to know if there is such an elliptic curve and a distortion map for any other $m$.

REPLY [4 votes]: The elliptic curve E would have to have complex multiplication.  If End(E) = Z then no such map will exist.
The word "distortion map" was introduced for elliptic curves over finite fields. Now the answer depends on wording.  You wrote "for some point T".  That case is easy.  The case "for all non-zero points T" is harder, and needs E to be supersingular.
The relevant literature is:

E.R. Verheul, Evidence that XTR is more secure than supersingular
elliptic curve cryptosystems, Journal of Cryptology
S. D. Galbraith and V. Rotger, Easy decision Diffie-Hellman groups, London Mathematical Society Journal of Computational Mathematics, Vol. 7 (2004) 201-218.<|endoftext|>
TITLE: Cone of a morphism in an abelian category when considered as a morphism in derived category. Connection between 4-term exact sequences and distinguished triangles.
QUESTION [6 upvotes]: Let $\mathcal{A}$ be an abelian category and let 
$$
0 \rightarrow E \rightarrow F \rightarrow G \rightarrow 0
$$
be a short exact sequence. Then in $D(\mathcal{A})$, the derived category of $\mathcal{A}$ we have a distinguished triangle
$$
E \rightarrow F \rightarrow G \rightarrow E[1].
$$ 
Moreover, the above short exact sequence determines an element of Ext$^1(G,E)$ and therefore a map $G[-1] \rightarrow E$ in $D(\mathcal{A})$. The axiom TR2 tells us that the cone of this map is exactly $F$.
I would like to understand if one can generalise this picture. 
Consider a morphism $f: A \rightarrow B$ in $\mathcal{A}$ such that both Ker$(f)$ and Coker$(f)$ are non-zero. $f$ can be also thought of as a morphism in $D(\mathcal{A})$. What is its cone then? Do we have a distinguished triangle 
$$
A \rightarrow B \rightarrow \textrm{Coker}(f) \oplus \textrm{Ker}(f)[1] \rightarrow A[1]?
$$
On the other hand, we have the following exact sequence in $\mathcal{A}$ 
$$
0 \rightarrow \textrm{Ker}(f) \rightarrow A \rightarrow B \rightarrow \textrm{Coker}(f) \rightarrow 0.
$$
It corresponds to an element in $\textrm{Ext}^2(\textrm{Coker}(f), \textrm{Ker}(f))$ and hence to a map $\textrm{Coker}(f)[-2] \rightarrow \textrm{Ker}(f)$ in $D(\mathcal{A})$. Can one write down the cone of this map using objects $A$ and $B$? My only guess is that it would be $A \oplus B[-1]$, but I don't know whether it is correct or how to prove it.

REPLY [2 votes]: I'll expand on Steve's comment, but I'm sure you must know this stuff already.
if $f: A \to B$ you take the cone $D$ and cohomology long exact sequence tells you that the cohomology of $D$ is $K = \text{ker } f$ in degree minus one and $C = \text{coker } f$ in degree zero.
Explicitly a cone is given by the mapping cone construction for the category of complexes.
The complex is given by $A[1] \oplus B$ but with the differential twisted by $f$ (actually, as these are complexes concentrated in a single degree, the differential is $f$).
The abelian category being hereditary should be equivalent to saying that mapping cones split, in the sense that they can be rewritten as a direct sum of $K$ and $C$ as you wrote.
As for your second question, maybe this helps.
Consider the inclusion $K \to [A \to B]$ where I'm thinking of the second guy as the complex where $A$ sits in degree zero, $B$ in degree one and the differential is $f$ (in other words it's the mapping cone of $f$ shifted by minus one).
Take the cone $E$. The merciful lord of cohomology long exact sequences then tells us
$$ 0 \to K \to K \to H^0(E) \to 0 \to C \to H^1(E) \to 0 $$
and, as the first map is an isomorphism, $E$ is actually $C[-1]$.
Is this enough?<|endoftext|>
TITLE: How much  does a quantum oracle to find a needle in a haystack really cost? 
QUESTION [11 upvotes]: Among the basic algorithms of quantum computations Lov Grover's result on quantum search stands out, both in regards to its intrinsic interest, and for its undisputable elegance. 
Grover's algorithm enables one to search an unsorted database of N elements in 
$O(N^{1/2})$ time, which is pretty remarkable.
However, I do have one perplexity, which perhaps someone here can dissipate. 
The starting point of Grover is this one (from the wiki entry):
Consider an unsorted database with N entries. The algorithm requires an N-dimensional state space H, which can be supplied by n=log2 N qubits. Consider the problem of determining the index of the database entry which satisfies some search criterion.

Let f be the function which maps
  database entries to 0 or 1, where
  f(ω)=1 if and only if ω satisfies the
  search criterion. We are provided with
  (quantum black box) access to a
  subroutine in the form of a unitary
  operator, $U_{f}$, which acts as:
$U_{f}|\omega \rangle =-|\omega \rangle$
and 
$U_{f}|x \rangle = |x \rangle $ $\forall x \neq
> \omega$

Now, my problem with Grover is this: 
IF you have your unitary oracle, THEN you can find your needle in the haystack. But, suppose a real life scenario, in which you have your database and you have your $f$ above as given input data. 
You still need a preliminary step  to create your operator, FROM $f$. 
If there is a simple way to write a program which carries out this preliminary step, you are cool. But if this step turned out to be expensive, you must add this cost in the total bill (and that in turn might erode your quantum benefits). 
Assuming you have $f$, how complicated is it to produce  $U_{f}$?

REPLY [8 votes]: $U_f$ is easy to create from a circuit that computes $f$. (I'm assuming you have a circuit that computes $f$. If you have a Turing machine, convert it to a circuit in the standard way.)
Now given a circuit (using AND, OR, NOT gates) of G gates that computes $f$, i.e., accepts a binary string $x$ as input and outputs the 1 bit answer $f(x)$, this can be converted to a quantum circuit that uses a Toffoli gate and NOT gates to simulate the AND and OR gates. This will create a unitary matrix that on input $|x\rangle|0^k\rangle$ gives you $|f(x)\rangle|\text{junk}(x)\rangle$. Using the standard "uncomputation trick", you can now get a circuit that performs $|x\rangle|b\rangle \mapsto |x\rangle |b \oplus f(x)\rangle$. Finally, if you put in the state $\frac{1}{\sqrt{2}}(|0\rangle-|1\rangle)$ for $|b\rangle$, you'll get a circuit that exactly implements $U_f$.
Note that this circuit for $U_f$ has increased the size and depth of the original circuit for $f$ by only a constant factor. Thus if $f$ had an efficient circuit, then so does $U_f$.<|endoftext|>
TITLE: What fraction of a sphere's volume lies within a cone?
QUESTION [5 upvotes]: Let $B \subset \mathbb{R}^n$ be a collection of $n$ (not necessarily independent) unit vectors which we will label $v_1,\ldots, v_n$ for convenience. The cone $K_B \subset \mathbb{R}^n$ associated to $B$ is the non-negative linear span of $B$, i.e., $$K_B = \lbrace r_1v_1 + r_2v_2 + \ldots + r_nv_n~|~r_j \geq 0 \rbrace. $$ Let $\mathbb{S}^n$ denote the unit $(n-1)$-sphere defined as usual by $$\mathbb{S}^n = \lbrace v \in \mathbb{R}^n ~|~ \|v\| = 1\rbrace.$$
Question:

Is there a nice formula known for the ratio $$\angle B = \frac{\text{Vol}(K_B\cap\mathbb{S}^n)}{\text{Vol}(\mathbb{S}^n)}?$$

Where $\text{Vol}$ refers to $(n-1)$ dimensional volume and nice means "directly involving the coordinates of the vectors in $B$"? The motivation comes from the trivial case $n=2$: when $B = \lbrace v_1, v_2\rbrace \subset \mathbb{R}^2$ then the fraction of the unit circle's perimeter lying within the cone spanned by $v_1$ and $v_2$ can immediately be recovered from the inner product (which of course directly involves coordinates):
$$ \angle B = \frac{1}{2\pi} \cos^{-1}(v_1\cdot v_2).$$
I assume this is an extremely well-studied problem, but all my google searches so far have only yielded high school trigonometry so I am obviously missing some keywords. All help is appreciated!

REPLY [2 votes]: As I wrote in KvanTTT's question on math.stackexchange: Solid angle between vectors in n-dimensional space, see Ribando's paper:
https://link.springer.com/content/pdf/10.1007%2Fs00454-006-1253-4.pdf
There seems to be no closed formula, but rather a Taylor series expression.<|endoftext|>
TITLE: What is the role of $\sum (-1)^p[\wedge^pT^*M]$ in the K-theory $K(M)$
QUESTION [5 upvotes]: I apologize for the vague title. Let $M$ be a compact smooth manifold, then we have $T^*M$ and hence $\wedge^pT^*M$ as vector bundles on $M$. There for we have 
$$
\sum (-1)^p[\wedge^pT^*M] \in K(M).
$$
It seems to have close relation with the Euler class of $M$.
My question is: Is there any study on $\sum (-1)^p[\wedge^pT^*M]$? What properties do we know? Can I have some references?

REPLY [5 votes]: I think the main topological significance of the element you identify is its close relationship with the Thom isomorphism $K(M) \cong K(T^*M)$.  I would imagine that there is also some connection with the Euler characteristic as well, perhaps by taking Chern characters, but this is probably secondary to its role in the Thom isomorphism.
The idea is the same for any compact Hausdorff space $X$ and any vector bundle $V$ over $X$, though in the case where $X = M$ is a manifold and $V = T^*M$ I believe the result becomes a version of Poincare duality in K-theory.  Recall that tensor product of bundles defines a product $K(X) \otimes K(Y) \to K(X \times Y)$.  Pulling back along the diagonal map $X \to X \times X$ one has a product $K(X) \otimes K(X) \to K(X \times X) \to K(X)$ and hence a ring structure on $K(X)$.  If $V$ and $W$ are vector bundles over $X$ then pullback along the diagonal map also induces a product $K(V) \otimes K(W) \to K(V \times W) \to K(V \oplus W)$ (where $V \times W$ is the product of spaces and $V \oplus W$ is the direct sum of bundles).  Taking $W$ to be the zero bundle over $X$, the map $K(V) \otimes K(X) \to K(V)$ gives $K(V)$ the structure of a $K(X)$-module.
Now view $\bigwedge^p V$ as a vector bundle over $V$ and consider the element $\lambda_V = \sum_p (-1)^p [\bigwedge^p V]$ in $K(V)$.  Some care must be taken as usual due to the fact that $V$ is not itself compact, but there are a number of ways to make this work out.  The product with $\lambda_V$ defines a map $\varphi: K(X) \to K(V)$, and the Thom isomorphism theorem in K-theory asserts that this map is an isomorphism.  To do computations one often pulls back along the zero section $i: X \to V$, and one finds that $i^* \circ \varphi$ is the map $K(X) \to K(X)$ given by multiplication by $\sum_p (-1)^p [\bigwedge^p V]$, thought of as an element of $K(X)$.
References: More detail can be found in the first section of Atiyah and Singer's "Index of Elliptic Operators I".<|endoftext|>
TITLE: If a non-trivial zero of the zeta function existed off the critical line, would infinitely many zeros exist with the same real part?
QUESTION [7 upvotes]: It is known that there exist infinitely many non-trivial zeros of the Riemann zeta function in the critical strip. Also, we know that infinitely many zeros are on the critical line - more than 1/3 among all, asymptotically - as well as most of the zeros live near the critical line.
In 1984, the theorem of Rademacher and Hlawka showed that the ordinates of the nontrivial zeros of the zeta function are uniformly distributed regardless of the abscissas.
Now, suppose that there exists one non-trivial zero having its real part (say A) other than 1/2. Then, let us call x=A "line A." (the complex plane defined as z=x+iy)
In that case when we take it true,
1) Might infinitely many zeros exist on the line A?
2) If 1) is possible, would they be distributed uniformly?

REPLY [8 votes]: 1) Yes, this is still possible 
2) No. We know that they have to distributed with low density, i.e. the number of zeros $z$ with $\Im z < T$ with $\Re z > \sigma >1/2$ is bounded by $T^{4\sigma(1-\sigma) + \epsilon}$ for $\epsilon >0$. So no uniform distribution is possible, since the gaps between consecutive zeros has to grow to infinity.
For sharper results in this direction, I suggest the first chapter of Joern Steuding "Universality of L-functions". This book actually explains pretty good what is going on in the critical stipe off the line $\Re s = 1/2$.<|endoftext|>
TITLE: Good notions of "perfect number" for rings of integers more general than ${\Bbb Z}$?
QUESTION [10 upvotes]: It seems natural to me to generalize the notion of perfect number to rings of integers more general that ${\Bbb Z}$.  I'll want to think of number fields concretely, as subfields of ${\Bbb C}$.  For a mutatis mutandis definition, a given algebraic integer must have
a distinguished finite set of divisors.  That makes two tricky parts:

figuring out what to say if the group of units is richer than $\{-1,1\}$
making do without the order if the field is not a subfield of ${\Bbb R}$.

Do definitions occur in the literature?  Do nontrivial examples exist?

REPLY [23 votes]: The interesting question seems to me not what a perfect number is but what an arithmetic function is (generalizing the divisor function, the sum-of-divisors, totient, etc.) over a number field.
One answer is as follows. My opinion is that arithmetic functions should not be thought of as having domain $\mathbb{N}$ but as having domain the poset of nonzero ideals in $\mathbb{Z}$ (which can be identified with $\mathbb{N}$ under division, but my point is that the additive structure of $\mathbb{N}$ does not enter into this). Dirichlet convolution
$$f \ast g = \sum_{d | n} f(d) g (n/d)$$
can then be seen as a special case of the multiplication in a subalgebra of the incidence algebra of the poset of ideals (the subalgebra which assigns the same number to an interval $[a, b]$ as it does to the interval $[1, \frac{b}{a}]$). The assignment of a Dirichlet series
$$\sum \frac{f(n)}{n^s}$$
to an arithmetic function merely instantiates this algebra as an actual algebra of functions. Now, the poset of nonzero ideals in $\mathbb{Z}$ is a product of posets, one for each prime, and the arithmetic functions which are multiplicative (in the sense that $f(mn) = f(m) f(n)$ if $(m, n) = 1$) are just the ones that are built from corresponding functions on each factor. In the Dirichlet series picture this is reflected in the existence of an Euler product
$$\sum \frac{f(n)}{n^s} = \prod_p \left( \sum_{k \ge 0} \frac{f(p^k)}{p^{ks}} \right).$$
The generalization to number fields is as follows. In any Dedekind domain $D$, we may consider the incidence algebra of the poset of nonzero ideals in $D$. This algebra has a subalgebra of functions which assign the same number to an interval $[I, J]$ as to the interval $[1, I^{-1} J]$, and this gives us the appropriate generalization of Dirichlet convolution
$$(f \ast g)(K) = \sum_{IJ = K} f(I) g(J)$$
where $I, J, K$ are ideals. When $D$ is the ring of integers in a number field, we can define norms $N(I)$ of ideals and also assign Dirichlet series. This assignment
$$\sum_I \frac{f(I)}{N(I)^s}$$
is now not faithful as there can be different ideals with the same norm, but it is still true that the multiplicative functions (the ones such that $f(IJ) = f(I) f(J)$ if $I, J$ generate the unit ideal) admit an Euler product
$$\sum_I \frac{f(I)}{N(I)^s} = \prod_P \left( \sum_{k \ge 0} \frac{f(P^k)}{N(P)^{ks}} \right)$$
where the product is now over all prime ideals in $D$.
This is all a long-winded way of setting the stage for what I actually wanted to say:

For number fields, an arithmetic function now manifestly has different domain and codomain: it takes input an ideal and returns a number.

In particular, to me the correct generalization of the sum-of-divisors function is the function $\sigma(I)$ which sums the norms of ideals dividing a given ideal $I$, and it no longer makes sense to ask whether this is equal to $I$ or $2I$. Of course there are other interesting questions to ask regarding comparing different functions, e.g. one might ask whether $\sigma(I) = c N(I)$ for some constant $c$ (this reduces to the perfect number question over $\mathbb{Q}$ taking $c = 2$, but I think from this perspective it is now clear that there is nothing special about $c = 2$).<|endoftext|>
TITLE: Quasi-unipotent monodromy for general families
QUESTION [7 upvotes]: This must be a naive question, but I'm wondering about the definition of quasi-unipotent monodromy for general families, not only 1-parameter families. The problem is that usually, in the books of algebraic geometry, quasi-unipotent monodromy is only discussed over a disc $\Delta^{\ast}$, i.e. for a 1-parameter family. In this case we know that for a fibration $f: X \rightarrow \Delta$, with monodromy representation $\rho : \pi_{1}(\Delta^{\ast}) \rightarrow Aut(H^{i}(X_{0}))$, the image $\rho (T)$ of a generator $T$ of $\pi_{1}(\Delta^{\ast})$ is a quasi-unipotent matrix. What is the correct generalization of this to arbitrary families? For example, in the multi-parameter case, it could happen that there are several generators $T_{i}$ such that each $\rho(T_{i})$ is a quasi-unipotent matrix but, for example, $\rho(T_{1})\rho(T_{2})$ is not quasi-unipotent. So is the possible generalization that all of the matrices in the monodromy group (image of the monodromy representation) should be quasi-unipotent or does it suffice that the image of the generators be quasi-unipotent?

REPLY [15 votes]: Quasi-unipotency is a well defined notion at any point of the discriminant. If we have a proper family $f : X \to S$ of varieties with a smooth total space and a smooth base, and if $p \in D \subset S$ is a point of the discriminant, then we say that the local monodromy of the family near $p$ is quasi-unipotent if we can find a small analytic neighborhood $p \in U \subset S$ of $p$ in $S$, so that if $o \in U - D$ is a base point, then the monodromy representation $mon : \pi_{1}(U-D,o) \to GL(H^{i}(X_{o},\mathbb{C})$ has an image whose Zariski closure $G$ is a quasi-unipotent linear algebraic group (that is, the quotient of $G$ by its unipotent radical is a finite-group). 
In general it is rare for the local monodromy  to be quasi-unipotent. If $p$ happens to be a very singular point of the discriminant, then the local monodromy tends to be big and is often as big as it can be, and not quasi-unipotent at all. However, if $p$ is at worst a normal crossing singularity of $D$, then the local monodromy is quasi-unipotent.<|endoftext|>
TITLE: To find an element of a  $\Pi^1_1$  set
QUESTION [6 upvotes]: I don't known the correct credit for the following: every non-empty $\Pi^1_1$ set of reals contains some $X \in L_{\alpha}$ for some $X$-recursive $\alpha$. (Addison-Kondo?)
So, my question is: what is the least $\beta$, s.t. $\mathcal{P}(\omega) \cap L_\beta$ is a basis for all non-empty $\Pi^1_1$ sets?

REPLY [4 votes]: The least such ordinal is the least ordinal which cannot be a $\Delta^1_2$-well-ordering over natural numbers. 
Let $$\delta^1_2=\mbox{ supremum of the }\Delta^1_2 \mbox{
wellorderings of } \omega,$$ and $$\delta=\min\{\alpha\mid
L\setminus L_{\alpha}\mbox{ contains no }\Pi^1_1 \mbox{
singleton}\}.$$
We claim that $\delta=\delta^1_2$.
$\mathbf{Proof}$:
If $\alpha<\delta$, then there is a $\Pi^1_1$ singleton $x \in
L_{\delta}\setminus L_{\alpha}$. Since $x\in L_{\omega_1^x}$ and
$\omega_1^x$ is a $\Pi^1_1(x)$-wellordering, it must be that
$\alpha<\omega_1^x<\delta^1_2$. So $\delta\leq \delta^1_2$.
If $\alpha<\delta^1_2$,  there is a $\Delta^1_2$ wellordering
relation $R\subseteq \omega\times \omega$ of order type $\alpha$. So
there are two arithmetical relations $S, T\subseteq
(\omega^{\omega})^2\times \omega^2$ so that
$$R(n,m)\Leftrightarrow \exists f \forall g S(f,g,n,m),
\mbox{ and}$$
$$\neg R(n,m)\Leftrightarrow \exists f \forall g T(f,g,n,m).$$
Define  $\Pi^1_1$ sets $$R_0=\{(h,\langle n,m\rangle)\mid h(0)=0\wedge \exists f\forall g 
(S(f,g,n,m)\wedge \forall n(f(n)=h(n+1)))\}$$ and $$R_1=\{(h,\langle n,m\rangle)\mid h(0)=1\wedge \exists f\forall g 
(T(f,g,n,m)\wedge \forall n(f(n)=h(n+1)))\}.$$ By $\Pi^1_1$-uniformization Theorem,  they both can be uniformized by $\Pi^1_1$ partial functions $p_{R_0}:\omega\to \omega^{\omega}$ and $p_{R_1}:\omega\to \omega^{\omega}$. Let $p=p_{R_0} \cup p_{R_1}$. Then $p$ is a $\Pi^1_1$  total function and can viewed as a $\Pi^1_1$-singleton. Then $R$ is recursive in $p$ and so $\alpha<\omega_1^p<\delta$. 
Thus $\delta^1_2=\delta$.<|endoftext|>
TITLE: Reference on the question mark cell complex
QUESTION [5 upvotes]: The question mark complex is a finite spectrum whose cohomology looks like a "question mark" (when drawn as a module over the Steenrod algebra): that is, there is an element in dimension zero $a_0$, an element $a_2 = \mathrm{Sq}^2 a_0$, and an element $a_6 = \mathrm{Sq}^4 a_2$. It can be constructed by starting with $\Sigma^{-2} \mathbb{CP}^2$, which maps to $S^2$ (thanks to the cofiber sequence $S^1 \to S^0 \to \Sigma^{-2} \mathbb{CP}^2 \to S^2$) and lifting the map $\nu: S^5 \to S^2$ to $\Sigma^{-2} \mathbb{CP}^2$ (which can be done since $\eta \nu = 0$). Then, one takes the cofiber of $S^5 \to \Sigma^{-2}\mathbb{CP}^2$.
Does anyone know any good references on this? I'd like, ideally, to see a few example computations done with it; as it is I don't really have much intuition for how to work with it.

REPLY [5 votes]: This is hardly a "canonical example", but one place I've seen the question mark complex $Q$ in action is in Hovey and Sadofsky's paper Invertible spectra in the $E(n)$-local stable homotopy category.  There they compute that the even part of the Picard group of the $E(1)$-local stable category at $p = 2$ is generated by (the $E(1)$-localization of) $Q$.  As a warning, the exposition at the tail of the paper where this appears is a bit skeletal; you'll be working out the details of the computations and digging up references on your own.
But, in turn, they do cite Hopkins' Minimal atlases of real projective spaces for some facts about $Q$, which I have not read but looks to be useful.  He seems to call the complex $N_2$ in section 7, for reasons he explains pertaining to the cohomology of $bo\langle i \rangle$.<|endoftext|>
TITLE: Why $G\to G/H$ is faithfully flat?
QUESTION [5 upvotes]: Some questions about algebraic groups.
Let $G$ be an affine algebraic group over algebraically closed field $k$. 
Questions: Let $H$ be a closed subgroup of $G$, then (as I learnt from some paper) the quotient $\pi\colon G\to G/H$ is faithfully flat, why? reference? When it is locally trivial, especially when $G$ is linear algebraic group?
Thank you very much!

REPLY [3 votes]: If you are really looking for a reference, here is one: Groupes algébriques, Demazure-Gabriel, Chapter III, §3, Proposition 2.5 p. 328. Note that the fact that the base is a field is not important. It might be any scheme. The important assumptions are 

the subgroup $H$ has to be flat over the base (which of course holds if it is smooth, which always hold over a field in char 0),
the quotient $G/H$ (defined as the fppf sheafification of the presheaf quotient) has to be a scheme (i.e. representable). This holds when $G$ is affine over a base field, by theorem 5.4, also in Ch. III, §3.

P.S. I am aware that this book is in french, and that the notation used in it makes it quite hard to browse through without spending too much time. Note, however, that it contains an index for the notation at the end. It is a very thorough reference for this type of questions.
P.P.S. As for the "locally trivial" part of the question, do you mean Zariski locally trivial (i.e. there is a Zariski open subset U of G/H over which as a scheme, the morphism becomes $U\times H \to U$)?<|endoftext|>
TITLE: Is the Mostowski collapse natural? 
QUESTION [9 upvotes]: The Mostowski collapse lemma  (see here for a quick ref) is one of the key basic tools in the set-theory arsenal. I wonder if the collapse is natural, in the functorial sense. 

More precisely, is this a
  reflection from the large category of
  well-founded models of ZF to the
  subcategory of transitive models?

My taste would say yes, but I have not thought it through ((apologies if the answer turns out to be trivial).
MOTIVATION: still thinking a bit about the MULTIVERSE Category. If the answer is affirmative, then it makes good sense to simply work with the subcategory of transitive models of ZF, which is certainly more manageable, and simpler to ponder. 
ADDENDUM TO THE MOTIVATION: on a quick  after-thought, I partially retract what I just said: there could still be some interest in considering the larger category of not necessarily well-founded models. In this case, perhaps someone could provide some speculations as to this larger cat and what can be found there (exotic models)

REPLY [13 votes]: You didn't say what the morphisms in your categories are supposed to be; Trevor assumed you meant elementary embeddings, but you could also have meant mere embeddings, or something else.  Nevertheless, unless you make a very strange choice of morphisms, the answer to your question is yes.  Not only is the Mostowski collapse a reflection, it's an equivalence of categories.  The transitive models constitute a skeleton of the category of all well-founded models; that is, every well-founded model $M$ is isomorphic to exactly one transitive model.  Better yet, the isomorphism is uniquely determined by $M$.  (All this information is part of the full statement of Mostowski's collapsing theorem.)  So, from a category-theoretic point of view, it makes no difference whether you work with arbitrary well-founded models or with only the transitive models.  Note, though, that in some situations, non-transitive well-founded models arise naturally, for example as elementary substructures of transitive ones, and in such cases your desire to work only with transitive models would require you to immediately apply the Mostowski collapse as soon as such a model enters your considerations.

REPLY [5 votes]: If I understand the definition correctly at https://en.wikipedia.org/wiki/Reflective_subcategory, the question boils down to showing that every elementary embedding $f: B \to A$ between wellfounded models uniquely factors through the transitive collapse of $B$.
This is true:
It factors through the transitive collapse of $B$ because the transitive collapse map is an isomorphism.
Uniqueness follows from the fact that isomorphic wellfounded models (in particular, $B$, its transitive collapse, and the range of $f$) are uniquely isomorphic.<|endoftext|>
TITLE: What is known about the distribution of average edge-degrees for 3-manifold triangulations (with the number of 3-simplices less than a fixed constant)?
QUESTION [6 upvotes]: This is my first question on mathoverflow! It relates to a project I'm undertaking with a student.
Work by Tamura (extending results by Luo and Stong) shows that for any closed 3-manifold $M$ and any rational number $4.5 < r < 6$ there is a triangulation $T$ of $M$ for which that the average edge-degree $\mu(T)$ is $r$. Here the degree of an edge $e$ is the number of 3-simplices having $e$ as a face.
Now, fix a closed 3-manifold $M$ and consider the (necessarily finite) set of all triangulation of $M$ containing at most $K$ 3-simplices. Call this set $\mathcal{T}_K(M)$.

QUESTION: Does anyone know of results concerning the "distribution" of the average edge-degree $\mu(T)$ for $T\in \mathcal{T}_K(M)$? That is, what is known about the fraction of triangulations from $\mathcal{T}_K(M)$ for which the edge-degree lies in a given small interval $[r - \epsilon, r+\epsilon] \subset (4.5,6)$? Results concerning any closed 3-manifold $M$ would be fine.

I suspect this is extremely difficult to answer in a precise way. However, I'd be very interested in knowing any asymptotic ($K\rightarrow \infty$) or approximate results as well. Thanks for your help!
NOTE: For the project we plan to use the Metropolis algorithm to sample from $\mathcal{T}_K(M)$ where $M$ is the 3-torus. Using these samples we hope to empirically estimate the distribution in question.

REPLY [6 votes]: Below is the data from Ben Burton's census of triangulations of $S^3$ with up to 9 tetrahedra. An Euler characteristic argument shows that the average edge degree is equal to $6\left(\frac{T}{V+T}\right)$, where $T$ is the number of tetrahedra and $V$ is the number of vertices, so the numbers below are all of that form. The data are given in (average edge degree, frequency) pairs. As Bruno guessed, it looks like the distribution weights heavily towards 6.
1 tetrahedron: 

(2.0000000000000000, 1)
(3.0000000000000000, 1)

2 tetrahedra:

(2.0000000000000000, 1)
(2.3999999999999999, 1)
(3.0000000000000000, 1)
(4.0000000000000000, 3)

3 tetrahedra:

(2.5714285714285716, 1)
(3.0000000000000000, 2)
(3.6000000000000001, 9)
(4.5000000000000000, 20)

4 tetrahedra:

(2.6666666666666665, 2)
(3.0000000000000000, 4)
(3.4285714285714284, 16)
(4.0000000000000000, 48)
(4.7999999999999998, 128)

5 tetrahedra:

(2.7272727272727271, 1)
(3.0000000000000000, 4)
(3.3333333333333335, 23)
(3.7500000000000000, 110)
(4.2857142857142856, 468)
(5.0000000000000000, 1297)

6 tetrahedra:

(2.7692307692307692, 1)
(3.0000000000000000, 5)
(3.2727272727272729, 36)
(3.6000000000000001, 199)
(4.0000000000000000, 1103)
(4.5000000000000000, 4931)
(5.1428571428571432, 13660)

7 tetrahedra:

(2.7999999999999998, 1)
(3.0000000000000000, 3)
(3.2307692307692308, 39)
(3.5000000000000000, 301)
(3.8181818181818183, 2186)
(4.2000000000000002, 13380)
(4.6666666666666667, 62657)
(5.2500000000000000, 169077)

8 tetrahedra:

(2.8235294117647061, 1)
(3.0000000000000000, 3)
(3.2000000000000002, 51)
(3.4285714285714284, 446)
(3.6923076923076925, 3870)
(4.0000000000000000, 28826)
(4.3636363636363633, 180128)
(4.7999999999999998, 829753)
(5.3333333333333333, 2142197)

9 tetrahedra:

(2.8421052631578947, 1)
(3.0000000000000000, 3)
(3.1764705882352939, 50)
(3.3750000000000000, 567)
(3.6000000000000001, 6046)
(3.8571428571428572, 54876)
(4.1538461538461542, 422860)
(4.5000000000000000, 2612407)
(4.9090909090909092, 11673471)
(5.4000000000000004, 28691150)<|endoftext|>
TITLE: What is the relationship between motivic cohomology and the theory of motives?
QUESTION [63 upvotes]: I will begin by giving a rough sketch of my understanding of motives.
In many expositions about motives (for example, http://www.jmilne.org/math/xnotes/MOT102.pdf), the category of motives is defined to be a category such that every Weil cohomology (viewed as a functor) factors through it. This does not define the category uniquely, nor does it imply that it exists.
There are two concrete candidates that we can construct. The category of Chow motives, which is well-defined, is trivially a category of motives. However, it has some bad properties. For example, it is not Tannakian. The second candidate is the category of numerical motives. It too is well-defined, however it is only conjectured that it is category of motives (i.e., that every Weil cohomology factors through this category). This conjecture is closely related to (or perhaps even equivalent to?) Grothendieck's standard conjectures. That would be desirable, because the category of numerical motives is very well-behaved.
Furthermore, the original motivation for motives is that Grothendieck has proven that if the category of numerical motives is indeed a category of motives, then the Weil conjectures are correct.
So far, even though I a murky on many of the details, I follow the storyline.
Question
Where does "motivic cohomology" (in the sense of, for example, http://www.claymath.org/library/monographs/cmim02.pdf) fit into this story?
I know that motivic cohomology has something to do with Milnor K-theory, but that is more or less where my understanding of the context of motivic cohomology ends. If motives are already an abstract object that generalizes cohomology, what does motivic cohomology signify? What is the motivation for defining it? What is the context in which it arose?

REPLY [71 votes]: Classically, Grothendieck's motives are only the pure motives, meaning abelian-ish things which capture the (Weil-cohomology-style) $H^i$ of smooth, projective varieties.  To see the relationship with motivic cohomology, one should extend the notion of motive so that non-pure (i.e. "mixed") motives are allowed, these mixed motives being abelian-ish things which capture the $H^i$ of arbitrary varieties.  The main novelty with mixed motives is that the (conjectural) abelian category of them is not semi-simple -- in fact every mixed motive should be a (generally non-trivial) iterated extension of pure motives, these extensions essentially coming from compactification and resolution of singularities, as in the story of mixed Hodge structures.
Then once one thinks of mixed motives, a natural direction of study (or speculation, as the case may be...) is that of determining all possible extensions (or iterated extensions) between two motives.  And that's what motivic cohomology is, essentially:  the study of these Ext groups.  More formally, every variety $X$ should determine an object $C(X)$ in the bounded derived category of mixed motives, collecting together all the various mixed motives $H^i(X)$, and the $(i,j)^{th}$ motivic cohomology of $X$ is (up to twisting conventions) the abelian group of maps from the unit object to $C(X)$ \ $[i](j)$ (the $j^{th}$ Tate twist of the $i^{th}$ shift of $C(X)$) in the derived category of mixed motives.
Now, there are a few points to make here.  The first is that, though the above motivation and definition of motivic cohomology rely on an as-yet-conjectural abelian category of mixed motives, one can, independently of any conjectures, define a triangulated category which, as far as anyone can tell, behaves as if it were the bounded derived category of this conjectural abelian category.  The most popular such definition, because of its simplicity and relative workability, is Voevodsky's.  So the basic theory and many basic results on motivic cohomology are unconditional.
Another thing to say is that, as always, matters with motives are illuminated by considering realization functors.  Let me single out the $\ell$-adic etale realization, since its extension from pure to mixed motives is straightforward (unlike for Hodge structures): any mixed motive, just as any pure motive, yields a finite-dimensional $\ell$-adic vector space with a continuous action of the absolute Galois group of our base field.  It then "follows" (in our conjectural framework... or actually follows, without quotation marks, in Voevodsky's framework) that the $(i,j)^{th}$ motivic cohomology of X maps to the abelian group of maps from the unit object to $C^{et}(X)$ \ $[i](j)$ in the bounded derived category of $\ell$-adic Galois representations.  But this abelian group of maps is just the classical (continuous) $\ell$-adic etale cohomology $H^i(X(j))$ of the variety $X$, making this latter group the natural target of an $\ell$-adic etale "realization" map from motivic cohomology.
So here comes the third point:  note that this is the etale cohomology of $X$ itself, not of the base change from $X$ to its algebraic closure.  So this etale cohomology group mixes up arithmetic information and geometric information, and the same is true of motivic cohomology in general.  (Think especially of the case $X=pt$: the motivic cohomology of a point admits a generally nontrivial realization map to the $\ell$-adic Galois cohomology of the base field.)  For example, it is expected (e.g. by Grothendieck -- see http://www.math.jussieu.fr/~leila/grothendieckcircle/motives.pdf for this and more) that for an abelian variety $A$ over an ``arithmetic'' base field $k$, the most interesting part of the motivic cohomology $H^{(2,1)}(A)$ (again my twists may be off...), by which I mean the direct summand which classifies extensions of $H^1(A)$ by $H^1(G_m)$, should identify with the rationalization of the abelian group of $k$-rational points of the dual abelian variety of $A$, the map being given by associating to such $k$-rational point the mixed motive given as $H^1$ of the total space of the corresponding $G_m$-torsor on $A$.  And in this case, the above "realization" map to $\ell$-adic etale cohomology is closely related to the classical Kummer-style map used in the proof of the Mordell-Weil theorem.
So in a nutshell: motivic cohomology is very related to motives, since morally it classifies extensions of motives.  But it is of a different nature, since it is an abelian group rather than an object of a more exotic abelian category; and it's also quite different from a human standpoint in that we know how to define it unconditionally.  Finally, motivic cohomology realizes to Galois cohomology of a variety itself, rather than to the base change of such a variety to the algebraic closure.
Hope this was helpful in some way.<|endoftext|>
TITLE: What's the difference between ZFC+Grothendieck, ZFC+inaccessible cardinals and Tarski-Grothendieck set theory?
QUESTION [20 upvotes]: Say that "U" is the axiom that "For each set x, there exists a Grothendieck universe U such that x $\in$ U", where Grothendieck universes are defined in the usual way (or, if that's unclear, in this way). Also, say that "Ca" is the axiom that "For each cardinal κ, there is a strongly inaccessible cardinal $\lambda$ which is strictly larger than κ."
It's known that ZFC+U and ZFC+Ca are completely equivalent and prove the same sentences. A sentence is a theorem of ZFC+U iff it's a theorem of ZFC+Ca.
In addition to the above, there's also Tarski-Grothendieck set theory, which can be found here. The axioms of TG are

The axiom stating everything is a set
The axiom of extensionality from ZFC
The axiom of regularity from ZFC
The axiom of pairing from ZFC
The axiom of union from ZFC
The axiom schema of replacement from ZFC
Tarski's axiom A

Tarski's axiom A states that for any set $x$, there exists a set $y$ containing, $x$ itself, every subset of every member of $y$, the power set of every member of $y$, and every subset of $y$ of cardinality less than $y$.
These three axioms from ZFC are then implied as theorems of TG:

The axiom of infinity
The axiom of power set
The axiom schema of specification
The axiom of choice

My question is as follows: is TG also completely equivalent to ZFC+U and ZFC+Ca, equivalent in the same sense that something is a theorem of TG iff it's a theorem of the other two? Is TG just an axiomatization of ZFC+U/ZFC+Ca which removes redundant axioms and allows them to just be theorems? Or is there some subtle difference between TG and ZFC+U/ZFC+Ca, in that there's some sentence which TG proves that's undecidable in ZFC+U/ZFC+Ca or vice versa?
In other words, instead of typing ZFC+Grothendieck, can I just type TG and be referencing a different axiomatization of the exact same thing?

REPLY [19 votes]: Yes.  Assume ZFC.  If there is a proper class of inaccessible cardinals, then Tarski's Axiom A holds because whenever $\kappa$ is inaccessible, the rank initial segment $V_\kappa$ of $V$ is a Tarski set.  Conversely, if Tarski's Axiom A holds then for every set $x$ there is a Tarski set $y$ with $x \in y$.  We will show that $|y|$ is an inaccessible cardinal greater than $|x|$, proving the existence of a proper class of inaccessible cardinals.
To show that the cardinality $\kappa$ of $y$ is a strong limit cardinal, given $\zeta < \kappa$ we take a subset $z$ of $y$ of size $\zeta$.  We have $z \in y$ because $y$ contains its small subsets.  Then we have $\mathcal{P}(z) \in y$ because $y$ is closed under the power set operation.  Finally $\mathcal{P}(\mathcal{P}(z)) \subset y$ because $y$ contains all subsets of its elements.  This shows that $2^{2^{\zeta}} \le \kappa$ and therefore that $2^{\zeta} < \kappa$.
To show that the cardinality $\kappa$ of $y$ is regular, notice that if $\kappa$ is singular then by the closure of $y$ under small subsets we can get a family of $\kappa^{cof(\kappa)}$ many distinct sets in $y$, contradicting the fact that $\kappa^{cof( \kappa)} > \kappa$ (which is an instance of Koenig's Theorem.)<|endoftext|>
TITLE: Weakly homogeneous trees under AD
QUESTION [7 upvotes]: If AD$_\mathbb{R}$ holds and $\kappa < \Theta$ then every tree $T$ on $\kappa$ is weakly homogeneous (Martin–Woodin, "Weakly homogeneous trees.")  I recall hearing that the hypothesis can be weakened to "AD holds and there is a Suslin cardinal above $\kappa$."  Is this correct?  If so, does anyone know where a proof can be found, or have the idea of the proof?
It may help to keep in mind that the main consequence of $AD_\mathbb{R} + \kappa < \Theta$ used in Martin–Woodin is the existence of a normal fine measure on $\mathcal{P}_{\omega_1}(\mathcal{P}(\kappa^{<\omega}) \cup \mathrm{meas}(\kappa^{<\omega}))$, where $\mathrm{meas}(\kappa^{<\omega})$ denotes the set of countably complete measures on $\kappa^{<\omega}$ (equivalently, on $\kappa^n$ for some $n<\omega$.)
In other words, the proof uses that $\omega_1$ is $\mathcal{P}(\kappa^{<\omega}) \cup \mathrm{meas}(\kappa^{<\omega})$-supercompact.  If we only assume "AD holds and there is a Suslin cardinal above $\kappa$" then because the measures are essentially ordinals one can show that $\omega_1$ is $\mathrm{meas}(\kappa^{<\omega})$-supercompact,  but I see no way to show that $\omega_1$ is $\mathcal{P}(\kappa^{<\omega})$-supercompact.  So perhaps a more substantial modification to the Martin–Woodin proof is required.

REPLY [6 votes]: I hope it's okay to post an answer to my own question.  I am essentially repeating Woodin's proof that he just showed me.  Any errors were probably introduced by me.
Recall that under AD all measures on ordinals are countably complete and ordinal-definable.
By the coding of measures theorem of Kechris assuming AD and that there is a Suslin cardinal above $\kappa$, there are fewer than $\Theta$ many measures on $\kappa$ (or $\kappa^{<\omega}$ for that matter.)
So we may push forward Martin's measure on $\mathcal{P}_{\omega_1}(\mathbb{R})$ to get a fine, countably complete measure $U$ on $\mathcal{P}_{\omega_1}(\operatorname{meas}(\kappa^{<\omega}))$, where $\operatorname{meas}(\kappa^{<\omega})$ is the set of all measures on $\kappa^{<\omega}$.
Fix a tree $T$ on $\omega\times \kappa$. As usual, given a real $x \in \omega^\omega$ we let $T_x = \lbrace s \in \kappa^{<\omega} : (x \restriction |s|, s) \in T\rbrace$.
Claim: $U$-almost all $\sigma$ witness the weak homogeneity of $T$.
Proof: For each $\sigma$ we define a game $G_{\sigma}$, closed for Player I, for which Player I has a winning strategy iff $\sigma$ does not witness the weak homogeneity of $T$.


I plays: $(x_0, \alpha_0, \beta_0)$, $(x_1, \alpha_1,\beta_1),\ldots$


II plays: $\mu_0$, $\mu_1,\ldots$


Let $x$, $\vec{\alpha}$, $\vec{\beta}$, and $\vec{\mu}$ denote the resulting sequence of moves.


Rules for I: $\vec{\alpha} \in [T_x]$ and the sequence $\vec{\beta} \in \mathrm{Ord}^\omega$ continuously witnesses that the tower $\vec{\mu}$ is illfounded.


Rules for II: $\vec{\mu}$ is a tower of measures in $\sigma$ concentrating on $T_x$.


If both players follow the rules until the end, we say that player I wins.
If $\sigma$ does not witness that $T$ is weakly homogeneous, say $x \in p[T]$ but there is no wellfounded tower of measures in $\sigma$ concentrating on $T_x$, then there is a continuous witness to the illfoundedness of towers of measures in $\sigma$ concentrating on $T_x$.  (This is proved by an argument similar to what follows but using a fine, countably complete measure on the set of subsets of $\kappa^{<\omega}$.)
So if $\sigma$ does not witness that $T$ is weakly homogeneous, then player I has a winning strategy in $G_\sigma$. (The $x$ and $\vec{\alpha}$ are fixed in advance and the $\vec{\beta}$ comes from $\vec{\mu}$ via the continuous witness mentioned above.)  Assume toward a contradiction that for $U$-almost every $\sigma$, player I has a winning strategy in $G_\sigma$. The game is closed, the moves are ordinals and measures, and all measures are ordinal-definable, so for such $\sigma$ player I has a winning strategy $F(\sigma)$ of playing the least move leading to a subgame where he or she still has a winning strategy.


Define the integer $x_0$ to be the one played by $F(\sigma)$ on the first turn for $U$-almost every $\sigma$.


Define a measure $\mu_0$ on $\kappa$ by $A \in \mu_0 \iff \forall^*_U \sigma\; (\alpha^{\sigma}_0 \in A)$ where $\alpha^{\sigma}_0$ is the ordinal $\alpha_0$ played by $F(\sigma)$ on the first turn.


Define the integer $x_1$ to be the one played by $F(\sigma)$ on the second turn against $\mu_0$ for $U$-almost every $\sigma$.


Define a measure $\mu_1$ on $\kappa^2$ by $A \in \mu_1 \iff \forall^*_U \sigma\; ((\alpha^{\sigma}_0,\alpha^{\sigma}_1)  \in A)$ where $\alpha^{\sigma}_1$ is the ordinal $\alpha_1$ played by $F(\sigma)$ against $\mu_0$ on the second turn.


Continuing in this way, we get a real $x \in \omega^\omega$ and a sequence of measures $\vec{\mu}$.  One can easily check that $\vec{\mu}$ is a tower of measures.
Each $\mu_i$ concentrates on $T_x$ because $(\alpha_0^\sigma,\ldots,\alpha_i^\sigma) \in T_x$.
It is a wellfounded tower because if $A_i \in \mu_i$ for all $i<\omega$ then by countable completeness of $U$ there is a $\sigma$ such that $(\alpha_0^\sigma,\ldots,\alpha_i^\sigma) \in A_i$ for all $i<\omega$.
However, by countable completeness of $U$ there is a $\sigma$ such that $\vec{\mu}$ is a legal play by player II against player I's winning strategy $F(\sigma)$, so player I's moves $\beta^\sigma_i$ continuously witness the illfoundedness of $\vec{\mu}$.  Contradiction.<|endoftext|>
TITLE: computing Morley rank using parameters from an arbitrary model
QUESTION [6 upvotes]: One of the ways to define the Morley rank of a definable set is with respect to a model, say $M$, i.e. a set has rank $\alpha+1$ if there are infinitely many definable subsets with parameters in $M$ of rank $\alpha$ (and a similar clause for limit ordinals). One then shows that once $M$ is $\aleph_0$-saturated then considering definable subsets with parameters in elementary extensions of $M$ doesn't change the rank.
However, in some theories, e.g. algebraically closed fields of fixed characteristic, any model will do, i.e. Cantor-Bendxson rank and Morley rank coincide.
What is known about this phenomenon, i.e. in which theories Cantor-Bendixson equals Morley rank in any (not necessary $\aleph_0$-saturated) model?

REPLY [5 votes]: The distinction between Morley rank as defined by arbitrary formulas and by definable families  of formulas is essential. $\aleph_1$- categoricity in particular implies the rank can defined by
definable families. Since my 1973 [?] article in the transactions AMS or Shelah's book or 
say Pillay's geometric model theory book<|endoftext|>
TITLE: Does the normal ordered product on differential operators lift to $U\left(\mathfrak{gl}_n\right)$?
QUESTION [8 upvotes]: Let $n\in\mathbb N$. Let $k$ be a commutative ring in which $1,2,3,\ldots$ are invertible. Let $\Omega$ denote the $k$-algebra of polynomial differential operators on $n$ variables $x_1$, $x_2$, ..., $x_n$ over $k$. (The multiplication in this algebra is the composition of differential operators.)
Let $M:k\left[x_1,x_2,...,x_n\right]\otimes k\left[\dfrac{\partial}{\partial x_1},\dfrac{\partial}{\partial x_2},...,\dfrac{\partial}{\partial x_n}\right]\to\Omega$ be the $k$-linear map which sends every $P\otimes Q$ to $P\cdot Q$. Clearly, $M$ is an isomorphism of $k$-modules, but not of $k$-algebras (unless $k=0$ or $n\leq 1$). This allows us to define a commutative multiplication $\boxdot$ on $\Omega$ by letting
$A\boxdot B = M\left(M^{-1}\left(A\right)\cdot M^{-1}\left(B\right)\right)$ for all $A\in\Omega$ and $B\in\Omega$.
This $\boxdot$ is called the normal(ly?) ordered product on $\Omega$. (When $A$ and $B$ are atomic terms, one often writes $:AB:$ for $A\boxdot B$, but when $A$ and $B$ are composite terms, $:AB:$ can mean something slightly different.)
The $k$-linear map $\ell:\mathfrak{gl} _ {n}\to \Omega$ which sends every elementary matrix $E_{i,j}$ to $x_i\dfrac{\partial}{\partial x_j}$ is a Lie algebra homomorphism. Thus, it gives rise to a $k$-algebra homomorphism $L:U\left(\mathfrak{gl} _ {n}\right)\to \Omega$. This homomorphism $L$ is generally not injective (I guess it's like a noncommutative Segre embedding: it sends $E_{i,j}E_{k,l}$ to the same differential operator as $E_{i,l}E_{j,k}$ if $i$, $j$, $k$, $l$ are pairwise distinct). Hence the following question: Is there a (commutative?) multiplication $\boxdot$ on $U\left(\mathfrak{gl} _ {n}\right)$ such that any $A\in U\left(\mathfrak{gl} _ {n}\right)$ and $B\in U\left(\mathfrak{gl} _ {n}\right)$ satisfy $L\left(A\right)\boxdot L\left(B\right) = L\left(A\boxdot B\right)$ ?
Note that (by a theorem of Sylvester from 1867, in P92 from the second tome of his Collected Works, but not the one I asked about in that thread) the image of $L$ is closed under $\boxdot$, and actually is the $k$-subalgebra of $\Omega$ generated by the image of $\ell$ under $\boxdot$-multiplication. This speaks in favor of the existence of a $\boxdot$ on $U\left(\mathfrak{gl} _ {n}\right)$ (but is not a proof yet, since $L$ is not injective).

REPLY [3 votes]: Yes, the multiplication $\boxdot$ on $U\left(\mathfrak{gl}_n\right)$ exists. Moreover, Alexander Chervov's conjecture is true: There is a commutative multiplication $\boxdot$ on $U\left(\mathfrak{gl}_n\right)$ such that, for every $m \in \mathbb{N}$, the $k$-algebra homomorphism
$$
U\left(\mathfrak{gl}_n\right) \to \mathcal{D}_{n,m},\ E_{i,j} \mapsto \sum\limits_{k=1}^{m} x_{i,k} \dfrac{\partial}{\partial x_{j,k}}
$$
(where $\mathcal{D}_{n,m}$ is the ring of all differential forms in the $nm$ variables $x_{i,k}$ with $1 \leq i \leq n$ and $1 \leq k \leq m$) sends the multiplication $\boxdot$ to the normal ordered product $\boxdot$ on $\mathcal{D}_{n,m}$ (which normal ordered product is defined similarly to the one above on $\Omega$).
For a proof, see Theorem 4.6 in my note On the PBW theorem for pre-Lie algebras. Unfortunately, the proof is really long; chiefly responsible for this are folklore lemmas which I couldn't easily find in the literature and annoying computations with differential operators which I had to do in all their gory detail in order not to get wrong. (As the introduction says, ignore the proofs except maybe that of Theorem 4.7 (b).)
Note that the notations in my note differ from those in my OP above; in particular, $\Omega$ means something different.<|endoftext|>
TITLE: Bijection between irreducible representations and conjugacy classes of finite groups
QUESTION [33 upvotes]: Is there some natural bijection between irreducible representations and conjugacy classes of finite groups (as in case of $S_n$)?

REPLY [4 votes]: It appears  that similar question has been asked at sci.math.research  Tue, 19 Oct 1999. The answer by G. Kuperberg is quite interesting. Hope no one don't mind if I put it here:

As Torsten Ekedahl explained, it is sometimes the wrong question, but
  in modified form, the answer is sometimes yes.
For example, consider A_5, or its central extension Gamma = SL(2,5).
  The two 3-dimensional representations are Galois conjugates and there
  is no way to choose one or the other in association with the conjugacy
  classes.  However, if you choose an embedding pi of Gamma in SU(2),
  then there is a specific bijection given by the McKay correspondence.
  The irreducible representations form an extended E_8 graph where two
  representations are connected by an edge if you can get from one to
  the other by tensoring with pi.  The conjugacy classes also form and
  E_8 graph if you resolve the singularity of the algebraic surface
  C^2/Gamma.  The resolution consists of 8 projective lines intersecting
  in an E_8 graph.  If you take the unit 3-sphere S^3 in C^2, then the
  resolution gives you a surgery presentation of the 3-manifold
  S^3/Gamma. The surgery presentation then gives you a presentation of
  Gamma itself called the Wirtinger presentation.  As it happens, each
  of the Wirtinger generators lies in a different non-trivial conjugacy
  class. In this way both conjugacy classes and irreps. are in bijection
  with the vertices of E_8.<|endoftext|>
TITLE: Free modules over integers
QUESTION [16 upvotes]: After the course of linear algebra I'm more familiar with vector spaces rather than modules so my question may seem to be silly but I think it's quite natural for someone who thinks of
modules as 'vector spaces over a ring': which of the following is free module (free means: having a basis, all of them are over ring $\mathbb{Z}$): 
a) $\mathbb{Z}^{\infty}$-all sequences of integers,
b) $\mathbb{Z}^{\mathbb{R}}$-all functions $f: \mathbb{R} \to \mathbb{Z}$,
c) the set of all functions $f: \mathbb{R} \to \mathbb{Z}$ with at most countable support?

REPLY [21 votes]: None of these are free.   For (a), there is a theorem of Specker ("Additive Gruppen von Folgen ganzer Zahlen" Portugaliae Math. 1950) that covers this group and lots of its subgroups (the so-called monotone subgroups).  I believe that for this particular group, the result may already be in a 1937 paper of Baer.  Since the groups in (b) and (c) have subgroups isomorphic to the group in (a), and since subgroups of free abelian groups are free, it follows that (b) and (c) aren't free either.<|endoftext|>
TITLE: Automorphisms of a certain digraph defined on the set of primes? [Edited]
QUESTION [11 upvotes]: Define a digraph $P=(V,E)$ with $V$ equal to the set of prime numbers and an arrow from $p$ to $q$ if $p|q-1$ (definition motivated by Pratt's primality certificates).
Does $P$ indeed admit only the trivial automorphism (as seems reasonable to guess)? 
Edit:  In light of Gjergji Zaimi's heuristic suggesting a very large automorphism group, perhaps the better question asks what can one say, otherwise unconditionally, about hypothetical nontrivial automorphisms of $P$ and the permutations of ${\Bbb N}$ that they induce.

REPLY [8 votes]: I would expect this digraph to have a lot of automorphisms, but I cannot prove it unconditionally.
First, notice that $2$ is the only prime with no incoming edges. The primes which have incoming edges only from $2$ are the Fermat primes. You can keep going defining new collection of primes with incoming edges from a specific set of primes that were already considered.
Here is a conjecture: For any primes $2=p_1\le p_2 \le \cdots \le p_k$ there are infinitely many primes $q$ so that $q=1+p_1^{a_1}\cdots p_k^{a_k}$. (This seems like a reasonable conjecture generalizing the infinitude of Fermat primes.)
Assuming this conjecture, for any collection of primes previously considered there is an infinite set of primes whose incoming edges come exactly from this set of primes. This implies that in this graph, for example, the Fermat primes can not be distinguished. But there are of course many more automorphisms.
I must say, however, that for all I know, there are only finitely many Fermat primes, or even better, for every set of primes $S$ there are only finitely many primes so that $q-1$ has $S$ as its set of prime factors. If, moreover, the number of such primes is distinct for ever $S$, then the graph turns out to be rigid (no automorphisms). But deciding unconditionally which is the case should be very hard.<|endoftext|>
TITLE: Do the "funny" tensor product and the cartesian product satisfy any algebraic "laws"?
QUESTION [11 upvotes]: Suppose that $X$ and $Y$ are two categories.  Let $\operatorname{Funny}(X,Y)$ denote the category whose objects are functors $X\to Y$ and whose morphisms are unnatural transformations $F\to G$, where an unnatural transformation $F\to G$ is given by an $\operatorname{Ob}(X)$-indexed family of arrows $\gamma_x:F(x)\to G(x)$, and that's it.
It is easy to see that its left adjoint gives a symmetric monoidal closed structure with unit object the terminal category.  We call this monoidal product the funny tensor product and denote it by $\Box$.  
What I'd like to know is if the funny tensor product satisfies any algebraic "laws" with the ordinary cartesian product, by which I mean, if we have some kind of sequence of tensor products and cartesian products together with parentheses, for example, something like the sequence  $$(((A\Box B)\times C)\Box D)\times E,$$ there is a way to either rearrange the parentheses or rewrite the expression with fewer or no parentheses.  
For instance, we can think of distributivity as an algebraic "law" between two operations.  We can see associativity as an algebraic "law" between an operation and itself.  However, I have never seen a nontrivial "law" between two associative operations having the same unit.  Is there an extended form of the Eckmann-Hilton argument which shows that no nontrivial relationship of this type can ever be satisfied unless the two operations are equal (given that they are both associative and share a unit)?

REPLY [13 votes]: In a recent article

Mark Weber, Free Products of Higher Operad Algebras, Theory and Applications of Categories, Vol. 28, No. 2, 2013, pp. 24–65, journal, arXiv:0909.4722.

there is a nice description of $□$ in terms of a pushout of the span $A_0\times B\leftarrow A_0\times B_0\rightarrow A\times B_0$.  This works as well for the presheaf category of graphs (directed multigraphs, if you like).  This pushout description yields canonical functors $A□(B\times C)\rightarrow(A□B)\times C$ that are easily seen not to be isomorphisms (both categories or graphs have the same objects, but different morphisms).  In fact, $□$ as tensor and $\times$ as "par" provide a "linearly distributive structure" on both, the category of small categories and the category of small graphs.  The latter notion is due to Robin Cockett and Robert Seely and was initially called "weak distributivity" (Weakly distributive categories, JPAA 114 Issue 2 (1997) pp 133--173, doi:10.1016/0022-4049(95)00160-3, plus corrected version (pdf) on Robert Seely's web page).<|endoftext|>
TITLE: Urge/reason for inventing  interior product ( Grassmann algebra )
QUESTION [7 upvotes]: Hello everyone, 
I wanted to lecture on Grassmann and his works , and I have been reading the collected works of Grassmann " Die Lineale Ausdehnungslehre ". There Grassmann introduced something called " Interior product " ( Left and Right interior products ) . So I was completely stuck up there, the Bourbaki papers don't speak on the reason or urge in creating such manipulations, but its understood by someone who really mastered them. 
Can anyone suggest me a good definition of the left and right interior products and explain the purpose of introducing them along with the intuition ? 
I would be really honored to hear that.  
Thank you.

REPLY [9 votes]: Grassmann's original motivations came from mechanics and geometry. Here is what he wrote in the foreword to his 1844 book "Die Lineale Ausdehnungslehre, ein neuer Zweig der Mathematik (Linear Extension Theory, a new branch of mathematics)":
"While I was pursuing the concept of product in geometry as it had been established by my father, I concluded that not only rectangles but also parallelograms in general may be regarded as products of an adjacent pair of their sides, provided one again interprets the product, not as the product of their lengths, but as that of the two displacements with their directions taken into account. When I combined this concept of the product with that previously established for the sum, the most striking harmony resulted; thus whether I multiplied the sum (in the sense just given) of two displacements by a third displacement lying in the same plane, or the individual terms by the same displacement and added the products with due regard for their positive and negative values, the same result obtained, and must always obtain."
More can be found at http://www-history.mcs.st-and.ac.uk/Extras/Grassmann_1844.html
There is a very accessible account in the  talk "Grassmann, Geometry and mechanics" by John Browne:
https://sites.google.com/site/grassmannalgebra/hermanngrassmann<|endoftext|>
TITLE: Homogeneous metrics of given volume element, on the n-sphere.
QUESTION [5 upvotes]: Let $\omega$ be an $n$-form on $S^n$, nowhere vanishing.
Is there a Riemannian metric $g$ on $S^n$, so that its volume form is $\omega$, and $(S^n,g)$ is homogeneous? Is it unique, and if not, what transformations of $S^n$ relates such metrics?
For example, for $n=1$, the unique metric with these properties is $\text{d} u^2$, where $u=\frac{1}{2\pi}\int\omega$.

REPLY [18 votes]: There is.  By Moser's Theorem, there exists a diffeomorphism $f:S^n\to S^n$ such that $f^*\omega = c\omega_{std}$ where $c>0$ and $\omega_{std}$ is the standard volume form homogeneous under $\mathrm{SO}(n{+}1)$.  Now use $f$ to transfer the right multiple of the standard metric (under which $S^n$ is homogeneous) back to one for which $\omega$ is the volume form.
About uniqueness:  (I forgot to answer this question on the first pass; also I am fixing a few errors in the cases below in this edit.)  The answer about uniqueness of homogeneous metrics with a given volume form depends on the dimension $n$.  Obviously, when $n=1$, it is unique.  It's also unique (up to volume-preserving diffeomorphism) when $n=2$ (and, in fact, for all even $n$).  However, it is not unique (up to volume-preserving diffeomorphism) when $n$ is odd. 
The reason has to do with this:  Suppose $g$ is a metric on $S^n$ that is homogeneous, and let $G$ be the identity component of the group of $g$-isometries. Then $G$ is compact, and $S^n = G/H$ where $H\subset G$ is a closed, connected subgroup of $G$ acting effectively on $S^n$.  It is known, from the work of Borel, which pairs such $(G,H)$ have the property that $G/H$ is diffeomorphic to $S^n$ for some $n$, so we know all of the possibilities and can work out the isometry groups in each case.  
When $n=2k$, the only possibility is $G = \mathrm{SO}(2k{+}1)$ with $H = \mathrm{SO}(2k)$.  The $G$-invariant metric $g$ is then uniquely determined by specifying its total volume. (While $S^6 = \mathrm{G}_2/\mathrm{SU}(3)$ would appear to be a counterexample, in this case $\mathrm{G}_2$ is not the full isometry group of the invariant metric, $\mathrm{SO}(7)$ is.)  
When $n = 4k{+}1>1$, we can have $G = \mathrm{SU}(2k{+}1)$ and $H = \mathrm{SU}(2k)$, and there is a $1$-parameter family of $G$-invariant metrics with a given total volume.  The full isometry group of such a metric is $\mathrm{U}(2k{+}1)$,
except that one element of the family (the metric of constant sectional curvature) has (connected) isometry group $\mathrm{SO}(4k{+}2)$.
When $n=4k{+}3$, we can have $G = \mathrm{Sp}(k{+}1)$ and $H=\mathrm{Sp}(k)$, and there is a $5$- (when $k=0$) or $6$- (when $k>0$) parameter family of $G$-invariant metrics on $S^n$ with a given total volume. Some of these are isometric though; one can get rid of $3$ of the parameters that way.  Some of these metrics have slightly larger isometry groups than $\mathrm{Sp}(k{+}1)$, such as $\mathrm{Sp}(k{+}1){\cdot}S^1$ or even $\mathrm{Sp}(k{+}1){\cdot}\mathrm{Sp}(1)$, and one, the metric of constant sectional curvature, has (connected) isometry group $\mathrm{SO}(4k{+}4)$. 
Finally, there is the exceptional case of $S^{15} = \mathrm{Spin}(9)/\mathrm{Spin}(7)$, which has a $1$-parameter family of inequivalent $\mathrm{Spin}(9)$-invariant metrics with the same total volume.   (Even though we also have $S^{15}=\mathrm{Sp}(4)/\mathrm{Sp}(3)$, these homogeneous metrics are not comparable, except at the constant curvature metric, which belongs to both families.)<|endoftext|>
TITLE: What finitely presented groups embed into GL_2? 
QUESTION [17 upvotes]: This is a naive question but I hope that the answers will be educational. When is it the case that a finitely presented group $G$ admits a faithful $2$-dimensional complex representation, e.g. an embedding into $\text{GL}_2(\mathbb{C})$? (I am mostly interested in sufficient conditions.) 
I think I can figure out the finite groups with this property (they can be conjugated into $\text{U}(2)$ and taking determinants reduces to the classification of finite subgroups of $\text{SU}(2)$ and an extension problem) as well as the f.g. abelian groups with this property (there can't be too much torsion). But already I don't know what finitely presented groups appear as, say, congruence subgroups of $\text{GL}_2(\mathcal{O}_K)$ for $K$ a number field. 
What can be said if you are given, say, a nice space $X$ with fundamental group $G$? I hear that in this case linear representations of $G$ are related to vector bundles on $X$ with flat connection.

REPLY [6 votes]: First, instead of subgroups of $GL(2,C)$ it suffices to consider subgroups of $SL(2, C)$. I will also restrict to subgroups which are not virtually solvable (since these should be easy to classify). Lastly, I will consider characterization of subgroups of $SL(2, C)$ "up to abstract commensuration" since the criterion is much cleaner in this setting. (Recall that two abstract groups $G_1, G_2$ are called "commensurable" if they contain isomorphic finite-index subgroups.) 
Definition. Let $p$ be a prime and $c$ an integer. A $p$-congruence structure of degree $c$ for a group $\Gamma$, is a descending chain of finite-index normal subgroups of $\Gamma$: 
$\Gamma=N_0\supset N_1 \supset ... \supset N_k ... $, such that:
(i) $\bigcap_{k=0}^\infty N_k= 1$; 
(ii)    $N_1/N_k$ is a finite $p$-group for every $k\ge 2$; 
(iii)   $d(N_i/N_j)\le c$ for all $j\ge i\ge 1$, where $d(H)$ denotes the minimal number of generators 
of a group $H$. 
Theorem. A finitely-generated (non-virtually solvable) group $\Gamma$ is commensurable to a 
subgroup of $SL(2,C)$ if and only if $\Gamma$ admits (for some prime $p$) a $p$-congruence structure of degree $c=3$. 
This theorem is an application of "A Group Theoretic Characterization of Linear Groups" by Alex Lubotzky (Journal of Algebra, 113 (1988), 207-214), combined with the classification of Lie algebras of dimension $\le 3$ over fields. Note that Lubotzky does not get a group-theoretic characterization of subgroups of $SL(n,C)$ for particular values of $n$ (even up to commensurability), however, for $n=2$ the situation is better than for the general $n$ since dimension of the Lie group is so small in this case.<|endoftext|>
TITLE: Reconstructing a word
QUESTION [15 upvotes]: Let $w(a,b)$ be a word in two letter alphabet. Let $$A=\left(\begin{array}{lll}x_1 & x_2 & x_3\\\ x_4 &x_5 & x_6\\\ x_7 & x_8 & x_9\end{array}\right), B=\left(\begin{array}{lll}y_1 & y_2 & y_3\\\ y_4 &y_5 & y_6\\\ y_7 & y_8 & y_9\end{array}\right)$$ where $x_i,y_i$ are commuting variables. Let $f_w=\mathrm{trace}(w(A,B))$, a polynomial in 18 variables. 
 Question.  Is it possible to reconstruct $w$ up to a cyclic shift from $f_w$? 
 Note that there exists a polynomial in one variable that encodes $w$: $x^{p_1}+...+x^{p_s}+x^{|w|}$ where $p_1,...,p_s$ are the places where $a$ occurs in $w$. Also note that for 2 by 2 matrices the answer is "no". For example if $w=abbaba$ and $w'=ababba$, then $f_w=f_{w'}$ for 2 by 2 matrices. The question is related to the study of the moduli space of representations (of degree 3) of the free group.
 Update  I think that as George suggested below, one can assume that $A=\mathrm{diag}(a,b,c)$ is a diagonal matrix (otherwise consider a conjugate of the pair $A,B$ over some algebraically complete field). After that the problem reduces to the following problem which seems longer but is in fact easier because we reduce the number of variables to from 18 to 3:
Pick a natural number $n\gg 1$. For every cyclic sequence $p$ (i.e. $p_{n+1}=p_1$) of $\{1,2,3\}$ of length $n$ consider a 9-vector 
$\phi(p)=($number of occurrences of 11, number of occurrences of 12, ..., number of occurrences of 33$)\in \mathbb{N}^9$.  The sum of coordinates of $\phi(p)$ is $n$, so we get a partition of $n$, and the number of different $\phi(p)$ is at most the number of partitions of $n$ into 9 parts, so less than $n^9$. Thus the map $\phi$ has a non-trivial kernel $\mathrm{Ker}(\phi)$ (i.e. the equivalence relation $p\equiv q$ iff $\phi(p)=\phi(q)$ ). Let $S$ be a preimage of a point in $\mathbb{N}^9$ under $\phi$. Let $v=(v_1,...,v_n)$ be a cyclic vector of natural numbers (including 0). For every $p\in S$ consider the monomial $m_p=a^sb^tc^u$ in  3 variables where $s$ is the sum of $v_i$ such that $p_i=1$, $t$ is the sum of $v_i$ such that $p_i=2$, $u$ is the sum of $v_i$ such that $p_i=3$. The sum of all the monomials $m_p$, $p\in S$, is a polynomial $f_S(v)$ in $a,b,c$. That polynomial is the coefficient of the monomial $\prod_{(i,j)} B[i,j]^{\phi(p)[i,j]}$ in $f_w$. 
 Question  Is the sequence $v$ determined by the sequence of polynomials $f_S(v)$ where $S$ runs over the equivalence classes of the partition $\mathrm{Ker}(\phi)$.

REPLY [4 votes]: This is not an answer, just an explanation of my comment.
Fix an integer $n$. Consider the Hilbert space $H=\ell^2(\{a,b\}^n)$ spanned by the words of length $n$ in $a$ and $b$. The symmetric group $S_n$ acts on $H$. For every permutation $\pi \in S_n$ consider $c(\pi)$, the number of cycles in $ \pi^{-1} \tau \pi \tau^{-1}$, where $\tau$ is the cycle $(1 2 \dots n)$. Consider the following operator on $H$:
$$T = \sum_{\pi \in S_n} 3^{c(\pi)} \pi.$$
The content of my comment was that for two words of length $n$, $f_w = f_{w'}$ if and only if $T(\delta_w - \delta_{w'})=0$. In particular, it would be sufficient to prove that the kernel of $T$ is just the kernel of $\sum_{k=1}^n \tau^k$ (= the functions on $\{a,b\}^n$ that sum to $0$ on each $\tau$-orbit) to answer positively Marks' question. Unfortunately I could not prove it.
Indeed, take $A$ and $B$ random with entries that are independant centered complex gaussians with variance $1$ (the real and imaginary parts are real gaussians centered with variance $1/2$). Then 
\[ \mathbb E[f_w(A,B) \overline{f_{w'}(A,B)}] = \langle T \delta_w, \delta_{w'}\rangle.\]
This formula implies that $T$ is a non-negative operator, and that $f_w \mapsto  \sqrt T \delta_\omega$ is an isometry from the subspace of $L^2$ spanned by the $f_w$'s to $H$, and hence that $f_w = f_{w'}$ iff $\sqrt T \delta_w = \sqrt T \delta_{w'}$ iff $T \delta_w =  T \delta_{w'}$.
I am not sure it is worth writing a precise proof of this equality: if one developps the LHS of the inequality, on gets a big sum of moments of gaussians. To simplify this expression, use some moment-cumulant transformation which yields to the RHS. I can at least say that $3^{c(\pi)}$ appears as the number of pairs of $n$-uples $(i_1,\dots,i_n)$ and $(j_1,\dots j_n)$ such that $(i_k,i_{k+1}) = (j_{\pi(k)},j_{\pi(k)+1})$.<|endoftext|>
TITLE: Does infinite-dimensional Brownian motion live in hyperplanes?
QUESTION [8 upvotes]: I'll begin this question with the finite-dimensional case, as a
warmup.
Let me say a continuous path $\omega : [0,1] \to \mathbb{R}^d$ is
hyperplanar if there exists a nonzero $x \in \mathbb{R}^d$ such
that $\omega(t) \cdot x = 0$ for all $t \in [0,1]$.
It is not hard to show:

Proposition.  Let $B_t$ be a standard Brownian motion in $\mathbb{R}^d$.  Almost surely, $(B_t : 0 \le t \le 1)$ is not hyperplanar.

To avoid possible confusion, let me emphasize that the hyperplane is
allowed to be random.  That is, I am claiming:
$$\mathbb{P}(\exists x \in \mathbb{R}^d\\, \forall t \in [0,1] : B_t \cdot
x = 0) = 0$$
which should not be confused with the weaker statement
$$\forall x \in \mathbb{R}^d :\mathbb{P}(\forall t \in [0,1] : B_t \cdot
x = 0) = 0.$$
One possible proof of the proposition is to choose $0 < t_1 < \dots <
t_d < 1$, and show by induction that, almost surely, $B_{t_1}, \dots,
B_{t_d}$ are linearly independent in $\mathbb{R}^d$.  (By the
induction hypothesis, $B_{t_1}, \dots, B_{t_{k-1}}$ span a
$k-1$-dimensional subspace of $\mathbb{R}^d$; by the Markov property
and the absolute continuity of the Gaussian distribution, $B_{t_k}$ is
almost surely not in this subspace.)

I am interested in the analogous statement for infinite dimensions.
Let $W$ be a real separable Banach space, and say a continuous path $\omega :
[0,1] \to W$ is hyperplanar if there exists a nonzero continuous
linear functional $f \in W^*$ such that $f(\omega(t)) = 0$ for all $t
\in [0,1]$.  Let $\mu$ be a non-degenerate Gaussian measure on $W$
(so that $(W,\mu)$ is an abstract Wiener space), and let $B_t$ be a standard Brownian motion in $W$.  That is, the process
$B_t$ has continuous sample paths and independent increments, starts
at 0, and the increments are distributed such that $(t-s)^{-1/2}(B_t -
B_s) \sim \mu$.

For infinite-dimensional $W$, what is the probability that $(B_t : 0 \le t \le 1)$ is hyperplanar?

As a start, I can show that the set of hyperplanar paths is analytic
in $C([0,1], W)$, and hence universally measurable, so the question
actually makes sense.  We can also use a scaling argument and the
Blumenthal zero-one law to see that the probability must be either 0
or 1.
I am not sure which way my intuition points here.  On the one hand, we
expect a Brownian motion to be pretty irregular and unconstrained,
suggesting the answer is 0 as in finite dimensions.  On the other
hand, an infinite-dimensional space has a lot of hyperplanes.  In principle, the answer could depend on the abstract Wiener space $(W,\mu)$.
Thanks!

REPLY [8 votes]: As suggested in my comment, here's a simple fact which applies to any probability measure $\mu$ on (the Borel σ-algebra of) a second countable topological space $X$. There is a unique minimal closed subset $S$ of $X$ with $\mu(S)=1$ -- the support of $\mu$ -- and, if $X_1,X_2,\ldots$ is an IID sequence of random variables, each with distribution $\mu$, then $\lbrace X_1,X_2,\ldots\rbrace$ is almost surely a dense subset of $S$.
Now, in the situation described in the question, choose a sequence $0\le t_0 < t_1 < t_2 < \cdots \le1$. Then, $X_k\equiv(t_k-t_{k-1})^{-1/2}(B_{t_k}-B_{t_{k-1}})$ is an IID sequence of random variables each with distribution $\mu$. Also, as it is non-degenerate, the support of $\mu$ is not contained in a closed proper subspace of the Banach space $W$.

REPLY [2 votes]: Shortly after posting this, I discussed it with Clinton Conley and we came up with what is essentially the same as George Lowther's argument.
The point is that, by the Hahn-Banach theorem, $\omega$ is hyperplanar iff the linear span of $\{\omega(t) : t \in [0,1]\}$ is not dense in $W$.  But with probability 1, this span is dense.  
$W$ is a separable metric space, so it has a countable basis $\{U_i\}_{i=1}^\infty$.  Choose an infinite sequence $0 < t_1 < t_2 < \dots < 1$ and set $\Delta_n = (t_{n+1}-t_n)^{-1/2}(B_{t_{n+1}}-B_{t_n})$ so that the $\Delta_i$ are iid with distribution $\mu$.  For each $i$, $\mu(U_i) > 0$ by non-degeneracy, so almost surely, one of the $\Delta_n$ lies in $U_i$.  In particular, the linear span of $\{B_t\}$ meets $U_i$.  Taking a countable intersection, almost surely, the linear span of $\{B_t\}$ meets every $U_i$ and hence is dense.
Edit: This characterization also makes it clear that the set of hyperplanar paths is much better than just analytic.  Indeed, for fixed $i$, the set of paths $\omega$ such that the linear span of $\{\omega(t) : 0 \le t \le 1\}$ meets $U_i$ is easily seen to be open.  The set of non-hyperplanar paths is thus $G_\delta$.  It is also dense (since a measure with full support gives it measure 1, or by the simpler argument that any hyperplanar path can be slightly perturbed to make it non-hyperplanar), and in particular comeager.  So this also answers the Baire category analogue of my question.<|endoftext|>
TITLE: Convergence of moments implies convergence to normal distribution
QUESTION [12 upvotes]: I have a sequence $\{X_n\}$ of random variables supported on the real line, as well as a normally distributed random variable $X$ (whose mean and variance are known but irrelevant). I know that the moments of the $X_n$ converge to the corresponding moment of $X$, that is, for every $k\ge1$,
$$
\lim_{n\to\infty} \mu_k'(X_n) = \mu_k(X).
$$
I need to conclude that the $X_n$ converge to $X$ in distribution.
I believe that this is a standard fact in probability, and I would like an excellent source (including a clear statement and proof) for this fact, to cite in a paper I'm writing. (The application is to number theory, which is why I added the probabilistic-number-theory tag.) I also believe that this conclusion holds for many, but not all, random variables $X$ and not just a normally distributed one; I'd be happy for a general statement or one that applies only to a normal variable.
Nominations for a good citing source, anyone?

REPLY [11 votes]: It is theorem 30.2 in Billingsley's Probability and Measure (I own a second Polish edition, so numbering may differ a little).
It's quite easy to prove it, once you estabilish Prokhorov's theorem; namely use boundedness of some moments to conclude that your sequence of distributions is tight and then it suffices to convey everyone that every convergent subsequence of $(X_n)$ converges to $X$ (because convergence in distribution is metrizable), which is easy, because the limit is characterized by its moments. Before that, one needs a lemma stating that convergence in distribution combined with convergence of moments implies that moments converge to the moments of the limit.<|endoftext|>
TITLE: Adjunction between Pro and Ind Completions
QUESTION [11 upvotes]: Hi,
Let $\mathcal{C}$ be a small category with finite limits and finite colimits. 
The Ind-completion $Ind(\mathcal{C})$ is a locally finitely presentable category which is the free cocompletion of $\mathcal{C}$ under directed colimits. That is, there is a full embedding $I : \mathcal{C} \hookrightarrow Ind(\mathcal{C})$ and for any functor $H : \mathcal{C} \to \mathcal{D}$ where $\mathcal{D}$ has directed colimits, there is an essentially unique directed colimit preserving extension $H^* : Ind(\mathcal{C}) \to \mathcal{D}$ with $H^* \circ I = H$.
The latter uses the fact that $\mathcal{C}$ has finite colimits. We can also construct the Pro-completion $Pro(\mathcal{C})$. Since $\mathcal{C}$ has finite limits it arises as the free completion of $\mathcal{C}$ under inverse limits and is dual to the locally finitely presentable category $Ind(\mathcal{C}^{op})$.
Then there is an adjunction $F \dashv U$ where $U : Pro(\mathcal{C}) \to Ind(\mathcal{C})$ is the unique inverse limit preserving extension of $\mathcal{C} \hookrightarrow Ind(\mathcal{C})$, likewise $F$ uniquely extends $\mathcal{C} \hookrightarrow Pro(\mathcal{C})$. For example if $\mathcal{C} = \mathsf{FinSet}$ then essentially $U : \mathsf{Stone} \to \mathsf{Set}$ is the usual forgetful functor and $F$ is the Stone-Cech compactification.
Questions:


Is it known whether $U$ and $F$ are always faithful and preserve both epis and monos?


Are any other general preservation properties known?


Aside from "Stone Spaces", what is a good source for the connection between Pro and Ind completions?


Any help much appreciated.

Here are some examples where $U$ and $F$ are both faithful and both preserve epis and monos.


If $\mathcal{C}$ is complete or cocomplete then it is essentially a complete lattice (since $C$ is small), so $Ind(\mathcal{C}) \cong Pro(\mathcal{C}) \cong C$ and $U$, $F$ define an equivalence.


If $\mathcal{C} = \mathsf{FinSet}$ then $U : \mathsf{Stone} \to \mathsf{Set}$ is the forgetful functor, whose left adjoint is the Stone-Cech compactification. 




If $\mathcal{C} = \mathsf{FinBA}$ then $U : \mathsf{CABA} \to \mathsf{BA}$ is the faithful forgetful functor, whose left adjoint could be described as the canonical extension of a boolean algebra.



If $\mathcal{C} = \mathsf{FinDL}$ then $U : \mathsf{Priestley} \to \mathsf{Poset}$ is the forgetful functor, whose left adjoint is the ordered Stone-Cech compactification.



If $\mathcal{C} = \mathsf{FinPoset}$ then $U : \mathsf{DADL} \to \mathsf{DL}$ is the faithful forgetful functor where $\mathsf{DADL}$ is the category of doubly algebraic distributive lattices with complete lattice morphisms. Its left adjoint could be described as the canonical extension of a distributive lattice.



If $\mathcal{C} = \mathsf{FinJSL}_\bot$ (finite join semilattices with bottom) then  $U : \mathsf{JSL}_\bot^{op} \to \mathsf{JSL}_\bot$ sends semilattices to their ideal completion and morphisms to the right adjoint of their extension.



If $\mathcal{C} = \mathsf{FDVect}(\mathbb{F})$ (finite dimensional vector spaces) then  $U : \mathsf{Vect}(\mathbb{F})^{op} \to \mathsf{Vect}(\mathbb{F})$ is the usual functor sending a vector space to its dual space.



If $\mathcal{C} = \mathsf{FinAb}$ (finite abelian groups) then  $U : \mathsf{Ab}(\mathsf{Stone}) \to \mathsf{TorAb}$ is the forgetful functor from the Stone topological abelian groups to abelian groups with torsion.


Admittedly I haven't checked all the details, but I'd also be interested in other such examples.

REPLY [2 votes]: No a answere but a specification and description of the issue (too long for a comment).
The free cocompletion property of $Ind(\mathcal{C})$ dont use the hypothesis "$\mathcal{C}$ has finite colimits". ANyway $Ind(\mathcal{C})$ is equivalent to the full subcategory $P_1(\mathcal{C})\subset \mathcal{C}^>$ of presheves that are coimits of a small, filtred diagram of representable. In this way the inclusion $i$ corresponds to the yoneda inclusion $h_-: \mathcal{C}\to P_1(\mathcal{C})$ and it preserves  limit, and finite colimits.
The inclusion $P_1(\mathcal{C})\subset \mathcal{C}^>$ create (small) filter colimits (if $P=\varinjlim_{i\in I}P_i$ is a filtrant colimit with $O_i\in P_1(P_1(\mathcal{C})$ then combining the comma categories $P_1(\mathcal{C})\downarrow P_i$ we
make a filtrant (small) diagram of $P_1(\mathcal{C})\downarrow P_i$
Is $F: \mathcal{C}\to \mathcal{D}$ where $\mathcal{D}$ as filter colimits we have a (iso)unique extentions $F': P_1(\mathcal{C})\to \mathcal{D}$ with $F'(P):=\varinjlim_{\ (X, x)\in \mathcal{C}\downarrow P} F(X)$, if $P$ has the Ind-representation $(X_i)_{i\in I}$ i.e. $P=\varinjlim_i h_{X_i}$ then the diagram of the  $h_{X_i}$ is a final diagram on the comma category $\mathcal{C}\downarrow P$ then $F'(P):=\varinjlim_i F(X_i)$
If $\mathcal{C}$ has finite colimits $P_1(\mathcal{C})\cong Cart(\mathcal{C}^{op}, Set)$ the latter is the the category cartesians presheaves i.e. that maps  finite colimits of $\mathcal{C}$ to finite limits in $Set$, of course the embedding $Cart(\mathcal{C}^{op}, Set)\subset \mathcal{C}^>$ create limits .
All above as a dual version for  $Proj(\mathcal{C}):=(Ind(\mathcal{C}^{op}))^{op}$, it is a free completion   of $\mathcal{C}$, and if $\mathcal{C}$ has finite limits $Proj(\mathcal{C})$ is equivalent to $Cart(\mathcal{C}, Set)^{op}$ (dual to the category of copresheaves that preserving finite limits)  it has (small) colimits and we have the embedding $\iota:=(h^-)^{op}: \mathcal{C}\to Cart(\mathcal{C}, Set)^{op} $.
Now the inclusion $h_-: \mathcal{C}\to Cart(\mathcal{C}^{op}, Set)$ and $\iota: \mathcal{C}\to Cart(\mathcal{C}, Set)^{op}$ induce for the universal properties of these two completions the functors $U: Cart(\mathcal{C}, Set)^{op}\to Cart(\mathcal{C}^{op}, Set)$ with $U(Q)=\varprojlim_{\ h^Y\to Q} h_Y$, and
$F: Cart(\mathcal{C}^{op}, Set)\to Cart(\mathcal{C}, Set)^{op}$ with 
$F(P)=\varprojlim_{\ h_X\to P} h^X$ (the limit is in $\mathcal{C}^{<}$ the copresheaves category)
we have the natural isomorphisms:
$Cart(\mathcal{C}, Set)^{op}(F(P), Q)=Cart(\mathcal{C}, Set)(Q, F(P))=$
$\mathcal{C}^{<}(\varinjlim_{\ h^Y\to Q}h^Y, \varprojlim_{\ h_X\to P}h^X)=$
$\varprojlim_{\ h_X\to P} \varprojlim_{h^Y\to Q}\mathcal{C}^<(h^Y, h^X)\cong$
$\varprojlim_{\ h_X\to P} \varprojlim_{h^Y\to Q}\mathcal{C}^>(h_X, h_Y)\cong$
$\mathcal{C}^>(\varinjlim_{\ h_X\to P} h_X, \varprojlim_{h^Y\to Q}h_Y) \cong$
$Cart(\mathcal{C}^{op}, Set)(P, U(Q))$
then $U$ is a adjoint to $F$.<|endoftext|>
TITLE: A name for Radon-Nikodym derivatives that are bound away from zero and infinity
QUESTION [6 upvotes]: Dear Mathoverflow, 
I would like to know if the nomenclature of mathematics has a name for Radon-Nikodym derivatives that are bounded away from zero and infinity almost everywhere. As in for equivalent measures $\mu, \nu$, there exists constants $c,C$ such that 
$$ 0 < c \leq \frac{d\nu}{d\mu}(x) \leq C < \infty$$
for $\mu$-almost every $x$.
Such measures could be called boundedly equivalent. But if there already exists a name, I'd like to use it.
Another possibility is to say the measures are correlated. Intuitively the condition means $\mu$ and $\nu$ either both give small or large values to the same $x$. However this is word already has a lot of meaning in maths - perhaps it is best not to add more.
Also, I'm open to suggestion if someone would like to offer a better name.

REPLY [5 votes]: Good question. I am not aware of any standard terminology, although this condition is quite natural and appears pretty often. I would rather call such measures uniformly equivalent. As for "correlated" - as you say, it would create wrong connotations.<|endoftext|>
TITLE: Computations and applications of Khovanov's functor valued invariant of tangles
QUESTION [6 upvotes]: Soon after his famous paper "A categorification of the Jones polynomial", Khovanov
introduced a "bordered" version.  His theory assingns to every oriented even tangle
a complex of (H_n,H_m)-bimodules, where H_n is a certain family of rings and n,m\geq 0
depend on the number of points on the top (resp. bottom) of the tangle.
This paper can be found here: http://arxiv.org/pdf/math/0103190.pdf
It is natural to ask if this relative version can be used as a tool for computations.
So my first question is:
Q1 Are there any examples in the literature of explicit computation regarding the chain complex or the homology associated to some families of tangles?
My second question is about possible applications of this invariant.
Q2 Are there any examples in the literature where this theory has been used?
Regarding the second question I should mention that the work of Smith and Abouzaid
which leads to a proof of the equivalence between Khovanov homology and Symplectic Khovanov homology actually makes use of this relative theory.

REPLY [4 votes]: I recall discussing this with you in Budapest last summer, and I was curious about the same thing. Here are my thoughts a year later:
I haven't seen any computational examples in the literature (not that they don't exist), for Q1, but in my experience any computation using Bar-Natan's formal-pictures theory (dotted cobordism version) can be translated to Khovanov's $\mathcal{H}^n$-theory. The module associated to a tangle is the projective $\mathcal{H}^n$-module whose generators are the generators of the Bar-Natan complex; in other words, you just view a Bar-Natan generator as representing all possible outside closures (and choices of signs on the resulting circles). The differential is given by applying the usual 2d TQFT to the (closed-off in all possible ways) cobordisms in Bar-Natan's complex. When you make a simplification in Bar-Natan's complex, e.g. a Gaussian elimination of two generators $x$ and $y$ connected by an identity-cobordism component of the differential, there's a corresponding homotopy equivalence of the associated $\mathcal{H}^n$-modules.
There are tons of computations using Bar-Natan's theory, so if you want to see how an $\mathcal{H}^n$ computation works, you could probably take one of those and just follow through in $\mathcal{H}^n$-language. (There are probably equally good examples, but I can't resist linking to my own computations for pretzel knots that I was working on last summer, arXiv:1303.3303- you could probably take any part of that computation and apply the $\mathcal{H^n}$ functor to get a computation in that language.)
For Q2, there's a whole lot on the representation theory side of the picture. Stroppel's paper arXiv:math/0608234 (and several other related papers) present $\mathcal{H}^n$ as an idempotent truncation of a larger, quasi-hereditary algebra $A_{n,n}$, which is (if I'm saying it correctly) the endomorphism algebra of a certain block of a parabolic category $\mathcal{O}$ for $\mathfrak{gl}_{2n}$. There's a corresponding tangle theory that works over $A_{n,n}$ instead of $\mathcal{H^n}$ and still recovers Khovanov homology. The category in question is equivalent to a category of perverse sheaves on the Grassmannian $G(n,n)$, but now I'm way out of my depth! This material might be an interesting read, though, if you're looking for how $\mathcal{H^n}$ fits into the broader scheme of things.<|endoftext|>
TITLE: line bundles and universal covers
QUESTION [9 upvotes]: When dealing with some lifting problems, I came across the following problem, which probably has a well-known answer, but anyway:
Suppose I have a (locally contractible) connected topological group $G$, such that $\pi_1(G) \cong \mathbb{Z}$. Let $\tilde{G} \to G$ be its universal covering group. Let $P \to X$ be a principal $G$-bundle over a compact Hausdorff space $X$. The obstruction of lifting $P$ to a principal $\tilde{G}$-bundle lives in $H^2(X, \pi_1(G)) \cong H^2(X, \mathbb{Z})$ and is therefore given by a complex line bundle over $X$.

Is it possible to construct this line bundle directly from $P$ (without going through the classification of line bundles by their Chern classes)?

REPLY [6 votes]: I don't think there is a canonical construction of the associated line bundle, this will be only defined up to homotopy, in some sense.
Here is a slightly more geometric construction. Let $G'$ be the group obtained by substituting $\mathbb Z \subseteq \widetilde G$ with $\mathbb R$; that is, $G' = (\widetilde G \times \mathbb R)/\mathbb Z$, with $\mathbb Z$ embedded anti-diagonally. The natural projection $G' \to \widetilde G/\mathbb Z = G$ has kernel $\mathbb R$, hence $P$ will lift to a $G'$-principal bundle $P'$, in a unique fashion, up to isomorphism (but not canonically). Also, $G'$ contains $\widetilde G$ as a normal subgroup, and $G'/\widetilde G \simeq \mathbb R/\mathbb Z \simeq \mathrm S^1$; so $P'$ gives an $\mathrm S^1$ principal bundle, which is the desired obstruction.<|endoftext|>
TITLE: Characterising semi-definite positiveness on vectors with non-negative entries
QUESTION [8 upvotes]: My problem is to characterise (or find useful information on) the cone $C$ of $N\times N$ matrices $M$ ($N\geq 1$) such that $$V^t M V\geq 0$$ for every vector $V $ with non-negative entries. Is this cone of matrices familiar to anyone?
Remark 1: $C$ clearly contains the convex cone $S$ of semi-definite positive matrices, and that of matrices $P$ with non-negative entries. Even though it looks too easy, I could not prove that $C$ is not the convex hull of $S$ and $P$.
Remark 2: My original problem is to characterise the dual cone of $C$, containing the covariances of random vectors of $\mathbb{R}^N$ with a.s. non-negative entries.

REPLY [6 votes]: (After Noah Stein's answer) By definition, the dual cone $C^\star$ is spanned by matrices $v\otimes v$ with $v\ge0$. The following counter-example is due to Hall. The $5\times5$ symmetric matrix
$$S=\begin{pmatrix}  4 & 0 & 0 & 2 & 2 \\\\ 0 & 4 & 3 & 0 & 2 \\\\ 0 & 3 & 4 & 2 & 0 \\\\ 2 & 0 & 2 & 4 & 0 \\\\ 2 & 2 & 0 & 0 & 4 \end{pmatrix}$$
has non-negative entries and is positive semi-definite. Therefore, it belongs to $(S\cup P)^\star$. Yet, it cannot be written as the sum of $v_j\otimes v_j$ where the vectors $v_j$ are non-negative. Therefore $A\not\in C^\star$. By duality, this proves that $C$ is not the convex hull of $S\cup P$.
The example is analyzed in details in Exercise 347 of my list.<|endoftext|>
TITLE: Which polynomials are Fricke polynomials ?
QUESTION [5 upvotes]: Let me recall the definition which seems the most standard of Fricke polynomials. 
Let $G$ be the free group with two generators $u,v$. It is not very hard to prove that there exists a unique application $t : G \rightarrow \mathbb{Z}[X,Y,Z]$
such that 
(1) $t(u)=X,t(v)=Y,t(uv)=Z$,
(2)  $t$ is the character (i.e. the trace) of a representation $r : G \rightarrow \rm{SL}_2(K)$ where $K$ is some field containing $\mathbb{Z}[X,Y,Z]$.
A  Fricke polynomial is just an element of $\mathbb{Z}[X,Y,Z]$ which is in the mage of $t$.

What is the image of $t$ ?  


Edit: As requested, I give here the proof of the statement about $t$. I believe that this statement is due in some form to Fricke in the nineteenth century: 
If $t:G \rightarrow K$ is
the trace of a representation $r: G \rightarrow {\rm{SL}}_2(K)$, then one has $t(gh)+t(gh^{-1}) = t(g)t(h)$ for all $g,h$ in $G$ (since this formula holds when $t$ is the trace, and $g,h$ are matrices in ${\rm SL}_2(K)$, as easily checked). Now this formula, and an easy induction on the length
of a word $g$ in $G$ shows that $t(g)$ can be expressed as a polynomial (with integral coefficients) in $t(u)$, $t(v)$ and $t(uv)$. Hence a $t$ satisfying (1) and (2) is unique if it exists. 
For the existence, let $U$ and $V$ be two matrices in ${\rm SL}_2(\mathbb{C})$ such that tr$U$,
tr$V$, and tr$UV$ are algebraically independent (this is really easy). Then see $\mathbb{Z}[X,Y,Z]$ as a subring of $\mathbb{C}$ by sending $X$ on tr$U$, $Y$ on tr$V$, $Z$ on tr$UV$.
Let $r$ be the representation $G \rightarrow \rm{SL}_2(\mathbb{C})$ sending $u$ on $U$ and $v$ on $V$, and $t=$tr $r$. Then by definition $t$ satisfies (2), with $K=\mathbb{C}$, one has $t(u)=X$, $t(v)=Y$, $t(uv)=Z$ by construction and $t$ takes value in the subring $\mathbb{Z}[X,Y,Z]$ of $\mathbb{C}$ by what we have said earlier. Hence the existence of $t$.

REPLY [2 votes]: I do not think that there is a complete answer to this question. However, one can give some necessary conditions (which show that any answer must be complicated).
One can show that a triple $(x,y,z) \in \mathbb C^3$ comes from traces of matrices in $SU(2)$ (i.e. there exist $u,v \in SU(2)$ such that $(x,y,z)=(t(u),t(v),t(uv))$) if and only if $(x,y,z) \in \mathbb [-2,2]^3$ (this is obviously necessary) and $$x^2 + y^2 +  z^2 - xyz \leq 4.$$
Hence, a necessary condition on a polynomial $p \in \mathbb Z[X,Y,Z]$ to be a Fricke polynomial is that for all $(x,y,z) \in [-2,2]^3$ we have the implication
$$x^2 + y^2 +  z^2 - xyz \leq 4 \quad \Rightarrow \quad p(x,y,z) \in [-2,2].$$
Indeed, the word evaluated on $u$ and $v$ will give again a matrix in $SU(2)$ and hence its trace is in the interval $[-2,2]$.
The semi-algebraic set $$S:=[-2,2]^3 \cap \lbrace (x,y,z) \in \mathbb R^3 \mid x^2 + y^2 +  z^2 - xyz \leq 4 \rbrace$$ is a spectrahedron (that means it can be defined by a linear matrix inequality) and is called the Elliptope $E_3$.<|endoftext|>
TITLE: Who proved that the plane partition generating function is valid?
QUESTION [6 upvotes]: I know Major Macmahon conjectured the formula $$ \prod_{m=1}^\infty \frac{1}{(1-q^m)^m}=1 + \sum_{n=1}^\infty PL(n)q^n$$
but who was the first to prove it?

REPLY [8 votes]: The answer is MacMahon himself, who proved this in his book Combinatory Analysis as a corollary of a more general theorem about plane partitions.  See Sections IX and X.
There is some additional historical information in the Notes to Chapter 7 of Richard Stanley's book Enumerative Combinatorics, volume 2.<|endoftext|>
TITLE: Who discovered the asymptotic formula for the number of partitions of n into distinct parts?
QUESTION [5 upvotes]: Who was the first to develop the asymptotic formulae for the distinct parts version of $p(n)?$

REPLY [7 votes]: According to Dickson, History of the Theory of Numbers, Volume 2, page 162, "G. H. Hardy and S.Ramanujan proved that the logarithm of the number $p(n)$ of partitions of $n$ is asymptotic to $\pi\sqrt{2n/3}$, and the logarithm of the number of partitions of $n$ into distinct positive integers is asymptotic to $\pi\sqrt{n/3}$." The reference is given as Proc London Math Soc 16 (1917) 131.

REPLY [6 votes]: Look, by Euler's theorem the number of partitions $p_{dist}(n)=p_{odd}(n)$.  Since the number of parts of a random odd partition (i.e. into odd part sizes) is about $O(\sqrt{n}\log n)$, removing 1 from each part gives an even partition of roughly the same size.  This gives
$$p_{odd}(n) \approx p_{even}(n) = p(n/2),
$$
when defined appropriately.  This shows that you really don't need a separate new asymptotic formula for $p_{dist}(n)$ if rough approximation is ok.  While informal, this argument can be made completely formal, and has been done a few times, I think. 
UPDATE: I came across Hua Luogeng's paper "On the number of partitions of a number into unequal parts" (1942), which gives an analogue of Rademacher's formula for $p(n)$.<|endoftext|>
TITLE: Minimal degree of polynomial vanishing on the variety of small degree. 
QUESTION [15 upvotes]: My question is assume that we know that the degree of some irreducible variety is small does it possible to conclude that there exists polynomial of small degree vanishing on this variety. 
Let us make the question more concrete: 
Let $V\subset A^{2n}$ be an irreducible algebraic variety of dimension $n$ and degree $d$ (say $2^n$).
Let $I \triangleleft \mathbb{C}[x_1,\ldots, x_{2n}]$ be an ideal of all polynomials vanishing on $I$. My question is what is the best upper bound on the  minimal degree of polynomials in $I$. 
The best upper bound that I know is $2^n$, while I think that it is not the true. 
For example if $V$ is a complete intersection variety then $I$ must have polynomial of degree $2$.

REPLY [4 votes]: You can have a look at the article "Direct methods for primary decomposition" by Eisenbud, Huneke and Vasconselos (Inventiones 110, 1992), available on Eisenbud's webpage. In proposition 3.5, they prove (and say it is long known) that a homogeneous equidimensional ideal $I$ of degree $d$ is generated up to radical by forms of degree at most $d$. 
See also the remark following the proposition for an example showing it is sharp: $V$ is the union of $d$ skew lines that all meet a common line $L$. They also mention a conjecture in the case where $I$ is prime: up to a radical it should be generated by form of degree at most $d-$ codim $I + 1$.<|endoftext|>
TITLE: Knot theory question: bridge number vs. min generators of fundamental group of complement
QUESTION [12 upvotes]: Given a knot in the 3-sphere in Bridge Position you can find a presentation for the fundamental group of the complement (a Wirthinger presentation) containing $n$ generators and $n-1$ relators, where $n$ is the number of bridges in the presentation. 
The bridge index of the knot is the minimal number of bridges it takes to present the knot.  Is that the same as the minimal number of generators among all presentations of the fundamental group of the knot complement?  
I imagine this is well known but after a while of Googling around and looking through Neuwirth's "Knot Groups" book I haven't found where this question is addressed.

REPLY [9 votes]: It is not a problem, but this is a problem.
A question posed by Cappell and Shaneson asks whether the bridge number of a given link equals the minimal number of meridian generators of its group.
http://ccms.claremont.edu/topology-seminar/Bridge-numbers-links-and-meridian-ranks-link-groups<|endoftext|>
TITLE: When tensor reflects exact sequences?
QUESTION [14 upvotes]: Is there a characterization of modules $N$ for which the functor $N\otimes-$ reflects exact sequences?

REPLY [8 votes]: As Alex Becker pointed out, $N$ is faithfully flat iff $N\otimes_R -$ preserves and reflects exact sequences. Since $N$ is flat iff $N\otimes_R -$ preserves exact sequences, you might think $N$ is faithful iff $N\otimes_R -$ reflects exact sequences. This is wrong, as we'll show below. These faithfully flat modules are extremely important, and probably the better place to do work if you can manage it. I think the concept of a module $N$ such that $N\otimes_R -$ reflects exact sequences (but $N$ is not faithfully flat) is an interesting one. However, this concept doesn't seem to appear in T.Y. Lam's Lectures on Modules and Rings, and that's my bible for these types of questions. So it seems the community hasn't found this concept interesting yet. Since no one seems to have coined a phrase for these yet, let's call them reflecting modules (not to be confused with reflexive modules).
When seeking to prove a module is flat you can use that $N\otimes_R -$ is always right exact, so all you need to show is that $N\otimes_R -$ preserves monomorphisms. To show $N$ is reflecting is harder. For example, you must show that if $N\otimes_R g$ is an epimorphism, then $g$ is an epimorphism. You must also show that if $N\otimes_R f$ is a monomorphism then $f$ is a monomorphism. These two follow if $N\otimes_R -$ is a faithful functor (this explains the terminology). According to Theorem 7.1 in Theory of Categories by Barry Mitchell, if $T:C\rightarrow D$ is faithful functor between exact categories which have zero objects, and if $T$ preserves the zero objects, then $T$ reflects exact sequences. Since $R$-mod is an exact category with a zero object, this tells us that $N$ is reflecting if $N\otimes_R -$ is faithful. We'll show the converse below. Consider the following two properties:
(1) For any left $R$-module $M$, $N\otimes_RM=0 \Rightarrow M=0$
(2) A homomorphism of left $R$-modules $\phi:M'\rightarrow M''$ is zero if the induced homomorphism $N\otimes_R M' \rightarrow N\otimes_R M''$ is zero. Note that this property is exactly saying $N\otimes_R -$ is faithful.
When $N$ is flat, these are equivalent, and also equivalent to the fact that $N$ is faithfully flat. Even without the flatness hypothesis, the theorem above shows (2) implies $N$ is reflecting. We can also see that (2) implies (1) because if $N\otimes_R M = 0$ then the map $N\otimes_R 0 \rightarrow N\otimes_R M$ is zero, so the map $0\rightarrow M$ is zero, proving $M=0$. Next, (1) implies (2) because if $N\otimes_R \phi = 0$ then $0=im(N\otimes_R \phi)=N\otimes_R im(\phi)$ so by (1), $im(\phi)=0$. Finally, $N$ reflecting implies (1) because knowing $N\otimes_R 0 \rightarrow N\otimes_R M\rightarrow N\otimes 0$ is exact implies $0\rightarrow M\rightarrow 0$ is exact, so we see $N\otimes_R M=0\Rightarrow M=0$. This completes the characterization:

$N$ is reflecting iff $N\otimes_R -$ is faithful iff $N\otimes_R M=0 \Rightarrow M=0$

(EDIT: Thanks to Martin for pointing out a flaw in the previous version. Here is an argument, now superfluous and a bit fishy, that $N$ reflecting implies (2): suppose $N\otimes_R \phi$ is zero. Then, because $N\otimes_R ker(\phi) = ker(N\otimes_R \phi)$, we see $0\rightarrow N\otimes_R M' \rightarrow N\otimes_R ker(\phi)\rightarrow 0$ is exact, so $0\rightarrow M'\rightarrow ker(\phi)\rightarrow 0$ is exact, i.e. $\phi=0$. This is fishy because it's not clear to me that $N\otimes ker(-) = ker(N\otimes -)$, though it does seem to work for $im(-)$.)
It may also be equivalent to say that for all maximal ideals $I\subset R$, if $N\otimes_R R/I=0$ then $R/I=0$, but I haven't checked this.
EDIT (in response to requests from Will Sawin and Zhen Lin for examples): I'll end by remarking on a confusing bit of terminology and giving some examples. There is a notion of what it means for a module $N$ to be faithful, but it is not equivalent to the above. One says $N$ is faithful if $Ann_R(N)=0$ (this appears in Lam). Note that this website erroneously calls a module faithful if it satisfies (1) above, but the link is still useful as it gives some properties of such modules and examples of modules which satisfy (1) but are not flat. An additional example is $N=\mathbb{Z}\oplus \mathbb{Z}/2$ over $R=\mathbb{Z}$. This seems to show that if $M$ is faithfully flat and $L$ is any module, then $N = M\oplus L$ will be reflecting but not necessarily flat. It's easy to see that $N$ is faithful iff for all nonzero $a\in R$, $aN \neq 0$. So (1) implies $N$ is faithful, because we can take $M=(a)$ for each $a$ and use the fact that $N\otimes_R (a) = aN$.
Note that $N$ faithful does not imply that $N\otimes -$ is a faithful functor. For example, $\mathbb{Q}$ is faithful as a $\mathbb{Z}$-module (since it's torsion free), but $\mathbb{Q}$ clearly fails (1) because $\mathbb{Q}\otimes \mathbb{Z}/2=0$. Faithfully flat implies faithful and flat, but not conversely (again because of $\mathbb{Q}$). Indeed, a module can even be projective and faithful, but fail to be faithfully flat. Consider $P=\mathbb{Z}\oplus \mathbb{Z}\oplus \dots$ as a module over $R = \mathbb{Z}\times \mathbb{Z}\times \dots$. It's faithful because it's a subring, it's projective (e.g. by the dual basis lemma), but it's not faithfully flat because for any maximal ideal $m$ of $R$ containing $P$, $Pm=P$.<|endoftext|>
TITLE: notation for formal Laurent series
QUESTION [7 upvotes]: I've found a few articles that write the ring of formal Laurent series in $t$ as $R((1/t))$, but what's the underlying meaning of $\cdot ((\cdot))$?
A mathematician of my acquaintance swears that $R((t))$, not $R((1/t))$, should be used to denote the ring of formal Laurent series in $t$.  We can't decide who's right without knowing what $\cdot((\cdot))$ means.  (We both agree that $R[[t]]$ denotes the ring of formal power series in $t$ with coefficients in $R$.)

REPLY [7 votes]: I've seen $k((x,y))$ used to mean two different things: the field of fractions of $k[[x,y]]$ and also
$k((x,y)) = \{\sum_{i \geq n} \sum_{j \geq m} f_{i,j}x^iy^j : f_{i,j} \in k; n,m \in \mathbb{Z}\}$
These rings are distinct, e.g. the second one does not contain $\sum_{i \geq 0}y^{-(i+1)}x^i = (y-x)^{-1}$.
In one variable the distinction disappears, and then the only difference between $R((t))$ and $R((1/t))$ is whether the powers of $t$ go to $\infty$ or $-\infty$.<|endoftext|>
TITLE: When is an acylic chain complex contractible
QUESTION [17 upvotes]: When is an acyclic chain complex contractible?  
I know an acyclic chain complex of free modules over a PID (or field) are always contractible, but what about over a more complicated ring, like a graded algebra over Z/p (for instance the mod p Steenrod Algebra)?  
EDIT:  I want to assume the chain complex is bounded below, but not necessarily that the ring is commutative.  I suspect that it isn't always true for a non-commutative ring, but I don't really have a counter example.  
Thanks everybody!

REPLY [19 votes]: There is a useful characterization in Brown: Cohomology of Groups, Prop. 0.3: 

A chain complex $C$ over any  ring is contractible iff it is acyclic and each short exact sequence $0 \to \ker(d_n) \to C_n \to \operatorname{im}(d_n) \to 0$ splits. 

This immediately explains the OP's PID example: If $C$ is a complex of free modules over a PID, then $\operatorname{im}(d_n) \le C_{n-1}$ is also free and thus the sequence splits. 
This observation can be axiomized as follows: A ring (with unit) is called hereditary, if each submodule of a projective module is again projective. As a corollary: 

Each acyclic chain complex of projective modules over a hereditary ring is contractible. 

An example of an non-commutative hereditary ring is given by the upper-triangular matrices over a field. 
BTW: Tom's remark also follows easily from the criterion: Let all $C_n$ be free and $C_n=0$ for $n < 0$. Since $C_0$ is free, the short exact sequence $0 \to \ker(d_1) \to C_1 \to C_0 \to 0$ splits. Hence $\ker(d_1)=\operatorname{im}(d_2)$ is a direct summand of a free module and therefore projective. By induction then all of the short exact sequences split.<|endoftext|>
TITLE: Are Conway's omnific integers the Grothendieck group of the ordinals under commutative addition?
QUESTION [30 upvotes]: This is a question in two parts.
Say that $\mathbf{On}$ is the proper class of all ordinal numbers in ZFC. We can define a binary operator over $\mathbf{On}$ which corresponds to the commutative version of ordinal addition; this has been called "Hessenberg addition" and "natural addition" before. It's also the operation you get by restriction of the $+$ operation from Conway's surreals to the subchain of ordinals (e.g. surreals with empty right set). I'll use the $+$ symbol for this operation over the ordinals.
$\langle\mathbf{On},+\rangle$ is a commutative monoid, which hence admits the notion of constructing a Grothendieck group $\mathrm{K}(\mathbf{On})$. The group $\langle\mathrm{K}(\mathbf{On}),+\rangle$ hence adds expressions such as $\omega$, $\omega-1$, $\omega^\omega - \omega^2 + 5$, etc. to the ordinals.
Question 1: is $\mathrm{K}(\mathbf{On})$ equivalent to Conway's "omnific integers" $\mathbf{Oz}$? In Conway's "On Numbers and Games," he defines an omnific integer $x$ as one which can be represented as a surreal number $\left \{ x-1 \mid x+1 \right \}$. Are these two classes isomorphic to one another?
It's also noteworthy that the field of fractions $Quot(\mathbf{Oz})$ is the full field $\mathbf{No}$ of surreal numbers. We can further turn $\mathrm{K}(\mathbf{On})$ into a ring $\langle\mathrm{R}(\mathbf{On}),+,\times\rangle$ by defining a new commutative operation called $\times$, called the "Hessenberg product", "Hausdorff product" or "natural product" of ordinals, which is commutative, associative, has an identity of 1, and distributes over the Conway normal form of the ordinal. A good definition for the Hessenberg product can be found on pages 24-25 of Ehrlich 2006.
Question 2: even if $\mathrm{K}(\mathbf{On})$ isn't isomorphic to $\mathbf{Oz}$, is $Quot(\mathrm{R}(\mathbf{On}))$ isomorphic to $\mathbf{No}$?
I'm tempted to answer in the negative for #1, as $\sqrt{\omega}$ is in $\mathbf{Oz}$, but is it in $\mathrm{R}(\mathbf{On})$? That is, given $\mathrm{K}(\mathbf{On})$ and ordinary commutative multiplication, is it the case that $\omega$ becomes a perfect square?
(Also, a last note - I'm aware that $\mathbf{On}$ is a proper class. I'm not sure what foundational issues arise specifically in the above question, but I don't care how you want to handle them - NBG set theory, Grothendieck universes, whatever.)

REPLY [18 votes]: There is an obvious extension of Cantor normal form to the Grothendieck group of the ordinals. Then the standard argument that $\sqrt{x}$ does not lie in the ring $\mathbb Z[x]$ applies to $\sqrt{\omega}$. Specifically, $\sqrt{\omega}$ must have a Cantor normal form $a + b \omega + $ higher-order terms, which squares to $a^2 + 2ab \omega +$ higher-order terms. For this to equal $\omega$ we need $a^2=0$ but $2ab=1$, which is of course impossible.
So your first question has a negative answer.
For your second, the equation $a^2=b^2\omega$ is equally problematic. Again apply the standard argument:
Write $a=k \omega^x +$ higher-order terms, and $b=l \omega^y + $ higher-order terms. Then the lowest term of $a^2$ and $b^2\omega$ must be equal, so $\omega^{x+_H x}=\omega^{y+_H y+_H 1}$, so $x+_H x = y+_H y+_H 1$, which cannot be because the $1$s coefficient of the first expression must be even while the $1s$ coefficient of the second expression must be odd.<|endoftext|>
TITLE: What is the "correct" definition of creation of limits?
QUESTION [19 upvotes]: There are several definitions of what it means for a functor $F$ to create limits of a certain type.
There is the definition in MacLane's CWM:

Definition 1: A functor $F:\mathcal{C}\to \mathcal{D}$ creates limits of type $J$ if for every diagram $D\in \mathcal{C}^J$ such that $FD$ has a limit, there exists a unique cone $k$ for $D$ such that its image through $F$ is the limit of $FD$, and moreover $k$ is the limit of $D$.

There is the definition in the nLab:

Definition 2: A functor $F:\mathcal{C}\to \mathcal{D}$ creates limits of type $J$ if for every diagram $D\in \mathcal{C}^J$ such that $FD$ has a limit, we have that $D$ has a limit, and moreover $F$ preserves and reflects limits.

There is also the definition in the category theory notes for a course taught by Eugenia Cheng:

Definition 3: A functor $F:\mathcal{C}\to \mathcal{D}$ creates limits of type $J$ if for every diagram $D\in \mathcal{C}^J$ such that $FD$ has a limit, there exists a cone for $D$ such that its image through $F$ is the limit of $FD$, and moreover $F$ reflects limits.

What are the advantages or disadvantages of one over the other? Surely there was something that motivated the modern definitions (2 and 3) to change the older definition (1) 
I'm aware of the remark on the nLab page on creation of limits: that probably answers why the need for a new definition. However, a functor that satisfies definition 2 also satisfies definition 3, but I don't see why a functor that satisfies definition 3 should preserve limits. So the question remains: why prefer definition 2 over definition 3 or viceversa?

REPLY [5 votes]: Definition. Let $F:\mathcal{C}\to\mathcal{D}$ be a functor, with $S:\mathcal{I}\to\mathcal{C}$ a diagram of shape $\mathcal{I}$ in $\mathcal{C}$. We say that $F$ creates the limit of $S$ iff $F\circ S$ having a limit $$\pi':\Delta L'\Rightarrow F\circ S$$ in $\mathcal{D}$ implies that there exists a unique (up to iso) source $$\hat L=\{\pi_I:L\to S(I)\}_{I\in{\bf Ob}_\mathcal{I}}$$ in $\mathcal{C}$ such that $$F(\hat L)\cong\pi'$$ as cones, and further that $\pi:\Delta L\Rightarrow S$ is a limit of $S$ in $\mathcal{C}$. We say that $F$ creates limits of shape $\mathcal{I}$ iff $F$ creates the limit of all functors $S:\mathcal{I}\to\mathcal{C}$, and that $F$ creates limits iff $F$ creates limits of shape $\mathcal{I}$ for all 'small' categories $\mathcal{I}$.

This definition is very similar to the one found in The Joy of Cats by Adamek, Herrlich, and Strecker (p. 227 definition 13.17), but they required uniqueness on the nose for $\hat L$ and that $F(\hat L)=\pi'$.
The discussion and lemmas in their book after their definition establish that it implies all other kinds of 'things you want a functor to do to limits' (c.f. the diagram in remark 13.38, p. 232, reproduced below), with the exception of preservation.

The definition above also implies all the good behavior one could ask for (except preservation), and is 'respected by equivalence' better than the one in JoC since it makes no reference to uniqueness on the nose or equality -- equivalences create limits in both senses of the definition though.
EDIT: Unless I'm mistaken, their definition is satisfied by the forgetful functors they list but not equivalences, since $\hat L$ is only unique up to iso and we only have $F(\hat L)\cong\pi'$ for the obvious source $\hat L$ corresponding to $\pi'$ by essential surjectivity followed by fullness. Accordingly it seems this definition is better than the one in JoC, as it is satisfied by equivalences which should certainly 'create limits' however we define the term.
I would argue that the lattice of nice implications above (that includes properties explicitly listed in the definitions you reference) make both versions of this definition more satisfying than the other options available. This definition doesn't imply preservation of limits, though, so if preservation of limits matches our intuition for what 'creating a limit' should mean we ought add that $F$ preserves limits to the above definition.
As a note, the subtle difference between the Adamek, Herrlich, and Strecker definition and most other definitions encountered is that we don't a-priori impose the cone coherence conditions on the source in the domain when we assert that it is unique; it must be unique as a source, not as a cone. We then 'create' the cone coherence conditions using $F$; this matches nicely with what 'creating' a limit should mean, in my opinion.

What follows is a proof that equivalences satisfy the above definition of 'creating limits'.

Proof Let $F:\mathcal{C}\simeq\mathcal{D}$ be an equivalence with $S:\mathcal{I}\to\mathcal{C}$ be a diagram of shape $\mathcal{I}$ in $\mathcal{C}$, and suppose that $F\circ S$ has a limit $\pi':\Delta L'\Rightarrow F\circ S$. Since $F$ is essentially surjective there exists some object $L\in\mathcal{C}$ and an isomorphism $u:F(L)\cong L'$, so $\pi'\circ u:\Delta F(L)\Rightarrow F\circ S$ is also trivially a limit of $F\circ S$. Further, since $F$ is full we obtain a source $$\hat L=\{\pi_I:L\to S(I)\}_{I\in\mathcal{I}}$$ in $\mathcal{C}$ with $F(\pi_I)=\pi'_I\circ u$ for all objects $I\in\mathcal{I}$, thus $F(\hat L)\cong\pi'$ as cones since $u$ was an iso and a morphism of cones by the preceding equations for all $I$. If any other source $\hat L''=\{\pi'':L''\to S(I)\}_{I\in\mathcal{I}}$ satisfies $F(\hat L'')\cong\pi'\cong F(\hat L)$ then there exists a unique isomorphism of cones $$w':F(L)\to F(L''),$$ and since $F$ is full this arrow is the image of an arrow $w:L\to L''$ which is unique satisfying $F(w:L\to L'')=w':F(L)\to F(L'')$ by faithfulness of $F$. We then have that $$F(\pi''_I\circ w)=F(\pi''_I)\circ F(w)=F(\pi''_I)\circ w'=F(\pi_I)\implies\pi''_I\circ w=\pi_I,$$ since $F$ is faithful, and $w$ is an iso since $w'$ is with $w^{-1}:L''\to L$ the unique arrow such that $F(w^{-1}:L''\to L)=w'^{-1}:F(L'')\to F(L)$, so $\hat L''\cong\hat L$ and since $\hat L''$ was an arbitrary source $\hat L$ is the unique source up to isomorphism satisfying $F(\hat L)\cong\pi'$. We further have that $\pi:\Delta L\Rightarrow S$ is a limit of $S$; it is a cone since

commutes for all arrows $f:I\to I'\in\mathcal{I}$ since $\pi'$ is a cone to $F\circ S$, thus

commutes for all arrows $f:I\to I'\in\mathcal{I}$ since $F$ is faithful. This cone is further terminal, since any other cone $'\pi:\Delta'L\Rightarrow S$ gives rise to a cone $F('\pi):\Delta F('L)\Rightarrow F\circ S$ which induces a unique morphism of cones $'u:F('L)\to F(L)=L'$ with $F(\pi_I)\circ{'u}=\pi'_I\circ {'u}=F({'\pi_I})$, and since $F$ is full there exists an arrow $v:{'L}\to L\in\mathcal{C}$ which is unique satisfying $F(v:{'L}\to L)={'u}:F({'L})\to F(L)$ since $F$ is faithful, and $v$ is also a morphism of cones since $$F(\pi_I\circ v)=F(\pi_I)\circ F(v)=\pi'_I\circ{'u}=F({'\pi_I})\implies\pi_I\circ v={'\pi_I}$$ for all objects $I\in\mathcal{I}$, again by faithfulness of $F$.

It is not immediately apparent how to modify the above proof to get that equivalences satisfy the JoC definition of creating limits. We might be able to get around this (non-canonically in a universe with choice) by using the fact that two categories are equivalent iff they have isomorphic skeletons, then choosing a skeleton of $\mathcal{D}$ that already contains the limit $\pi':\Delta L'\Rightarrow F\circ S$ and a skeleton of $\mathcal{C}$ isomorphic to this skeleton -- we should then have that the JoC definition is satisfied as well, unless I'm mistaken.<|endoftext|>
TITLE: Alternating sums of alternate Stirling numbers 
QUESTION [5 upvotes]: Does anybody know of any identities or combinatorial interpretations for alternating sums of alternate Stirling numbers?
I am particularly interested in expressions of the form:
$$\pm\sum_{k}(-1)^k|s(n,2k)|=\mp|s(n,2)|\pm|s(n,4)|\mp|s(n,6)|\pm\ldots $$
or:
$$\pm\sum_{k}(-1)^k|s(n,2k+1)|=\pm|s(n,1)|\mp|s(n,3)|\pm|s(n,5)|\mp\ldots, $$
where $|s(n,k)|$ is an unsigned Stirling number of the first kind.  
However, I would be happy with any related identities or information (including an argument as to why sums like this might not have nice formulae or count anything interesting).  As far as I can tell these do not appear on the OEIS, or in any of the literature.

REPLY [13 votes]: The generating function here is
$$\sum_{n\geq 0}s(n,k) x^n y^k=\sum_{n\geq 0} \frac{x^n}{n!}y(y-1)\cdots (y-n+1)=e^{y\log(1+x)}.$$
If we put $y=i$ the coefficient of $x^n$ becomes $A_n+iB_n$ where $A_n$ and $B_n$ are your sequences. It is pretty clear from here that the exponential generating function for $A_n$ is 
$$\mathfrak {Re}\left(e^{i\log(1+x)}\right)=\cos(\log(1+x)),$$ 
and for $B_n$ it is 
$$\mathfrak{Im}\left(e^{i\log(1+x)}\right)=\sin(\log(1+x)).$$ Also these are A009454 and A003703.<|endoftext|>
TITLE: Actions on Sⁿ with quotient Sⁿ
QUESTION [16 upvotes]: What is known about isometric actions on $\mathbb S^n$ such that the quotient space is homeomorphic to $\mathbb S^n$?

Comments.

I am mostly interested in (maybe trivial) properties of such actions for large $n$.
I see that the orientation preserving part of Coxeter's group has this property. 
Originally I thought that any such action is generated by rotations around $\mathbb S^{n-2}$'s; now I see that there are other examples for $\mathbb S^3$; thanks to Lee Mosher.

REPLY [12 votes]: I think I have completely answered the question in the following form:
Theorem. For a finite subgroup $\Gamma < O(n)$ the quotient space $S^{n-1}/\Gamma$ is homeomorphic to $S^{n-1}$ if and only if $\Gamma$ has the form
\begin{eqnarray*}
  \Gamma = \Gamma_{ps} \times P_1 \times \ldots \times P_k
\end{eqnarray*}
for a pseudoreflection group $\Gamma_{ps}$ and Poincaré groups $P_i<SO(4)$, $i=1,\ldots,k$, such that the factors act in pairwise orthogonal spaces and such that $n>5$ if $k=1$. 
cf. http://arxiv.org/abs/1307.4875.
Here, a pseudoreflection group is understood in the sense of Mikhailova and a Poincaré $P$ group comes from the binary icosahedral group in $SU(2)$, i.e. $S^3/P$ is Poincaré's homology sphere.<|endoftext|>
TITLE: Prescribed values for the uniform density
QUESTION [7 upvotes]: Strauch & Tóth [1] Georges Grekos [3][4] showed that for any choice of upper and lower density, there is some subset of $\mathbb{N}$ with the chosen densities, provided the lower is no more than the upper and both are in [0, 1].
Mišík [2] extended this to show that for any choice of upper density, lower density, upper logarithmic density, and lower logarithmic density there is a subset of $\mathbb{N}$ with the chosen densities, provided they follow the necessary inequalities $0\leqslant\underline d\leqslant\underline\delta\leqslant\overline\delta\leqslant\overline d\leqslant 1.$
Is there a similar result with the uniform densities?
$$
\underline{u}(A)=\lim_s\frac1s\liminf_n\sum_\stackrel{a\in A}{\ n < a\leqslant n+s}1
$$
$$
\overline{u}(A)=\lim_s\frac1s\limsup_n\sum_\stackrel{a\in A}{\ n < a\leqslant n+s}1
$$
An ideal result would combine all three density types with the inequality
$$
0\leqslant\underline{u}\leqslant\underline{d}\leqslant\underline{\delta}\leqslant\overline{\delta}\leqslant\overline{d}\leqslant\overline{u}\leqslant1
$$
but I'm looking for any published results on the topic.
[1] Strauch and Tóth, Asymptotic density of $A\subset N$ and density of the ratio set $R(A)$. Acta Arith. 87 (1998), pp. 67-78. eudml
[2] Ladislav Mišík, Sets of positive integers with prescribed values of densities, Mathematica Slovaca 52:3 (2002), pp. 289-296.dml.cz
[3] Georges Grekos, Sur la répartition des densités des sous-suites d'une suite donnée, PhD thesis, Université Pierre et Marie Curie (1976).
[4] Georges Grekos, Répartition des densites des sous-suites d'une suite d'entiers, J. Number Theory 10:2 (1978), pp. 177-191 (in French). doi:10.1016/0022-314X(78)90034-3

REPLY [5 votes]: OK. I think you can do this pretty easily by hand. 
First you need a way to generate sequences with very uniform density. The Sturmian sequences are perfect for this. A Sturmian sequence with parameter $\alpha$ has the property that sub-blocks of length $N$ have density converging to $\alpha$ uniformly in $N$.
Now: start by interspersing a Sturmian with parameter $\underline\delta$ with one of parameter $\bar\delta$. How to do the interspersing? Have one from $2^{n!}$ to $2^{(n+1)!}$. Then switch to the other between $2^{(n+1)!}$ and $2^{(n+2)!}$ etc. This switching is slow enough to guarantee that the sequence that you obtain has the prescribed $\underline\delta$ and $\bar\delta$. These are also the upper and lower densities for the time being.
Next we'll modify the sequence to obtain densities $\underline d$ and $\bar d$.
Alternately splice in segments of Sturmian parameter $\underline d$ and $\bar d$ between $2^{4^i}$ and $2^{4^i+2^i}$. This won't affect the upper and lower logarithmic densities (because $2^i/4^i\to 0$). 
But looking at these segments, you obtain a sequence with upper and lower densities $\underline d$ and $\bar d$ (the upper and lower uniform densities are also $\underline d$ and $\bar d$). 
Finally, we'll perturb things as in my comment to get the uniform densities we want.
For the segments between $2^n$ and $2^n+n$, insert alternately segments of the Sturmian sequences with densities
$\underline u$ and $\bar u$. These segments are so sparse, they will have no effect on the upper and lower densities or the logarithmic densities. They are enough to guarantee that you get the uniform densities you want.
And Robert's your mother's brother.<|endoftext|>
TITLE: Distance metric between two sample distributions (histograms)
QUESTION [10 upvotes]: Context: I want to compare the sample probability distributions (PDFs) of two datasets (generated from a dynamical system). These datasets depend on a set of parameters, and I want a concise way to evaluate the distance between the two PDFs over several different parameter regimes, ideally by a single number. For a fixed parameter regiume, my two sample PDFs are given by the vectors $x$ and $y$, where $x_i$ is the relative frequency of samples which lie in the $i$th bin.
One method I've seen is the Kolmogorov-Smirnov statistic, which is the maximum vertical distance between the cumulative distribution functions of the two datasets. This would work for my purposes, but I'm starting to think that the chi-squared distance will be better (at the very least I had heard of it). It is given by: $d(x,y) = \frac{1}{2}\sum_i \frac{(x_i-y_i)^2}{x_i+y_i}$. But this doesn't seem to make sense in the case where $x_i=y_i=0$, which seems like a fairly reasonable occurrence.
Any recommendations?
Edit: My first instinct was just to take the $L^2$ ($l^2$) norm of the difference of the two PDFs (sample PDFs). But then I thought the $L^1$ norm might be more suitable for a probability distribution. After looking a little further, I stumbled on the K-S and Chi-Squared distances.

REPLY [3 votes]: For particle distributions, you can, e.g., use the Wasserstein distance or OSPA (its counterpart for densities with different numbers of particles), see [1]. These two are based on the linear assignment problem (one-to-one assignment of the particles, if the numbers are equal) and therefore do not have a continuous gradient. If you don't want a a hard assignment of the particles, you can use the LCD distance from [2].
[1] J. R. Hoffman and R. P. S. Mahler. Multitarget Miss Distance and its Applications. In Proceedings of the Fifth International Conference on Information Fusion (Fusion 2002), July 2002.
[2] U. D. Hanebeck. Optimal Reduction of Multivariate Dirac Mixture Densities. at -Automatisierungstechnik, 63(4), 2015.<|endoftext|>
TITLE: $\Theta$ and the Hartogs of $2^\mathbb R$
QUESTION [7 upvotes]: Let $a,b$ be sets, we write $a\leq^\ast b$ if either $a=\varnothing$ or there exists a surjection $f\colon b\to a$. With the axiom of choice this is a linear ordering equivalent to the usual ordering of cardinals $a\leq b$ if there is an injection from $a$ into $b$).
Without the axiom of choice this order need not be linear, or even a partial order. Indeed there are models where $a < b$ (so trivially $a\leq^\ast b$) but also $b\leq^\ast a$.
Let $\Theta(a)$ denote the least ordinal $\alpha$ such that $\alpha\nleq^\ast a$. This is a dual notion to $\aleph(a)$ which is the least ordinal $\alpha$ such that $\alpha\nleq a$. It is not very hard to see that both these ordinals are in fact cardinals, and that they exist for every set in $\mathrm{ZF}$. In the case where $a=\mathbb R$ we simply write $\Theta$ for $\Theta(\mathbb R)$.
It is consistent that for some set $a$ it holds $\aleph(a)<\Theta(a)$. For example if $V$ is a Solovay model then $\aleph(\mathbb R)=\aleph_1\leq^\ast\mathbb R$.
We also know that if $a\leq^\ast b$ then $a\leq 2^b$, where the injection is simply the sending a point in $a$ to its fiber under a fixed surjection from $b$. It follows, if so, that $\Theta(a)\leq\aleph(2^a)$. Even in $\mathrm{ZFC}$ one can see that it is consistent that there is an equality, and it is consistent that there is no equality (e.g. $a=\omega$ and take a model in which $\frak c=\aleph_1$ and another where it is is $\aleph_2$).

Question: Suppose that $V$ is a model of $\mathrm{ZF+AD}$, we know that if $\alpha\leq^\ast\mathbb R$ then $2^\alpha\leq^\ast\mathbb R$, and therefore $\alpha^+\leq^\ast\mathbb R$. Is it consistent that $\Theta<\aleph(2^\mathbb R)$?
Equivalently, suppose $\alpha<2^\mathbb R$, can we find a surjection from $\mathbb R$ onto $\alpha$?

REPLY [4 votes]: Yes, it is consistent with AD.  In $L(\mathbb{R})$, if AD holds then $\Theta$ injects into $\mathcal{P}(\mathbb{R})$, or equivalently, into $2^{\mathbb{R}}$. The OD sets of reals are Wadge-cofinal and $\Theta$ is regular, so we can define an injection $F : \Theta \to \mathcal{P}(\mathbb{R})$ by induction.  Let $F(\alpha)$ be the $<_{OD}$-least OD set of reals with Wadge rank greater than the Wadge ranks of all the sets of reals $F(\gamma)$, $\gamma < \alpha$.
Going further, we can determine the Hartogs number $\aleph(2^{\mathbb{R}})$ in all models of $AD + V=L(\mathcal{P}(\mathbb{R}))$.  The proof splits into two cases corresponding to the length of the Solovay sequence being a successor ordinal or a limit ordinal.
If the length of the Solovay sequence is a successor ordinal, this means that there is a set of reals $A$ such that every set of reals is OD from $A$ and a real. In this case $\aleph(2^{\mathbb{R}})$ has the largest possible value, namely $\Theta(2^{\mathbb{R}})$.
Given a surjection $F : \mathcal{P}(\mathbb{R}) \to Z$ for any set $Z$ we can construct an injection $Z \to \mathcal{P}(\mathbb{R})$.  Let $G(z)$ be the set of pairs $(x,y)$ of reals such that $y$ is in the least $OD_{A,x}$ set of reals $B$ with $F(B) = z$ (if it exists.)
On the other hand, if the length of the Solovay sequence is a limit ordinal then $\aleph(2^{\mathbb{R}})$ is equal to $\Theta$; that is, to $\Theta(\mathbb{R})$.
We have $\aleph(2^{\mathbb{R}}) \ge \Theta$ on general grounds like you said in the question.  But any function $F : \Theta \to \mathcal{P}(\mathbb{R})$ is OD from a set $A$ of reals because $V=L(\mathcal{P}(\mathbb{R}))$, so the range of $F$ consists of $OD_A$ sets of reals.  The Solovay sequence has limit length, so the range of $F$ cannot be all of $\mathcal{P}(\mathbb{R})$.<|endoftext|>
TITLE: Does the moduli space of smooth curves of genus g contain an elliptic curve
QUESTION [27 upvotes]: Let $M_g$ be the moduli space of smooth projective geometrically connected curves over a field $k$ with $g\geq 2$. Note that $M_g$ is not complete.
Does $M_g$ contain an elliptic curve?
The answer is no if $g=2$. In fact, $M_2$ doesn't contain any complete curves.
Note that one can construct complete curves lying in $M_g$ for $g\geq 3$. There are explicit constructions known. 
Probably for $g>>0$ and $k$ algebraically closed, the answer is yes.
What if  $k$ is a number field?

REPLY [14 votes]: Here is how the function field version of Shafarevich's conjecture (=Arakelov-Parshin Theorem) implies that there are no elliptic curves or (at most) twice punctured rational curves in $M_g$:
(See Noam's comment to Felipe's answer)
Suppose there exists a smooth non-isotrivial family $f:X\to C$ of curves of genus $g$ for some fixed $g>1$ parametrized by a curve $C$. Call such a family admissible, let $m\in\mathbb N$ fixed and consider the set of numbers
$$
D_m=\left\{ 
\deg (f_*\omega_{X/C}^m) \mid f \text{ is an admissible family }
\right\}
$$
By Shafarevich's conjecture (=Arakelov-Parshin Theorem) this set is finite and hence bounded for any given $m$. On the other hand, it is well-known that for $m\gg 0$ the line bundle $\det (f_*\omega_{X/C}^m)$ is ample, but we only need that it is not trivial and hence has a non-zero degree.
Now assume that $C$ admits an endomorphism of degree $>1$, say $\sigma:C\to C$. Then the base change $f_\sigma:X_\sigma\to C$ of any admissible family $f:X\to C$ is still admissible, but 
$$
\deg ({f_\sigma}_*\omega_{X_\sigma/C}^m) = \deg\sigma \cdot \deg (f_*\omega_{X/C}^m),
$$
which would mean that if non-empty, then $D_m$ could not be bounded, therefore if $C$ admits such an endomorphism, then $D_m$ has to be empty.    
If $C$ is an elliptic curve, or a rational curve minus (at most) two points, then it admits such an endomorphism, so they cannot parametrize smooth non-isotrivial families of curves of genus $>1$.

Remark
The boundedness of (the analogous set in arbitrary dimension) $D_m$ is sometimes called weak boundedness. The above argument shows that 
"Weak Boundedness" implies "Hyperbolicity".  This statement, in a somewhat more general form, is contained in Thm 0.8/0.9 of Logarithmic vanishing theorems and Arakelov-Parshin boundedness for singular varieties. Compositio Math. 131 (2002), no. 3, 291–317<|endoftext|>
TITLE: What is the origin of the term magma?
QUESTION [10 upvotes]: Wikipedia credits Bourbaki with coining it, but doesn't provide a source. Does anyone happen to know the motivation for using this term?

REPLY [4 votes]: Wikipedia also says that "magma" is used by Serre in his book Lie algebras and lie groups: 1964 Lectures given at Harvard University.  This seems to be the case (at least for the 1992 Springer reprint that I have access to.)  Is this the earliest use in print of "magma" in this sense?<|endoftext|>
TITLE: Irrationality proof technique: no factorial in the denominator
QUESTION [19 upvotes]: Jonathan Sondow elegantly proves the irrationality of e in his aptly titled A Geometric Proof that e Is Irrational and a New Measure of Its Irrationality (The American Mathematical Monthly, Vol. 113, No. 7 (Aug. - Sep., 2006), pp. 637, http://www.jstor.org/stable/27642006).
In his argument, he constructs a sequence of nested intervals $I_n$ for every $n \geq 1$, each of the form $[k/n!, (k+1)/n!]$, such that $\bigcap I_n = \{e\}$, with $e$ lying strictly between the endpoints of each $I_n.$ From this, we conclude that $e$ cannot be written as a fraction with denominator $n!$ for any $n \geq 1.$
Fact: Every rational number $p/q$ can be written as a fraction with a factorial in its denominator: $p/q = p(q-1)!/q!$. 
Thus, we conclude that $e$ is irrational.
The reason this proof technique works so well with $e$ is, of course, related to the Maclaurin series for the exponential function, $e^x.$
That any rational number can be written in lowest terms is employed in other irrationality proofs (e.g., the classic proof for that of $\sqrt{2}$) but I had not seen the above fact drawn upon before reading this particular paper.
My question is: are there other examples of real numbers (which are not related to $e$ in some trivial way) whose irrationality can be proved using the Fact above?

REPLY [5 votes]: For any interval $I_n$ of the form you define, there are $n+1$ intervals that could be $I_{n+1}$ nested in it. Choose one of those intervals, then repeat the process. As long as you choose an interval other than the first infinitely many times and an interval other than the last infinitely many times, you get an irrational number by this proof.
Or in other words, we have:
$$\sum_{m=n+1}^\infty \frac{m-1}{m!} = \sum_{m=n+1}^\infty \left(\frac{1}{(m-1)!} - \frac{1}{m!}\right) = \frac{1}{n!}$$
so in user22202's answer, for any sequence of integers $a_m$ such that $0 \leq a_m \leq m-1$, and $a_m>0$ infinitely often and $a_m < m-1$ infinitely often, we have
$$ \sum_{m=0}^\infty \frac{a_m} {m!}$$
is irrational.<|endoftext|>
TITLE: What is the simplest oscillatory integral for which sharp bounds are unknown?
QUESTION [8 upvotes]: I have either heard or read that sharp asymptotics and bounds for oscillatory integrals of the form
$ \int e^{i \lambda \Phi(x)} \psi(x) dx \quad \lambda \to \infty $
are unknown when the critical points for the phase function are not isolated.  If this impression is correct, what are the simplest / most important integrals of this form for which the optimal decay rate and asymptotic have not been proven?  E.g. are there examples with $\Phi$ being a polynomial?  (I would also appreciate recommendations of references for estimating multidimensional oscillatory integrals if anyone has them.)

REPLY [7 votes]: As soon as the Hessian is not full rank, the problem becomes quickly messy:

if the hessian has rank $n-1$, then one can treat the one direction separately since we have explicit bound for a one-dimensional integral where the taylor expansion of $\Phi$ near a critical point $x_0$ looks like $(x-x_0)^p$ for any $p\ge 2$, the other directions will always give you $\lambda^{-\frac{1}{2}}$.
when the rank is less, then one must first identify those directions where the phase isn't  quadratic, and look at the next terms in the expansion. V.I. Arnold then classifies the simple jets of functions in terms of their corresponding maximal decay in $\lambda$ in the following paper: 

V.I. Arnold, Remarks on the stationary phase method and coxeter numbers, Russian Math.
Surveys, 28 (1973), p. 19
See also J.J. Duistermaat, Oscillatory integrals, Lagrange immersions and unfolding of singularities, CPAM vol XXVII, 207-281 (1974)
The classification is algebraic and does not rely on estimating integrals, so it never tells you how to obtain the estimate corresponding to that optimal decay. For the simpler classes of degenerate critical points, Popov has worked on estimating the oscillatory integrals:
D.A. Popov, Estimates with constants for some classes of oscillatory integrals, Russian Math. Surveys, 52, pp. 73–145.
D.A. Popov, Remarks on uniform combined estimates oscillatory integrals with simple singularities, Izv. Math., 72, pp. 793–816.
These papers helped me for the following problem, where the oscillatory integral I had to study had an interesting degenerate behavior.
F. Monard-G. Bal "Inverse transport with isotropic time-harmonic sources", SIAM J. Math. Anal., Vol. 44, No. 1, pp. 134-161 (2012).
Along the way, another paper I found interesting for direct estimates: 
G.I. Arkhipov, A.A. Karatsuba, and V.N. Chubarikov, Trigonometric integrals, Izv. Akad.
Nauk SSSR Ser. Mat., 43 (1979), pp. 971–1003 (in Russian); Math. USSR-Izv., 15 (1980),
pp. 211–239 (in English).
(I think they also have a multidimensional counterpart).<|endoftext|>
TITLE: What is the functor tensor product?
QUESTION [7 upvotes]: I'm familiar with the tensor product of modules, but I've also come across functor tensor product (in emily riehls paper on homotopy limits), what are they, and how are they (if they are) related to traditional tensor products? (Emily shows that they can be defined as a particular coend, but that doesn't really provide any intuition for me).

REPLY [3 votes]: Familiar examples of modules can be expressed in terms of categories. For example:

If $\mathcal{G}$ is a one-object groupoid, then functors $\mathcal{G}^\circ \to \mathbf{Set}$ are the same thing as right actions of the group $\hom_{\mathcal{G}}(*,*)$ on sets.
If $\mathcal{R}$ is a one-object additive category, then additive functors $\mathcal{R} \to \mathbf{Ab}$ are the same thing as left modules over the ring $\hom_{\mathcal{R}}(*,*)$

Generalizing, we can view any functor $\mathcal{C} \to \mathcal{D}$ as a left action of $\mathcal{C}$ (on objects of $\mathcal{D}$). Similarly, functors $\mathcal{C}^\circ \to \mathcal{D}$ are right actions. The most common example is actions on sets; i.e. $\mathcal{D} = \mathbf{Set}$. 
Let there be a presheaf $P : \mathcal{C}^\circ \to \mathbf{Set}$, a functor $F : \mathcal{C} \to \mathbf{Set}$, and a set $S$.
As with other sorts of modules, there is a "hom-tensor" adjunction: there is a functor $\otimes_\mathcal{C} : \mathbf{Set}^{\mathcal{C}^\circ} \times \mathbf{Set}^{\mathcal{C}} \to \mathbf{Set} $ such that there is a natural bijection
$$\hom_\mathbf{Set}(P \otimes_\mathcal{C} F, S) \cong \hom_{\mathbf{Set}^{\mathcal{C}^\circ}}(P, \hom_{\mathbf{Set}}(F-, S))$$
We can also characterize the tensor product by the facts. Let $\mathbf{y} : \mathcal{C} \to \mathbf{Set}^{\mathcal{C}^\circ}$ be the yoneda embedding. Then,

$P \otimes_\mathcal{C} F$ preserves colimits in both variables
There is a natural bijection $\mathbf{y}(C) \otimes_\mathcal{C} F \cong F(C)$

One of the important intuitions about presheaves are that they are formal colimit diagrams. If $P$ is written as a colimit of representables, then we can see that
$$ P \otimes_\mathcal{C} F \cong \left( \operatorname{colim}_j \mathbf{y}(C_j) \right) \otimes_\mathcal{C} F  \cong \operatorname{colim}_j F(C_j) $$
This meshes with other tensor product intuitions; for example, you can imagine $P$ as a "free" colimit, and $- \otimes_\mathcal{C} F$ specializes the formal generators $\mathbf{y}(C_j)$ to the special values $F(C_j)$.
The story generalizes almost unchanged to functors $F : \mathcal{C} \to \mathcal{M}$ for a locally small, cocomplete $\mathcal{M}$. You still have a functor $\otimes_\mathcal{C} : \mathbf{Set}^{\mathcal{C}^\circ} \times \mathcal{M}^{\mathcal{C}} \to \mathbf{Set} $  and a hom-tensor adjunction
$$\hom_\mathcal{M}(P \otimes_\mathcal{C} F, M) \cong \hom_{\mathbf{Set}^{\mathcal{C}^\circ}}(P, \hom_{\mathcal{M}}(F-, M))$$
you still have the properties

$P \otimes_\mathcal{C} F$ preserves colimits in both variables
There is a natural bijection $\mathbf{y}(C) \otimes_\mathcal{C} F \cong F(C)$

and $P \otimes_\mathcal{C} F$ can be interpreted as a colimit computation in $\mathcal{M}$.
These stories can be generalized to the enriched setting; my impression that it's easier to see how various constructions should go when they're phrased in terms of homs and tensors rather than in other terms.<|endoftext|>
TITLE: A strange question about closed geodesics on a closed manifold
QUESTION [10 upvotes]: I'm studying a particular kind of curve evolution on Riemannian manifolds. It would help me
to know the answer to the following kinda weird question:
Does there exist a closed Riemannian manifold $M$ and a pair of distinct closed geodesics 
$\gamma, \alpha$ in $M$ satisfying the follow properties?
(1) $\gamma$ and $\alpha$ are of the same length. (Call the length $l$.)
(2) $\gamma$ is isolated in the space of loops of length $l$.
(3) For any $\varepsilon>0$, there exists a path of loops, each of length between $l$ and $l+\varepsilon$, starting at $\gamma$ and ending at $\alpha$.
I suspect (and hope) that the answer is no.
Details: The loop space I'm working with is $C^0(S^1,M)$ with the compact-open topology.
In (2), I mean that $\gamma$ has a neighborhood in the loop space in which the only loops of length $l$ are reparameterizations of $\gamma$.
The path in (3) of course has to be a path of rectifiable loops. 
Thanks,
Dmitri

REPLY [6 votes]: Is it a homework problem?
Define the distance between curves as 
$$d(\gamma,\gamma')=\inf_h\sup_x|\gamma'(x)-\gamma\circ h(x)|,$$
where $h:\mathbb S^1\to\mathbb S^1$ is reparametrization.
Fix small $\delta$ so that if $d(\gamma,\gamma')\le \delta$ 
and $\mathop{\rm length}\gamma=\mathop{\rm length}\gamma'$ 
then $\gamma'=\gamma\circ h$ for some $h$.
Fix $\varepsilon>0$ a choose the path $\gamma_t$.
Choose the smallest value $t$ such that $d(\gamma,\gamma_t)=\delta$
and set $\beta_\varepsilon=\gamma_t$.
You may assume that $\beta_\varepsilon$ has unit-speed parametrization,
in particular they all uniformly Lipschitz.
Pass to partial limit as $\varepsilon\to 0$ and that is it.<|endoftext|>
TITLE: Did Hermite really prove "Hermite's Theorem" on number field discriminants?
QUESTION [15 upvotes]: Hermite's theorem, as it is typically called, is that there are only finitely many number fields of bounded (equivalently, fixed) discriminant.
The usual proof (see Neukirch's Algebraic Number Theory for example) proceeds as follows. First, one proves the Minkowski bound: $|Disc(K)| > (C + o(1))^{deg \ K}$ where $C > 1$. This reduces the problem to fields of fixed degree. For fields $K$ of degree $n$ with $Disc(K) < X$, one can then obtain bounds on the coefficients of the minimal polynomial of $K$, and in particular there are only finitely many possibilities.
However, Minkowski's work was more than thirty years after Hermite's. I looked at Hermite's original paper and although I confess to not having read it in detail, it seems to be essentially the proof I described above. In particular he only claims on the first page to prove the theorem for fields of a fixed degree, and there is nothing I found in the paper which looks like it applies to all degrees. 
So did Hermite actually prove the result that bears his name, or has he been given credit for the jazzed-up version, which apparently could only have been proved thirty years later?
And regardless of whether he actually did, might he have reasonably done so other than coming up with Minkowski's bound on his own?
Thank you!

REPLY [3 votes]: What Rhett Bulter says is also detailed in a chapter entitled 'From Hermite to Minkowski' in a book of Scharlau and Opolka (From Fermat to Minkowski, Lectures on the Theory of Numbers and Its Historical Development, Springer, 1984).<|endoftext|>
TITLE: Acyclic categories related to structures in algebraic topology
QUESTION [11 upvotes]: An acyclic category (also called loopfree category or scwol (small category without loops)) is a small category where only identity morphisms have inverses, and any morphism from an object to itself is the identity.
Every poset P can be regarded as an acyclic category by identifying the set of objects with the elements of P. And saying there is a morphism from x to y, if and only if $x\le y$. Hence we can regard acyclic categories as a generalization of posets.
Posets play a crucial role in combinatorial algebraic topology, e.g. in form of intersection lattices related to hyperplane arrangements, or face posets of simplicial complexes. I'm looking for examples where we get acyclic categories as generalized posets encoding information about some structure (e.g.: Is there some kind of generalized hyperplane arrangement which yields an intersection category).
I'm aware of the face category of a polytopal complex and the salvetti categori of a complex, toric arrangement, but I'd be glad if anyone knew some further applications for acyclic categories in the field of algebraic topology

REPLY [6 votes]: Ralph Cohen, John Jones, and Graeme Segal found an interesting "construction" of a topological acyclic category $C(f)$ from a Morse-Smale function $f : M\to \mathbb{R}$ in a preprint in early 90's. Objects are critical points and morphism spaces are given by moduli spaces of gradient flows.
See also papers by Tanaka and Qin.<|endoftext|>
TITLE: Random pseudoprimes vs. primes
QUESTION [9 upvotes]: (Edit. What I called "pseudoprimes" are known as "Cramér random primes" in the literature,
of which I was unaware.)

Say that a set $S$ of natural numbers is a set of pseudoprimes if they
are (a) random in a sense made explicit below, and (b) the
number in $S$ less than $n$ grows as $n / \log_e n$.
They are, in a sense, random numbers distributed like the primes.
To be more specific, $S$ is created by generating random numbers
$u_2, u_3, u_4, \ldots$, 
uniformly distributed in $[0,1]$, 
and including $n$ in $S$ iff $u_n < 1 / \log_e n$.
(This method was suggested by two MSE users in response to an
MSE question on prime-like distributions.)
Here are the first twenty numbers in various pseudoprime sets generated by this process:
$$3, 8, 10, 11, 12, 13, 15, 28, 35, 39, 44, 50, 53, 65, 66, 67, 68, 70, 78, 86$$
$$8, 27, 29, 31, 32, 33, 41, 50, 52, 55, 57, 62, 63, 71, 72, 74, 78, 82, 83, 92$$
$$5, 28, 31, 34, 44, 45, 46, 48, 50, 53, 59, 64, 68, 69, 70, 72, 76, 78, 85, 95$$
One can imagine other ways of defining pseudoprimes, but let me stick to this simple defintion.
Here is their cumulative distribution up to $n=10^6$:

          


My question is:

Which of the many unsolved problems concerning primes could be proved for pseudoprimes?

For example:

Are there almost surely an infinite number of twin pseudoprimes in any pseuodprime set $S$? 
(See the graph below for the cumulative distribution.)
Are there almost surely an infinite number of Sophie Germain pseudoprimes?
Can you walk to infinity via a pseudoprime Ulam-spiral?
(This was my original motivation.)
Do the Hardy–Littlewood conjectures hold for pseudoprimes? 

One can extend this list indefinitely.  I am seeking to understand to what extent the $1/\log n$ distribution
determines the properties of the primes.
Thanks for insights!

REPLY [2 votes]: This is a bit of a meta-answer; regarding actual facts regarding the specific questions I do not have anything to add to the existing answer; still this or that might be of interest. 
As commented the way of studying problems on primes via considering them as random sets with a certain (changing) density is a common tool known as Cramér's model.
And, there is also a refinement where one takes 'local' obstructions, that is congruence conditions such as 'almost all primes are odd' into account.
To test a conjecture on the distributions of prime numbers against this 'random model' is quite standard. And, indeeed, if something would not hold for this random model of the primes it typically would be discarded as a plausible conjecture. 
In some sense this is an oversimplification, since the model in the question is not that precise (not including local obstructions/congruence conditions) and one could imagine (though I do not have a concrete example handy) a conjecture on the primes that would hold in a more refined model but not in the state one; the issue being that with the model in the question one has a density of $1/ \log n$ whereas with more refined models (and for the primes using the so-called W-trick, ie, sieving for small primes) one can push-up the primes to be a subset of about relative density $\log\log n / \log n$ in certain arithmetic progressions. 
Yet, essentially all the predictions/conjectures on the numer of twin-primes, primes tuples and related things indeed all stem (explicitly or implicitly) from such a refined random model. 
Conversely, results on such questions are even (in particular recently) 'almost exactly' proved along these lines. One establishes that the primes are 'sufficiently (pseudo)random' for heuristics based on random models to apply, or one can argue differently. (Needless to say this is infinitely simpler said like this than actually done.)
Two text I can recommend related to this are the following blog post by Tao and, in particular the introduction to, 'Linear equations in primes' by Green and Tao<|endoftext|>
TITLE: sigma-algebra generated by OD sets
QUESTION [6 upvotes]: Assume $V=L(\mathbb{R})$ and the Axiom of Determinacy. Is every set of reals generated by ordinal-definable sets of reals under the operations of countable union and intersection?
The class of sets generated in this way is Wadge-cofinal and not wellorderable (it contains $\lbrace x\rbrace$ for every $x \in \mathbb{R}$) so there don't seem to be obvious limitations on its extent.
This question came up when I was trying to answer Asaf Karagila's "bonus question" here:
Generating family for the Lebesgue $\sigma$-algebra

REPLY [7 votes]: This does not answer the question, but for every set
$A\subset\mathbb{R}$ in $L(\mathbb{R})$, you can get
$A\times\{r\}$ in your algebra, for some real $r$, and this does not use AD.
The reason is that if $V=L(\mathbb{R})$, then every set is
definable from an ordinal and a real. So if $A\subset\mathbb{R}$,
then $A=\{ x\in\mathbb{R}\mid \varphi(x,\alpha,r)\}$ for some
definition $\varphi$ and parameters $\alpha\in\text{Ord}$,
$r\in\mathbb{R}$. For any finite binary sequence $t$, let
$B_t=\{(x,s)\mid t\subset s,\varphi(x,\alpha,s)\}$, which is an
ordinal definable subset of the plane. Finally, observe that
$A\times\{r\}=\{(x,r)\mid
\varphi(x,\alpha,r)\}$ is precisely $\bigcap_{t\subset r} B_t$, a
countable intersection of OD sets.<|endoftext|>
TITLE: When is a Banach Algebra $C^\star$
QUESTION [5 upvotes]: I know that if there are enough Hermitian elements in a Banach algebra, then the Banach algebra is stellar. In particular, I'm interested in the two spaces $B(L^1(S^1,\Sigma,\mu))$ the space of bounded linear operators on Lebesgue integrable functions of the circle and $B(ba(\Sigma))$ the space of bounded linear operators on finite, finitely-additive Borel measures. I know about the results that having enough Hermitian elements is sufficient, but I'm not quite sure how to apply them.
The issue comes up because I am trying to bound the inverse of a Hermitian element in terms of its spectral radius. From my reading, we have an equality for $C^\star$ algebras and an inequality for Banach algebras.

REPLY [2 votes]: This is not an answer to the question posed in your first paragraph (which I think is asking for more than you need, and more importantly, more than you can really hope for). However, for the specific purpose outlined in your second paragraph, the following paper might be helpful:

H. König, A functional calculus for Hermitian elements of complex Banach algebras.
  Arch. Math. (Basel) 28 (1977), no. 4, 422–430.

IIRC, one has a C^1-functional calculus for Hermitian elements in Banach algebras (this is proved by taking the Fourier transform of your C^1 function after introducing a smooth cutoff outside the support of the spectrum of your Hermitian element). So if the spectrum of your $x$ is contained in $[a,b]$ for $0 \lt a$ then $\Vert x^{-1}\Vert$ should hopefully be bounded above by some universal constant times $a^{-2} = \rho(x^{-1})^2$. CAVEAT: I have not checked this in detail!
Update: I've just remembered that there are theorems to the effect that if $E$ is a Banach space, $A(E)$ the algebra of approximable operators, and $X$ is a reflexive Banach space, then an injective algebra homomorphism $A(E)\to B(X)$ must arise from some embedding of $E$ into $X$ as a closed, complemented subspace. In particular, if there is an injective HM $B(E) \to B(H)$ for some Hilbert space then $E=H$. So the two Banach algebras mentioned at the start of your question can't possibly be $C^\ast$-algebras.<|endoftext|>
TITLE: Matrix-tree theorem via supersymmetry (i.e. Grassman algebras)
QUESTION [8 upvotes]: The matrix-tree theorem states the number of spanning trees of a graph $G$ is equal to a modified determinant of the adjacency matrix or "graph Laplacian", $\Delta_G$:
$$\#\{ \text{spanning trees of }G  \}  = \lambda_1 \lambda_2 \dots \lambda_n = \lim_{\epsilon \to 0}\frac{ \det(\Delta_G - \epsilon I)}{\epsilon}.$$
Here $\lambda_1, \dots, \lambda_n$ are the eigenvalues.
Apparently, Matrix-Tree can be used to compute effective resistances between points in Electrical networks or show the number of distinct labelled trees of size n.
Grassman algebra (or supersymmetry) seems to be a great bookkeeping device which does a lot of the linear algebra for you.  Maybe the proof of that result can be simplified here.
"Fermions" and 1-forms are similar objects.  We can say $\psi_1\psi_2= - \psi_2 \psi_1$ or $ v \wedge  w = - w \wedge v $.  For fermions there's something called Berezin integration or Grassman integration.
$$ \int (a \psi + b ) d\psi= a \hspace{0.333in}\text{and}\hspace{0.333in} \int e^{\phi^T A \psi}\; d\phi d\psi = \det A$$
I am not sure what the analog is for 1-forms in the cotangent bundle of a manifold $T^1(\mathbb{R}^n)$.

Between the papers [arXiv:math.CO/0306396][6] by A Abdesselam and [arXiv:math-ph/0107005][7] by Brydges and Imbrie two approaches for proving this result arise:

Grassmann integration (above)
Forest-Root formula (below)

The Forest-Root formula says, for any compactly supported function:
$$ f(\mathbf{0}) = \sum_{(F,R)} \int_{\mathbb{C}^N} f^{(F,R)} (\mathbf{t}) \left(-\frac{d^2 z}{\pi} \right)^N $$
Here the variables $t_i = |z_i|^2$ and $t_{ij} = |z_i - z_j|^2$ and sum is over all possible roots and forests.  Apparently it can be put even more concisely as:
$$\int_{\mathbb{C}^n} f(\tau) = f(0) $$
(for compactly supported functions) where the integral is over some ``supersymmetric" measure, $\tau = z \overline{z} + \frac{dz d\overline{z}}{2\pi i}$.
It also seems these kinds of "free-fermion" calculations lead to many generalizations of matrix-tree that I won't get into (including relations to Branched Polymers and Diffusion Limited Aggregation).  Personally, I wonder what the cotangent bundle picture looks like for these.

The proofs in the two above papers are left as exercises, which more general results proven.  Mainly, I just would like to see the proofs the no-frills Matrix-Tree theorem from either of these two starting points

REPLY [4 votes]: Check out this talk by Alan Sokal, and references therein (on the title page).<|endoftext|>
TITLE: What are some interesting problems in the intersection of Algebraic Number Theory and Algebraic Topology?
QUESTION [30 upvotes]: I'm a beginning graduate student and while my background is primarily in algebraic number theory, I've found myself a bit smitten with the subject of algebraic topology recently after only having read some of Hatcher (Chapters 1 and 2 and I'm currently reading Chapter 3- and this is currently the extent of my topology background). 
I'm trying to get an idea of what sorts of problems lay at the intersection of these two fields, for now just so I have a direction of what sort of background I might want to be learning in the next year or so.
I realize this question is rather broad, but does algebraic number theory have any nice applications to topological problems?

REPLY [6 votes]: There are applications of the theory of cyclotomic fields to free actions of finite groups on $S^n$. The existence of such actions is tied to the class groups of certain cyclotomic fields $\mathbb{Q}(\mu_N)$ and their maximal real subfields $\mathbb{Q}(\mu_N)^+$.
You can find a brief introduction to this concept on p265 in Lang's Units and Class Groups in Number Theory and Algebraic Geometry: http://projecteuclid.org/euclid.bams/1183548780
Here are some of the relevant references he cites:
J. MILGRAM, Odd index subgroups of units in cyclotomic fields and applica-
tions, Springer Lecture Notes no. 854 (1981).
J. MILGRAM, Patching techniques in surgery and the solution of the compact
space form problem.
D. KUBERT, The 2-primary component of the ideal class group in cyclotomic
fields.
C. T. C. WALL, Classification of hermitian forms VI, Ann. of Math. 103 (1976)
pp. 1-80.<|endoftext|>
TITLE: Why are injective modules more complicated than projective modules?
QUESTION [22 upvotes]: For beginners in homological algebra, it is a fact of life that injective modules seems to be more mysterious than projective modules. For example, for finitely generated modules over a noetherian ring, projective resolution can be taken as resolution by free modules of finite rank, but I don't see how one can easily write down injective resolutions.
I'm wondering if there is a deep reason behind this. What makes injective modules so complicated?

REPLY [12 votes]: I'll give you my more pedestrian reason. The proofs of existence of injective  resolutions require the axiom of choice, in one form or another. Translation:  these proofs  are not constructive, so there are no general algorithms for producing such objects.  This becomes a painful issue in concrete  situations.     This  has similarities with another famous existence result, the Hahn-Banach theorem  which postulates the existence of  continuous linear functionals with certain properties. It is particularly useful for existence theorems for PDE's. Unfortunately it gives you no guide for finding those solutions.<|endoftext|>
TITLE: A Nisnevich cover which is not Zariski
QUESTION [5 upvotes]: The Nisnevich topology on $Sch$ is a Grothendieck topology strictly finer than the Zariski topology, and the etale topology is strictly finer than the Nisnevich topology.
Colin McLarty asked me for an example of a Nisnevich cover which is not a Zariski cover. The standard example I have seen in several places is rather of the distinction between etale and Zariski. Namely, a family of etale covers $\{A^1 - \{0\} \stackrel{(-)^2}{\to} A^1, A^1 - \{a\}\hookrightarrow A^1\}$ of the affine line $A^1$ over a field $k$ indexed by elements $a\in k^\times$, such that a cover is only Nisnevich if $a$ has a square root in $k$. I'm looking for something that will separate Zariski from Nisnevich.
Examples inspired by arithmetic or geometry are both good.

REPLY [6 votes]: My standard example is an $n$-gon of $\mathbb P^1$'s covering the nodal cubic. For some reason, I especially like the case $n=2$.

Here's the affine version, which always takes me a bit to work out. It's two parabolas joined at two points covering the nodal cubic:
$$
\def\spec{\mathrm{Spec\,}}
\spec k[s,t]/(t^2-(s^2-1)^2) \to \spec k[x,y]/(y^2-x^2(x+1))
$$
given by $x\mapsto (s^2-1)$ and $y\mapsto st$.



This map is étale. To see it is Nisnevich, you have to check that the residue field extensions of the generic point are isomorphisms (this is where the squaring map  on $\mathbb A^1-0$ fails to be Nisnevich). You can see that by restricting to the two components, giving the normalization maps
$$
k[x,y]/(y^2-x^2(x+1)) \to k[s,t]/(t-(s^2-1)) \cong k[s]
$$
given by $(x,y)\mapsto (t,st)$ (so $s=y/x$), and 
$$
k[x,y]/(y^2-x^2(x+1)) \to k[s,t]/(t+(s^2-1)) \cong k[s]
$$
given by $(x,y)\mapsto (-t,st)$ (so $s=-y/x$).<|endoftext|>
TITLE: Is there a typical example of Nisnevich covers?
QUESTION [9 upvotes]: There is a popular (and I think helpful) example of etale covers, namely covers of Riemann surfaces with ramification points removed.  Is there a similarly accessible example to motivate Nisnevich covers?

REPLY [13 votes]: Well, a representative example is where you take some arbitrary etale cover Y of X which splits over a closed subvariety Z of X, then form the Nisnevich cover of X consisting of the open complement X - Z together with the open subscheme Y' of Y where you remove all but one of the copies of Z lying above Z.  For instance Z could be a point, and our field could be algebraically closed.
The intuition I find helpful is that descent for the Nisnevich topology is meant to be an easier-to-precisely-phrase consequence of the principle "X is gotten by gluing X-Z to a tubular neighborhood of Z, along the punctured tubular neighborhood of Z".  The idea being that, in the situation of the previous paragraph, the tubular neighborhoods of Z in X and Z in Y should be the same, Y --> X being etale.
As an example of this intuition, note that if X is a smooth variety of dimension n (over an algebraically closed field for simplicity) and x is a point of X, then by choosing n independent parameters at x you can find a Zariski neighborhood X' of x and an etale map X' --> A^n which, together with A^n - 0 ---> A^n, makes a Nisnevich cover of A^n with intersection equal to X'-x.  This corresponds (ish) to the fact that the tubular neighborhood of x in X should be the same as the tubular neighborhood of 0 in A^n.<|endoftext|>
TITLE: Can formal power series become polynomial often, when composed with polynomials?
QUESTION [13 upvotes]: Let $F$ be a finite field. Let $F[X]$ and $F[[X]]$ denote the ring of polynomials and power series over $F$, respectively. I'm trying to show a statement like the following:
Fix a $d > 0$. Let $g\in F[[X]]$. If there exists a set $C\subseteq F[X]$ of polynomials (with no constant term) of degree at most $k$ such that for all $c\in C$, $g(c)$ -- $g$ composed with $c$ -- is a degree $kd$ polynomial, and $|C|$ is "large" (some function of $k$, $d$, and $|F|$), then $g$ must actually be a polynomial. 
I'm trying to beat the bound that one might be able to get via Schwartz-Zippel, where $|C| > kd |F|^{k-1}$ (where $kd \ll |F|$). 
What bounds on $|C|$ can we get?
Thank you,
Henry

REPLY [6 votes]: I claim that, if $g(x) \in k[[x]]$ is not a polynomial, then $g \circ c$ is a polynomial for at most $|F|^{k/2}$ polynomial $c$ of degree $\leq k$. We will always use the letter $c$ to represent a polynomial with $c(x)=0$.
Case 1: $g$ is transcendental over the field $k(x)$. In this case, I claim that $g \circ c$ is a polynomial only when $c$ is $0$. Proof: Suppose that $g(c(x)) = h(x)$ for some polynomial $h$. Then there is a polynomial relation $F(c(x), h(x))=0$ for some polynomial $f$. Set $G(x) := F(x, g(x))$. By hypothesis, $G \neq 0$, but $G(c(x))=0$. The only way this can happen is if $c=0$.
Case 2: $g$ is algebraic over $k(x)$. Let $F(x,g(x))=0$ be the minimal polynomial relation between $x$ and $g$. Let $A$ be the ring $k[x,y]/F(x,y)$. At this point I really, really want to use the language of algebraic geometry. If you aren't happy with this, I'll try to convert into commutative algebra, but it will be harder to write and, in my opinion, harder to read. My goal is the following claim:

Claim: If there is any nonzero $c$ such that $g \circ c$ is polynomial, then the ring $A$ is a subring of $k[u]$ for some $u \in \mathrm{Frac}(A)$.

Example: Let $g = (1+x)^{3/2}$. Then $F(x,y) = y^2 - (1+x)^3$. Letting $u = \frac{y}{1+x}$, we have $A \subseteq k[u]$, since $x=u^2-1$ and $y=u^3$.
Proof: $\mathrm{Spec}(A)$ is an algebraic curve. Since $F$ is the minimal polynomial relation, it is irreducible. Let $X$ be the normalization of $\mathrm{Spec}(A)$.
Let $h=g \circ c$. Now, $t \mapsto (c(t), h(t))$ is a map $\mathbb{A}^1 \to \mathrm{Spec}(A)$ and, since $c$ is nonconstant, is a nonconstant map. Since $\mathbb{A}^1$ is normal, this lifts to a nonconstant map $\mathbb{A}^1 \to X$. So $X$ has genus zero, and has at most one puncture. Since $X$ is affine, it has at least one puncture. So $X \cong \mathbb{A}^1$. In other words, the normalization of $A$ is $k[u]$, so we get an embedding $A \subseteq k[u]$ for some $u \in \mathrm{Frac}(A)$. $\square$
From now on, we will assume that there is some nonzero $c$ for which $g \circ c$ is polynomial. So we may assume that there is an embedding $A \subseteq k[u]$, with $\mathrm{Frac}(A) = k(u)$, and we fix such an embedding. Let $x = a(u)$ and $g(x) = b(u)$, for some polynomials $a$ and $b$. If $a$ has degree $1$, then $u$ is a linear function of $x$ and we can write $g$ as a polynomial in $x$, contradicting your hypothesis. So $\deg a \geq 2$. I claim that there are only $|F|^{k/\deg a}$ values of $c$ for which $g \circ c$ is a polynomial. Specifically, the polynomials $c$ of the form $a(\phi(t))$, where $\phi$ is a polynomial of degree $k/\deg a$.
Why is this? Well, if $c(t) = a(\phi(t))$, then $g(c(t)) = b(\phi(t))$ and is thus a polynomial. Conversely, if we have some $c$ such that $g \circ c$ is polynomial then, as discussed above, we get a map from the normalization of $A$ to $k[t]$. Take $\phi$ to be the image of $u$.<|endoftext|>
TITLE: A possible mistake in Kac's "Infinite Dimensional Lie Algebras"
QUESTION [5 upvotes]: I have a paperback 3rd edition and on page 65 you can find Proposition 5.8. My question is about part (c):

If $A$ is of indefinite type, then
  $$ \overline{X} = \{ h \in \{ \frak h_{\mathbb{R}} \}| \left \langle \alpha, h \right \rangle \geq 0 \text{ for all } \alpha \in \Delta_+^{im} \}, $$
  where $\overline{X}$ denotes the closure of $X$ in the metric topology of $\frak h_{\mathbb{R}}$.

Some context: $A$ is a generalized Cartan matrix, $\frak h_{\mathbb{R}}$ is a real form of the Cartan subalgebra in the Kac-Moody algebra associated to $A$ and finally $X$ is the Tits cone. 
Question: if one reads the proof given by Kac, it is written for $X$ and not $\overline{X}$. Is there a mistake or am I missing something here?

REPLY [11 votes]: After taking a closer look at the proof by Kac of Prop. 5.8 c), I can see that it's too sketchy to be followed easily.   Here the generalized Cartan matrix is assumed to be of indefinite type, which I have little experience with.  But the basic steps in the proof might be organized as follows:
In the background is the infinite root system $\Delta= \Delta_+ \cup \Delta_-$ along with a further partition $\Delta_+ = \Delta_+^{re} \cup \Delta_+^{im}$.  Here $\Delta_+$ consists of $\mathbb{Z}^+$-linear combinations of the fixed simple  roots $\alpha_1, \dots, \alpha_n$. The $\alpha_i$  are linearly independent and lie in the dual space of $\mathfrak{h}_\mathbb{R}$ where $\mathfrak{h}$ is the finite dimensional Cartan subalgebra.  The Weyl group  $W$ is generated by all reflections $r_\alpha$ with $\alpha \in \Delta_+$.  
The Tits cone $X$ is the image under $W$ of $C:= \{ h \in \mathfrak{h}_\mathbb{R} | \langle \alpha_i, h \rangle \geq 0 \text{ for all } i\}$.  The candidate for its metric closure is $X' := \{h \in \mathfrak{h}_\mathbb{R} | \langle \beta, h \rangle \geq 0 \text{ for all } \beta \in \Delta_+^{im}\}$.   Being defined by inequalities, $X'$ is closed.
By Prop. 5.2 a), $\Delta_+^{im}$ is $W$-invariant (so $X'$ is).  Obviously $X' \supset C $, so $X' \supset \overline{X}$.  
In the reverse direction, consider just those $h \in X'$ for which $\langle \alpha_i, h \rangle \in \mathbb{Z}$ for all $i$.   These elements of $X'$ are dense in the metric topology, so it's enough to show they all lie in $X$ (where they will form a dense subset of $\overline{X}$).
Use Thm. 5.6 c) to find $\beta \in \Delta_+^{im}$ such  that the "Cartan integers" $\langle \beta, \alpha_i^\vee \rangle <0$ for all $i$.  (So $\beta= \sum_i b_i \alpha_i$ with all $b_i>0$.)
In turn, for all $\gamma \in \Delta_+^{re}$, 
$$r_\gamma (\beta) = \beta - \langle \beta, \gamma^\vee \rangle \gamma = \beta + s \gamma$$ with $s$ larger than the sum of coefficients of $\gamma$ because $\langle \beta, \alpha_i^\vee \rangle <0$ for all $i$.  Thanks to Prop. 5.2 c), all $\beta +s \gamma \in \Delta_+^{im}$. Since  $h \in X'$, it follows that $\langle \beta + s\gamma, h \rangle \in \mathbb{Z}^+$.  In particular, only finitely many such $\gamma$ exist with $\langle \gamma, h \rangle \leq -1$.    
But  Prop. 3.12 c) characterizes $X$ as the set of all $h$ for which  only finitely many $\gamma \in \Delta_+$ satisfy $\langle \gamma, h \rangle <0$.
Combined with the special choice of $h$,  we get  $h \in X$ as desired.  
ADDED: To fill in details of the argument I've automatically tended to think in terms of Zariski-density and Zariski-closure, but something else must be going on here to deal with the metric topology.  This is the point at which I'm doubtful about the strategy used by Kac.   But given the brevity of the argument it's probably necessary to look further into the surrounding material for some kind of insight.  (Maybe it's just a question of pointing to the fact that both $X$ and $X'$ are cones?   It would be easier if the author of the book told us what he was thinking about.)<|endoftext|>
TITLE: CW structures on unitary sphere of a banach/Hilbert space
QUESTION [5 upvotes]: Does the  unit sphere in a Banach  space carries a structure of  a CW complex? What about Finsler Manifolds?

REPLY [12 votes]: The answer is no in infinite dimension (and yes in the finite dimensional case, of course). Because the sphere in a Banach space is Baire. But an infinite dimensional CW-complex is not, being a countable union of finite dimensional skeletons.<|endoftext|>
TITLE: Isoperimetry and Poincaré Inequality
QUESTION [9 upvotes]: What are the known relations between isoperimetric and Poincaré inequalities on manifolds?
For example, for manifolds with a lower bound on Ricci curvature, the Cheeger-Buser inequality relates the isoperimetric constant to the first eigenvalue of the Laplacian. Does this imply a Poincaré inequality for such manifolds?

REPLY [14 votes]: Everything here is for closed Riemannian manifolds.
If you have a lower bound on Cheeger's isoperimetric constant $h(M)$, then Cheeger's inequality
$\lambda_1(M)\geq \frac{h(M)^2}{4}$ gives you a lower bound on the first eigenvalue of the Laplacian.
Thanks to the variational characterization of $\lambda_1(M)$, this is exactly equivalent to an upper bound on the constant $C_P$ in the $L^2$ Poincaré inequality
$$\int_M f^2 dV \leq C_P \int_M |\nabla f|^2 dV,$$
for all smooth functions $f$ with $\int_M f dV=0$.
If you have a lower bound for the Ricci curvature, and also an upper bound for $C_P$ (equivalently, a lower bound for $\lambda_1(M)$), then Buser's inequality gives you a lower bound for $h(M)$.
See for example Wikipedia's page.
On the other hand a result of Yau shows that $h(M)$ is equal to the reciprocal of the constant $C_P'$ in the $L^1$ Poincaré inequality
$$\int_M |f| dV \leq C_P' \int_M |\nabla f| dV,$$
for all smooth functions $f$ with $\int_M f dV=0$.
If you are interested in obtaining bounds on $h(M)$ or $\lambda_1(M)$ in terms of geometric quantities, there are the following basic results: $h(M)$ can be bounded below in terms of a lower bound for the volume of $M$, an upper bound for its diameter, and a lower bound for its Ricci curvature (Croke). On the other hand $\lambda_1(M)$ can be bounded below in terms of an upper bound for its diameter and a lower bound for its Ricci curvature (Li-Yau). Of course these bounds also depend on the dimension of the manifold.
Finally, if you are intersted in the relation with Sobolev inequalities, you should look at this paper of Peter Li.<|endoftext|>
TITLE: congruences for Fourier coefficients of modular forms
QUESTION [6 upvotes]: Are there other good articles on congruences for Fourier coefficients of modular forms beside Swinnerton-Dyer's article in "Modular Functions of One Variable III"?
I am looking for generalisations and other explicit cases than the Ramanujan $\tau$-function and the $5$ other modular forms in [Swinnerton-Dyer].

REPLY [2 votes]: Also, have a look at Ken Ono's The Web of Modularity: 
Arithmetic of the Coefficients of Modular Forms and Q-Series. This book is full of what you are looking for.<|endoftext|>
TITLE: Recent impressive combinatorial developments in probability theory
QUESTION [7 upvotes]: In the preface to the second edition of Daniel Stroock's book "Probability Theory: An Analytic View", there is this striking claim (on p. xv)

... I suspect that, for at least a decade, the most important developments in the field will have a strong combinatorial component ...

I have several questions

What have been the most striking and impressive developments along that line in the past decade?
Is there an overwhelming agreement in the research community about that statement?
What are the most promising avenues of exploration?
Was the progress occurring throughout probability theory and some specific fields such as stochastic processes or statistics or was it confined to areas that were fundamentally combinatorial in nature such as random graph theory?

REPLY [7 votes]: A whole body of results in probability with strong combinatorial flavour are around 2-dimensional stochastic models. Some of this progress started 15 years ago but much was achieved in the last decade. Much of this research has combinatorial flavour. This includes conformal invariance for planar percolation on the triangular grid; The stochastic Lowevner equations, SLE, and their relations with Brownian motion, crirical percolation, loop-erased random walks, the Ising model, and other models. These relations allowed the computation of many critical exponents of 2D models. You can add to that the recent results on self avoiding random walks (again in 2D), and the computation of critical probabilities for 2D Potts model.

REPLY [4 votes]: Another important development in probability theory with strong combinatorial flavour and  relations to mathematical physics is around random partitions. This is closely related to random surfaces and random matrices. The origin are again older but much was achieved in the last decade or so.<|endoftext|>
TITLE: A combinatorial question
QUESTION [5 upvotes]: This possibly easy question is related to  this one.  Let $s_1,...,s_n$ be a sequence of natural numbers (some of them may be equal to 0). Consider the following sequence of multisets of 2-vectors (each vector is counted with its multiplicity) of natural numbers $A_1=\{ (s_1,s_2),...(s_n,s_1)\}$, $A_2=\{(s_1+s_2,s_3),...,(s_n+s_1,s_2)\}$, ...,$A_n=\{(s_1+...+s_n,0)\}$. 
Note that each multiset is closed under taking cyclic shift on the set of indices $1,2,...,n$.
 Question Does this sequence of multisets determine the sequence $s_1,...,s_n$ up to a cyclic shift?
In particular, is the Prouhet-Morse-Thue sequence reconstructible (up to a cyclic shift)?
 Example  If we encode every pair $(i,j)$ by a monomial $a^ib^j$ in commuting variables $a,b$, and interpret the multiplicity as a coefficient, we encode every multiset as a polynomial in $a,b$ over $\mathbb{Z}$. Here are the first 7 polynomials corresponding to the Prouhet-Morse-Thue word of length $8$ p_8="10010110":
$$3a+1+3b+ab\\\
3a+b+3ab+{a}^{2}\\\
2ab+2a+2{a}^{2}b+2{a}^{2}\\\
2{a}^{2}+2\,ab+2{a}^{2}b+2\,{a}^{3}\\\
3{a}^{2}b+3{a}^{3}+{a}^{3}b+{a}^{2}\\\
3{a}^{3}b+3{a}^{3}+{a}^{4}+{a}^{2}b\\\\
4{a}^{4}+4{a}^{3}b$$
Does this sequence of polynomials determine the word $p_8$ up to a cyclic shift?  Update  Answer for $p_8$ is "yes" (computed using Maple).

REPLY [5 votes]: Unless I made mistake a counterexample is formed by the binary $m$-sequence of length 7 $s=(1,0,0,1,0,1,1)$ and its reversal $\tilde{s}=(1,1,0,1,0,0,1)$ that is not a cyclic shift of $s$. Both lead to the sequence of generating functions $1+2a+2b+2ab$, $2a+b+a^2+2ab+a^2b$, $a+a^2+2ab+a^3+2a^2b$, $a^2+ab+2a^3+2a^2b+a^3b$, $2a^3+2a^2b+a^4+2a^3b$ and $3a^4+4a^3b$.
The $m$-sequences are examples of de Bruijn -sequences. That is binary sequences of length $2^n-1$ such that every sequence of $n$ bits (with the exception of $n$ zeros) occurs exactly once in the cycle. This is, of course, then a natural source for an eventual counterexample as the condition is automatically satisfied for $A_j, j\lt n.$ Length 7 is the shortest, where not all de Bruijn sequences are cyclic shifts of each other. 
An $m$-sequence is generated by a linear feedback shift register that has a primitive polynomial of degree $n$ from the ring $\mathbb{F}_2[x]$ as a feedback polynomial. If you decimate an $m$-sequence with a decimation exponent $d$, $\gcd(d, 2^n-1)=1,$ by cyclically taking every $d^{th}$ member, you get another $m$-sequence. Therefore for larger $n$ the number of non-cyclically equivalent $m$-sequences increases. For example, the sequence $s$
is generated by the feedback polynomial $x^3+x^2+1$, or equivalently by the given first 3 bits and the recurrence relation $$s_n=s_{n-3}+s_{n-2}\pmod2$$
with subscript arithmetic done modulo seven. The reversed sequence $\tilde{s}$ is similarly generated by the reciprocal polynomial $x^3+x+1$.
I dare not guess yet, whether all $m$-sequences of a given length give rise to the same sequence of multisets.<|endoftext|>
TITLE: Two Concepts of Monotonicity
QUESTION [5 upvotes]: Let $K$ be a closed convex subset in $\mathbb{R}^n$ and $F: K\rightarrow \mathbb{R}^n$. We say that

$F$ is strongly monotone on $K$ if there exists $\gamma>0$ such that

$$
\langle F(y)-F(x), y-x\rangle\geq \gamma\|y-x\|^2, \quad \forall x,y\in K.
$$

$F$ is strongly pseudomonotone on $K$ if there exists $\gamma>0$ such that
$$
\langle F(x), y-x\rangle\geq 0 \Longrightarrow \langle F(y), y-x\rangle\geq \gamma\|y-x\|^2
$$
for all $x,y\in K$.

It is easily to verify that strongly monotone implies strongly pseudomonotone.
The converse is not true in general. For example, in one-dimensional case 
$$
F(x)=(2-x), \quad K=[0,1],
$$ 
the mapping $F$ is strongly pseudomonotone but not strongly monotone on $K$.
$\textbf{Question:}$ Can we find a mapping $F: K\rightarrow \mathbb{R}^n (n\geq 2)$ such that $\text{int}K\ne \emptyset$ ($K$ has a nonempty interior) and $F$ is strongly pseudomonotone but not strongly monotone on $K$. It is interesting to find $\textbf{an affine mapping}$ as in the above example.
I am grateful to all your comments and helping.

REPLY [2 votes]: You can find an example here:
Modified projection method for strongly pseudomonotone variational inequalities
PD Khanh, PT Vuong, Journal of Global Optimization 58 (2), 341-350<|endoftext|>
TITLE: Subgroups of $GL_n(\mathbb Z)$ with finite coinvariants
QUESTION [5 upvotes]: Is there a finite index torsion-free subgroup $G$ of $GL_n(\mathbb Z)$, where $n\ge 3$,
such that the coinvariants group $\mathbb Z^n_G$ is finite? 
Here $G$ acts on $\mathbb Z^n$ in the standard way, and $\mathbb Z^n_G$  by definition is the quotient of $\mathbb Z^n$ by the subgroup generated by the set $\{gz-z: g\in G, z\in\mathbb Z^n\}$. 
If $G=GL_n(\mathbb Z)$, then $\mathbb Z^n_G$ is finite because G contains $-I_n$, but then $GL_n(\mathbb Z)$ isn't torsion-free.

REPLY [6 votes]: van der Kallen gave a nice answer for the situation at hand, but I thought I'd give a somewhat more general one.  The question is equivalent to showing that $(\mathbb{R}^n)_G=0$ for some finite-index torsion-free subgroup $G$ of $\text{GL}_n(\mathbb{Z})$.  The representation $\mathbb{R}^n$ is a nontrivial irreducible representation of $\text{SL}_n(\mathbb{R})$, so this follows from the following more general result (just take $\Gamma$ to be a torsion-free lattice, for instance the level $3$ principle congruence subgroup).
LEMMA : Let $V$ be a nontrivial finite-dimensional irreducible representation of $\text{SL}_n(\mathbb{R})$ over $\mathbb{R}$ and let $\Gamma$ be any lattice in $\text{SL}_n(\mathbb{R})$.  Then $V_{\Gamma} = 0$.
To see this, observe that using the Borel density theorem (which says that $\Gamma$ is Zariski dense in $\text{SL}_n(\mathbb{R})$), we can get that $V$ is also a nontrivial irreducible $\Gamma$-representation.  Now, $V_{\Gamma} = V/K$ where $K$ is spanned by the set $\{x-g(x)\text{ $|$ }x \in V, g \in \Gamma\}$. Clearly $K$ is a nontrivial $\Gamma$-subrepresentation of $V$, so by the irreducibility of $V$ we must have $K=V$.<|endoftext|>
TITLE: Strong monotone limits and dense subalgebras of von Neumann algebras, again
QUESTION [6 upvotes]: Edit: I just realized that this question is related to Andreas Thom's very interesting question here. I think the question below is more crude...
Michael's question here reminded me of the first lemma of this paper of Kadison, establishing that those $C^{*}$-algebras $\mathcal{A}\subset B(\mathcal{H})$ for which the set of self-adjoint elements $\mathcal{A_{s.a.}}$ is strong-operator closed under taking monotone increasing limits are, in fact, von Neumann algebras. I've often wondered about the following

Question: What are necessary and sufficient conditions on a $*$-subalgebra $\mathcal{A}\subset B(\mathcal{H})$ such that strong operator closure of $\mathcal{A_{s.a.}}$ under monotone increasing nets guarantees that $\mathcal{A}$ is closed in the strong operator topology (i.e. is actually a von Neumann algebra)?

I'd like to know if there are weaker conditions than $\mathcal{A}$ being a $C^{\ast}$-algebra for which Kadison's conclusion holds. For example, conditions including things like "$\mathcal{A}$ is closed under continuous functional calculus" and such, which are essentially equivalent to $\mathcal{A}$ being a $C^{*}$-algebra in the (singly-generated) abelian case, but not in general.
(My impression is that this question is really tough, but I hope I'm wrong. It would be nice even to have an expert's digression on precisely why this question should be tough...as I'd learn some new things from that insight!) 
It is also possible to ask about analogues of Pedersen's "up-down" theorem, which says that any self-adjoint element in the strong closure of a $C^{\ast}$-algebra $\mathcal{A}$ on a separable Hilbert space $\mathcal{H}$ is the strong limit of a monotone decreasing sequence of self-adjoint elements each of which is a strong limit of a monotone increasing sequence of self-adjoint elements. Can one weaken the $C^{\ast}$-condition on $\mathcal{A}$ and still get this result? (I've tried to modify the argument in Pedersen's $C^{*}$-algebras and their automorphism groups but if I remember correctly this seems to use the definition of a $C^{*}$-algebra in an essential way. Is there a way around this?!)

REPLY [3 votes]: This is not an answer. Recently, I have been thinking questions related to yours, and I posted some of them on the board, so you may have seen them. I also posted related references in my comments there, but I think that you have already read those papers. It’s possible to ask a more general question: “Which subsets $S\subset\mathbb{B}(\mathcal{H})_{sa}$ have the property $S^m=S_m=S$?” or “Which real linear subspaces $S\subset\mathbb{B}(\mathcal{H})_{sa}$ have the property $S^m=S$?”  It would be interesting if these simple conditions implied $S$ being strongly closed.<|endoftext|>
TITLE: Overview of Arakelov intersection theory and the Arakelov Chow ring
QUESTION [8 upvotes]: I'm looking for a reference that gives an overview of the most important properties of Arakelov intersection theory (on arithmetic varieties of arbitrary dimension) and that describes basic properties of the Arakelov Chow ring. There's a similar MO question asking about
survey articles on (classical) intersection theory, so I guess that I'm asking the same question, but for Arakelov intersection theory.

REPLY [7 votes]: A good reference in my humble opinion is Bost's paper in Bourbaki:
Théorie de l'intersection et théorème de Riemann-Roch arithmétiques
Séminaire BOURBAKI. Novembre 1990. 43ème année, 1990-91, n° 731
Another reference would be Soule's book on Arakelov geometry "Lectures on Arakelov geometry" written with Abramovich, Burnol and Kramer. 
Finally, I know you didn't ask this, but in the case of arithmetic surfaces there are more references (besides Faltings' "Calculus on arithmetic surfaces" and Arakelov's original paper). For example, Deligne's paper "le determinant de la cohomologie" and  R. de Jong's Ph.D. thesis: http://www.math.leidenuniv.nl/~rdejong/publications/thesis.pdf 
Also, Moret-Bailly's paper "Metriques permises" in Szpiro's 1985 Asterisque is wonderful.<|endoftext|>
TITLE: Symmetric group action on squarefree polynomials
QUESTION [12 upvotes]: The following dynamical system on polynomials comes mostly from idle curiosity, but I hope it is of some interest.
Background Fix some natural number $n$. Let $P$ be the quotient of the polynomial ring $\mathbb{Z}[x_1,\ldots,x_n]$ by the ideals $\lbrace(x_j^2=x_j)~|~1\leq j \leq n\rbrace$.
Let $\Sigma(n)$ denote the symmetric group on $n$ elements and note that it acts on $P$ by permuting the variables. The polynomials fixed by all $\sigma \in \Sigma(n)$ are sums and products of the symmetric polynomials and we can enumerate them by monomial degree. That is, the subring $P' \subset P$ of those polynomials is generated by the elementary symmetric polynomials $$S_1 = x_1 + \ldots + x_n,\ S_2 = \sum_{i<j}s_is_j,\ \dots\ ,\ S_n = x_1x_2\ldots x_n.$$ 
Consider the function $s:P \to P'$ defined as follows: $s$ takes a polynomial $p \in P$ to $$ \prod_{\sigma \in \Sigma(n)}\sigma(p).$$ Also consider a function $e:P' \to P$ defined by sending the elementary symmetric polynomial $S_j$ to its corresponding $x_j$ and extending the domain to all of $P'$ via $\mathbb{Z}$-linearity.
The "dynamical system" I'm interested in comes from the self-map $f = e\circ s: P \to P$. 
Example Consider $n = 2$ and $p = x_1$. Then, $s(p) = x_1x_2 = S_2$ and so $f(p) = e(S_2) = x_2$. Then, $f(f(p)) = e \circ s(x_2) = e(x_1x_2) = x_2 = f(p)$. Thus, $p$ lies in the basin of attraction of the fixed point $x_2$. On the other hand, if you start with $p = 2x_1+x_2$ then the orbit of $p$ is $$2x_1 + x_2 \to 2x_1 + 5x_2 \to 10x_1 + 29x_2 \to \ldots $$

Question What can be said about the behavior of $f:P \to P$ in terms of fixed points, periodic orbits, etc. for arbitrary $n$?

Update Contrary to what was stated in the prior versions of this question, neither $s$ nor $e$ is a ring homomorphism.

REPLY [2 votes]: Interesting things happen if you reduce this dynamical system modulo some integer $N$.
This makes sense because your map $s$ is compatible with reduction modulo $N$, and your map $e$ is $\mathbf{Z}$-linear, so $f$ induces a map $f_N : P_N \to P_N$ with $P_N = P \otimes \mathbf{Z}/N\mathbf{Z}$.
In the case $N=2$, what is the map $f_2$? Note that $P_2 = \mathbf{F}_2[x_1,\ldots,x_n]/(x_1^2-x_1,\ldots,x_n^2-x_n)$ can be identified with the ring of functions $\mathbf{F}_2^n \to \mathbf{F}_2$ (every such function is polynomial!). This ring is in bijection with the powerset of $\mathbf{F}_2^n$ : we can map any subset of $\mathbf{F}_2^n$ to its characteristic function. Now, the map $s_2: P_2 \to P_2$ has a very natural interpretation : it maps a subset $A$ of $\mathbf{F}_2^n$ to the subset $\cap_{\sigma \in \Sigma(n)} \sigma(A)$. Also, by definition, the image of $e_2 = e \otimes \mathbf{F}_2$ is contained in the set of affine hyperplanes of $\mathbf{F}_2^n$ (together with $\emptyset$ and $\mathbf{F}_2^n$).
So we want to determine which affine hyperplanes are fixed by $f_2$. This is not obvious at first sight, since the map $e_2$ doesn't seem to have a nice interpretation like $s_2$. But luckily, some linear algebra over $\mathbf{F}_2$ shows that there are only $6$ possibilities for $s_2(H)$ where $H$ is an affine hyperplane of $\mathbf{F}_2^n$. Namely $s_2(H)=H$ when $H$ is symmetric, and $s_2(H)$ can only be $\emptyset$, $\{0\}$, $\{1\}$ or $\{0,1\}$ otherwise. Here $0$ resp. $1$ denotes the constant vector $(0,\ldots,0)$ resp. $(1,\ldots,1)$. It's possible to compute the image by $e_2$ of these subsets, and after some computations we find that the only fixed points of $f_2$ are the polynomials $0$, $1$, $x_n$, together with $1+x_1+\cdots+x_{n-1}$ when $n$ is odd.
Now let's go back to $\mathbf{Z}$. Obviously every fixed point $p$ of $f$ must reduce to a fixed point of $f_2$ mod $2$. Conversely, we have a kind of Hensel lemma : every fixed point of $f_2$ lifts to a unique fixed point of $f_{\mathbf{Z}_2} : P \otimes \mathbf{Z}_2 \to P \otimes \mathbf{Z}_2$. Here $f_{\mathbf{Z}_2}$ is defined by just replacing the base ring $\mathbf{Z}$ by $\mathbf{Z}_2$ in the definition of $f$.
This lifting property is a consequence of the following fact : $f$ is contracting for the $2$-adic topology on every ball $p_0+2P$ with $p_0 \in P$. Indeed, assume $p=p_0+2q$ with $q \in P$. Then
\begin{equation*}
s(p) = \prod_{\sigma \in \Sigma(n)} p_0^\sigma + 2q^\sigma \equiv s(p_0)+2\sum_{\sigma} q^\sigma \cdot \prod_{\tau \neq \sigma} p_0^\tau \pmod{4}
\end{equation*}
Now, observe that the stabilizer $H$ of $\overline{p_0} \in P_2$ in $\Sigma(n)$ is isomorphic to $\Sigma(m) \times \Sigma(n-m)$ for some $0 \leq m \leq n$, so in particular $\operatorname{card}(H) \geq 2$. This implies that $\prod_{\tau \neq \sigma} \overline{p_0}^\tau = s(\overline{p_0})$ in $P_2$. Furthermore $\sum_{\sigma} \overline{q}^\sigma = 0$ since the stabilizer of $\overline{q} \in P_2$ has even order. It follows that $s(p) \equiv s(p_0) \pmod{4}$. The same argument shows that $p \equiv p_0 \pmod{2^m}$ implies $s(p) \equiv s(p_0) \pmod{2^{m+1}}$, so $s$, and thus $f$, is contracting.
It follows immediately that the only fixed points of $f$ in the balls $2P$, $1+2P$ and $x_n+2P$ are $0$, $1$ and $x_n$ respectively. It remains to consider the fixed point $1+x_1+\cdots+x_{n-1}$ with $n \geq 3$ odd. The corresponding fixed point of $f_{\mathbf{Z}_2}$ is obtained as the limit of $f^m(1+x_1+\cdots+x_{n-1})$ in $P \otimes \mathbf{Z}_2$ when $m \to \infty$.
EDIT This question is still surprising me. I thought this fixed point would not belong to $P$, but it does. In fact $1-x_1+x_2 -\cdots +x_{n-1} \in P$ is a fixed point of $f$ for any odd $n \geq 3$. This is because of the following identity in $P$
\begin{equation*}
\prod_{\sigma \in \Sigma(n)} 1-x_{\sigma(1)}+x_{\sigma(2)}- \cdots +x_{\sigma(n-1)} = 1-S_1+S_2-\cdots+S_{n-1}
\end{equation*}
which is true because it's true for any choice of $x_1,\ldots,x_n \in \{0,1\} \subset \mathbf{Z}$ (both sides are $0$ except when $x_1=\cdots=x_n=0$).
To sum up, for $n \geq 2$ the only fixed points of $f$ are $0$, $1$, $x_n$, together with $1-x_1+x_2-\ldots+x_{n-1}$ for odd $n$.
By the same method, one might determine the periodic orbits of $f$. Since $f$ is contracting on each ball, each $f$-orbit reduces mod $2$ to a $f_2$-orbit with the same period. The only non-trivial orbit of $f_2$ is $\{1+x_1,1+x_1+\cdots+x_n\}$. Again, this non-trivial orbit lifts to a unique orbit in $P \otimes \mathbf{Z}_2$, but I don't know whether this orbit is in $P$. I guess the next step would be to determine all preperiodic points of $f$.
This method also suggests to study the analogous dynamical system on $P = \mathbf{Z}[x_1,\ldots,x_n]/(x_1^p-x_1,\ldots,x_n^p-x_n)$ for any prime $p$.<|endoftext|>
TITLE: what is the maximum number of rational points of a curve of genus 2 over the rationals
QUESTION [19 upvotes]: Conjecturally, there exists an integer $n$ such that the  number of rational points of a genus $2$ curve over $\mathbf{Q}$ is at most $n$. (This follows from the Bombieri-Lang conjecture.)
We are very far from proving the existence of such an integer, let alone find an explicit value which works.
My question is:
What is the best known lower bound for $n$?
One way to obtain a lower bound $m$ for $n$ is to prove the existence of a curve of genus $2$ over $\mathbf{Q}$ with at least $m$ rational points.

REPLY [11 votes]: Here is some more information.

The curve that establishes the current record is obtained from a K3 
surface $S$ that was found by Noam Elkies. $S$ is a double cover of 
${\mathbb P}^2$ ramified above a smooth sextic $B$ that has lots of 
tritangent lines and also higher degree curves meeting it with even 
intersection multiplicity in all points.
Pulling back any rational line $L$ in ${\mathbb P}^2$ to $S$ that is not 
tangent to $B$ gives a genus 2 curve $C$ on which all the tritangents induce 
pairs of rational points, and the same is true for the higher degree
curves when they intersect $L$ in rational points. Of course, there can
be additional rational points on $C$ that do not arise in this way.
I found the record curve by a systematic search through rational lines
of relatively small height. The previous record was 588 points, due to
Keller and Kulesz. Their curve is of a special form and has 12 automorphisms
defined over $\mathbb Q$; the 588 points come in 49 orbits. By contrast,
the new record curve has minimal automorphism group (only the hyperelliptic
involution).
The result of Caporaso, Harris and Mazur needs the weak Lang conjecture
for varieties of arbitrarily large dimension. (I have seen Bombieri protest
against the name `Bombieri-Lang Conjecture', saying that he only ever
made the conjecture for surfaces.) We know essentially nothing in this
direction beyond what is covered by Faltings' result on subvarieties of
abelian varieties (or can be deduced from that).
What is perhaps more convincing is the conjecture that there should be
a bound in terms of the genus $g$ and the rank $r$ for the number of 
(geometric) points
on a genus $g$ curve $C$ mapping into a rank $r$ subgroup of its Jacobian
(under some embedding given by a base-point on $C$). This follows from
the Zilber-Pink conjecture for families of abelian varieties. This of
course implies a bound for the number of rational points on a genus 2 curve
in terms of the Mordell-Weil rank of its Jacobian. If these ranks are
bounded, then one would expect a bound on the number of rational points.
But this is another open question. Perhaps the data in
this paper might be interesting.
Such bounds do exist (and are even explicit) when the rank $r$ is sufficiently small compared to $g$; concretely for $r \le g-3$ (which 
unfortunately does not tell us anything about the case $g = 2$). See this and this paper. For hyperellptic curves, the bound can
be taken to be $33(g-1) + 8rg \pm 1$  ($+$ when $r = 0$, $-$ otherwise).<|endoftext|>
TITLE: What is known about first cohomology of the units in a number field?
QUESTION [12 upvotes]: Let $K/Q$ be a finite Galois extension with Galois group $G$. Let $U\subset K^\times$ be the group of units. I am interested in any available information about $H^1(G,U)$.
Motivation: in the theory of fusion categories one is interested in "d-numbers": an algebraic number
$\alpha$ is a d-number if for any Galois conjugate $\beta$ of $\alpha$ the ratio $\frac{\alpha}{\beta}$ is a unit. Let us look at d-numbers contained in the number field $K$. It is clear that d-numbers form a group under multiplication; this group contains two obvious subgroups: units and rational numbers. An exact sequence $1\to U\to K^\times\to K^\times/U \to 1$ and Hilbert theorem 90 show that the quotient of d-numbers in $K$ by the units and rational numbers is precisely $H^1(G,U)$.
In the theory of fusion categories one is mainly concerned with the case when $K/Q$ is abelian and
totally real. Using the properties of Herbrand quotient one shows that if $K/Q$ is cyclic (and real) of degree $n$ then the order of $H^1(G,U)$ is $n$ if $K$ contains a unit of norm $-1$ (this is always the case if $n$ is odd) and $2n$ otherwise. I suspect that group $H^1(G,U)$ is cyclic or direct sum of two cyclics in these cases but I don't see how to prove this. I don't know how to extend this computation to more general extensions (say, to biquadratic).
Finally, the computation of norm of a d-number gives a map from $H^1(G,U)$ to positive rationals modulo $|G|-$th powers. What can be said about image of this map? This seems to be nontrivial even for quadratic fields not containing a unit of negative norm.

REPLY [9 votes]: $\newcommand\Q{\mathbf{Q}}
\newcommand\OL{\mathcal{O}}
\newcommand\I{\mathcal{I}}
\newcommand\Z{\mathbf{Z}}
\newcommand\eps{\epsilon}
\newcommand\p{\mathfrak{p}}
\newcommand\PP{\mathfrak{P}}
\newcommand\Hom{\mathrm{Hom}}
\newcommand\R{\mathbf{R}}
\newcommand\q{\mathfrak{q}}
\newcommand\Gal{\mathrm{Gal}}$
 Summary: I think that it's difficult problem and that there won't be a "nice" answer in general.
Let $\I_K$ denote the group of invertible fractional ideals.
There is a tautological exact sequence
$$1 \rightarrow K^{\times}/\OL^{\times} \rightarrow \I_K
\rightarrow C_K \rightarrow 0.$$
 Taking cohomology gives the following sequence:
$$0 \rightarrow (K^{\times}/\OL^{\times})^{G}
\rightarrow \I^{G}_K
\rightarrow C^{G}_K \rightarrow H^1(G,K^{\times}/\OL^{\times}),$$
which can naturally be modified to yeild the sequence:
$$0 \rightarrow (K^{\times}/\OL^{\times})^{G}/\Q^{ > 0}
\rightarrow \I^{G}_K/\Q^{ > 0}
\rightarrow C^{G}_K \rightarrow H^1(G,K^{\times}/\OL^{\times}).$$
You are interested in the order of $ (K^{\times}/\OL^{\times})^{G}/\Q^{ > 0}$.
As Will Sawin noted, the second term is simply  $\prod \Z/e_p$. Thus one wants
to understand  the classes $I \in C^G_K$ (that is,  the so-called   ambiguous 
classes)
which are actually  strongly ambiguous , that is, $\sigma I = I$ as  ideals , not
just ideal classes. If one defines $S_K \subset C^G_K \subset C_K$ to denote the subset of strongly
ambiguous classes, then one "formally" has an answer to your question, namely,
$$\frac{1}{|S_K|} \cdot \prod e_p.$$
OTOH, this is really a proof by definition,  so much content so far, although it gives a "name" to some of the objects to connect you with the literature.
The strongly ambiguous classes can also be described (given the exact sequence above)
as the kernel of the  map
$$C^G_K \rightarrow H^1(G,K^{\times}/\OL^{\times})
\hookrightarrow H^2(G,\OL^{\times})$$
What makes things  much   easier when $K$ is cyclic is that one can essentially determine
$C^G_K$ (by the ambiguous class number formula, which only exists for cyclic extensions), and also $S_K$. In fact, for a cyclic extension 
$S_K$ and $C^G_K$  are (almost) the same group.
Let $C^{+}_K$ denote the group of invertible fractional ideals $I_K$ modulo the following relation:
$[I] \sim [J]$ if and only if $I = (\alpha) J$ for some $\alpha$ with $N(\alpha) > 0$. In particular, $C^{+}_K$
is a quotient of the narrow class group, and surjects onto the class group. For example, $C^{+}_K =  C_K$
if there exist units of norm $-1$. Suppose that $K/\Q$ is cyclic. I claim that the image of  $(C^{+}_K)^G$ 
in $C^{G}_K$ lands in $S_K$. By assumption, $[\sigma I] \sim [I]$  in $C^{+}_K$, so $\sigma I = (\alpha) I$ for some $\alpha$ with positive norm.
Clearly $N(\alpha)$ is a unit, so $N(\alpha) = 1$. Since $K/\Q$ is cyclic, by Hilbert 90 there exists a $\beta \in K^{\times}$
such that $\alpha = \beta/\sigma \beta$. (This version of Hilbert 90 only makes sense for cyclic extensions, which is one
of the difficulties in the general case.) Replacing $I$ by $J = (\beta) I$ we deduce that $\sigma J = J$ and $[I] = [J]$ in
$C_K$. When $K/\Q$ is not cyclic, however, then it's much trickier to get a handle on the
ambiguous classes --- I don't think that there will be a nice answer in general.
The simplest possible non-cyclic case is the case of biquadratic extensions. Suppose that $K$ is totally real.
Then $K$ contains three subfields $K_1$, $K_2$, and $K_3$. The unit group of $K$ is, up to finite index,
generated by $U:=\{\eps_1, \eps_2, \eps_3\}$. Kubota 
(Uber den bizyklischen biquadratischen Zahlkörper) classified the possible $\Gal(K/\Q)$-module
structure of $\OL^{\times}_K$ (there are $8$ or so different types, of indices ranging from $2^0$ to $2^3$). One could simply compute the corresponding
cohomology groups of each type, and see what one gets.
It's not clear that the answer will be any more precise than a list of cases.
(Even in the case of real quadratic fields the answer depends on the existence of unit of norm
$-1$, which is itself a notoriously fickle condition.)
Yet another way to explain why the cyclic case is not typical is that, since (at least for $p$ odd) the units tensor $\Z_p$ are annihilated by the norm map, they form a module under $$\Z_p[x]/(x^n-1,1+x+x^2+ \ldots + x^{n-1}) = \Z_p[\zeta_n],$$
 which essentially a direct sum of PIDs.
I guess it depends on exactly what you are interested in doing, but it might be useful 
to consider the following approach. Assuming $K/\Q$ is abelian, one can easily compute
the genus class field $L/K$.  The field $L$  is the largest field such that $L/\Q$ is totally real, abelian, and $L/K$ is unramified everwhere.
The point of working with $L$ is that one knows that 
the strongly ambiguous classes in $\OL_K$ (and in $\OL_L$) become (are) principal in $\OL_L$, so the transgression map
is injective, and
$$(L^{\times}/\OL^{\times}_{L})^{G_L}/\Q^{ > 0} \simeq
\I^{G}_L/\Q^{ > 0} \simeq \prod \Z/e_p.$$
(The $e_p$ are the same for $L$ and $K$ since $L/K$ is everywhere unramified.)
If $L = \Q(\zeta_N)$, then one has explicit generators for this group, namely,
$1 - \zeta$ for each $p^n$th root of unity for $p^n \| N$. 
Note that, in some sense, this gives a "complete" description of the $d$-integers in $\Q^{\mathrm{ab}}$:
The $d$-numbers in $K$ are
the $G_K$-invariants of the finitely generated group
$$\Z[\zeta_N]^{\times} \times \prod_{p^n \| N} (1 - \zeta)^{\Z}.$$
 Concerning the image of the norm map, I think you are again out of luck.
To explain why, consider what is close to  the simplest possible non-trivial  example, namely,
 $K = \Q(\sqrt{6p})$, where
$p \equiv 1 \mod 4$ is prime. Write $(2) = \p^2_2$, $(3) = \p^2_3$, and
$(p) = \p^2_p$ respectively.
The field $K$ does not contain a unit of norm $-1$, so we know that the
$d$-numbers generate the group $(\Z/2\Z)^2$.  Since $\p_2 \p_3 \p_p$ is principal, it follows
that  exactly  one of $\p_2$, 
$\p_3$, and $\p_p$ will be principal.
The genus class field of $K$ is 
$$F  =  \Q(\sqrt{6},\sqrt{p}).$$
It follows that if $C_K$ is the class group of $K$, then
$C_K/2C_K$ is cyclic, and hence the $2$-part of the class group is
cyclic. Via the Artin map, we can identify the primes ideals which lie in $C_K[2]$ as exactly the primes which
split completely in the genus field. Since the genus field is given explicitly, we may compute that

$\p_2$ splits  in $F/K$ if and only if $p \equiv 1,17 \mod 24$,
$\p_3$ splits in $F/K$ if and only if $p \equiv 1,13 \mod 24$,
$\p_p$ splits in $F/K$ if and only if $p \equiv 1,5  \mod 24$.

In particular, if $p$ lies  outside  one of these equivalence classes, then
the image of the corresponding $\p$ is non-trivial in $C_K/2 C_K$, and hence $\p$ is not principal.
It follows that:

If $p \equiv 17 \mod 24$, then the $d$-numbers are generated by $2$ and $6p$,
If $p \equiv 13 \mod 24$, then the $d$-numbers are generated by $3$ and $6p$,
If $p \equiv 5 \mod 24$, then the $d$-numbers are generated by $6$ and $6p$,

This leaves open the case when 
$q \equiv 1 \mod 24$. The computation above merely shows that
$\p_2$, $\p_3$, and $\p_p$ all become trivial in $C_K/2C_K$. Since they cannot all be principal,
it follows that when $p \equiv 1 \mod 4$, there must be a surjection
$C_K \rightarrow \Z/4\Z$. By class field theory, this corresponds to the existence of an unramified extension
$E/K$ with $\Gal(E/K) = \Z/4\Z$. We may construct $E$ explicitly as follows.
Since $\Z[\sqrt{6}]$ has class number one, and $p \equiv 1 \mod 24$ splits this field, there exists
a $\pi \in \Z[\sqrt{6}]$ with $N(\pi) = p$. (For local reasons the sign will be positive.)
The choice of $\pi$ will be unique up to a sign and the fundamental unit $\eps = 5 + 2 \sqrt{6}$.  If we explicitly write
$\pi = A + B \sqrt{6}$, then we have
$$A^2 - 6 B^2 =  p.$$
The congruence on $p$ forces $B$ to be even and $A$ to be odd. After possibly multiplying by a unit and by $-1$, we 
may assume that $A \equiv 1 \mod 4$ and $B \equiv 0 \mod 4$. This determines $\pi$ up to squares, and $E$
is identified with the Galois closure of $\Q(\sqrt{\pi}) = \Q(\sqrt{A + B \sqrt{6}})$.
In particular, $E$ is the splitting field of
$$X^4 - 2 A X^2 + p.$$
A necessary condition (and sufficient if the $2$-part of the class group has order $4$) for $\p_p$ to be prime is
that the residue degree of $p$ in $E$ is $1$, or equivalently that $(2 A/p) = 1$.
Since $p \equiv 1 \mod 24$, this is the same as saying that $(A/p) = 1$. In fact, (proof omitted, because this
answer is already too long and the argument is a somewhat tedious calculation of ring class fields  and Kummer extensions)
this is equvalent to $(6/p)_4 = 1$. So this leads to the following criterion:

If $p \equiv 1 \mod 24$, and the quartic residue $(6/p)_4 = -1$, then  there are no $d$-numbers of norm $p$. 
If
$p \equiv 1 \mod 24$, and $8 \nmid h_K$, then there do exist $d$-numbers of norm $p$.

One could go on, giving similar criteria for $\p_2$ and $\p_3$, but it will just get worse (at least for $4  \| h_K$
one can give some sort of classical criteria due to the existence of governing fields, probably for
$8 \| h_K$ and certainly for $16 \| h_K$ there won't be any non-tautological criterion.)
All in all, the "best" cases are when $K$ is its own genus field, or at least, for all $p$ dividing $[K:\Q]$, the $p$-class field of $K$ is the $p$-genus field of $K$.
If this doesn't help, perhaps you could say more precisely you want to prove about dimensions (or otherwise) of fusion categories?<|endoftext|>
TITLE: Moishezon manifolds with vanishing first Chern class
QUESTION [17 upvotes]: Suppose $M$ is a Moishezon manifold with $c_1(M)=0$ in $H^2(M,\mathbb{R})$. Does it follow that $K_M$ is torsion in $\mathrm{Pic}(M)$?
This is true whenever $M$ is Kähler (and therefore projective) and was proved independently by Bogomolov, Fujiki and Lieberman. It is also a well-known consequence of Yau's solution of the Calabi Conjecture. Also when $\mathrm{dim}M=2$ then $M$ is automatically projective, so the question is really about dimensions $3$ or more.
The only examples that I know of non-Kähler Moishezon manifolds with $c_1(M)=0$ are obtained by applying a Mukai flop to a projective hyperkähler manifold, and so they have holomorphically trivial canonical bundle. They are described here.
Are there other simple examples of such manifolds?
The same question can also be asked for compact complex manifolds bimeromorphic to Kähler (i.e. in Fujiki's class $\mathcal{C}$).

REPLY [2 votes]: This problem is solved affirmatively in Theorem 1.5 in this paper. 
The idea is that after some blowups we obtain a compact Kähler manifold whose canonical bundle is effective after twisting by a numerically trivial line bundle. A seminal result of Simpson, recently extended to the Kähler case by Wang, then shows that the manifold has nonnegative Kodaira dimension (this is related to the proof of the abundance conjecture when the numerical dimension of the canonical bundle is zero).
Once we know that there is a nontrivial pluricanonical section, it is easy to conclude that this section must be never vanishing.<|endoftext|>
TITLE: cone angle at infinity for product of cones
QUESTION [5 upvotes]: Let $A=\lim_{r \rightarrow +\infty} \frac{Vol(B(o,r))}{\omega_{n} r^{n}}$ for any Riemannian manifold $(\mathbb{M}^{n},g)$ with nonnegative Ricci curvature. Here $\omega_{n}$ is the volume of unit ball in $\mathbb{R}^n$. We call $A$ the cone angle at infinity or asymptotic volume ratio, and the manifold is cone-like if $A>0$. I got a direct question with this definition. Let $(\mathbb{M}^{n},g)$ be a metric product of two manifolds $(\mathbb{M}_1^{k},g_1,A_1)$ and $(\mathbb{M}_2^{n-k},g,A_2)$ both with nonnegative curvature and cone-like. Does the cone angle of the product manifold only depends on $k$, $n$, $A_1$ and $A_2$. It sounds like the new cone angle will be some average of $A_1$ and $A_2$, but I failed to get a clean formula even assuming product of two rotationally symmetric cones in $R^3$ (like surface of evolution). Is there something simple I miss? Thanks!

REPLY [5 votes]: I claim that $A=A_1A_2$.
Indeed, the distance in the product is given by the standard Pythagorean formula, therefore
$$
 B(o,r) = \bigcup_{x\in B_1(o_1,r)} B_2(o_2,\sqrt{r^2-|o_1x|^2})
$$
where $B$, $B_1$ and $B_2$ denote metric balls in $M$, $M_1$ and $M_2$, $o=(o_1,o_2)$ and $|o_1x|$ denotes the Riemannian distance from $o_1$ to $x$. Hence by Fubini
$$
 Vol(B(o,r)) = \int_{B_1(o_1,r)} Vol_2(B_2(o_2,\sqrt{r^2-|o_1x|^2})) \  dVol_1(x)
$$
where $Vol_1$ and $Vol_2$ are the volume forms in $M_1$ and $M_2$.
Using the co-area formula, this can be rewritten as
$$
 Vol(B(o,r)) = \int_0^r dt \int_{S(t)} Vol_2(B_2(o_2,\sqrt{r^2-t^2})) \  dVol^{k-1}_1(x) .
$$
where $S(t)$ denotes the sphere of radius $t$ centered at $o_1$ in $M_1$ and $Vol_1^{n-1}$ its $(k-1)$-dimensional volume form. For most values of $t$, the $Vol_2(...)$ under the integral equals $A_2$ times the similar Euclidean volume, with a small relative error. Hence
$$
 Vol(B(o,r)) \sim A_2 \int_0^r dt \int_{S(t)} \omega_{n-k}(r^2-t^2)^{(n-k)/2} \  dVol^{k-1}_1(x) 
$$
(you need Bishop-Gromov inequality here, to estimate the volumes that are not among those "most"). The right-hand side equals to
$$
A_2 \int_0^r  \omega_{n-k}(r^2-t^2)^{(n-k)/2} \  Vol^{k-1}_1(S(t)) \ dt .
$$
The growth of volumes of spheres in $M_1$ is the same as that of balls, i.e.
$$
  Vol^{k-1}_1(S(t)) \sim k\omega_k t^{k-1} .
$$
Therefore
$$
Vol(B(o,r)) \sim A_1A_2 \int_0^r \omega_{n-k}(r^2-t^2)^{(n-k)/2} k\omega_k t^{k-1} \  dt .
$$
By the same argument applied to $\mathbb R^n$, the latter integral equals the volume of the Euclidean $r$-ball, hence the result.<|endoftext|>
TITLE: Dual of a Specht module
QUESTION [7 upvotes]: For a partition $\mu$ of $n$, let $S^{\mu}$ be the associated Specht module, defined over $\mathbb{Z}$.  For any field $k$, we can tensor $S^{\mu}$ with $k$ to get a representation $S^{\mu}_k$ of the symmetric group $S_n$ over $k$.  Finally, let $A_n \cong \mathbb{Z}$ be the sign representation of $S_n$ over $\mathbb{Z}$ and let $A_{n,k} \cong k$ be the sign representation over $k$.
In James's book on the representation theory of the symmetric group, he proves that $S^{\mu}_k \otimes A_{n,k} \cong (S^{\mu'}_k)^{\ast}$, where $\mu'$ is the conjugate partition and the $\ast$ indicates that we are taking the dual.  My question is whether this is true over $\mathbb{Z}$ in the sense that there is a nondegenerate bilinear pairing
$$\omega : S^{\mu} \times S^{\mu'} \rightarrow \mathbb{Z},$$
and assuming this is true, how can we calculate $\omega(e_t,e_s)$, where $t$ (resp. $s$) is a tableau of shape $\mu$ (resp. $\mu'$) and $e_t$ (resp. $e_s$) is the associated polytabloid.
If the above is false, I'm still interested in explicit pairings
$$S^{\mu}_k \times S^{\mu'}_k \rightarrow k.$$
I've had trouble extracting them from James's book.

REPLY [11 votes]: Yes, this works over $\mathbb Z$, and the pairing can be explicitly realised with polytabloids. See Section 4 of my paper "On the structure of Specht modules", J. London Math. Soc. 67 (2003) 85–102. (In retrospect, I wish I'd given this paper a more helpful title.)
Briefly: start by defining a map
$[,]: M^\mu\times M^{\mu'}\longrightarrow\mathbb Z$
as follows.  Fix a $\mu$-tableau $r$. Now given a $\mu$-tabloid $S$ and a $\mu'$-tabloid $T$, we want to define $[S,T]$. If there are two numbers in the same row of $S$ and in the same row of $T$, then set $[S,T]=0$. If not, then there is a unique $\mu$-tableau $v$ such that $v$ is in $S$ and $v'$ (the transpose of $v$) is in $T$. Set $[S,T]$ to equal the sign of the permutation that takes $r$ to $v$.
Now for the map
$S^\mu\times S^{\mu'}\longrightarrow\mathbb Z$
you just take the above map $[,]$, restrict to $S^\mu\times S^{\mu'}$, and divide by the product of the hook lengths in $\mu$.<|endoftext|>
TITLE: Equivalence between statements of Hodge conjecture
QUESTION [10 upvotes]: Dear everyone, 
I was unable to obtain the equivalence between the two statements of the Hodge conjecture. I searched for some previous questions that others asked here, to check whether someone has asked the same thing previously. But I didn't find any such instance, that is why I am asking. 
We know that Hodge conjecture gives some relation between the topological cycles and algebraic cycles. But I have read two different variations of the same conjecuture. I number my pointers. 

A fantastic description given by Prof.Dan Freed (here), which an undergraduate student can also understand. 
A bit tough description given by Prof.Pierre Deligne (here), with lot of technical terms and constructions. 

So I was befuddled in asking myself that how can one obtain equivalence between the both statements. 
Dan Freed's Version : 

He considers a Topological cycle ( boundary less chains that are free to deform ) on a projective manifold. Then he says that the topological cycle is homologous to a rational combination of algebraic cycles, if and only if the topological cycle has a rotation number Zero. 

P.Deligne's Version : 

On a projective non-singular algebraic variety over $\mathbb{C}$ , and Hodge Class is a rational combination of classes $\rm{Cl(Z)}$ of algebraic cycles. 

So now I have the following queries for my learned friends.

How can one explain that both the statements are equivalent to each other ?  One speaks about the rotation number and another doesn't even speak about it. How can one say that both the statements are valid ?  I infact know that both the statements are valid ( as both the speakers are seminal mathematicians ) But how ? 
So can anyone explain me what the Rotation number has to do with the Hodge Conjecture ? I obtained some information about the rotation number from Wiki. But I am afraid , to decide whether Freed is speaking about the same rotation number ( given in wiki ) in his talk ? or something different ? 

I would be really honored to hear answers for both of them . Thank you one and all for sparing your time reading my question.

REPLY [13 votes]: The rotation number in question has to do with the behaviour of a differential form $\omega$ on the manifold $X$ under rotation $e^{i\theta}$ of tangent vectors : a complex $k$-form $\omega$ has rotation number $p-q$ iff $$\omega(e^{i\theta}v_1,\dots,e^{i\theta}v_k)=e^{i(p-q)\theta}\omega(v_1,\dots,v_k)$$ (with $p+q=k$, of course). 
This should explain the terminology, although it is not much in use today. Such a form is called a $(p,q)$-form, their space is denoted  $\Omega^{p,q}(X)$, and the subspace of $H^{p+q}(X,\mathbb{C})$ (de Rham cohomology) of classes having a (closed) $(p,q)$-form representative is denoted $H^{p,q}(X)$. 
Hodge theory then gives a decomposition $H^k(X,\mathbb{C})=\bigoplus_{p+q=k} H^{p,q}(X)$. 
Now a topological cycle $C$ (or its class) of (real) codimension $2p$ has a Poincaré dual de Rham cohomology class $[\omega]$, for a closed $2p$ form $\omega$, such that intersection of cycles of dimension $2p$ with $C$ coincides with integration of $\omega$.
"The rotation number of $C$ is zero" means that one can choose $\omega$ of type $(p,p)$.
On the other hand, a cohomology class in $H^{2p}(X,\mathbb{Q})$ is said to be Hodge if under the Hodge decomposition
$$H^{2p}(X,\mathbb{Q})\otimes\mathbb{C}\simeq \bigoplus_{i=-p}^{p} H^{p-i,p+i}(X)$$
it belongs to $H^{p,p}(X)$.
I hope this clarifies the equivalence of the two statements you found.<|endoftext|>
TITLE: Isotopy in 3-manifolds
QUESTION [8 upvotes]: Assume $\Sigma_1$ and $\Sigma_2$ are two embedded compact surfaces 
(say orientable) in an orientable 3-manifold $M$. Assume $\Sigma_1$ and $\Sigma_2$ 
are homotopic in $M$. Then are they isotopic?

REPLY [9 votes]: If $\Sigma_1 \hookrightarrow M$ is an embedded $\pi_1$-injective surface, then any homotopic embedded surface will be isotopic to $\Sigma$. As Ryan and Allen point out, this is due to Waldhausen for incompressible surfaces of genus $>0$, and to Laudenbach for 2-spheres, together with the Poincare conjecture. 
If $\Sigma_1$ is not incompressible in $M$, then there exists $\Sigma_2$ which is homotopic to $\Sigma_1$ but not isotopic. The point is that one may compress $\Sigma_1$ to get a surface $\Sigma'\hookrightarrow M$ which has smaller genus, and then reembed the 1-handle (in the same homotopy class) in a knotted fashion to get a non-isotopic surface $\Sigma_2$. Misha observed this in the comments on Ryans question for tori in $S^3$, but it holds more generally. 
There's an intermediate case of $\Sigma_1\hookrightarrow M$ which is incompressible and not $\pi_1$-injective. By the loop theorem, this can only occur if $\Sigma_1$ is 1-sided in $M$, which implies that the surface is non-orientable, so does not fall under the purview of your question.  I'm not sure if homotopy implies isotopy in this case - I suspect there are 1-sided Heegaard surfaces which are homotopic but not isotopic, but I don't know examples off the top.<|endoftext|>
TITLE: The direct sum of injective modules need not be injective
QUESTION [12 upvotes]: The Bass-Papp Theorem asserts that a commutative ring $R$ is Noetherian iff every direct sum of injective $R$-modules is injective. Thus every non-Noetherian ring carries a counterexample. 
If
$$
I_1 ⊊I_2⊊\cdots⊊I_n⊊\cdots
$$
is an infinite properly ascending chain of ideals of $R$, then for all $n$ let $E_n=E(R/I_n)$ be an injective envelope of $R/I_n$, and let $E=\bigoplus_{n\ge 1} E_n$. Then $E$ is a direct sum of injective modules and $E$ is not itself injective.
How to prove that $E$ itself is not injective?

REPLY [17 votes]: The standard proof that I am aware of is actually explicit in this regard.
As you note, assume that $I_1\subsetneq I_2\subsetneq\cdots$ is an infinite ascending chain of ideals of $R$, let $E(R/I_n)$ be the injective envelope of $R/I_n$ for each $n$, and let $$E=\bigoplus_{n=1}^{\infty}E(R/I_n)$$
be the their direct sum.
Let $I=\bigcup\limits_{i=1}^{\infty} I_n$. 
For each $n$, let $f_n$ be the composition of the embedding $I\hookrightarrow R$ with the canonical map $R\to E(R/I_n)$ (map to the quotient, then embed into the envelope). 
Since we have a map from $I$ to each $E(R/I_n)$, we obtain a map $f\colon I\to \mathop{\prod}\limits_{n=1}^{\infty}E(R/I_n)$ by the universal property of the product. In fact, the image of $f$ lies in the direct sum, since for every $x\in I$ there exists $n\in\mathbb{N}$ such that $x\in I_m$ for all $m\geq n$, hence the image of $x$ is $0$ in $R/I_n$.  So we have a map $I\to E$. 
I claim that $f$ does not extend to a module homomorphism $\overline{f}\colon R\to E$ (which it would necessarily do if $E$ were injective). Assume to the contrary that we have an extension $\overline{f}\colon R\to E$. Being a module homomorphism with domain the free module of rank $1$, it is completely determined by $\overline{f}(1)$, and so it has the form $\overline{f}(x) = xe$ for all $x\in R$, where $e=\overline{f}(1)\in E$. 
Now, let $n_0$ be a positive integer such that the $m$th component of $e$ is $0$ for all $m\geq n_0$. Let $x\in I_{n_0}\setminus I_{n_0-1}$. When we map $x$ to $R/I_{n_0}$, we obtain a nonzero element; hence the $n_0$th component of $f(x)$ must be nonzero (since $R/I_{n_0})$ embeds into $E(R/I_{n_0})$). But $f(x) = \overline{f}(x) = xe$, and the $n_0$th component of $e$ is $0$, hence so is that of $f(x)$, a contradiction.
The contradiction arises from the assumption that  we can extend the map $f\colon I\to E$ to a module homomorphism $R\to E$. Hence no such extension exists, so $E$ is not injective.<|endoftext|>
TITLE: Number of distinct values taken by $\alpha$ ^ $\alpha$ ^ $\dots$ ^ $\alpha$ with parentheses inserted in all possible ways, $\alpha\in\mathbf{Ord}$
QUESTION [16 upvotes]: Let $\alpha\in\mathbf{Ord}$ and $n\in\mathbb{N}^+$.
Let $F_\alpha(n)$ be the number of distinct values taken by ordinal exponentiation $\underbrace{\alpha \hat{\phantom{\hat{}}} \alpha \hat{\phantom{\hat{}}} \dots \hat{\phantom{\hat{}}} \alpha}_\text{n times}$ with parentheses inserted in all possible ways. 
The case $\alpha<\omega$ (i.e. natural numbers) was studied in "The Nesting and Roosting Habits of The Laddered Parenthesis", by R. K. Guy and J. L. Selfridge. Here I am interested in a case when $\alpha\ge\omega$.
Questions: Does the sequence $F_\alpha(n)$ depend on $\alpha$ in this case? Is there a simple closed form formula, recurrence relation or generating function for this sequence? What can be said about its asymptotic growth? What is $\lim\limits_{n\to\infty}\frac{F_\alpha(n+1)}{F_\alpha(n)}$? Do odd and even numbers occur infinitely often in the sequence?

$F_\omega(n)$ is given in A199812: $1, 1, 2, 5, 13, 32, 79, 193, 478, 1196, 3037, 7802, 20287, 53259, 141069, 376449, 1011295, \dots$
It is bounded by Catalan numbers $\frac{(2n-2)!}{n! (n-1)!}$. The first difference occurs at $n=5$ because among $14$ possible parenthesizations there are only $13$ distinct ordinals:
$(((\omega^\omega)^\omega)^\omega)^\omega < ((\omega^\omega)^\omega)^{\omega^\omega} < ((\omega^\omega)^{\omega^\omega})^\omega < ((\omega^{\omega^\omega})^\omega)^\omega < (\omega^{\omega^\omega})^{\omega^\omega} < (\omega^\omega)^{(\omega^\omega)^\omega} < $
$(\omega^{(\omega^\omega)^\omega})^\omega < \omega^{((\omega^\omega)^\omega)^\omega} < \pmb{(\omega^\omega)^{\omega^{\omega^\omega}} = \omega^{(\omega^\omega)^{\omega^\omega}}} < (\omega^{\omega^{\omega^\omega}})^\omega < \omega^{(\omega^{\omega^\omega})^\omega} < \omega^{\omega^{(\omega^\omega)^\omega}} < \omega^{\omega^{\omega^{\omega^\omega}}}$

REPLY [10 votes]: The sequence $F_{\alpha}(n)$ indeed does not depend on $\alpha$ when $\alpha \geq \omega$. This question is closely related to this other question, and I will refer to my answer in that question.
Let $S_\alpha(n)$ be the set of ordinals defined by $\alpha ^ {\alpha ^ {\cdots ^ \alpha}}$ with all possible arrangements of parentheses.  Since $\alpha = \alpha^{\alpha^0}$ and $(\alpha^{\alpha ^ \beta})^{\alpha^{\alpha^\gamma}} = \alpha^{\alpha ^ {\beta + \alpha^\gamma}}$, every element of $S_\alpha(n)$ is of the form $\alpha^{\alpha ^ \beta}$.  So we define $T_\alpha (n) = \lbrace \beta | \alpha^{\alpha^\beta} \in S_\alpha(n) \rbrace$.  We observe that $T_\alpha(0) = \lbrace 0 \rbrace$ and $T_\alpha(n) = \lbrace \beta + \alpha^\gamma | \beta \in T_\alpha(i), \gamma \in T_\alpha(j), i + j = n - 1 \rbrace$.
From my answer to the other question, we observe that $T_\alpha(n)$ is contained in the set $E(\alpha)$, which I refer to as the "exponential polynomials over the base $\alpha$".  For each element $\beta$ of $E(\alpha)$, we can write $\beta$ in the iterative normal form to the base $\alpha$, which I describe as follows:
0 is represented as 0
$\beta$ is represented as $\alpha^{\beta_1} + \alpha^{\beta_2} + \ldots + \alpha^{\beta_n}$, where $\beta_1 \geq \beta_2 \geq \ldots \geq \beta_n$ and each $\beta_i$ is itself represented in iterative normal form.
We claim that, for any n and any $\alpha, \beta \geq \omega$, $T_\alpha(n)$, when written in iterative normal form, is the same set as $T_\beta(n)$, except the base $\beta$ is replaced everywhere it appears by $\alpha$.  If we let $f(\gamma)$ be the ordinal obtained by replacing the base $\beta$ by $\alpha$ everywhere in the iterative normal form of $\gamma$, our claim amounts to $T_{\alpha}(n) = f(T_\beta(n))$.  From our above expression for $T_\alpha(n)$, we see that it suffices to show that $f(\gamma + \beta^\delta) = f(\gamma) + \alpha^{f(\delta)}$ for any $\gamma, \delta \in E(\beta)$.
Let $\gamma = \beta^{\gamma_1} + \beta^{\gamma_2} + \ldots + \beta^{\gamma_n}$.  Let $m$ be the largest index such that $\gamma_m \geq \delta$.  Then we claim that
$\gamma + \beta^\delta = \beta^{\gamma_1} + \beta^{\gamma_2} + \ldots + \beta^{\gamma_m} + \beta^\delta$
(note that this is in iterative normal form, provided that $\gamma_i$ and $\delta$ are.)
First we prove:
Lemma.  If $\beta, \mu, \nu$ are any ordinals such that $\mu < \nu$, then $\beta^\mu + \beta^\nu = \beta^\nu$.
Proof:  Let $\beta = \omega^{\beta_1} + \omega^{\beta_2} + \ldots + \omega^{\beta_n}$ be the Cantor Normal Form for $\beta$.  Then the leading term of the Cantor Normal Form of $\beta^\mu$ will be $\omega^{\beta_1 \mu}$, i.e.
$\beta^\mu = \omega^{\xi_1} + \omega^{\xi_2} + \ldots + \omega^{\xi_m}$ where $\xi_i \leq \beta_1 \mu$.
Since $\omega$ is a principal ordinal, for $\mu < \nu, \omega^\mu + \omega^\nu = \omega^\nu$.  Therefore
$\beta^\mu + \beta^\nu = \omega^{\xi_1} + \omega^{\xi_2} + \ldots + \omega^{\xi_m} + \beta^\nu = \beta^\nu$, as desired.  (end of proof of Lemma)
Therefore, since $\gamma_i \leq \delta$ for $m+1 \leq i \leq n$, we have
$\beta^{\gamma_{m+1}} + \ldots + \beta^{\gamma_n} + \beta^\delta = \beta^\delta$
and therefore
$\gamma + \beta^\delta = \beta^{\gamma_1} + \beta^{\gamma_2} + \ldots + \beta^{\gamma_m} + \beta^\delta$.
We have
$f(\gamma) = \alpha^{f(\gamma_1)} + \alpha^{f(\gamma_2)} + \ldots + \alpha^{f(\gamma_n)}$
$f(\gamma + \beta^\delta) = \alpha^{f(\gamma_1)} + \alpha^{f(\gamma_2)} + \ldots + \alpha^{f(\gamma_m)} + \alpha^{f(\delta)}$,
and, provided that $m$ is the largest index such that $f(\gamma_m) \geq f(\delta)$, the same argument as above shows that
$f(\gamma) + \alpha^{f(\delta)} = \alpha^{f(\gamma_1)} + \alpha^{f(\gamma_2)} + \ldots + \alpha^{f(\gamma_n)} + \alpha^{f(\delta)} $ $= \alpha^{f(\gamma_1)} + \alpha^{f(\gamma_2)} + \ldots + \alpha^{f(\gamma_m)} + \alpha^{f(\delta)} = f(\gamma + \beta^\delta)$.
So we are done if we can show that for all $\gamma, \delta \in E(\beta)$, $\gamma < \delta \Leftrightarrow f(\gamma) < f(\delta)$.
This follows from the fact that we can describe the comparison relation < using the same inductive rules, regardless of base.  The inductive rules are as follows:
0 is less than any ordinal besides 0.
If $\gamma = \beta^{\gamma_1} + \beta^{\gamma_2} + \ldots + \beta^{\gamma_m}$, and $\delta = \beta^{\delta_1} + \beta^{\delta_2} + \ldots + \beta^{\delta_n}$, with the $\gamma_i$ and the $\delta_i$ weakly decreasing, than $\gamma < \delta$ if and only if one of the following two conditions hold:

For some $i$, $\gamma_i < \delta_i$ and for all $1 \leq j < i, \gamma_j = \delta_j$.
$m < n$ and for all $1 \leq i \leq m$, $\gamma_i = \delta_i$.

It is easy to see that if the second condition holds, then $\gamma < \delta$.  So assume the  first condition holds.  Since $\beta \geq \omega$ and $\delta_i \geq \gamma_i + 1$,
$\beta^{\delta_i} \geq \beta^{\gamma_i + 1} \geq \beta^{\gamma_i} \omega > \beta^{\gamma_i} m \geq \beta^{\gamma_i} + \beta^{\gamma_{i+1}} + \ldots + \beta^{\gamma_m}$, so
$\beta^{\delta_1} + \beta^{\delta_2} + \ldots + \beta^{\delta_n} \geq \beta{\delta_1} + \beta^{\delta_2} + \ldots + \beta^{\delta_i} = \beta^{\gamma_1} + \beta^{\gamma_2} + \ldots + \beta^{\gamma_{i-1}} + \beta^{\delta_i} > \beta^{\gamma_1} + \beta^{\gamma_2} + \ldots + \beta^{\gamma_{m}}$.
Going the other way, observe that, if neither conditions 1 nor 2 hold for $\gamma < \delta$, nor conditions 1 nor 2 hold for $\delta < \gamma$, we must have $m = n$ and $\gamma_i = \delta_i$ for $1 \leq i \leq m$, so $\gamma = \delta$.  So if $\gamma < \delta$, either conditions 1 or 2 must hold.
We have shown that $T_\alpha(n)$ and $T_\beta(n)$ are the same ordinals except for a change of base from $\alpha$ to $\beta$.  It follows that $S_\alpha(n)$ and $S_\beta(n)$ are the same except for the same change of base.  So $F_\alpha(n) = F_\beta(n)$ for any $\alpha, \beta \geq \omega$.

As for the asymptotics of $F_\omega(n)$, all I can prove is that $F_\omega(n) \geq 2^{n-2}$.  Take any binary string of n-2 symbols.  Starting from $\omega^\omega$, apply the function $\alpha \rightarrow \alpha^\omega$ if the $i$th symbol is a 0, and apply the function $\alpha \rightarrow \omega^\alpha$ if the $i$th symbol is a 1.  Defining $\beta$ by $\alpha = \omega^{\omega^\beta}$, the two operations affect $\beta$ by $\beta \rightarrow \beta + 1$ and $\beta \rightarrow \omega^\beta$ respectively.  It's not hard to see that two distinct binary strings lead to two different final ordinals.
I conjecture that $\lim_{n \rightarrow \infty} \frac{F_\omega(n+1)}{F_\omega(n)} = 4$, but am nowhere close to a proof.<|endoftext|>
TITLE: K-theory of compact Lie groups
QUESTION [7 upvotes]: The complex K-theory of a compact connected Lie group $ G $ is computed by Hodgkin in the case that $ G $ has torsion-free fundamental group. The result is that $ K^*(G) $ is an exterior algebra in the fundamental representations of $ G $. Note that every finite dimensional representation gives a canonical $ K^1 $-class. 
What happens if $ \pi_1(G) $ has torsion? Are there any general results analogous to Hodgkin's theorem? I am specifically interested in the case of the projective unitary group $ PU(n) = U(n)/center $. In the case $ n = 2 $ this amounts to $ SO(3) $, and identification with $ \mathbb{R}P^3 $ shows that $ K^0 $ has a torsion subgroup $ \mathbb{Z}/2\mathbb{Z} $ in this case. I would expect to find a torsion subgroup $ \mathbb{Z}/n\mathbb{Z} $ inside $ K^0 $ in general - is this true?

REPLY [3 votes]: Here are some relevant references. First, three papers on different methods to compute the K-theory of compact Lie groups with finite cyclic fundamental group. (The first one also discusses the projective unitary groups as example.)

R.P. Held and U. Suter. On the unitary K-theory of compact Lie groups with finite fundamental group. Quart. J. Math. Oxford Ser. (2) 24 (1973), 343–356. (link to Quart J. Math)
L. Hodgkin. The equivariant Künneth theorem in K-theory. In: Topics in K-theory. Lecture Notes in Math., Vol. 496. Springer 1975, 1–101.
H. Minami. Note on complex K-groups of compact Lie groups with fundamental group of prime order. Osaka J. Math. 35 (1998), no. 3, 547–551. (link to Osaka J. Math)

There is a further paper dealing specifically with projective unitary groups

T. Petrie. The K theory of the projective unitary groups. Topology 6 (1967), 103–115 (link to Topology)

There are further papers by Minami dealing with the computations for specific projective groups ($PE_6$, $PE_7$, $PSp(n)$, $PO(4l+2)$).<|endoftext|>
TITLE: Why do we distinguish the continuous spectrum and the residual spectrum?
QUESTION [6 upvotes]: As we know, continuous spectrum and residual spectrum are two cases in the spectrum of an operator, which only appear in infinite dimension. 
If $T$ is a operator from Banach space $X$ to $X$, $aI-T$ is injective, and $R(aI-T)$ is not $X$. If $R(aI-T)$ is dense in $X$, then $a$ belongs to the continuous specturm, it not, $a$ belongs to the residual spectrum. 
1) I want to know why do we care about whether $R(aI-T)$ is dense.
2) It seems as we just emphasize this difference for the operator, but not for the Banach algebra, why?

REPLY [4 votes]: For $\lambda$ in the continuous spectrum, while there may not be genuine eigenvectors, there are approximate eigenvectors.
That point is related to the fact that (as Bob Israel mentions) normal operators on Hilbert spaces, which do have a spectral theorem, do not have residual spectrum. It is only slightly frivolous to observe that an operator on Hilbert space with residual spectrum could not admit a reasonable spectral theorem. 
(By "reasonable spectral theorem" we need not require decomposition into actual eigenvectors, but something much weaker, than can be described in various ways.)<|endoftext|>
TITLE: Rain droplets falling on a table
QUESTION [9 upvotes]: Suppose you have a circular table of radius $R$. This table has been left outside, and it begins to rain at a constant rate of one droplet per second. The drops, which can be considered points as they fall, can only land in such a way such that they impact the surface of the table. Once they strike the table, they form a puddle of radius $r$, centered at their point of impact. What is the expected number of droplets it takes to cover the table in water?
The answer should be left in terms of $R$ and $r$. 
I have tried decomposing the problem by considering only the 1-dimensional case with line segments, but even its solution has eluded me. A potential starting point could be the discrete case of marbles falling into buckets.

REPLY [3 votes]: The expected number of droplets it takes is the sum of the probabilities that $n$ droplets don't suffice for $n \ge 0$. For the $1$-dimensional version of the problem, you can determine this probability by inclusion-exclusion. 
Geometrically, the set of $n$ impact locations corresponds to a point in a simplex with $n+1$ vertices by using barycentric coordinates, and the density is uniform. Two of the vertices are special because they correspond to the first and last segments, which have to have length at most $r$ instead of $2r$ for all points to be covered. (You can avoid this complication by covering a circle instead.) The probability that something is left uncovered is the normalized volume of the points within $R-2r$ of at least one regular vertex (by the $L^1$ distance), or within $R-r$ of one of the two special vertices. 
$$P(n~ \text{cover})= \sum_{i=0}^{n-1} \sum_{j=0}^2 (-1)^{i+j}{n-1 \choose i}{2\choose j} \max(0,(1-jr/R-2ir/R)^n)$$
so, the expected number of droplets it takes to cover a line segment is
$$\sum_{n=0}^\infty \bigg(1-\sum_{i=0}^{n-1} \sum_{j=0}^2 (-1)^{i+j}{n-1 \choose i}{2\choose j} \max(0,(1-jr/R-2ir/R)^n)\bigg).$$
For a particular value of $r/R$, a fixed number of terms are nonzero so this series can be simplified. For example, if $r/R = 1/3$, then the expected value is 
$$\sum_{n=0}^\infty \bigg(1 - (1^n - (n-1)(\frac13)^n - 2(\frac23)^n  + (\frac13)^n)\bigg) =\frac{15}{4}.$$<|endoftext|>
TITLE: 6-regular bipartite graphs with no 8-cycles
QUESTION [8 upvotes]: I'm looking for simple 6-regular bipartite graphs with no 8-cycles, as small as possible.  It doesn't matter if there are 4-cycles or 6-cycles, provided there are no 8-cycles.  Such graphs must exist since the girth can be arbitrarily high, but what smaller examples are there?  There are certainly none on less than 46 vertices.

REPLY [2 votes]: Hi Brendan.  I just happened upon this while searching for some related material, and I'm not sure if you're still interested in finding such a graph.  I'm relatively certain I can construct one on 7812 vertices, following a few old ideas used by Lazebnik, Ustimenko and me in a series of papers on extremal graphs (in the sense of Turan).  In fact the 7812-vertex graph I mention has girth 10, and its construction, as well as a proof that the girth is 10, should be easy to follow, and/or present.  Let me know if you're still interested.  I'll be looking back here every so often.  Or just drop me an email. Regards.<|endoftext|>
TITLE: What is known about size-restricted power set axioms?
QUESTION [6 upvotes]: What is known about ZF without powerset but with an axiom "every set
has a set of all its countable subsets"?
This seems stronger than positing that the set of natural numbers has
a powerset, though I do not know a proof that it is.
More generally, for any definable cardinal $\alpha$, what about the
axiom "every set has a set of all smaller-than-$\alpha$ subsets"?

REPLY [7 votes]: Isn't $H(\mathfrak c^+)$, the collection of sets whose transitive closures have cardinality at most $\mathfrak c=2^{\aleph_0}$, a model of your theory (but not the power set axiom)?  The point is that a set of cardinality $\mathfrak c$ has only $\mathfrak c$ countable subsets.
EDIT: Colin actually asked whether this theory is stronger than ZF minus power set plus "the set of natural numbers has a power set".  To get a model where $\omega$ has a power set but some sets don't have a set-of-all-countable-subsets, replace $\mathfrak c^+$ in my original answer with $\kappa^+$ where $\kappa$ is a singular cardinal, greater than or equal to $\mathfrak c$, and of cofinality $\omega$.  So the proposed model consists of all sets whose transitive closures have size at most $\kappa$.  From $\kappa\geq\mathfrak c$, it follows that the model contains the actual power set of the (von Neumann) natural numbers.  From the fact that $\kappa$ has cofinality $\omega$, it follows that a set of size $\kappa$ has strictly more than $\kappa$ countable subsets.  Therefore, some sets in the model, for instance $\kappa$ itself, don't have a set-of-all-countable-subsets in the model.  (Note that each individual countable subset of $\kappa$ is in the model, so nothing other than the genuine set-of-all-countable-subsets of $\kappa$ can serve as the set-of-all-countable-subsets of $\kappa$ in the sense of the model.)
(By the way, I'm aware that, in this situation, we actually have strict inequality $\kappa>\mathfrak c$ because $\mathfrak c$ can't have cofinality $\omega$.  I ignored that fact above, because it's not needed for the argument.)<|endoftext|>
TITLE: When does a homomorphism factor through a free group?
QUESTION [21 upvotes]: Let $f\colon\thinspace G\to H$ be a surjective homomorphism of finitely generated groups. Are there any methods to decide whether $f$ factors through a free group? That is, does there exist a free group $F$ and homomorphisms $g\colon\thinspace G\to F$ and $h\colon\thinspace F\to H$ such that $f=h\circ g$?
I know of one necessary condition, given by cohomology. If $f$ factors through a free group then $f^*\colon\thinspace H^i(H;M)\to H^i(G;M)$ is zero for all $i>1$ and all $H$-modules $M$ (since free groups have cohomological dimension one).
This question was inspired by Tom Goodwillie's answer to my earlier question on cohomological dimension of group homomorphisms.

REPLY [14 votes]: $f: G \to H$ factors through a free group iff there is a subgroup $N \le \operatorname{ker}(f)$ and a free group $F$ such that $G = N \rtimes F\;\;$ ($N$ normal ).
Proof: $(\Rightarrow)$ Let $G \xrightarrow{g} E \xrightarrow{} H$ be a factorization of $f$ with $E$ free and let $N$ be the kernel of $g$. Clearly $N \le \operatorname{ker}(f)$ and as a subgroup of a free group, $F := G/N$ is itself free. Thus we have an extensions $1 \to N \to G \to F \to 1$ that splits since $F$ is free. 
$(\Leftarrow)$ Suppose $N\le \operatorname{ker}(f)$ and $G = N \rtimes F$ with $F$ free. Hence $N$ is normal in $G$ and because $N \le \operatorname{ker}(f)$, $f$ has a factorization $G \to G/N=F \to H$ through a free group. 
Remark: That $f$ is surjective wasn't used. But if we know that $f$ is surjective, we can conclude that the rank of $F$ is greater or equal than the minimal number of generators of $H$.<|endoftext|>
TITLE: Grothendieck Topologies versus Pretopologies
QUESTION [12 upvotes]: The wikipedia article(s) as well as the nlab article(s) about Grothendieck topologies and Grothendieck pretopologies are careful to differentiate the two very emphatically and to point out that distinct pretopologies can give rise to the same topology on a category. My (almost certainly trivial) questions are the following:
1) What are the key differences between Grothendieck topologies and Grothendieck pretopologies, and
2) What are some nice examples of distinct pretopologies that give rise to the same topology?

REPLY [10 votes]: I think of pretopologies (and coverages and other variations of this idea) as convenient ways of presenting a topology, just as groups (and other algebraic structures) can be conveniently presented by generators and relations, or (since I'm a set theorist) a complete Boolean algebra can be conveniently presented by giving a dense subposet (called a notion of forcing).  Usually (though not always), the important thing is the bigger, more abstract object --- the Grothendieck topology, the group, or the complete Boolean algebra --- on which the interesting constructions can be based.  Then the choice of a particular presentation is, in principle, irrelevant, but it can be very useful for working with the object and understanding it, especially because the presentation is often much smaller than the abstract object.<|endoftext|>
TITLE: Proof of a combinatorial identity (possibly using trigonometric identities)
QUESTION [10 upvotes]: For integers $n \geq k \geq 0$, can anyone provide a proof for the following identity?
$$\sum_{j=0}^k\left(\begin{array}{c}2n+1\\\ 2j\end{array}\right)\left(\begin{array}{c}n-j\\\ k-j\end{array}\right) = 2^{2k} \left(\begin{array}{c} n+k\\\ 2k \end{array}\right)$$
I've verified this identity numerically for many values of $n$ and $k$, and suspect it to be true.
I found similar identities in http://www.math.wvu.edu/~gould/Vol.6.PDF, most notably:
$$\sum_{j=0}^k\left(\begin{array}{c}2n\\\ 2j\end{array}\right)\left(\begin{array}{c}n-j\\\ k-j\end{array}\right) = 2^{2k} \frac{n}{n+k}\left(\begin{array}{c} n+k\\\ 2k \end{array}\right)$$
which is Eq. (3.20) in the above link, and
$$\sum_{j=0}^k\left(\begin{array}{c}2n+1\\\ 2j+1\end{array}\right)\left(\begin{array}{c}n-j\\\ k-j\end{array}\right) = 2^{2k} \frac{2n+1}{n-k}\left(\begin{array}{c} n+k\\\ 2k+1 \end{array}\right)$$
which is Eq. (3.34) in the above link.  The derivations of these two identities seem to rely on trigonometric identities, which I've been having trouble reconstructing.

REPLY [3 votes]: In terms of hypergeometric series, the sum is 
$\binom nk {}_2F_1(-n-1/2, -k; 1/2;1)$, so the identity is a case of the Chu-Vandermonde theorem. It may be obtained from the more "usual-looking" case of Vandermonde's theorem 
$$\sum_j \binom{k-1/2}{k-j}\binom {n+1/2}{j} = \binom{n+k}{k}
$$
by multiplying both sides by $\binom nk/\binom{k-1/2}{k}$ and simplifying.
Similarly, the other two identities are also cases of Vandermonde's theorem. The sums are
$\binom nk {}_2F_1(-n+1/2, -k; 1/2;1)$ and $(2n+1)\binom nk {}_2F_1(-n+1/2, -k; 3/2;1)$
Incidentally, there is a fourth identity to go along with these three:
$$\sum_{j=0}^k \binom{2n+2}{2j+1}\binom{n-j}{k-j} = 2^{2k+1}\binom{n+k+1}{2k+1}.
$$
All four are equivalent to cases of Vandermonde's theorem.<|endoftext|>
TITLE: Serre's theorem about regularity and homological dimension
QUESTION [23 upvotes]: One of the nicest results I know of is (Auslander-Buchsbaum-)Serre's theorem asserting that a (commutative!) local ring is regular iff it has finite global dimensional.
I'd like to ask a somewhat vague question: 

what is the history and what was the context of this result? 

By this I mean: presumably the above characterization did not come out of thin air (or just out of Serre's mind!), and there was a buildup of ideas which lead to such an elegant and, I guess, surprising characterization of regularity. Nowadays, such a thing appears almost natural to our minds brought up in the nice set-up constructed by the founders of homological algebra and tended to by a few generations already, but I suspect it was less «evident» at the time.

REPLY [45 votes]: In 1953-54 E. Artin asked me to describe homological algebra to him. In addition to Ext and Tor, I decided to show him the "homological proof" of the Hilbert Basis Theorem and indicated that the same proof showed that a regular local ring had finite global dimension. Artin then mentioned the open localization problem, and I said that if the converse of the theorem I just showed him were true, the localization result would be trivial. He asked if I could prove the converse, I said no, and we both agreed that it would be nice to prove it. The question of factoriality in regular local rings also came up in that conversation, and so I set myself the goal of proving those two theorems. I persuaded Auslander to join me in that project. When Auslander and I had almost all the results, and an outline of the string of inequalities needed, Eilenberg asked to see them, we wrote them up for him, and he went to Paris with them. It was there that Serre saw the outline of our project, and he beat us to the final proof by around a week (or the time it took an air mail letter to travel from Paris to Princeton). This may clear up the questions initially posed.<|endoftext|>
TITLE: Hexagonal rooks
QUESTION [22 upvotes]: Suppose you have a triangular chessboard of size $n$, whose "squares" are ordered triples $(x,y,z)$ of nonnegative integers that add up to $n$.  A rook can move to any other point that agrees with it in one coordinate -- for example, if you are on $(3,1,4)$ then you can move to $(2,2,4)$ or to $(6,1,1)$, but not to $(4,3,1)$.
What is the maximum number of mutually non-attacking rooks that can be placed on this chessboard?
More generally, is anything known about the graph whose vertices are these ordered triples and whose edges are rook moves?

REPLY [2 votes]: This has a connection with additive permutations, asked about here:
Are there enough additive permutations? .
http://oeis.org/A002047 has references to a similar problem involving a hexagonal board and counting the number of placements. (The Bennett and Potts reference calls the piece a 'brook'.) This number also counts the permutations.  A recent ArXiv paper http://arxiv.org/abs/1510.05987 appears to provide an asymptotic estimate.  Note that placement on a hexagonal board gives one for the triangle, but there may be placements in a triangle corner that are not realizable in the hexagon.
Gerhard "Cutting Corners For More Results" Paseman, 2017.03.13<|endoftext|>
TITLE: Number Fields Arising from Newforms
QUESTION [9 upvotes]: It is well-known that, given a normalized eigenform $f=\sum a_n q^n$, its coefficients $a_n$ generate a number field $K_f$. 
In their 1995 paper "Fermat's Last Theorem", Darmon, Diamond, and Taylor remark that, at the time of writing, very little was known about what sort of number fields could arise as some $K_f$. They do claim, however, that $K_f$ must be totally real or CM. This claim is made just before Lemma 1.37, on page 40 of the copy I linked to.
This is probably standard knowledge among experts, but I'm having trouble finding a reference, so my questions are:
1) Can someone please provide a reference for this claim?
2) Is this still the state of the art, or do we now know more about what types of fields can appear as $K_f$ for some $f$? What if we restrict our attention to weight $k=2$?
Thank you!

Edit: In my question, I originally just wrote "modular form" instead of "normalized eigenform". Thanks to @Stopple for pointing this out! Also, I originally claimed the paper was published in 2007, but Kevin Buzzard pointed out it was published in 1995. Thanks Kevin!

REPLY [7 votes]: For (1), see Ribet's wonderful article Galois representations attached to eigenforms with Nebentypus (http://dx.doi.org/10.1007/BFb0063943). It's proposition 3.2.<|endoftext|>
TITLE: Asymptotics for the coefficients of a rational function 
QUESTION [7 upvotes]: Let $(a_n)$ be a sequence of non-negative real numbers and assume that the resulting power series defines a rational function 
$$\sum_{n=0}^\infty a_n x^n = \dfrac{f(x)}{(1-x^{k_1})\cdots (1-x^{k_d})}$$
where $k_1,...,k_d>0$ are integers and $f(x)$ is a real polynomial s.t. $f(1) \neq 0$. 
It is not hard to show that 
$$\frac{f(1)}{k_1 \cdots k_d}\le \limsup \frac{a_n}{n^{d-1}} \cdot (d-1)! \le f(1)$$
As a special case we obtain for example $\limsup = f(1)$ if $k_1=\cdots k_d = 1$ (this is in fact not only the limsup but even the limit of the sequence). 
Questions: 1) Are there known formulas or better estimates for the $\limsup$ above in terms of $f$ and $k_1,...,k_d$ ? 
2) Are there particular techniques, that can be used to obtain good estimates (the one above is simply based on the binomial series for $(1-x)^{-d}$). 
Background: Such rational functions occur as Poincaré series of graded Noetherian algebras where $a_n$ is the dimension of the subspace of homogeneous lements of degree $n$. I'm trying to relate this quantity to the rational function.

REPLY [8 votes]: We have the partial fraction decomposition $$\dfrac{f(x)}{(1-x^{k_1}) \ldots (1 - x^{k_d})} = \text{polynomial}(x) + \sum_\omega \sum_{j=1}^{d(\omega)} b_{\omega,j} ( \omega-x)^{-j}$$
where $\omega$ are the roots of $(1 - x^{k_1})\ldots(1 - x^{k_d})$ and $d(\omega)$ is the multiplicity of the root $\omega$.  Then $(\omega-x)^{-j} = \sum_{n=0}^\infty {{j+n-1} \choose {j-1}} \omega^{-j-n} x^n$.  The asymptotics will be dominated by the $\omega$'s with highest multiplicity, namely the $\gcd(k_1, \ldots, k_d)$'th roots of unity, which have multiplicity $d$ (assuming $f(\omega) \ne 0$).  In particular if $\gcd(k_1,\ldots,k_d) = 1$,
we have $$a_n \sim \frac{f(1)}{k_1 \ldots k_d} {{d+n-1} \choose {d-1}} \sim \frac{f(1) n^{d-1}}{(d-1)!\; k_1 \ldots k_d}$$  But if $\gcd(k_1,\ldots,k_d) = g > 1$ we also have to consider the other $g$'th roots of unity:
$$ a_n \sim \sum_{\omega^g=1} \dfrac{f(\omega) n^{d-1}}{\omega^{n + k_1 + \ldots + k_d} (d-1)!\; k_1 \ldots k_d}$$  
EDIT: For example, with $f(x) = x + c(x-1)$, $d=2$, $k_1 = 2$, $k_2 = 4$ I get
$ a_n \sim  \dfrac {1 - (2c+1)(-1)^n}{8} n$
which, by the way, shows that your upper bound is wrong since $c$ can be arbitrary without affecting $f(1)$.<|endoftext|>
TITLE: Orbits of the projective special linear group on $\mathbf{Q} \cup \{\infty\}$ 
QUESTION [8 upvotes]: The group $\mathrm{PSL}_2(\mathbf{Q})$ of fractional linear transformations $x \mapsto (ax+b)/(cx+d)$ such that $a,b,c,d \in \mathbf{Q}$ and $ad-bc = 1$ acts on $\mathbf{Q} \cup \lbrace \infty\rbrace$. The weakest version of my question is the following:

Is there a short proof that no element of $\mathrm{PSL}_2(\mathbf{Q})$ acts transitively on $\mathbf{Q} \cup \lbrace \infty\rbrace$?

It seems obvious to me that no element of $\mathrm{PSL}_2(\mathbf{Q})$ can have finitely many orbits on $\mathbf{Q} \cup \lbrace\infty\rbrace$, but I have not been able to prove this.

Is this stronger claim true, and if so, does it have a short proof?

Some remarks and motivation: (1) The subgroup of $\mathrm{PSL}_2(\mathbf{Q})$ generated by the maps $x \mapsto x+1$ and $x \mapsto x/(x+1)$ acts transitively on $\mathbf{Q} \cup \lbrace \infty\rbrace$; these elements are used to generate the Calkin–Wilf tree of rational numbers.
(2) By conjugating by elements of $\mathrm{PGL}_2(\mathbf{Q})$ it suffices to look at maps of the form $g : x \mapsto 1/(b-x)$. If $|b| > 2$ then $g$ is conjugate in $\mathrm{PGL}_2(\mathbf{R})$ to a map of the form $x \mapsto \lambda x$. It follows that the iterates of $g$, starting at the initial value $x_0$ are of the form
$$ \frac{1}{b-\frac{1}{b-\ldots\frac{1}{b-x_0}}} $$
and converge to the continued fraction $1/(b-1/(b- \ldots ))$. The iterates of $g^{-1}$ behave similarly. So in this case each orbit of $\langle g \rangle \le \mathrm{PSL}_2(\mathbf{Q})$ has precisely two limit points in $\mathbf{R}$, namely the two roots of $x^2-bx+1$. Clearly this implies that $\langle g \rangle$ has infinitely many orbits on $\mathbf{Q} \cup \lbrace \infty\rbrace$. 
However if $|b| < 2$ then I have not been able to make this approach work.
(3) If $\mathbf{Q}$ is replaced with the finite field $\mathbf{F}_p$ then
is not too hard to show that no element of $\mathrm{PSL}_2(\mathbf{F}_p)$ acts transitively on $\mathbf{F}_p \cup \lbrace \infty \rbrace$: see Lemma 8.2 in this paper with John Britnell. This leads to a proof of the result for $\mathbf{Q}$ by reduction mod $p$.

REPLY [11 votes]: If your matrix is scalar it acts trivially on the projective line, so assume it's not scalar; hence it's conjugate in $\text{GL}_2(\mathbf{Q})$ to a companion matrix $A=\pmatrix{0 & -1 \cr 1 & z}$. 
If $z$ is integer with $|z|\ge 3$ then the dynamics of $A$ on the real projective line $\mathbf{P}^1(\mathbf{R})$ has a repelling point $a_-$ and a attractive point $a_+$; in particular, if $K$ is an infinite compact subset of $\mathbf{P}^1(\mathbf{R})\smallsetminus (a_-,a_+)$, then its intersection with each $A$-orbit $\{A^nx:n\in\mathbf{Z}\}$ is finite and hence $K$ intersects infinitely many $A$-orbits. Since $K$ can be chosen to be contained in $\mathbf{P}^1(\mathbf{Q})$, it follows that $A$ admits infinitely many orbits on $\mathbf{P}^1(\mathbf{Q})$. If $|z|=2$ a similar argument works (with $a_+=a_-$). If $|z|<2$ then $A$ has finite order.
Assume now $z$ is not an integer. Let $p$ be a prime divisor of the denominator of $z$. Let $|z|_p$ be the norm of $z$ in $\mathbf{Q}_p$. An easy calculation shows that the function $x\mapsto -1/(x+z)$ maps $\mathbf{Z}_p$ into itself and is $(1/|z|_p^2)$-contracting on 
$\mathbf{Z}_p$,
so it admits a unique fixed point which I denote by $a_+$. Thus $A$ admits a north-south dynamics on $\mathbf{P}^1({\mathbf{Q}_p})$, with repelling and attracting points $a_-$ and $a_+$. So the same argument as in the real case, using density of $\mathbf{P}^1({\mathbf{Q}})$ in $\mathbf{P}^1({\mathbf{Q}_p})$ shows 
that $A$ has infinitely many orbits. I assumed (as you do) that $A$ has determinant 1 but the argument carries over with minor modification for arbitrary $A$.<|endoftext|>
TITLE: Diophantine theory of homogeneous cubic polynomials
QUESTION [10 upvotes]: Arithmetic of quadratic forms over $\mathbb{Z}$ (or lattices theory) has received much attention and there are many applications in broad area of mathematics (such as intersection forms on fourfolds). I now wonder whether or not a similar theory for cubic forms can be developed. I recently found a beautiful theorem about binary cubic forms: 

Theorem (B. N. Delone and D. K. Faddeev,W.-T. Gan, B. H. Gross, and G.
  Savin) There is a canonical
  bijection between isomorphism classes
  of cubic rings and the set of
  $GL_{2}(\mathbb{Z})$-equivalence
  classes of integral binary cubic
  forms. Under this bijection, the
  discriminant of a cubic ring is equal
  to the discriminant of the
  corresponding binary cubic form.

I don't know how useful this theorem is because I don't know how difficult to classify cubic rings. My question is, are there any classification theory when the number of variable is small? I would appreciate it if anyone could give me a reference for recent development of cubic forms.

REPLY [8 votes]: Let me begin with a historical comment. The correspondence between cubic rings 
and binary cubic forms that you mentioned is not due to Delone and Faddeev, but 
rather to F. Levi 
(Kubische Zahlkörper und binäre kubische Formenklassen,
  Leipz. Ber. 66, 26-37 (1914); this article presents the results of
Levi's thesis, which was supervised by Weber in 1911). Actually Delone and
Faddeev credit Levi in their book not in the chapter where the material is
presented but in the preface.
BTW, Levi was Jewish and had to emigrate from Germany in 1936, when he went to
the University of Calcutta and apparently was elected president of the Indian
Mathematical Society for a few years. In 1949 he went to the Tata Institute,
and he returned to Germany in 1952.
Now for your question: there is a theory of cubic forms analogous to that of 
binary quadratic forms, which was developed by Eisenstein but abandoned after 
the success of Dedekind's ideal theory. The theory deals not with arbitrary 
cubic forms but only those that can be written as products of three linear 
factors (decomposable forms), and more specifically norm forms. For a modern 
account of the arithmetic of binary cubic forms you may want to look at 
Hoffman and Morales, Arithmetic of binary cubic forms,
Enseign. Math. 46 (2000) 61-94.<|endoftext|>
TITLE: Finite measure on the power set
QUESTION [6 upvotes]: Let $X$ be an uncountable set, and let $\Omega$ be the power set of $X$, viewed as a $\sigma$-algebra. Does there exist a positive $\sigma$-additive measure of finite total mass on $(X, \Omega)$ such that each point of $X$ has measure zero?

REPLY [3 votes]: Just to complement Andres's excellent answer with another reference, you can find a nice summary of the status of this question, as well as further references, in chapter 1.12(x) of Bogachev's monograph "Measure Theory I". 
The (very) short summary is that in all concrete cases the answer is no.<|endoftext|>
TITLE: On the difference between a projective chain complex and a level-wise projective chain complex
QUESTION [7 upvotes]: Let R be an associative ring with a unit, and consider the standard projective model structure of non-negatively graded (left) R-module, $Ch_R$. A map $f:M\to N$ in $Ch_R$ is a weak equivalence if it induces isomorphisms $H_kM\to H_kN$ on all homologies, and a cofibration if each $f_k:M_k\to N_k$ is a monomorphism, with a projective R-module as its cokernel.
Let A be a projective R-module. The chain complex $(*)$ $0\to A\to A\to0$ is a projective object in the world of chain complexes, $Ch_R$. So is any direct sum of such chain complexes. It is a known result (see [DS95, 7.10] for instance) that any acyclic object in $Ch_R$ which is level-wise projective is isomorphic to a direct sum of chains complexes as in $(*)$, with $A$'s projective.
We have here two notions, of (i) a chain complex which is level-wise project, and of (ii) a projective object in $Ch_R$. My gut's feeling is that (i) doesn't imply (ii). In the other way around I'm not sure. I would be glad to have an answer in both directions.   :-)
[DS95] = Dwyer & Spalinski, Homotopy theories and model categories.

REPLY [9 votes]: Edit: I put my answer into a broader perspective. 
In the following [We] refers to Weibel's "An Introduction to Homological Algebra". 
Recall that a chain complex $C$ is split exact if it is acyclic and $Z_n$ (the cycles) is a direct summand of $C_n$ for each $n$ [We, Def. 1.4.1, Ex. 1.4.2]. Moreover, let $Ch$ be the category of (unbounded) chain complexes over an abelian category with enough projectives. Then, by [We, Ex. 2.2.1] and my answer in  
$\qquad$https://mathoverflow.net/questions/103056/when-is-an-acylic-chain-complex-contractible 
we have: 

For a chain complex $P$ the following are equivalent: 

$P$ is a projective object in $Ch$ 
$P$ is a split exact complex of projectives 
$P$ is a contractible complex of projectives


Also from the link we obtain the following examples where $Ch_R$ denotes the category of unbounded chain complexes of modules over the ring $R$: 

If $R$ is hereditary (e.g. PID's, Dedekind domains), then the projective objects of $Ch_R$ are exactly the acyclic complexes of projective $R$-modules.
If $R$ is any ring with unit and the acyclic complex $P$ of projective $R$-modules is bounded below, then $P$ is a projective object in $Ch_R$. 

However, not all acyclic complexes of projective or free modules are projective objects in $Ch_R$. A counter-example (due to Dold) is given in [We, Example 1.4.2]: 

Over $R=\mathbb{Z}/4$ the following complex is exact 
$$\cdots \to \mathbb{Z}/4 \xrightarrow{2} \mathbb{Z}/4 \xrightarrow{2} \mathbb{Z}/4 \to \cdots $$
But it's no projective object in $Ch_R$ since $Z_n = \mathbb{Z}/2$ can't be a direct summand of $C_n = \mathbb{Z}/4$. 


Added: Let $Ch_b \subseteq Ch$ be the subcategory of chain complexes that are bounded below. In contrast to $Ch$ the following holds in $Ch_b$: 

The projective objects in $Ch_b$ are exactly the acyclic chain complexes (bounded below) of projectives. 

Proof: Let $P$ be an acyclic chain complex of projectives that is bounded below. We have already seen in example 2 above (compare also [We, Ex. 1.4.1 2.]) that $P$ is projective in $Ch$. Since $P \in Ch_b$ it's a projective object in $Ch_b$  
Now let $P \in Ch_b$ be a projective object. The same proof as in $Ch$ shows that each $P_i$ is projective (consider objects of the abelian category as chain complexes concentrated in a single degree). Also the same proof as in $Ch$ can be used to see that $P$ is acyclic: The mapping cone of $id: P[1] \to P[1]$ yields a short exact sequence 
$$0 \to P[1] \to \operatorname{cone}(id_{P[1]}) \to P \to 0.$$
But $P, P[1] \in Ch_b$ and by definition $\operatorname{cone}(id)_i = P_i\oplus P_{i+1}$ whence it is also bunded below.  So the short exact sequence is in $Ch_b$ and splits since $P$ is a projective object. Hence $id_{P[1]}$ is nullhomotopic. So in particular, $P[1]$  and thus $P$ is acyclic. q.e.d.

REPLY [5 votes]: Here is a general nonsense fact: if $F \dashv U : \mathcal{A} \to \mathcal{B}$ is an adjunction and $U$ preserves epimorphisms, then $F$ preserves projective objects. Epimorphisms in $\textrm{Ch}(R)$ are precisely the levelwise epimorphisms in $\textrm{Mod}(R)$, so if we take $\mathcal{A} = \textrm{Mod}(R)$, $\mathcal{B} = \textrm{Ch}(R)$, and set $U(M)$ to be $M$ considered as a chain complex concentrated in degree $n$, and $F(K_\bullet) = K_n$, we have an adjunction $F \dashv U$ satisfying the above hypotheses, and therefore $F$ preserves projective objects. Thus,
Proposition. A projective chain complex is necessarily levelwise projective. 　◼
The converse is false, as you anticipated. Consider the following chain complexes:
$$P_\bullet = L_\bullet = (R^3 \to R^3) \text{ with differential given by }
\begin{pmatrix}
0 & -1 & 1 \newline
1 & 0 & -1 \newline
-1 & 1 & 0
\end{pmatrix}$$
$$K_\bullet = (R^3 \to R^4) \text{ with differential given by } \begin{pmatrix}
0 & -1 & 0 \newline
1 & 0 & -1 \newline
-1 & 1 & 0 \newline
0 & 0 & 1
\end{pmatrix}$$
Let $f_\bullet : K_\bullet \to L_\bullet$ be the map given in degree $1$ by the identity and in degree $0$ by the matrix
$$f_0 = \begin{pmatrix}
1 & 0 & 0 & 1 \newline
0 & 1 & 0 & 0 \newline
0 & 0 & 1 & 0 \newline
\end{pmatrix}$$
It is clear that $f_\bullet$ is an epimorphism. Geometrically, $P_\bullet$ is the simplicial chain complex of a triangle $S^1$, $K_\bullet$ is the simplicial chain complex of the interval $[0, 1]$ subdivided into 3 segments, and $f$ is the quotient map that identifies the endpoints of $[0, 1]$. Intuitively, we expect that there is no morphism $g : S^1 \to [0, 1]$ such that $f \circ g = \textrm{id}$, and this turns out to be true in the chain complex world as well... provided $R$ is not the trivial ring. Indeed, if $f_0 \circ g_0 = \textrm{id}$ and $f_1 \circ g_1 = \textrm{id}$, then by elementary algebra we must have $g_1 = \textrm{id}$ and
$$g_0 = \begin{pmatrix}
a & b & c \newline
0 & 1 & 0 \newline
0 & 0 & 1 \newline
1 - a & -b & -c
\end{pmatrix}$$
for some $a, b, c$, but if $g_\bullet$ is to be a chain map, $a, b, c$ must satisfy
$$\left\lbrace\begin{aligned} b & = c \newline
a & = 1 + c \newline
a & = b \newline
b & = c 
\end{aligned}\right.$$
which implies $0 = 1$. So $P_\bullet$ cannot be projective in $\textrm{Ch}(R)$, even though it is degreewise projective.<|endoftext|>
TITLE: smooth homotopy on exotic R^4
QUESTION [5 upvotes]: Take an exotic $\mathbb{R}^4$ i.e. $V = (\mathbb{R}^4,d)$ such that $V$ is not diffeomorphic to $\mathbb{R}^4$ with standard metric.
Is it true (obvious?) that any two smooth maps $f_1, f_2: S^k \to V$ are equivalent via smooth homotopy?
edited: (for any k)

REPLY [9 votes]: Yes, since smooth maps which are (continuously) homotopic are always smoothly homotopic. See Kosinski's "Differential Manifolds", Theorem III.2.5 and Corollary III.2.6.<|endoftext|>
TITLE: A variation on Pisot–Vijayaraghavan numbers
QUESTION [5 upvotes]: Suppose a non-real algebraic integer $\alpha$ has, aside from itself and its complex conjugate $\bar\alpha$, all its algebraic conjugates of norm less than 1.  Then the fractional
parts of $\Re(\alpha^n)$ will cluster about 0, 1/2 and 1 as $n$ grows large. 
Do such numbers have a name and/or a literature?

REPLY [5 votes]: These numbers have been called complex Pisot numbers in at least a couple of papers. But they seem to have been investigated far less than their real counterparts.
Chamfy [1] and Garth [2] found the few smallest (in modulus) complex Pisot numbers.
Solomyak and Xu [3] showed that some of the dynamical properties of Pisot numbers continue to hold in the complex case, in particular, complex Bernoulli convolutions associated to complex Pisot numbers are singular. See Definition 2.2, Theorem 2.3 and the following remarks in [3].
References:
[1] Chamfy, Christiane. Fonctions méromorphes dans le cercle-unité et leurs séries de Taylor. (French) Ann. Inst. Fourier. Grenoble 8 1958 211--262.
[2] Garth, David. Complex Pisot numbers of small modulus. C. R. Math. Acad. Sci. Paris 336 (2003), no. 12, 967--970.
[3] Solomyak, Boris; Xu, Hui. On the `Mandelbrot set' for a pair of linear maps and complex Bernoulli convolutions. Nonlinearity 16 (2003), no. 5, 1733--1749.<|endoftext|>
TITLE: If $X$ is a smooth and proper stack, does it admit a smooth and proper atlas?
QUESTION [7 upvotes]: Fix a ground scheme $S$ (a field say).
By atlas for an algebraic stack I mean a smooth and surjective morphism $Y \to X$ from a scheme (or algebraic space or affine scheme) $Y$.
If the stack $X$ is smooth then all atlases are smooth schemes over $S$.
What about when $X$ is smooth and proper? Can I find a proper atlas?
If in general the answer is no, what about for Deligne-Mumford stacks?

REPLY [11 votes]: The answer is "no".  Consider a smooth, proper $1$-dimensional Deligne-Mumford stack over $\mathbb{C}$ which has coarse moduli space $\mathbb{P}^1$ and which has a single "stacky" point (with any nontrivial stabilizer group at that point).  
Edit.  A more precise definition of the stack is as follows.  Let $m$ be an integer, $m>1$.  Let $A$ be $\mathbb{A}^2 \setminus \{(0,0)\}$ with coordinates $x$ and $y$.  Let $\mathbb{G}_m$ act on $A$ by $t\ast (x,y) = (tx,t^m y)$.  The stack $Q_m$ is the quotient stack $[A/\mathbb{G}_m]$.  The quotient morphism $A\to Q_m$ is a smooth atlas, implying that $Q_m$ is smooth (since $A$ is smooth).  In fact, this morphism is naturally a $\mathbb{G}_m$-torsor; denote the associated invertible sheaf on $Q_m$ by $\mathcal{L}$.  It is not hard to see that this is an $m$-torsion invertible sheaf; the $\text{m}^{\text{th}}$ tensor power of $A$ as a $\mathbb{G}_m$-torsor over $Q_m$ is the quotient stack $[(\mathbb{G}_m\times A)/\mathbb{G}_m]$ where the $\mathbb{G}_m$-action is $t\cdot (u,(x,y)) = (t^mu, (tx,t^my))$.  This admits a "trivializing" morphism to $\mathbb{G}_m$, $(u,(x,y)) \mapsto u/x^m = u/y$, each defined on the appropriate open $\mathbb{G}_m\times D(X)$ or $\mathbb{G}_m\times D(y)$.  The only nontrivial stabilizer group is $\mathbf{\mu}_m$ acting on the $\mathbb{G}_m$-orbit $\{0\}\times \mathbb{G}_m$ inside $A$.  So $Q_m$ is a Deligne-Mumford stack.  Finally, $Q_m$ is finite over its coarse moduli space $\mathbb{P}^1$, where the $\mathbb{G}_m$-invariant morphism $A\to \mathbb{P}^1$ is just $(x,y) \mapsto [x^m,y]$.<|endoftext|>
TITLE: Can a string's sophistication be defined in an unsophisticated way?
QUESTION [29 upvotes]: This question is about sophistication, a way of measuring the amount of "interesting, non-random information" in a binary string, which was proposed by Kolmogorov and others in the 1980s.  I'll define all the needed concepts below, but for further reading, I recommend this paper by Antunes and Fortnow, this PhD thesis by Antunes, or this paper by Gacs, Tromp, and Vitanyi.
Given an n-bit string x, recall that K(x), the Kolmogorov complexity of x, is the length in bits of the shortest program p (in some fixed universal programming language) such that p()=x: that is, p halts and outputs x when given a blank input.  Given a set S ⊆ {0,1}n, one can also define K(S) to be the length in bits of the shortest program that outputs the 2n-bit characteristic sequence of S.  Finally, one can define K(x|S) to be the length in bits of the shortest program p such that p(S)=x: that is, p halts and outputs x when fed the characteristic sequence of S as input.
The "problem" with Kolmogorov complexity is that it's maximized by random strings, which are intuitively not very "complex" at all.  This motivates the following alternatives to K(x):
Given an n-bit string x and a constant c>0, the oxymoronically-named naïve sophistication of x, or NSophc(x), is the smallest possible value of K(S), over all sets S ⊆ {0,1}n such that x∈S and K(x|S) ≥ log2|S| - c.  Intuitively, NSoph measures the minimum number of bits needed to specify a set of which x is an incompressible or Kolmogorov-random element.  I call it "naïve" because it's the first measure I would think of that's sort of like Kolmogorov complexity but small for random strings (small because for random strings, one can take S={0,1}n, whence NSophc(x)=O(1)).
Meanwhile, the coarse sophistication of x or CSoph(x), defined by Antunes, is the smallest possible value of 2K(S)+log2|S|-K(x), over all sets S ⊆ {0,1}n such that x∈S.  Intuitively, CSoph measures the minimum number of bits needed to specify x via a "two-part code," where the first part specifies a set S containing x, the second part gives the index of x in S, and a penalty gets applied both for K(S) (the length of the first part of the code) and for K(S)+log2|S|-K(x) (the amount by which the total code length exceeds K(x)).  Despite the unwieldy definition, Antunes amasses evidence that CSoph is in various ways the "right" measure of the non-random information in a string.
My question is now the following:
Let c=O(1).  Is NSophc(x), my "unsophisticated kind of sophistication," always close to CSoph(x), Antunes' "sophisticated kind of sophistication"?  Or can there be a large gap between the two?  If so, how large?
Here's what I know about this question:

CSoph(x) ≤ 2NSophc(x)+c.  To see this: let the set S minimize K(S) subject to x∈S and K(x|S) ≥ log2|S| - c.  Then CSoph(x) ≤ 2K(S)+log2|S|-K(x) ≤ 2NSophc(x)+log2|S|-K(x) ≤ 2NSophc(x)+log2|S|-K(x|S) ≤ 2NSophc(x)+c.

NSophc(x) can be about twice as large as CSoph(x).  To see this: first, as observed by Antunes, if x is an n-bit string, then CSoph(x) never exceeds n/2+o(n).  (For we can always achieve that bound by setting S={x} if K(x)≤n/2, or S={0,1}n if K(x)>n/2.)  Second, as discussed by Gacs, Tromp, Vitanyi, it's possible to construct what Kolmogorov called "absolutely non-random objects," meaning n-bit strings x such that K(x|S) ≤ log2|S| - O(1) whenever K(S) ≤ n - clog(n).  For these strings, we clearly have NSophc(x) ≥ n-O(log n) if c=O(1).  Combining now yields the result.

As a final note, NSoph and CSoph are both different from the "ordinary sophistication" Soph, which is defined as follows: Sophc(x) is the smallest possible value of K(S), over all sets S ⊆ {0,1}n such that x∈S and K(S) + log2|S| ≤ K(x)+c.  Intuitively, Sophc(x) measures the minimum number of bits needed for the first part of a near-minimal two-part code specifying the string x.  One can observe the following (I'll give details on request):

NSophc(x) ≤ Sophc(x)
CSoph(x) ≤ Sophc(x)+c
There exist strings x for which Sophc(x) is very large but NSophc(x) and CSoph(x) are both very small.

I'll also observe that NSophc(x), CSoph(x), and Sophc(x) are all upper-bounded by the Kolmogorov complexity K(x) (or rather, by K(x)+c).
Update: Sorry, just minutes after writing this post, I think I see the answer to one direction of my problem!  Consider the "absolutely non-random objects" x discussed above.  These objects satisfy K(x) ≥ NSophc(x) ≥ n-O(log n).  But precisely because their Kolmogorov complexity is so large, they should also satisfy CSoph(x)=O(log n), achieved by setting S={0,1}n.  On the other hand, I still don't know whether CSoph(x) can ever be larger than NSophc(x) (only that, if so, it's never more than a factor of 2 larger).  And I'd still be extremely interested if anyone could answer that question.

REPLY [9 votes]: Hi Scott,
are you asking whether CSoph(x) may be much more NSoph_c(x) for every constant c? This question looks strange as NSoph_c(x) increases, as c decreases. Thus the question reduces to the case c=0 (or may be you allow negative values?). Which question precisely did you have in mind? Note also that Kolmogorov complexity K(x)  itself is defined up to an additive constant term (BTW, I assume that you meant the plain complexity and not prefix one). Thus you should specify the quantifier over K(x). 
One reading of your question is the following: 
Is it true that for all c, all K and all d there is a x 
with CSoph(x) > NSoph_c(x) + d
Another reading is the following: 
Is it true that for all c there is K such that for all d there is a x 
with CSoph(x) > NSoph_c(x) + d
Kolia.
P.S. May be you will be interested also in the following result from our joint with Paul Vitanyi paper from FOCS 2002: if we allow logarithmic changes of c than soph and NSoph coincide (with logarithmic precision):
soph_{c+O(\log K(x))}(x) <  Nsoph_c(x) +O(\log K(x)).<|endoftext|>
TITLE: Geometric information on transferred structure
QUESTION [7 upvotes]: Let $(M,g)$ be a Riemannian manifold, let $\Omega^*(M)$ denote the cochain complex of differential forms on $M$ and $H^*(M)$ its cohomology considered as a chain complex with trivial differential. We have a map $f:\Omega^*(M) \to H^*(M)$ given by projecting a form to its harmonic part and taking its cohomology class and we have a map $h: H^*(M) \to \Omega^*(M)$ which sends a cohomology class to its harmonic representative. Then $fh$ is the identity and $hf$ is chain homotopic to the identity. Hence, we have the right setting to transfer the $C_{\infty}$-algebra structure of $\Omega^*(M)$ with structure maps $m_1=d$, $m_2=\wedge$, and $0=m_3=m_4=...$ to obtain an $C_{\infty}$-algebra structure on $H^*(M)$. We can get formulas for the structure maps of the transferred structure on $H^*(M)$ by taking sums over trees and putting the chain homotopies in the right internal edges, etc...
The topological information information obtained from this transferred structure is understood: up to homotopy, the transferred structure contains rational homotopy information. 
However, my question is the following: What kind of geometric information is contained in the transferred structure (which depends on the metric $g$ and it involves choices) up to isomorphism ?

REPLY [2 votes]: As far as I understand it, Theorem 7 of http://arxiv.org/abs/0811.1655 states that the $C_\infty$-algebra structures on $H^\ast(M)$ up to isomorphism correspond to rational homotopy types with the given cohomology algebra. This implies that the answer to your question is "any geometric information which can be read off from the rational homotopy type". This is a significant amount, see the book "Algebraic Models in Geometry" by Felix, Oprea and Tanré, or this Oberwolfach Report. 
(Of course, it may just be that your interpretation of "isomorphism" of $C_\infty$-algebras is stricter than Kadeishvili's.)<|endoftext|>
TITLE: Spin and SO groups associated to a degenerate symmetric bilinear form
QUESTION [5 upvotes]: In "Spin geometry" by Lawson and Michelsohn it is defined the Clifford algebra $Cl(g)$ associated to a symmetric bilinear form $g$ in general, including the degenerate case. But the rest of the book is devoted exclusively to the non-degenerate case.
Are there any references concerned with the spin group $Spin(g)$, the group $SO(g)$, their representations, corresponding to a degenerate symmetric bilinear form $g$?

REPLY [5 votes]: Some naive comments. Any real vector space $V$ with a symmetric bilinear form $g$ admits an orthogonal direct sum decomposition $V_0 \oplus V_1$ where $V_0$ consists of the vectors $v$ such that $g(v, -) = 0$ and $g$ is nondegenerate on $V_1$ (e.g. by the spectral theorem). An inspection of the defining relation
$$\frac{uv + vu}{2} = g(u, v)$$
of the Clifford algebra shows that $\text{Cl}(V,  g)$ is the (graded) tensor product $\Lambda(V_0) \otimes \text{Cl}(V_1, g)$. So this is not too bad.
The corresponding special orthogonal group is more complicated; $\text{SO}(g)$ consists of block matrices
$$\left[ \begin{array}{cc} A & B \\\ 0 & C \end{array} \right]$$
where $A \in \text{GL}(V_0), C \in \text{O}(V_1, g)$, $B$ is an arbitrary linear map $V_1 \to V_0$, and $\det(A) \det(C) = 1$. This does not seem like a very nice group to work with and I have no comment on what the corresponding spin groups might look like.<|endoftext|>
TITLE: Modern Proof of the Theorem of the Base
QUESTION [9 upvotes]: I am looking for a modern proof of the so-called "Theorem of the Base"--that the Neron-Severi rank of a smooth projective variety is finite.  One can prove this for varieties over $\mathbb{C}$ easily via transcendental methods---however, the most recent proof I can find which works e.g. in positive characteristic is due to Lang and Neron and is written in the language of Weil's foundations.  (Lang includes a similar proof in his book "Diophantine Geometry.")  

Does anyone know of a proof written in the language of schemes?

My motivation is that I'd like to have a brief reading seminar on this theorem with some graduate students at my institution---while it looks like the Lang/Neron paper is translatable, I think that this sort of translation is a serious burden for seminar participants.  So it would be nice to have a more modern reference.

REPLY [8 votes]: SGA 6, Exp. XIII<|endoftext|>
TITLE: Global dimensions of non-commutative rings
QUESTION [9 upvotes]: This is related to my previous question: When is a quantum affine space $\mathbb{A}^{n}$ Calabi-Yau? I now would like to know the global dimension of the ring $R=\mathbb{C}\langle x_1,\dots,x_n\rangle/I$, where I is the two-sided ideal generated by $x_ix_j=a_{ij}x_jx_i$ for some $a_{ij}=a_{ji}^{-1}\in \mathbb{C}$ for $1≤i,j≤n$. Recall that the global dimension $gl\dim(S)$ of a ring $S$ is defined to be the supremum of the set of projective dimensions of all left $S$-modules.  
In case your ring $R$ is defined by a quiver with relations, there seem some techniques. Are there another approach to compute $gl\dim(R)$? Moreover, are there any standard way to compute $gl\dim(S)$ when $S$ is non-commutative?  
There are many characterization of $gl\dim(R)$ when $R$ is commutative and I am aware of similar questions, such as Commutative Ring of Finite Global Dimension.

REPLY [7 votes]: There is another way to see this than constructing an explicit resolution.  This involves viewing $R$ as an iterated skew polynomial ring.
I am assuming that you want $a_{ii} = 1$; otherwise $x_i^2 = 0$ for all $i$.
Also, since this works over any field, I am just going to denote the base field by $k$.
Start with $R_1 = k[x_1]$.  Then let $\sigma_1$ be the $k$-algebra automorphism of $R_1$ defined by $\sigma_1(x_1) = a_{21} x_1$, and let $R_2$ be the skew-polynomial ring
$$
R_2 = R_1[x_2; \sigma_1].
$$
Thus $R_2$ is generated by $x_1$ and $x_2$ with the relation
$$
x_2 x_1 = \sigma_1(x_1)x_2 = a_{21} x_1 x_2.
$$
Then we continue this game.  Having constructed $R_i$, define
$$
R_{i+1} = R_i[x_{i+1}, \sigma_i],
$$
where $\sigma_i \in \mathrm{Aut}(R_i)$ is defined by $\sigma_i(x_j) = a_{i+1,j} x_j$ for $1 \le j \le i$.  Then for $j < i$ we have the relations
$$
x_i x_j = \sigma_{i-1}(x_j) x_i = a_{ij} x_j x_i,
$$
and hence your ring $R$ coincides with $R_n$.
This gives us two things.  First, there is an analogue of the Hilbert Basis Theorem for skew polynomial rings; if $A$ is left Noetherian then the skew polynomial ring $A[x;\sigma]$ is left Noetherian for any automorphism $\sigma$ of $A$.  You can find this in Section 1.2.9 of McConnell-Robson or Theorem 1.14 of Goodearl-Warfield.
The other fact is that there is an analogue of the (generalized) Hilbert Syzygy Theorem for skew polynomial rings over Noetherian rings.  This is in Section 7.9.10 of McConnell-Robson.  Explicitly, it says the following: if $A$ is left Noetherian with $\mathrm{l.gl.dim} \, A = n < \infty$, then $\mathrm{l.gl.dim} \, A[x;\sigma] = n+1$ for any automorphism $\sigma$ of $A$.
Starting with $\mathrm{l.gl.dim}k[x] = 1$ and iterating shows that each $R_i$ is both left and right Noetherian and has 
$$
\mathrm{l.gl.dim} R_i = \mathrm{r.gl.dim} R_i = i.
$$<|endoftext|>
TITLE: Why does $\aleph_{\omega}$ have more than $\aleph_{\omega}$ countable subsets?
QUESTION [5 upvotes]: I believe  $\aleph_{\omega}$ has more than $\aleph_{\omega}$ countable subsets but I do not see the proof.  I fear it is obvious, but not to me today.

REPLY [14 votes]: Let $\{A_\alpha:\alpha<\omega_\omega\}$ be all countable subsets of $\omega_\omega$. We build one that is not among them, giving the contradiction. Pick $x_0\in\omega_1$ wich is not in $\bigcup\{A_\alpha:\alpha<\omega\}$. Then choose $\omega_1\leq x_1<\omega_2$ which is not in $\bigcup\{A_\alpha:\alpha<\omega_1\}$, etc. Eventually we get a countable set 
$\{x_0,x_1,\dots\}$ different from each $A_\alpha$. (This is essentially the argument for proving Konig's inequality.)<|endoftext|>
TITLE: Create examples of deg=1 maps
QUESTION [5 upvotes]: My question is:
For given k-dimension manifold M, Is there any way to produce a manifold N so that there exist a map $f:M\rightarrow N$, $degf=1$
The trivial method I know:
The only trivial method I can image is when $M=M_1\#M_2$, then we can produce a deg=1 map from $M$ to $M_1$. Or for the torus, we can produce deg=1 map.
Topological Restrictiom I know:
if $f:M\rightarrow N$, $degf=1$ 
(1)then every betti number of M must bigger than N.
(2)$f_*\pi_1(M)$ is onto. 
edit: Another restriction is that the Gromov norm of $M$ must be at least as large as the Gromov norm of $N$.  add by Zare

REPLY [9 votes]: In dimension 2, Edmonds proved that every degree one map between closed orientable surfaces is homotopic to a "pinch map", which takes a subsurface which is bounded by one component, and sends it to a disk. This is a special case of the connect sum method you describe, so that is really the only method in 2D. 
In dimension 3, Yi Liu has proven that every closed orientable 3-manifold has a degree one map to finitely many other closed orientable 3-manifolds. I think it ought to be possible to make the proof into an algorithm. So given a closed manifold $M$, there should be an algorithm which produces finitely many manifolds $N$ such that there is a degree one map $M\to N$. It might also be possible to say something about the homotopy classes of such maps, but this hasn't been worked out. I should say also that there are many known 3-manifolds which have non-zero degree maps only to themselves or $S^3$. 
Unfortunately, the proof doesn't give a prescription of how to produce such maps. There is a version of the pinching map for Seifert manifolds, by extending a pinch over the base orbifolds to the fibrations. 
A related method for producing degree one maps is to observe that any knot complement has a degree one map to a solid torus which is the identity on the boundary torus. Thus, one can glue the knot complement and the solid torus to another manifold with torus boundary, and extend the degree one map as the identity over the manifold. This only works if you have a manifold with essential tori in it.
A third method allows you to specify the range, but not the domain. Take a 3-manifold $N$, and take a knot $K\subset N$ which is null-homotopic. Then one may perform a surgery on this knot with slope $\alpha$ to get a manifold $N_K(\alpha)$ which has a degree one map $N_K(\alpha)\to N$, by extending the meridian of the Dehn filling over a disk in $N$ which realizes the null-homotopy of $K$. This method is described in Section 3 of this paper.<|endoftext|>
TITLE: Maximal subfields in a division algebra over a local field
QUESTION [5 upvotes]: Hi, Let $A$ be a division algebra over a local field $F$ of dimension $n^2$ and $K$ be an extension of $F$ of degree $n$. Then if follows from COROLLARY 2 in page 225 of Weil's Basic Number Theory that $A$ is split over $K$. Now my question is 

Does $A$ contain a subfield isomorphic to $K$ ? Why?
Could you descirbe the general picture about "maximal subfields in central simple algebras" over a (not necessarily local) field ?  Or tell me some references.

Thank you !

REPLY [2 votes]: The answer to your first question can also be found in Weil's Basic Number Theory. See Corollary 3 on page 180, and note that if two central simple algebras of the same dimension are similar then they are isomorphic.<|endoftext|>
TITLE: Cartier divisors on singular curves
QUESTION [7 upvotes]: (1) Let $X$ be a projective (integral) curve over $\mathbb{C}$ and let $P$ be a singular point of $X$. Is there always a Cartier divisor whose support is exactly $P$ (set-theoretically)?
The following may be two related questions:
(2) Let $f: Y \to X$ be the normalisation of the curve $X$. Then we have a pushforward functor for Weil divisors from $Y$ to $X$. Is there some analogous contruction for Cartier divisors?
(3) Is every Weil divisor on a singular curve $\mathbb{Q}$-Cartier (i.e. a mutiple of the divisor is Cartier)?  A 2-dimensional analog seems to have been discussed in the following MO question:
Is every Weil divisor on an arithmetic surface Q-Cartier
I still wonder what is known in the 1-dimensional situation.

REPLY [7 votes]: Here is a part of an email I sent you because I couldn't log in MO. 
The answer to (3) is yes: there is a Cartier divisor $D$ on $X$ such that the associated Weil divisor ($0$-cycle for a  curve) is $[P]$. This idea is just to consider  the normalization $f : Y\to X$. Let $Q\in Y$ be a point lying over $P$. Then $[Q]\sim Z$ (rational equivalence) for some $0$-cycle $Z$ on $Y$ with support disjoint from $f^{-1}(P)$ (use e.g. Riemann-Roch). By pushforward, $[P]=f_*[Q] \sim f_*Z$. By definition of rational equivalence, this means that $[P]=f_*Z+[\mathrm{div}(h)]$ for some rational function $h$ on $X$. As the support of $f_*Z$ is disjoint from $P$, this means that in an open neighborhood $V$ of $P$, $[P]$ coincides with $[\mathrm{div}(h)]$. Define a Cartier divisor $D$ by ${D|}\_{V}=\mathrm{div}(h)$ and $D|_{X\setminus { P}}=1$. Then $[P]$ equals to  (the $0$-cycle associated to) $D$.  The idea of taking normalization can be found, say, in a paper of Colliot-Thélène (Inv. Math. 2005, p. 599).
This works over any algebraically closed field. If the base field is not necessarily algebraically closed, let $Q_1, \cdots, Q_r$ be the points of $f^{-1}(P)$, let $n$ be the gcd of the $[k(Q_i): k(P)]$. Then the above argument shows that $n[P]$ is associated to a Cartier divisor on $X$. And $n$ is the smallest possible.
In higher dimension, as you already known, this conclusion is false. However the following weaker result holds. Let $X$ be a noetherian scheme, let $E$ be a prime cycle. Then for some positive integr $n$, $nE$ is the cycle associated to a Cartier divisor (on an integral closed subscheme of $X$ containing $E$) in an open neighborhood of the generic point $P$ of $E$. The integer $n$ can be controlled by some invariants related to the singularity at $P$ (can take $n=1$ is $P$ is a regular point of $X$, but this is not necessary as we saw in the case of curves). See Theorem 4.5, §5, and Theorem 6.5 in this paper.<|endoftext|>
TITLE: Strictly order preserving maps into the integers
QUESTION [7 upvotes]: If $P$ and $P'$ are partial orders, a strictly order preserving map from $P$ to $P'$ is an $f:P\to P'$ satisfying that $x\lt y$ implies $f(x)\lt f(y)$ for all $x,y\in P$.
An interval in $P$ is a set of the form ${}[x,y]=\{z\mid x\le z\le y\}$, where $x\le y$. Note that points in an interval need not be comparable with one another. The height of an interval is the supremum of the lengths of $< $-chains of elements in the interval.
Clearly, if there is a strictly order preserving map from $P$ to $\mathbb Z$, then all intervals in $P$ have finite height. If $P$ is countable, this condition also suffices for the existence of such a map. This dates back to Erné in 1979. 
However, there are examples of uncountable $P$ with all intervals of finite height for which there is no strictly order preserving map into $\mathbb Z$. The examples I know have size continuum ($=|\mathbb R|$) or larger. A nice example due to Farley and Schröder can be found here.

Is there an example of size $\omega_1$, preferably one that does not require the axiom of choice?

REPLY [5 votes]: The construction of Farley and Schroeder that you linked to essentially gives an example of size the bounding number $\mathfrak b$.  So it's at least consistent with $\neg$CH to have an example of size $\aleph_1$.  Details: Take an unbounded family $\mathcal B$ of functions $\omega\to\omega$, and form the analog of the Farley-Schroeder example with just the functions in $\mathcal B$ at the bottom level instead of all functions.  If there were a strictly order-preserving map $f$ to $\mathbb Z$, then the paper shows how to obtain a $F$ that majorizes every $b\in\mathcal B$ up to an additive constant (namely $-f(b)$).  Then any $G$ that eventually majorizes the countably many functions $F+$ constant would also eventually majorize all the functions in $\mathcal B$, which contradicts the choice of $\mathcal B$.
EDIT: Here's a variant of the same construction that gives an $\omega_1$-sized example.  The poset will consist of a "low" copy of the set $\omega_1$ of countable ordinals, a "high" copy of the same set, and some finite chains joining elements in the low copy to elements in the high copy.  I'll write $\alpha-$ for the low copy of an ordinal $\alpha<\omega_1$ and $\alpha+$ for the high copy.  The main issue is how long to make the finite chains between low and high copies.  For each countable ordinal $\beta$, fix a one-to-one function $f_\beta:\beta\to\omega$.  Join $\alpha-$ to $\beta+$ by a chain of length $f_\beta(\alpha)$ if $\alpha<\beta$; don't join them by any chain if $\alpha\geq\beta$.  Clearly each interval in this poset is either one of the finite chains or a subset of one.  So I need only prove that there is no strictly order-preserving map $g$ from my poset to $\mathbb Z$.  
Suppose, toward a contradiction, that $g$ were such a map.  Since the low copy of $\omega_1$ is uncountable and the range of $g$ is countable, $g$ must be constant, say with value $a$, on an uncountable subset $\{\alpha-:\alpha\in X\}$ of the low copy.  Similarly, it is constant, say with value $b$, on $\{\beta+:\beta\in Y\}$ for some uncountable $Y$.  Fix some $\beta\in Y$ that is larger than the $\omega$-th element of $X$.  So there are infinitely many ordinals $\alpha\in X$ with $\alpha<\beta$.  These infinitely many $\alpha$'s are mapped by $f_\beta$ to infinitely many natural numbers, so we can find $\alpha\in X$ with $f_\beta(\alpha)>b-a$.  But then the chain between $\alpha-$ and $\beta+$ is longer than the difference between $g(\beta+)=b$ and $g(\alpha-)=a$, so $g$ cannot be strictly order-preserving.<|endoftext|>
TITLE: Global Definition of the Dolbeault Complex of a Vector Bundle
QUESTION [5 upvotes]: For an $2n$-dimensional complex manifold $M$, and a smooth vector bundle $E$ over $M$, it is well-known (see Voisin, Huybrechts) that there exists an operator $\overline{\partial}$, built locally from the usual anti-holomorphic derivative, that acts on $\Gamma^{\infty}(E) \otimes_{C^{\infty}} \Omega^{(0,\bullet)}(M)$ so as to give a complex
$$
\Gamma^{\infty}(E) \overset{\overline{\partial}}{\to} \Gamma^{\infty}(E) \otimes_{C^{\infty}} \Omega^{(0,1)}(M) \overset{\overline{\partial}}{\to} \cdots \overset{\overline{\partial}}{\to} \Gamma^{\infty}(E) \otimes_{C^{\infty}} \Omega^{(0,n)}(M)\overset{\overline{\partial}}{\to} 0.
$$
As I prefer global constructions, I began to wonder how one would construct this globally. To apply id $\otimes \overline{\partial}$ to $\Gamma^{\infty}(E) \otimes_{C^{\infty}} \Omega^{(0,\bullet)}(M)$ is of course not well-defined since we are tensoring over ${C^{\infty}(M)}$. So what can one do . . .?
P.S. Does such a complex exist in the purely real case, and if not, then why not?

REPLY [6 votes]: To get the Dolbeault complex, you need a choice of holomorphic structure on $E$, not just a smooth one. If $\mathcal{E}$ is the locally free sheaf of $\mathcal{O}_M$-modules corresponding to $E$, then $\mathcal{E}\subset \mathcal{A}^0(\mathcal{E})=\mathcal{C}^\infty_M\otimes_{\mathcal{O}_M}\mathcal{E}$ and
 $\overline{\partial}_E: \mathcal{A}^0(\mathcal{E})\to \mathcal{A}^{0,1}(\mathcal{E})$
is the unique morphism (of sheaves of $\mathbb{C}$ vector spaces) satisfying
$$ \overline{\partial}_E(f\sigma) = \overline{\partial}f\otimes \sigma + f\overline{\partial}_E(\sigma),$$
for any smooth function $f$ and $\sigma$ a smooth section of $E$, such that 
$\left. \overline{\partial}_E\right| _{\mathcal{E}}=0$. The first term in the Leibniz formula involves $\overline{\partial}=d^{0,1}$.
You can then extend the Dolbeault operator to
$\overline{\partial}_E: \mathcal{A}^{0,p}(\mathcal{E})\to \mathcal{A}^{0,p+1}(\mathcal{E})$,
$\overline{\partial}_E^2=0$, 
by imposing the Leibniz rule with the usual sign. This gives you the Dolbeault resolution
$$ 0\to \mathcal{E}\to \mathcal{A}^0(\mathcal{E})\to \mathcal{A}^{0,1}(\mathcal{E})\to\ldots$$
The complex you write is obtained by passing to global sections of $\mathcal{A}^{0,\bullet}(\mathcal{E})$.
You cannot do any of this without the holomorphic structure. Differently put, you need
 the total space of $E$ to be a complex manifold and  the projection $E\to M$ to be holomorphic. 
You cannot play the same game in the real case, but if you are willing to assume that $E$ carries a flat connection, then you can look at the de Rham resolution of the corresponding local system, as David explains.
ADDENDUM
The requirement that a (smooth) complex vector bundle $V$ admits a holomorphic structure
$\overline{\partial}_E$ is non-trivial. It can be phrased as follows:
$V$ admits a holomorphic structure if and only if it admits a connection, $D$, such that
$D^{0,1}\circ D^{0,1}=0$, i.e., a connection for which the $(0,2)$ component of the curvature vanishes.<|endoftext|>
TITLE:  Are all the theorems true?
QUESTION [6 upvotes]: The title sounds a bit philosophical, but it is
about mathematics. Let me explain. 
Consider a first order theory $T$, which is an extension 
of Peano Arithmetic. Call this theory "good" if it is 
consistent and satisfies the following
Property: 
For any $\phi\in\Sigma^0_{n+1}$ such that 
$T\vdash \phi$ there exists  $\psi\in\Pi^0_{n}$ such that 
$T\vdash\psi$ and $PA\vdash\psi\to\phi$.
Question 1. Is $ZFC$ "good"?
Question 2. The same for $ZFC+something$ (from the lot of proposed new axioms). 
Motivation. 
If $ZFC$ is not "good" then there (EDIT) may be theorems which can be proved in $ZFC$
despite they are false (in the standard model of PA).
I believe that $ZFC$ is "good". However, I would like 
to know if there is a formal proof. (Admittedly, I don't have a slightest 
idea what a proof may be like). By the way, "goodness" implies consistency, 
hence a proper proof requires some new axioms (a large cardinal, perhaps). 
(EDIT). As Andreas Blass pointed out correctly, 
even if a theory is not "good" in the above sense,
it does not yet follow that some of the theorems are wrong 
(an obvious fact which I have missed somehow). Still, 
the question if ZFC is "good" may be of some interest, in my opinion.
Question 3. Is "goodness" equivalent to consistency?
(I doubt this).
EDIT: (Clarification). In this question, the theory $T$ is supposed to be "good"
and at least as strong as $ZFC$. (Thus, the answer to Question 1 
must be yes). The question is, whether $T\vdash Con(T)\to Good(T)$, 
where $Good(T)$ is a formalization of "goodness"; note that 
$Good(T)\in \Pi^0_{2}$.
P.S. Is there a standard term for "good"?

REPLY [11 votes]: $\def\zfc{\mathrm{ZFC}}\def\pa{\mathrm{PA}}$First, there is no consistent recursively axiomatizable theory extending Robinson’s arithmetic which has the property of having existential witnesses as described by Sridhar Ramesh. Let $\pi=\forall x\,\theta(x)$ be a true but $T$-unprovable $\Pi^0_1$ sentence with $\theta$ bounded, which exists by Gödel’s theorem. Then $\exists y\,(\theta(y)\to\forall x\,\theta(x))$ is a tautology, but there is no $n\in\omega$ such that $T\vdash\theta(n)\to\forall x\,\theta(x)$: since $\pi$ is true, $\theta(n)$ is provable in Robinson’s arithmetic, hence $T$ would prove $\pi$.
In fact, an iteration of the same idea shows that the only consistent theory with the property of having existential witnesses is the true arithmetic $\mathrm{Th}(\mathbb N)$.
The situation with goodness is more complicated: there are good theories, such as any consistent theory axiomatizable over $\pa$ by a set of $\Pi^0_1$ sentences. Nevertheless, neither $\zfc$ nor any its recursively axiomatized extension is good.
Let $T=\zfc$, or more generally, let $T$ be any recursively axiomatizable extension of $\pa$ which proves the local $\Sigma^0_1$-reflection principle for $\pa$. Let $\Box_\pa$ denote the provability predicate for $\pa$, and $T_{\Pi^0_1}$ the set of all $\Pi^0_1$ theorems of $T$. By a theorem of Lindström, there exists a $\Pi^0_1$ sentence $\pi$ such that $\pa+\pi$ is a $\Sigma^0_1$-conservative extension of $\pa+T_{\Pi^0_1}$. $T$ proves the reflection principle
$$\tag{$*$}\Box_\pa(\neg\pi)\to\neg\pi$$
which can be written as a $\Sigma^0_2$ sentence, hence assuming $T$ is good, $(*)$ is provable in $\pa+T_{\Pi^0_1}$, and a fortiori in $\pa+\pi$. But then $\pa+\pi$ proves its own consistency, hence by Gödel’s theorem, it is inconsistent. By $\Sigma^0_1$-conservativity, $\pa+T_{\Pi^0_1}$ is also inconsistent, hence $T$ is inconsistent, contradicting its goodness.
Reference:
Per Lindström, On partially conservative sentences and interpretability, Proc. AMS 91 (1984), no. 3, pp. 436–443.<|endoftext|>
TITLE: Evaluating the average distance from a point in the unit disk to the disk
QUESTION [6 upvotes]: I am interested in finding the average Euclidean distance from a point $(x,y)\in\mathbb{D}_2$, the unit disk $\{(u,v):u^2+v^2\leq 1\}\subseteq\mathbb{R}^2$, to the disk $\mathbb{D}_2$. This amounts to essentially calculating the integral
$$\iint_{\mathbb{D}_2} \|(x,y)-\omega\|_2d\omega.$$
By rotational symmetry this gives an integral of the form
$$\iint_{\mathbb{D}_2}\|(a,0)-\omega\|_2d\omega$$
where $0\leq a \leq 1$. I believe this can be converted to an elliptic integral of the second kind where the integral over the angle $\theta$ appears in the form
$$\int_{0}^{2\pi}\sqrt{1-\frac{4ar}{(r+a)^2}\cos^2(\theta/2)}d\theta.$$
I am not sure how to proceed from here.

REPLY [11 votes]: Let me set
$$k^2=\frac{4ar}{(r+a)^2} <1. $$ 
If we replace $\theta$ with $2\theta$ we reduce this to an integral;
$$ 2 \underbrace{\int_0^\pi \sqrt{1-k^2\cos^2\theta} d\theta}_{=I}. $$
Now set
$$ x=\cos\theta $$
so that  $$ dx=-\sqrt{1-x^2} d\theta $$
and 
$$ I=\int_{-1}^1\frac{\sqrt{1-k^2x^2}}{\sqrt{1-x^2}} dx= 2\underbrace{\int_0^1 \frac{\sqrt{1-k^2x^2}}{\sqrt{1-x^2}} dx}_{=:E_2(k)}.  $$
The integral $E_2(k)$ is called  Jacobi's complete elliptic integral of the second kind. There is no simple formula for it but  you can have a look at the beautiful book by H. McKean and V. Moll, Elliptic Curves. Function Theory, Geometry Arithmetic Cambridge University Press,1997.<|endoftext|>
TITLE: Lattice-cube minimal blocking sets
QUESTION [6 upvotes]: Let $C_d(n)$ be the lattice cube consisting of the $n^d$ points with
each of its $d$ coorindates in $\lbrace 1,2,\ldots,n \rbrace$.
Define a blocking set for a lattice cube to be a set of points
in the cube that meets every axis-parallel line through a lattice
point
of any facet of the cube.
Let $b_d(n)$ be the minimum cardinality of a blocking set for
$C_d(n)$.

Q1. What is the function $b_d(n)$?

For $d=2$, selecting the $n$ points on the diagonal through $(1,1)$
and $(n,n)$ suffices to block the $n$ lines through each facet/side of the cube/square;
so $C_2(n)=n$.
Because a facet consists of $n^{d-1}$ points, and a line orthogonal to
the facet through each point must be blocked, $n^{d-1}$ is a lower bound on
$b_d(n)$.  A more specific version of Q1 is:

Q2. Can the lower bound be achieved?  Is $b_d(n) = n^{d-1}$? If the lower bound cannot
  always be achieved, when can it be achieved?

For $C_3(3)$, the $\mathbb{Z}^3$-cube consisting of $27$ points, here is a blocking set $B$ with $|B|=10=n^2+1$:

        

Perhaps these questions have already been explored?
If so, I would appreciate references.  Thanks!
My interest in the above questions derive from the 
more geometrical question:
What is the 
"minimal object'' that casts a (hyper)cube-shadow from parallel light rays in 
the $d$ coordinate directions? I defer this question until I understand the
combinatorial version.
Addendum.  As per Johan Wästlund and Eoin, indeed the lowerbound is achievable.
Here is a 9-point blocking set for $C_3(3)$:

REPLY [2 votes]: This is somewhat of an expanded comment on Eoin's answer and his follow-up question of whether such sets correspond to hyperplanes in the discrete torus.
In the $C_2(n)$ case minimal blocking sets correspond to permutation of $\{1...n\}$ (consider a permutation matrix; then a minimal blocking set can be formed by choosing the points corresponding to the positions of the 1's in this matrix). Conversely, every minimal blocking set in the $d=2$ case arises this way, so for $n\geq 4$ not every blocking set corresponds to a hyperplane on a torus.
In the $d=3$ case this can be extended to see that every minimal blocking set corresponds to a set of $n$ permutations whose permutation matrix sums to the all 1's matrix (think of stacking the permutation matrices up to form our cube); again it is clear that not all such sets can correspond to hyperplanes if $n \geq 4$.
For $d \geq 4$, one can apply an analogous stacking approach using sets in $C_{d-1}$ to generate all minimal blocking sets although it is not immediately obvious to me how to relate these sets to permutations easily.
For $n = 2$, it is easy to see that there are exactly two minimal blocking sets in any dimension and both always correspond to toric hyperplanes, and it seems that all minimal blocking sets are also toric hyperplanes for $n=3$ in any dimension due to how little room there is in $S_3$ to form such sets (this should follow from induction on the dimension; once one lays down the first $d-1$-dimensional layer and one point in the next layer, this should determine where all the remaining points must go in the last two layers). If anyone can easily formalize this idea for the $n=3$ case (or provide a counterexample) I would love to see it.<|endoftext|>
TITLE: compact quotient
QUESTION [11 upvotes]: Let X be a topological space that is not too bad (let's say "not too bad" = "compactly generated Hausdorff"), and let ∼ be an equivalence relation such that X /∼ is compact Hausdorff.
Does there exist a compact subspace A⊂X that meets every equivalence class of ∼? (This would then imply that A /∼ is homeomorphic to X /∼).

REPLY [5 votes]: Here is a counterexample, inspired by Henrik Rüping's answer, but somewhat simpler:
Take $X$ to be the disjoint union of $\quad X_1:=([0,1]\times [0,1])\setminus \{(0,0)\}$
and $\quad X_2:=(]0,1]\times [0,1])\cup ([-1,0]\times\{0\})$. Both $X_1$ and $X_2$ are given the subspace topology from $\mathbb R^2$.
The quotient space is $[0,1]^2$, with quotient map $\pi:X\to [0,1]^2$.
The map $\pi|_{X_1}$ is the inclusion $X_1\hookrightarrow [0,1]^2$.
The map $\pi|_{X_2}$ is a continuous bijection $X_2 \to [0,1]^2$ given by $\pi(-a,0)= (0,a)$ for $(-a,0)\in [-1,0]\times\{0\}$.<|endoftext|>
TITLE: An extension of Edelstein's Attraction Theorem?
QUESTION [5 upvotes]: Let $X$ be a metric space.  Recall that a function $f: X \rightarrow X$ is contractive if there exists $C \in (0,1)$ such that for all $x,y \in X$, $d(f(x),f(y)) \leq C d(x,y)$; a function $f$ is weakly contractive if for all $x \neq y \in X$, $d(f(x),f(y)) < d(x,y)$.
Also let us say that an attracting point for $f: X \rightarrow X$ is a point $L \in X$ such that for every $x_0 \in X$, the sequence of iterates $\{ (f \circ \cdots \circ f)(x_0)\}_{n=0}^{\infty}$ converges to $L$.  (Since $f$ is continuous, an attracting point is necessarily a fixed point.)  Finally, we say $f: X \rightarrow X$ is attractive if it has an attracting point.  
Recall the following two Attraction Theorems:


Banach Attraction Theorem: If $X$ is complete, then every contractive mapping $f: X \rightarrow X$ is attractive.
Edelstein Attraction Theorem: If $X$ is compact, then every weakly contractive mapping $f: X \rightarrow X$ is attractive.  


(Side Remark: Of course these theorems imply that $f$ has a fixed point and are usually called "fixed point theorems".  In the case of Banach's Theorem the stronger conclusion is well known: in fact it is well known that the sequences of iterates converge with exponential speed.  But in the case of Edelstein's Theorem, most of the literature I have found states it as a fixed point theorem only, despite the fact that this weaker conclusion is at the level of a homework or exam problem: consider $\min_{x \in X} d(x,f(x))$...One exceptional reference which treats this topic very nicely is Keith Conrad's note on contraction mappings.)
I started thinking about these results in the context of teaching a "Spivak Calculus course" this past academic year.  In the context of that course, functions were defined on intervals, not metric spaces, but not necessarily intervals that are closed and/or bounded.  Thus one is not always working in a complete metric space.  It is clear that in an incomplete case a contraction mapping need not have a fixed point: one need look no further than
$f: (0,1) \rightarrow (0,1), \ x \mapsto \frac{x}{2}$.
However this counterexample doesn't look "serious", because $f$ continuously extends to $[0,1)$ and has $0$ as an attracting point there.  Thinking over this phenomenon, I realized it was rather general:

$3$. Let $I \subset \mathbb{R}$ be an(y) interval, and let $f: I \rightarrow I$ be continuous.
  a) At least one of the following occurs:
  (i) $f$ has a fixed point in $I$.
  (ii) $\sup I$ is an attracting point for $f$.  (If $I$ is unbounded above, this means that every sequence of iterates diverges to $\infty$.)
  (iii) $\inf I$ is an attracting point for $f$.  (If $I$ is unbounded below...)
  b) If $f$ is moreover weakly contractive, then it has an attracting point in $[\inf I,\sup I]$. 

I expect many readers will find a proof immediately, but see $\S 10.4$ of these notes if you like.
The case $I = \mathbb{R}$ of Theorem 3a) was the subject of a short Monthly note of A.F. Beardon.
Is there an extension of Theorem 3 to $\mathbb{R}^n$?  If you look at the proof of Theorem 3, you see that it really uses the order-theoretic, rather than the metric properties of $\mathbb{R}^n$: when $I$ is a bounded interval the order-theoretic completion and the metric completions coincide, but not in the unbounded case.  So let's restrict to bounded sets and ask the following question:

Question 1: Let $X \subset \mathbb{R}^n$ be a bounded subset.  Does every weakly contractive map $f: X \rightarrow X$ have an attracting fixed point in $\overline{X}$ (closure in $\mathbb{R}^n$ = metric completion)?

Here is a reformulation / generalization of the question using the concepts of metric spaces.  If I have a contraction mapping $f$ on an incomplete metric space $X$, I can't expect it to have a fixed point, but I can ask for a fixed point in the unique extension of $f$ to a continuous mapping $\overline{f}: \overline{X} \rightarrow \overline{X}$ on the metric completion $\overline{X}$.  But it is easy to see that if $C$ is a contraction constant for $f$, it is also a contraction constant for $\overline{f}$, so the following result is not really any more general than Theorem 1.

$1'$ Every contractive map on a metric space extends to an attractive map on the metric completion.

As for 2., it is clear that the compactness hypothesis cannot just be thrown out entirely: for instance $f: \mathbb{R} \rightarrow \mathbb{R}, \ x \mapsto \sqrt{x^2+1}$ is weakly contractive but has no (finite!) fixed point.  However, recalling that a metric space is compact iff it is complete and totally bounded  there is a way to "remove only the completeness hypothesis".  So I ask:

Question 2: Let $X$ be a totally bounded metric space.  Must every weakly contractive map on $X$ have an attracting point (or even a fixed point) in the compact space $\overline{X}$?  

Note that bounded sets in $\mathbb{R}^n$ are totally bounded, so an affirmative answer to Question 2 indeed implies an affirmative answer to Question 1.
Finally, is a priori conceivable -- although I am rather doubtful about it -- that Question 2 may follow trivially from Theorem 2 in the same way that Theorem 1'. follows trivially from Theorem 1.  That is:

Question 3: Let $f$ be a weakly contractive map on a metric space $X$, and let $\overline{f}$ be the unique continuous extension to the metric completion $\overline{X}$.  Must $\overline{f}$ be weakly contractive?

REPLY [8 votes]: Q1.
Take $X=\{\\,z\in\mathbb C\mid 1<|z|<2\\,\}$ and $f(z)=-\tfrac{z}{\sqrt{|z|}}$.
It seems to answer the other questions.
You may take a spiral in X which is $f$-invariant. This produce a simply connected example.<|endoftext|>
TITLE: About the intrinsic definition of the Weyl group of complex semisimple Lie algebras
QUESTION [15 upvotes]: It may be a easy question for experts. 
The definition of the Weyl group of a complex semisimple Lie algebra $\mathfrak{g}$ is well-known: We first $\textbf{choose}$ a Cartan subalgebra $\mathfrak{h}$ and we have the root space decomposition. The Weyl group now is the group generated by the reflections according to roots.
Naively this definition depends on the choices of the Cartan subalgebra $\mathfrak{h}$. Of course we can prove that for different choices the resulting Weyl groups are isomorphic.
My question is: can we define the Weyl group intrinsically such that we don't need the do check the unambiguity. 
One thought is: we have the abstract Cartan subalgebra $\mathfrak{H}:=\mathfrak{b}/[\mathfrak{b},\mathfrak{b}]$ of $\mathfrak{g}$ (which is in fact not a subalgebra of $\mathfrak{g}$). Can we define the Weyl group along this way? Again is there any references for this?

REPLY [14 votes]: Sometimes when you define a group using an arbitrary choice of object and then show the choice of object doesn't matter, you could have defined a groupoid without making an arbitrary choice.  
For example, to define the fundamental group $\pi_1(X,x)$ of a path-connected space $X$ we need to choose a basepoint $x \in X$, but then we can show we get isomorphic groups no matter what basepoint we choose, with an isomorphism given by a homotopy class of paths between the basepoints.  To avoid this maneuver we can work with the fundamental groupoid of $X$, whose objects are points of $X$ and whose morphisms are homotopy classes of paths.  If $X$ is path-connected all objects in this groupoid are isomorphic, and thus the automorphism groups of all objects are isomorphic. The automorphism group of $x$ is just $\pi_1(X,x)$.  The fundamental groupoid is thus equivalent, as a category, to the one-object groupoid corresponding to the group $\pi_1(X,x)$.   But the advantage of the fundamental groupoid is that we can define it without choosing a basepoint, and it makes sense and works well even when $X$ is not path-connected.
Similarly, I think we can define the Weyl groupoid of a compact semisimple Lie group $G$ in a way that gives a groupoid equivalent to the usual Weyl group, but doesn't require a choice of maximal torus.  The idea should go like this.  The objects of the Weyl groupoid are maximal tori.  A morphism $f : T \to T'$ in the Weyl groupoid is a Lie group isomorphism of the form
$$   t \mapsto g t g^{-1}   \textrm{ for all } t \in T $$
for some $g \in G$.   If I did this right, the automorphism group of any object $T$ in the Weyl groupoid is the usual Weyl group 
$$  W_G(T) = N_G(T) / T ,$$
that is, the normalizer of $T \subset G$ modulo the centralizer of $T \subset G$, which is $T$ itself.  If this is true, the Weyl groupoid will be equivalent, as a groupoid, to the usual Weyl group $W_G(T)$ for any maximal torus $T$.<|endoftext|>
TITLE: Elementary submodels of V
QUESTION [7 upvotes]: Consider the claim:
(C) There is a transitive set $S \in V$ such that the structure $(S, \in)$ is an elementary submodel of $(V, 
\in)$.
Obviously, this claim cannot be a theoreom of ZFC, by Godel's 2nd Incompleteness Theorem. But does (C) follow from ZFC+CON(ZFC)? Or are large cardinals needed to prove (C)? More generally (and vaguely), what is the weakest assumption needed to prove (C)?

REPLY [13 votes]: A cardinal $\delta$ is correct if the theory $V_\delta\prec V$ holds, that is, if $V_\delta$ is an elementary substructure of $V$.  This theory is expressible as a scheme in first order logic in the language of set theory augmented by a constant symbol for $\delta$ as the following scheme of statements:
$$\forall \vec a\in V_\delta\ \ (\ \varphi(\vec a)\quad\iff\quad V_\delta\models\varphi[\vec a]\ \  ).$$
Note that any set $\langle S,{\in}\rangle$ such as you desire must have the form $S=V_\delta$ for some $\delta$, since it will the union of the $V_\alpha$ for $\alpha<\delta$ because by elementarity it will be right about these $V_\alpha$, and so this is your main case.
The property of being correct is not expressible as a single assertion in ZFC, unless inconsistent, since otherwise there could be no least reflecting cardinal, since if $\delta$ is the least reflecting cardinal and this were expressible, then $V$ would have a reflecting cardinal, but $V_\delta$ would not. A similar argument applies directly to the existence of a set $\langle S,{\in}\rangle$ as in your question, showing that $S\prec V$ is not expressible as a single assertion in the language of set theory.
Meanwhile, as a scheme, the existence of a reflecting cardinal is equiconsistent with ZFC, and in particular it does not imply Con(ZFC). To see this, observe that if there is a model of ZFC, then by the reflection theorem, we may easily produce models of any given finite subset of the theory. So by compactness the theory $V_\delta\prec V$ is also consistent. 
One may easily extend this theory to have a unbounded closed proper class $C$ of cardinals $\delta$ for which $V_\delta\prec V$, a theory known as the Feferman theory, and this also is equiconsistent with ZFC, with no increase of consistency strength, by the same argument. Feferman propsed this theory as a natural background set theory in which to undertake category theory, because it provides a robust universe concept (more robust than Grothendieck universes in that the universes all cohere into an elementary chain), but without the large cardinal commitment. 
If one adds to the theory $V_\delta\prec V$ that $\delta$ is inaccessible, then $\delta$ is a reflecting cardinal, whose existence is equiconsistent with Ord is Mahlo. 
If one desires a purely first-order concept, expressible as a single property in the language of set theory, then one may restrict to a less severe form of elementarity. For example, a cardinal $\delta$ is $\Sigma_2$-correct if $V_\delta\prec_{\Sigma_2} V$, that is, $V_\delta$ is elementary for formulas of complexity $\Sigma_2$, and $\Sigma_2$-reflecting if $\delta$ is also inaccessible. For example, the $\Sigma_2$-reflecting cardinals commonly arise in large cardinal arguments, and this is expressible in the language of set theory. More generally, one has the concept of a $\Gamma$-reflecting cardinal for any class $\Gamma$ of formulas, and if these have bounded complexity, then these also will be expressible. 
For many purposes, the $\Sigma_2$-correct cardinals carry much of the useful power of what you want from $V_\delta\prec V$, since one of the equivalent formulations is that $\delta$ is $\Sigma_2$ correct if and only if whenever there is $\theta$ with $V_\theta\models\sigma$ for any assertion $\sigma$, using parameters in $V_\delta$, then there is $\theta\lt\delta$ with $V_\theta\models\sigma$. In other words, any thing that happens in a $V_\theta$ above $\delta$ also happens below $\delta$.<|endoftext|>
TITLE: Fast-growing hierarchy and Turing machines
QUESTION [12 upvotes]: Is it possible to get an estimate of the size of a Turing machine computing $f_\alpha(n)$, for a given $\alpha$ (I am especialy interested in moderately large $\alpha$ like the ordinal of Fefferman-Schütte, or the small Veblen ordinal)? The idea is to get an idea of the size of the BusyBeaver function $BB(n)$ for moderate values of $n$, as the litterature usually only mention the exact known values (for $n\le 6$) and the fact that such values will probably never be known for $n=10$, say.

REPLY [2 votes]: Consider Gödel's speed-up theorem which discusses sentences like "this sentence is unprovable in theory T with less than $\phi$ symbols" where $\phi$ is some computable formula like TREE(3), and T is some consistent, effective theory like PA or ZFC with large cardinals (assuming those are consistent).  Call the above sentence S.
Per the speed-up theorem, S is both true and provable (see the wiki article).  The provability means that for reasonable theories, there is a relatively small $n$ such that an $n$-state Turing machine will eventually (by unbounded search) find the proof and halt, but that the running time will be (much) longer than $\phi$.  Of course BB(n) has to be even bigger than that.  With a suitable encoding, n=15 might be enough.  So BB(n) quickly dominates any computable formula that can be written down.
"BB(15) is uncomputable" could be interpreted something like: there is a 15-state TM that halts if and only if your favorite strong arithmetic theory T is inconsistent.  So if you believe that T really is consistent, that means you'll never be able to know what BB(15) is.<|endoftext|>
TITLE: got any tricks to build up t-structures on derived categories?
QUESTION [12 upvotes]: Are there any good tricks to construct a heart of a t-structure? (I'm thinking on the derived category of coherent sheaves of some variety)
I'll start with the only one I know. If $(T,F)$ is a torsion pair on an abelian category $A$ then you can form the tilt inside $D(A)$.
By definition, the tilt is complexes in $D(A)$ such that the minus one cohomology lies in $T$ and the zeroth cohomology lies in $F$ and all other cohomology vanishes.
Unfortunately this method is only good for hearts concentrated in two degrees.
An example of such a torsion pair is if you take $Ab$ the category of abelian groups, $T$ torsion groups and $F$ free groups.
In any abelian group you can find the largest torsion subgroup and the quotient will be torsion free.
This can be generalised to arbitrary integral domains and from that to arbitrary integral schemes, although the geometric meaning of the tilt (if there is one) escapes me.
Another example is Bridgeland's category of perverse coherent sheaves, which can be defined as a tilt.
I would be very interested in generalisations of Bridgeland's category. There is also a notion of perverse coherent sheaf by Bezrukavnikov, but from what I understand it is only useful with Artin stacks, as the jump in the dimension of the coarse space given by stabiliser groups is what allows interesting perversities. Comments on these latter perverse sheaves would also be very much appreciated, as I don't really understand the construction.

REPLY [20 votes]: This will be a short overview on techniques I am familiar with. For simplicity, I will talk about bounded t-structures, which are determined by their heart $\mathcal{A} = D^{\le 0} \cap D^{\ge 0}$; and on the bounded derived category of coherent sheaves $D^b(X)$ on a variety/stack.

Tilting is in principle extremely powerful: $\mathcal{A}_1$ is obtained by tilting from $\mathcal{A}_2$ whenever objects of $\mathcal{A}_1$ have only two cohomologies with respect to $\mathcal{A}_2$, e.g. $\mathcal{A}_1 \subset \langle \mathcal{A}_2, \mathcal{A}_2[1]\rangle$. (See e.g. Lemma 1.1.2 in Link)

Tilting can be iterated. As an example, using 1. it is a not too difficult exercise to see that Bezrukavnikov's t-structures of perverse coherent sheaves can be constructed by iterated tilting. When $D^b(X)$ is equivalent to a derived category of quiver representations, iterated tilting can often be descibed by iterated quiver mutations.

It is not very difficult to construct torsion pairs. For example, when $\mathcal{A}$ is Noetherian, then any subcategory $\mathcal{T} \subset \mathcal{A}$ that is closed under extensions and quotients is the torsion part of a torsion pair $(\mathcal{T}, \mathcal{F} = \mathcal{T}^{\perp})$. Alternatively, any notion of stability condition on a heart $\mathcal{A}$, say, induced via a slope function: $\mathcal{T}_{> \mu_0}$ is the extension-closed subcategory generated by stable objects $E$ with $\mu(E) > \mu_0$. (See Link, section 6.)

Given a semi-orthogonal decomposition of a triangulated category, one can construct a t-structure on the full category by gluing t-structures on the components - this is all in the original BBD.

It is much easier to construct (unbounded) t-structures in the unbounded derived category $D_{qc}(X)$ of quasi-coherent sheaves. (Any subcategory closed under [1], extensions and small coproducts is the $D^{\le 0}$-part of a t-structure.) Sometimes one can use this to construct bounded t-structures on $D^b(X)$ by showing that they restrict - see e.g. section 2 of Link. But in general t-structures on $D_{qc}(X)$ do not descend, and even if they do, it might be hard to prove.

As an example of the latter techniques, when $G$ acts freely on $X$, then $G$-invariant t-structures on $X$ are in 1:1-correspondence with t-structures on the quotient $X/G$ satisfying an additional assumption: tensoring with $f_* \mathcal{O}_X$ is right-exact.

Any derived equivalence $D^b(X) \cong D^b(\mathcal{A})$ induces a t-structure on $D^b(X)$ by pull-back - I guess you already knew that!


I realize that I talked mostly about tilting - I do think it's a very powerful method.<|endoftext|>
TITLE: "No Small Subgroups" Argument
QUESTION [10 upvotes]: What is the "no small subgroups" argument for $GL(n,\mathbb R)$? That is, how do we show that in $GL(n,\mathbb R)$ there exists a neighborhood of the identity which contains no subgroup other than the trivial one? I had some scribbling (for the $n=2$ case) but could not arrive at a clean proof.

REPLY [21 votes]: Here is Asaf's agrument expanded a bit.  It has the advantage of working for all Lie groups simultaneously.
Given a Lie group $G$ with Lie algebra $\mathfrak{g}$, consider the exponential map $\exp:\mathfrak{g}\rightarrow G$.  It is known that it  is a diffeomorphism on a small enough open set $U\subseteq\mathfrak{g}$.
Choosing an inner product on $\mathfrak{g}$, we may assume wlog that $U$ has the form $U = \{v\in \mathfrak{g} : \; |v| < \epsilon\}$ for some $\epsilon > 0$.  Let $V\subseteq U$ with $V = \{v\in\mathfrak{g} : \; |v| < \epsilon/2\}$.
I claim that $\exp(V)$ contains no nontrivial subgroups.  Indeed, suppose $H\subseteq \exp(V)$ is a subgroup and choose $g\in H$ so $g = \exp(v)$ for some $v\in V$.  I claim that $2v \in V$ as well.  To see this, notice that since $g^2 \in H\subseteq \exp(V)$, we must have $g^2 = \exp(w)$ for some $w\in V$.  Then $\exp(w) = g^2 = \exp(v)^2 = \exp(2v)$ which implies $w=2v$ since $\exp|_U$ is a diffeomorphism.  Thus, $2v \in V$.
But now can iterate this argument showing $2^n v \in V$ for all $n$.  Since $|2^n v| = 2^n |v|$, this implies $v=0$, i.e. that $g =e$ so $H$ is trivial.<|endoftext|>
TITLE: Geometric invariant theory for geometers
QUESTION [15 upvotes]: I am trying to learn "Geometric invariant theory" like it was introduced by Mumford. But I do not have a strong background in algebraic geometry since I work in geometric topology and geometry. 
So my question is if there is any nice reference where people explain geometric invariant theory from a geometric viewpoint. In particular, I am looking for a good reference where the analogies between algebraic geometry and differential geometry are pointed out.

REPLY [2 votes]: Read the survey in my article and go over the references therein. It is written with exactly similar intentions you have asked for.
Kalafat, Mustafa, Geometric invariant theory and Einstein-Weyl geometry, Expo. Math. 29, No. 2, 220-230 (2011); addendum ibid. 37, No. 1, 92-95 (2019). ZBL1250.14034, MR2787619.<|endoftext|>
TITLE: Eigenvalues of reverse bidiagonal matrices
QUESTION [6 upvotes]: Is there an easy way of determining if the eigenvalues of a real-valued reverse bidiagonal matrix are real. Basically I have two vectors $(a_1,...,a_n)$ and $(b_1,...,b_{n-1})$ that form the "reverse" diagonals of a matrix A. So that $A_{1,n}=a_1, ..., A_{n,1}=a_n$ and $A_{1,n-1}=b_1,...,A_{n-1,1}=b_{n-1}$ and all other $A_{i,j}=0$. What conditions must the vectors $a$ and $b$ satisfy to guarantee that $A$ has real eigenvalues? Thanks for help.

REPLY [4 votes]: One sufficient condition is that $b_i b_{n-i} \ge 0$ for $1 \le i < n/2$ and $a_i a_{n+1-i} \ge 0$ for $1 \le i < (n+1)/2$.  In fact, if $\ge$ is replaced by $>$, 
there is a diagonal matrix $U$ such that $U A U^{-1}$ is symmetric. 
On the other hand, if $a_i a_{n+1-i} < 0$ and $|b_i|, |b_{i-1}|, |b_{n+1-i}|, |b_{n-i}|$ are sufficiently small,  there  are non-real eigenvalues (close to $\pm \sqrt{a_i a_{n+1-i}}$).
Similarly if  $b_i b_{n-i} < 0$ and $|a_i|, |a_{i+1}|, |a_{n-i}|, |a_{n+1-i}|$ are sufficiently small, there are non-real eigenvalues close to $\pm \sqrt{b_i b_{n-i}}$.<|endoftext|>
TITLE: intensional equality in type theory
QUESTION [7 upvotes]: I want to know why we add an intensional equality in type theory to definitional equality ?
What is the aim with this intensional equality ?
thanks

REPLY [7 votes]: The key thing to notice is that the definitional equality is a judgement and not a proposition (so in particular, definitional equalities can not be part of your assumptions, nor can they be proved, they can only be checked by the type-checker).  The intensional equality, on the other hand, is a proposition, so it can be used as an assumption, and we can e.g. use induction to prove an intensional equality.  
The two equalities are closely connected, but do not generally agree: Martin-Löf's intensional type theory has decidable type checking, and hence also the definitional equality judgement is decidable, but you can convince yourself that the intensional equality is not decidable.
Note that definitional equality entails intensional equality, but not the other way around.<|endoftext|>
TITLE: What is the upper shriek in Grothendieck duality in the non-proper case? 
QUESTION [14 upvotes]: I'm trying to learn a little about Grothendieck duality. One version of the theorem states that if $f: X \to Y$ is a proper morphism of schemes, then the induced functor on derived categories $f_*: D^+(\mathrm{QCoh}(X)) \to D^+(\mathrm{QCoh}(Y))$ has a right adjoint $f^!$ (and under nice hypotheses, these will preserve the subcategories with coherent cohomology). The existence of an adjoint can be proved via adjoint functor arguments, even without assuming $f$ proper; this was done, e.g., by Neeman. (The point is that a triangulated  functor between nice triangulated categories (or,  stable $\infty$-categories) which preserves coproducts is a left adjoint.) 
However, in trying to identify $f^{!}$, we might want to be able to localize on $X$ and $Y$, and thus deal with the non-proper case. My understanding is that the upper-shriek functor $f^{!}$ there is not supposed to be the right adjoint to $f_{\ast}$: for example, for an open immersion it should be the upper-star $f^*$. 
In the topological version, one can define a $f_{!}$ functor for sheaves (push-forward with compact support) and $f^!$ is the right adjoint to $f_{!}$. Is there any "functorial" way to interpret $f^{!}$ when $f$ is not proper?

REPLY [18 votes]: Classically, the functor $f^!$ is indeed not a right adjoint in general. Clausen and I have recently found a way to make it a right adjoint in general, by enlarging the category of modules to that of solid modules, and constructing a general $f_!$ functor on solid modules directly. Solid modules are a version of "completed topological modules", but with excellent categorical properties. As an example, for $f: \mathbb A^1 =\mathrm{Spec} \mathbb Z[T]\to \mathrm{Spec}\mathbb Z$, one has
$$
f_! \mathcal O_{\mathbb A^1} = \left[\mathbb Z[T]\to \mathbb Z((T^{-1}))\right],
$$
the idea being that one wants to look at those functions $f\in \mathbb Z[T]$ that vanish to all orders at $\infty$, i.e. lie in the kernel of $\mathbb Z[T]\to \mathbb Z((T^{-1}))$. This kernel is of course trivial, but there's an interesting cokernel, so it gives an interesting functor on the derived level. And it's really necessary to consider $\mathbb Z((T^{-1}))$ not as an abstract $\mathbb Z$-module, but with its topological (or rather condensed) structure. See here for an account of this approach to coherent duality.
(I should say that a version of this is due to Deligne in the appendix to Hartshorne "Residues and Duality", working with pro-coherent sheaves.)<|endoftext|>
TITLE: Definability of the direct limit
QUESTION [7 upvotes]: Suppose $\kappa$ is a measurable cardinal and $U$ is a normal measure on $\kappa$. $M_1$ is the ultrapower of the universe $V$ constructed form $U$,$j_{01}$ is the induced elementary embedding, $M_2$ is the ultrapower of $M_1$ constructed from $j_{01}(U)$...We can iterate this process. We know $M_0,M_1,M_2,...$ and $j_{01},j_{02},j_{12},..$ are all classes of the universe, i.e. they are all definable. Also, they form a direct system, so we can define the direct limit $(M_{\omega},E_{\omega})$. $M_{\omega}$ and $E_{\omega}$ are both subset of the universe $V$, but my question is why they are definable subset of the universe?

REPLY [6 votes]: There are in fact many different definable ways to represent the direct limit. One particularly concrete representation of the direct limit is to use maximal threads, that is, maximal sequences $\langle x_i \mid j\leq i\lt\omega\rangle$, where $x_{i+1}=j_{i,i+1}(x_i)$, where $j_{i,i+1}:M_i\to M_{i+1}$ is the ultrapower of $M_i$ by $U_i=j_{0,i}(U)$, a normal measure on $\kappa_i$, where $j_{i,j}:M_i\to M_j$ is the corresponding composition of these embeddings. One then defines the $\in$ relation in $M_\omega$ by consulting the common coordinates, and this is well-defined. Alternatively, one may take the disjoint union of the $M_i$, and declare that any two such points are equivalent when they map to the same point in some later $M_i$. And there are many other representations of the direct limit, and these may all be defined by defining the constituent elements of the system used to define the direct limit, often taking the quotient by an equivalence relation, which is definable from the system and the measure.
The point is that any of these concrete representations of the direct limit can be defined from the measure $U$. With $U$ as a parameter, one may define any of the classes $M_i$ and also the embeddings $j_{i,j}:M_i\to M_j$, and furthermore, we may do so uniformly in $i$ and $j$. Thus, the point is that not only are the various $M_i$ and embeddings $j_{i,j}$ definable individually, but uniformly in the parameters $i$, $j$, and this is what it takes to define the direct limit using any of the usual representations of the direct limit. In short, we use the fact that we can define the entire system of embeddings $j_{i,j}:M_i\to M_j$ uniformly in order to know that we can build a representation of the direct limit of that system. 
But what is more, since the direct limit is well-founded, there is in fact a canonical way to represent the direct limit: the Mostowski collapse of your favorite representation. This provides $\langle M_\omega,{\in}\rangle$ as a transitive model of set theory, using the actual $\in$ relation. And the result will be a proper transitive class $M_\omega\subset V$, definable from $U$ as a parameter, with definable class embeddings $j_{i,\omega}:M_i\to M_\omega$.<|endoftext|>
TITLE: Does higher order arithmetic interpret the axiom of choice?
QUESTION [8 upvotes]: By second order arithmetic I mean the axiomatic theory $Z_2$, that is Peano arithmetic extended by second order variables with the full comprehension axiom, and not defined semantically using power set in ZF.  By third order arithmetic I mean that extended by third order variables and the comprehension axiom.  And so on.  Does each of these have an inner model which also satisfies the axiom of choice in each order, using constructibility?  If not, do such inner models exist if we also extend induction to a higher order axiom?  Is there a good reference on it?

REPLY [7 votes]: There is quite a bit of this in Simpson's book Subsystems of Second Order Arithmetic in the specific context of second-order arithmetic.  Here are three relevant results:

Corollary VII.5.11 (conservation theorems). Let $T_0$ be any one of the $L_2$-theories
   $\Pi^1_\infty\text{-CA}_0$,  $\Pi^1_{k+1}\text{-CA}_0$, $\Delta^1_{k+2}\text{-CA}_0$, $0 ≤ k < \infty$. Let $\phi$ be any $\Pi^1_4$ sentence. Suppose that $\phi$ is provable from $T_0$ plus $\exists X \forall Y (Y ∈ L(X ))$. Then $\phi$ is provable from $T_0$ alone.

Here $\Pi^1_\infty\text{-CA}_0$ has the full comprehension scheme for second order arithmetic, and hence also the full induction scheme. 

Theorem VII.6.16 ($\Sigma^1_{k+3}$ choice schemes). The following is provable in
  $\text{ATR}_0$. Assume $\exists X \forall Y (Y ∈ L(X ))$. Then: 

$\Sigma^1_{k+3}\text{-AC}_0$ is equivalent to $\Delta^1_{k+3}\text{-CA}_0$.
$\Sigma^1_{k+3}\text{-DC}_0$ is equivalent to $\Delta^1_{k+3}\text{-CA}_0$ plus $\Sigma^1_{k+3}\text{-IND}$. 
Strong $\Sigma^1_{k+3}\text{-DC}_0$ is equivalent to $\Pi^1_{k+3}\text{-CA}_0$.
$\Sigma^1_\infty \text{-DC}_0$ ($=\bigcup_{k < \omega} \Sigma^1_k\text{-DC}_0$ ) is equivalent to $\Pi^1_\infty\text{-CA}_0$.


and

Corollary IX.4.12 (conservation theorem). For all $k <\omega$, $\Sigma^1_{k+3}\text{-AC}_0$ (hence also $\Delta^1_{k+3}\text{-AC}_0$ ) is conservative over $\Pi^1_{k+2}\text{-CA}_0$ for $\Pi^1_4$ sentences.<|endoftext|>
TITLE: Is there a categorical treatment of dynamical systems?
QUESTION [33 upvotes]: Let $X$ be a set and $(T,\cdot)$ an abelian group. Is there a category of $T$-dynamical systems on $X$ which yields useful information about $X$ and $T$?
More precisely, is there a category whose objects are dynamical systems, i.e., $\phi:X \times T \to X$ such that $\phi(x,t + s) = \phi(\phi(x,t),s)$? And if so,

what are morphisms $\phi \to \phi'$ of $T$-dynamical systems on $X$?

I assume that in general this would involve imposing structure on $X$: for example, a topology so that one could consider homotopically perturbing $\phi$. Mostly, I am interested in asking when two such dynamical systems may be considered equivalent, and what it would take to have functors from $T$-dynamical systems on $X$ to $T'$-dynamical systems on $X'$, and to have natural transformations of those functors, etc.
I promise I've done (some) homework by looking at this. But note that this document
only provides candidates for equivalent dynamical systems which presumably only accounts for isomorphisms in the desired category rather than all morphisms.
This may be too basic a question for the folks here; in this case I will delete it.

REPLY [7 votes]: In my opinion another interesting reference is "An algebraic approach to chaos" (Appl. Cat. Struct.
4 (1996) 423-441) by Susan Niefield. Unfortunately I could find neither a free version nor any followup anywhere, which is a pity. She had very interesting approach to formulating  phenomena related to ergodicity and some of its topological analogs like topological transitivity in a localic way which permitted her to treat metric, topological and algebraic cases in parallel.<|endoftext|>
TITLE: What is a higher genus analogue of the Pontryagin product?
QUESTION [5 upvotes]: Given a compact oriented aspherical $3$--manifold $M$ with torus boundary $\partial M\simeq T^2$ (e.g. a knot complement), the condition that the images in $\pi_1 M$ of basis $x,y\in \pi_1 T^2$ under the inclusion generate the boundary of a compact oriented $3$-manifold with fundamental group $\pi_1 M$ (the $3$-manifold $M$ itself, to be precise) is algebraically encoded by the vanishing of their Pontryagin product $\langle x,y\rangle\in H_2 (\pi_1 M)$, which you can think of as the fundamental class of the torus under the inclusion. Recall that the Pontryagin product sends two commuting elements of a group to an element of $H_2$ of the group.

Question: Given a compact oriented aspherical $3$--manifold $M$ with boundary a closed surface $\partial M \simeq \Sigma$ of genus $g\geq 2$, how can I express the statement that $\Sigma$ bounds a $3$--manifold (the $3$-manifold $M$) on the level of fundamental groups? What condition must images under the inclusion of a basis for $\pi_1 \Sigma$ satisfy? How can I express the fundamental class of $\Sigma$ in terms of elements of $\pi_1$, if I know what $\Sigma$ is topologically?

I'm trying to identify some homomorphs of the fundamental group of a knotted theta, with peripheral data, and the answer to this question would, I presume, be something all such homomorphs would have to satisfy.

REPLY [3 votes]: An obvious necessary condition is that the pair $(M,\Sigma)$ satisfies Lefschetz duality. Since $M$ and $\Sigma$ are aspherical, this can be reformulated in terms of group cohomology.
Let $\Gamma=\pi_1(M)$ and $S=\pi_1(\Sigma)$. Assume that the inclusion $\Sigma\subset M$ is $\pi_1$-injective, so that $S$ is a subgroup of $\Gamma$. Then the pair $(\Gamma,S)$ is a Poincaré duality pair of dimension $3$. See Definition 4.6 of this survey by Davis, which refers to papers of Bieri and Eckmann. 
This should place some pretty serious restrictions on how the homology of $\Sigma$ sits in the homology of $M$, from which you might be able to extract a useful answer to your question.<|endoftext|>
TITLE: Why are Galois Representations so important in Number theory ? 
QUESTION [12 upvotes]: Dear everyone, 
Motivation :
From the past few days, I have been reading about the Galois Representations . I was really amazed to see that every seminal idea in the theory of elliptic curves have  Galois Representation as an ingredient !!  
I would be very happy listening to :

What made Galois representations so famous ? ( especially in number theory ), I was wondering, may be Galois representations are having some special symmetries
that can facilitate the problem solving more easily. 
What are the special properties of Galois representations ? 

Case Study  : 
To describe the application of Galois representation in a beautiful manner, I came accross a paper of Skinner and Urban , where they relate the ranks of Selmer Groups to 
the non-vanishing of $L$-functions. They use this Galois representations as a major ingredient, but due to extensive use of Algebraic Geometry , I was not able to understand the quintessence of the paper. It was so difficult to read. But on the other hand, 
I know how can one relate the volumes of lattices ( groups ) to the $L$-functions, using Siegel's formula. But I didn't come across any such track in that paper ( The word Siegel is not found in that paper ) . May be they have used some other different approach. I would be very happy in listening to that , as an application of Galois Representation. 
Any other good applications of Galois Representations are welcomed with high appreciation. 
My Background : 
I know number theory ( Mass formula and other things ) and rudimentary theory of Elliptic curves. 
Epilogue :
I thank everyone for sparing your time in answering / reading my questions and other questions at MO, in-spite of your hectic schedule. 
-Shanmukha.

REPLY [10 votes]: For abelian number fields, the Dedekind zeta function factors into Dirichlet $L$-functions. Hecke characters are one-dimensional representations. For possible generalizations to non-abelian field extensions, you will require higher dimensional Galois representations and Artin $L$ functions.
(Maass 1949) Maass wave forms of eigenvalue $1/4$ correspond to 2-dimensional Galois representations, see Bump "Autom.reps...." Chapter 1.9. Similar things happen for modular forms of weight one (Hecke 1925). In general, automorphic representations and Galois representations are expected to be in a certain correspondence (the $L$ functions and the root numbers should be the same).
A possible conceptual explanation for the importance of Galois representations delivers the Tannaka-Krein theorem. Roughly, this states that knowing the representation theory is equivalent to knowing the group. The group you want to understand is the absolute Galois groups (with a profinite topology) via its Galois representations, and understand the Galois representation via automorphic forms.
Perhaps one famous example is the Taniyama Shimura conjecture and consequently Fermat's last theorem: A certain construction with the elliptic curve gave a Galois representation, and the later was then shown to correspond to an automorphic form.<|endoftext|>
TITLE: Mordell-Weil group of the universal abelian scheme
QUESTION [10 upvotes]: Let $n>2$ and let $k$ be either $\bf Q$ or a finite field whose characteristic is prime to $n$. Let $A_{g,n}$ be the moduli scheme, which represents the functor, which with every $k$-scheme $S$ 
associates the set of principally polarized abelian schemes $\cal A$ over $S$, together with a symplectic isomorphism $({\bf Z}/n{\bf Z})^{2g}_S\simeq A[n]$.` This scheme is geometrically irreducible by Chai-Faltings. 
Let ${\widetilde A}_{g,n}\to A_{g,n}$ be the universal family and let $K$ be the function field of $A_{g,n}$.
My ${\bf question}$ is: is anything known about  ${\widetilde A}_{g,n}(K)$ ? $(\ast)$ 
A guess would be that ${\widetilde A}_{g,n}(K)\simeq {\widetilde A}_{g,n}[n](K)$.` 
Note that part of the difficulty of the question $(\ast)$ lies in the fact that I am asking for the structure of 
${\widetilde A}_{g,n}(K)$ and not for the structure of its subset ${\widetilde A}_{g,n}({A}_{g,n})$.` 
In the case $g=1$ (elliptic curves), these two sets coincide and the question should be easier to answer.
A final remark is that question $(\ast)$ is maybe not "the right one". It might make more sense to ask for the structure 
of the group of rational sections of the universal abelian scheme over the moduli stack of all abelian varieties 
(forgetting level structures and even polarizations) - but this group is not the Mordell-Weil group of a concrete abelian variety so I prefer to focus on the more down-to-earth question $(\ast)$.

REPLY [10 votes]: For $g=1$ there is a classical paper of Shioda (one of the two cited below) that proves that in char. zero, the group is what you expect but in char. p there are situations in which you get sections of infinite order.
On rational points of the generic elliptic curve with level N structure over the field of modular functions of level N. 
J. Math. Soc. Japan 25 (1973), 144–157
On elliptic modular surfaces. 
J. Math. Soc. Japan 24 (1972), 20–59. 
I seem to recall that in char zero the same is true for $g>1$ but I don't remember the reference, so I can't be sure. In char p, I don't know.
Added $g>1$ char zero:
Silverberg, Alice
Mordell-Weil groups of generic abelian varieties. 
Invent. Math. 81 (1985), 71–106.<|endoftext|>
TITLE: The monotone closure of a $C^*$-algebra
QUESTION [7 upvotes]: Related to Jon's question, I have two questions. Let $\mathcal{A}$ be a concrete $C^*$-algebra on a Hilbert space $\mathcal{H}$. For any selfadjoint subset $S$ of $\mathbb{B}(\mathcal{H})$, let $S^m$ denote the set of elements of $\mathbb{B}(\mathcal{H})_{sa}$ that can be obtained as the strong limits of monotone increasing nets from $S$.

Question 1. Is $((\mathcal{A}_{sa})^m)^m=(\mathcal{A}_{sa})^m$? (Maybe this is very basic.)Question 2. Does the $C^*$-algebra $C^*((\mathcal{A}_{sa})^m)$ generated by $(\mathcal{A}_{sa})^m$ in $\mathbb{B}(\mathcal{H})$ coincide with the strong closure of $\mathcal{A}$ in $\mathbb{B}(\mathcal{H})$?

For Question 2, I have been thinking that Pedersen's up-down-up theorem [Theorem 2 in American Journal of Mathematics 94 (1972), 955-962] might be useful, but I couldn't figure out.

REPLY [6 votes]: I can give you a partial answer to question 1, which you may already be aware of if you're familiar with Pedersen's "C*-algebras and their automorphism groups".  Specifically, if $A\subseteq A^{**}$, i.e. if we are considering the universal representation of $A$, and if $A$ is either unital or separable then you do indeed have $(A_\mathrm{sa}^m)^m=A_\mathrm{sa}^m=\overline{A_\mathrm{sa}^m}(=$ norm closure of $A_\mathrm{sa}^m$).  For if $A$ is unital then, by Pedersen 3.11.7, $A_\mathrm{sa}^m=\overline{A_\mathrm{sa}^m}$, while if $A$ is separable then we again have $A_\mathrm{sa}^m=\overline{A_\mathrm{sa}^m}$, by Corollary 3.25(a) of Lawrence G. Brown's "Semicontinuity and Multipliers of C*-algebras".  By Pedersen 3.11.5, $\overline{A_\mathrm{sa}^m}$ consists precisely of those $x\in A^{**}_\mathrm{sa}$ that are lower semicontinuous on $Q=A^{*1}_+$, which are immediately seen to be closed under taking supremums of bounded increasing nets.
If we consider $A$ with its atomic representation instead, then the above remarks still apply, as there is then a normal morphism $\pi$ from $A^{**}$ onto $A''$ which is faithful on $A_\mathrm{sa}^m$, by Pedersen 4.3.13 and 4.3.15.  Arbitrary representations can be faithful on $A$ but not on $A^m_\mathrm{sa}$, although in general you might perhaps try to use the predual $A''_*$ of $A''$, in the topology induced by $A$, as a replacement for $A^*$ with the weak* topology.  The only problem is that $A''^1_{*+}$ may not be compact in this topology so you would have to somehow adjust the proof of Pedersen 3.11.2, on which 3.11.5 relies.
Incidentally, it is a problem of Akemann and Pedersen from 1973 (still open as of 2014 according to Brown) whether $A_\mathrm{sa}^m=\overline{A_\mathrm{sa}^m}$ for arbitrary $A$ in its universal representation, so if you or Nik have a counterexample it would be quite important.<|endoftext|>
TITLE: Does there exist a wide but not full abelian subcategory of an abelian category?
QUESTION [5 upvotes]: Does there exist an example of an abelian category A and abelian subcategory B, where B is wide but not full (as a subcategory of A)?

REPLY [13 votes]: Take $\textrm{Mod-}A$, where $A$ is a finite dimensional local $k$-algebra ($k$ a field). Let $\mathcal C$ be the category whose objects are the same as those of $\textrm{Mod-}A$, but whose homomorphisms are replaced by all $k$-vector space homomorphsims, i. e.  $\textrm{Hom}_{\mathcal C}(X,Y) := \textrm{Hom}_k(X,Y)$. By construction, $\textrm{Mod-}A$ is a wide subcategory of $A$, and clearly it isn't full unless $A$ is equal to $k$. Moreover, $\mathcal C$ is equivalent to the category of vector spaces over $k$ (since every $k$-vector space can be given the structure of a $k$-module by letting $A$ act via $A/\textrm{Rad}(A)\cong k$). Hence $\mathcal C$ is abelian. Even kernels and cokernels in $\textrm{Mod-}A$ coincide with those in $\mathcal C$.<|endoftext|>
TITLE: Minimal blocking objects with shadows like a cube
QUESTION [14 upvotes]: This is a more geometric version of the previous question,
"Lattice-cube minimal blocking sets".  I will first specialize to $\mathbb{R}^3$, $d=3$.
View an $n \times n \times n$ cube $C_3(n)$ as formed of $n^3$ unit cubes glued face-to-face.
I would like to find a minimal blocking object $B$ inside $C_3(n)$,
which I define as a collection of the unit cubes in $C_3(n)$ with the following
two properties:
(a) The shadow of $B$ by parallel light rays in the three orthogonal directions
(parallel to the cube edges) is an $n \times n$ filled square.
(b) $B$ is a connected object, in the sense that its dual graph is connected.
Here the dual graph has a node for each unit cube in $B$ and an edge
between each pair of cubes that share a face.
$B$ is intended to be a minimal volume object that casts shadows like a cube.
The connectedness condition ensures one could build a physical model of $B$.
The previous MO question did not include the connected condition. There it was
shown that blocking sets of size $n^2$ are attainable, $n^{d-1}$ in
dimension $d$.
Certainly that lower bound is no longer achievable in general,
as can be seen with $C_2(2)$: 
a $2 \times 2$ square in dimension $d=2$ needs $3$ rather than $2=2^1$
unit squares to form a connected blocking object.
Exploring $n=3$ in $\mathbb{R}^3$, I have been unable to create a connected blocking
object with fewer than 15 cubes:

 
 
 
 
 
 


\[ \left[\begin{array}{ccc} 0 & 1 & 1 \\\\ 0 & 1 & 0 \\\\ 1 & 1 & 0 \end{array}\right] \hspace{0.25 in} 
\left[\begin{array}{ccc} 0 & 1 & 0 \\\\ 1 & 1 & 1 \\\\ 0 & 1 & 0 \end{array}\right] \hspace{0.25in}
\left[\begin{array}{ccc} 1 & 0 & 0 \\\\ 1 & 1 & 1 \\\\ 0 & 0 & 1 \end{array}\right]
\]
Fifteen seems excessive—more than half!.  Can anyone see a better solution?
In addition, I do not see how to generalize to $C_3(n)$, let alone to
$C_d(n)$, the same question in $d$ dimensions.
Addendum. Here is Joel's 13-cube blocker from his answer below:

REPLY [9 votes]: Lower bound:
Let there be $c$ cubes in a blocking configuration. Consider the graph whose vertices are cubes so that cubes are connected if they share a face. Any connected graph has at least  $c-1$ edges. By the pigeonhole principle, there is at least one direction with at least $(c-1)/d$ edges in that direction. When you project parallel to that axis, the image has size $n^{d-1}$, and the size of the image is also at most $c - (c-1)/d$ since at least one vertex on each of those edges is redundant. So, 
$$ n^{d-1} \le c - \frac{c-1}{d}  = c \frac{d-1}{d} + \frac 1d$$ 
$$ c \ge \bigg(\frac {d}{d-1}\bigg)n^{d-1} - \frac {1}{d-1}. $$
For $d=3$, $c \ge \frac 32 n^2 - \frac12$ (mentioned by Gjergji Zaimi in the comments). This is sharp for $n=3$ by the construction with $13$ cubes shown by Joel David Hamkins.
Upper bound ($d=3$):
Here is a construction of a connected blocking configuration with $\frac32n^2 + O(n)$ cubes related to Zack Wolske's constructions. We'll identify the cubes with lattice points. We start with the points $\lbrace(x,y,z) | x-y \equiv z \mod n, 0 \le x,y,z \lt n \rbrace$, illustrated for $n=7$.
|......X|  |x......|  |.x.....|  |..x....|  |...x...|  |....x..|  |.....x.|
|.....X.|  |......X|  |x......|  |.x.....|  |..x....|  |...x...|  |....x..|
|....X..|  |.....X.|  |......X|  |x......|  |.x.....|  |..x....|  |...x...|
|...X...|  |....X..|  |.....X.|  |......X|  |x......|  |.x.....|  |..x....|
|..X....|  |...X...|  |....X..|  |.....X.|  |......X|  |x......|  |.x.....|
|.X.....|  |..X....|  |...X...|  |....X..|  |.....X.|  |......X|  |x......|
|X......|  |.X.....|  |..X....|  |...X...|  |....X..|  |.....X.|  |......X|

This is made of two triangles of points, in the planes $x-y=z$ (marked X) and $x-y=z-n$ (marked x), which are separated by some distance. We'll first make the connections within the triangles, and then connect the triangles to each other.
To connect the bottom triangle, use $n-1$ extra points (marked A) to connect the base of the triangle in the plane $z=0$ to itself and to the points in the triangle with $z=1$. Then for $i = 2, ..., n-1$, use $\lceil (n-i-1)/2 \rceil$ points (marked O) to connect the points in the plane $z=i$ to each other and to the lower points in the triangle. These points have $y$ odd, and are just above an X in the layer below. 
|......X|  |x......|  |.x.....|  |..x....|  |...x...|  |....x..|  |.....x.|
|.....XA|  |......X|  |x.....O|  |.x.....|  |..x....|  |...x...|  |....x..|
|....XA.|  |.....X.|  |......X|  |x......|  |.x.....|  |..x....|  |...x...|
|...XA..|  |....X..|  |....OX.|  |.....OX|  |x.....O|  |.x.....|  |..x....|
|..XA...|  |...X...|  |....X..|  |.....X.|  |......X|  |x......|  |.x.....|
|.XA....|  |..X....|  |..OX...|  |...OX..|  |....OX.|  |.....OX|  |x.....O|
|XA.....|  |.X.....|  |..X....|  |...X...|  |....X..|  |.....X.|  |......X|

This has added $n-1$ A's (this is one of the few correct uses of apostrophes to indicate a plural), and $0+1+1+2+2+...+\lfloor (n-1)/2 \rfloor = \lfloor (n-1)^2/4 \rfloor$ O's (see A002620). 
We repeat the process upside down to connect the upper left triangle using $n-2$ A's and $\lfloor (n-2)^2/4 \rfloor$ O's. 
|......X|  |x......|  |.x.....|  |..x....|  |...x...|  |....x..|  |....Ax.|
|.....XA|  |O.....X|  |xO....O|  |.xO....|  |..xO...|  |...x...|  |...Ax..|
|....XA.|  |.....X.|  |......X|  |x......|  |.x.....|  |..x....|  |..Ax...|
|...XA..|  |....X..|  |....OX.|  |O....OX|  |xO....O|  |.x.....|  |.Ax....|
|..XA...|  |...X...|  |....X..|  |.....X.|  |......X|  |x......|  |Ax.....|
|.XA....|  |..X....|  |..OX...|  |...OX..|  |....OX.|  |.....OX|  |x.....O|
|XA.....|  |.X.....|  |..X....|  |...X...|  |....X..|  |.....X.|  |......X|

Finally, we connect the upper and lower triangles with $n-2$ Z's. There are many choices for how to do this. We'll put them in $z=1$, $y=n-2$, $1 \le x \le n-2$. 
|......X|  |x......|  |.x.....|  |..x....|  |...x...|  |....x..|  |....Ax.|
|.....XA|  |OZZZZZX|  |xO....O|  |.xO....|  |..xO...|  |...x...|  |...Ax..|
|....XA.|  |.....X.|  |......X|  |x......|  |.x.....|  |..x....|  |..Ax...|
|...XA..|  |....X..|  |....OX.|  |O....OX|  |xO....O|  |.x.....|  |.Ax....|
|..XA...|  |...X...|  |....X..|  |.....X.|  |......X|  |x......|  |Ax.....|
|.XA....|  |..X....|  |..OX...|  |...OX..|  |....OX.|  |.....OX|  |x.....O|
|XA.....|  |.X.....|  |..X....|  |...X...|  |....X..|  |.....X.|  |......X|

In total, this configuration contains $n^2$ X's, $2n-3$ A's, $n-2$ Z's, and $\lfloor (n-1)^2/4\rfloor +\lfloor(n-2)^2/4 \rfloor = {n-1 \choose 2}$ O's, a total of $\frac 32 n^2 + \frac 32 n - 4$. 
Therefore, 
$$ \frac 32 n^2 - \frac 12 \le \min |C_3(n)| \le \frac 32 n^2 + \frac 32 n - 4.$$<|endoftext|>
TITLE: Can $\omega_1$ be supercompact?
QUESTION [9 upvotes]: Is "ZF + $\omega_1$ is supercompact" consistent relative to "ZFC + there is a supercompact cardinal"?
In particular, if $\delta$ is supercompact, does it remain so in $V(\mathbb{R} \cap V[G])$ where $G \subset Col(\omega,<\delta)$ is $V$-generic?  This seems to be the case for measurability but I am having trouble proving it for supercompactness.  It seems likely that someone else has tried this, so I though I'd ask here.
The appropriate definition of supercompactness in ZF is the one in terms of normal fine measures, where normality is defined using diagonal intersections.
I am aware that $\omega_1$ has some amount of supercompactness under AD. I am interested in a more direct proof using forcing, which I hope will give (full) supercompactness.

REPLY [3 votes]: While might not be a full and satisfactory answer, you might be interested in the following paper by Spector:

Spector, M. Iterated extended ultrapowers and supercompactness without choice.,
  Ann. Pure Appl. Logic 54 (1991), no. 2, 179–194.<|endoftext|>
TITLE: Beyond Presburger Arithmetic
QUESTION [9 upvotes]: Do there exist known examples of predicates $P$ (possibly functional) such that
1) $P$ admits a first-order definition in the language ${\Bbb N}(+,\times,0,1)$;
2) $P$ admits no definition that does not involve both $+$ and $\times$;
3) the theory of ${\Bbb N}(+,P,0,1)$ is decidable?
(Or 
3') the theory of ${\Bbb N}(\times,P,0,1)$ is decidable? 
Or even better, both.)
Last I heard (a long time ago) no one can prove the undecidability of ${\Bbb N}(+,{\rm Prime}(),0,1)$, but Alan Woods showed that standard conjectures imply the definability of multiplication in this language.

REPLY [7 votes]: From a paper by Françoise Point, "On the expansion $(N, +; 2^x)$ of Presburger arithmetic," I learned of a much more general result of Semenov, “Logical theories of one-place functions on the set of natural numbers”, Izv. Akad. Nauk SSSR Ser. Mat., 47:3 (1983), 623–658.  
Semenov's result implies that Presburger arithmetic together with the function $f(x) = c^x$ is decidable for any fixed $c \geq 2$.  Of course, $f$ is characterized by the fact that $f(0) = 1$ and $f(x) = c \cdot f(x-1)$ for $x > 0$.  So $f$ is definable in $(N,+,\times,0,1)$.
Something about the sparseness of powers of $2$, it seems... but I haven't read the details.<|endoftext|>
TITLE: Invertible matrix
QUESTION [5 upvotes]: Let $K$ be a field s.t. $charac(K)\not= 2$. Let $A\in\mathcal{M}_n(K)$ be such that $rank(A)\geq n/2$. Can one find a matrix $B$ such that $B$ is similar to $A$ and $A+B$ is invertible ?

REPLY [3 votes]: Initially my assumption was that the question was primarily about finite fields $K$, but after reading the comments it appears to me that there is some confusion about the infinite case as well. So I'll prove the claim for $|K|=\infty$. I would guess the assertion is also true if $K$ is finite, but I don't know how to show it.
If one assumes $K$ to be algebraically closed, then it's very easy to see that one can find an appropriate $B$: Let $\lambda_1,\ldots,\lambda_n$ be the eigenvalues of $A$ (eigenvalues with multiplicity $m$ are supposed to appear $m$ times in that list), then for any permutation $\sigma\in S_n$ we can conjugate $A$ to a matrix of the form
$$
   \textrm{diag}(\lambda_{\sigma(1)}, \ldots, \lambda_{\sigma(n)}) + N
$$
where $N$ is a strictly upper triangular matrix.
(This should be clear but I'll give a short explantion anyhow: View $K^n$ as a $K[x]$ module where $x$ acts as $A$. Then finding a basis that conjugates $A$ into the form given above corresponds to finding a filtration $0= V_0 < V_1 <\ldots< V_n= K^n$ of the $K[x]$-module $K^n$ such that the quotient $V_i/V_{i-1}$ is isomorphic to the simple $K[x]$-module $K[x]/(x-\lambda_{\sigma(i)})$. But by the structure theorem on f. g. modules over PID's, $K^n$ is a direct sum of modules each having just one isomorphism type of simple modules as composition factors. So choosing such a filtration is no problem: in each step just pick a maximal submodule in the desired isotypic direct summand and leave the other summands unchanged.)
We may assume that $A$ is of the above shape with $\sigma=\textrm{id}$. So just choose a permutation $\sigma$ such that $\lambda_i+\lambda_{\sigma(i)} \neq 0$ for all $i$ (this can be done: just choose $\sigma$ to be the transposition that swaps each $\lambda_i$ which is zero with some $\lambda_i$ which is non-zero and fixes the rest; the rank condition implies that there are enough non-zero eigenvalues to pair with the zero eigenvalues; also $char(K)\neq 2$ is essential: it guarantees $\lambda_i\+\lambda_i\neq 0$ whenever $\lambda_i\neq 0$). Then $A+B$ can becomes
$$
\textrm{diag}(\lambda_1+\lambda_{\sigma(1)}, \ldots, \lambda_n+\lambda_{\sigma(n)}) + \textrm{something strictly upper triangular}
$$
and you are done.
As for the case of a non-algebraically closed (but still infinite) $K$: Note that the algebraically closed case implies that the rational function
$$
   det(A+TAT^{-1})=\frac{det(AT+TA)}{det(T)} \in K(T_{ij})
$$
is non-zero. But $GL(n,K)$ is Zariski dense in $GL(n,\bar K)$ (see here), and so the above rational funktion must be non-vanishing on $GL(n,K)$, which shows there is a $T\in GL(n,K)$ such that $det(A+TAT^{-1})\neq 0$.<|endoftext|>
TITLE: Primes that are the sum of a positive cube and a square in 13 ways < 107122676734733201
QUESTION [11 upvotes]: In order to try and add to the integer sequence at http://oeis.org/A173795 I am attempting to fill in a gap in a sequence of primes that are the sum of a positive cube and a square in n different ways.
To date, and as far as I am aware, for n=0 to 10 the smallest primes are known that fulfill this criteria.  The largest being 333413867957257 where n = 10.
Candidates for the smallest at n=11 and n=12 are due to Elkies and these are 4417190430889897 and 84658174289284249 respectively.
There then exists a gap at n=13 before 107122676734733201 fulfills the criteria for n=14.
My questions are:

Are 4417190430889897 and 84658174289284249 the smallest primes for n=11 and 12 respectively?
Is 107122676734733201 the smallest prime where n=14?
Is there a known prime < 107122676734733201 where n=13?

Kevin.

REPLY [4 votes]: Many thanks to Noam Elkies for taking the time and effort to resolve the first part of this question and for sending me the results by email.
His, summarized, answer is:
The smallest primes for $n=11$ and $n=12$ are respectively $p=1057543811051633$ and $p=1448734752622601$.
For $(n, p) = (12, 1448734752622601)$.  There are no other examples of $n>11$  for prime $p$ less than $7.5 * 10^{15}$.<|endoftext|>
TITLE: Non-classical specializations of Hida families
QUESTION [10 upvotes]: Let ${\mathbb T}$ denote the ordinary $\Lambda$-adic Hecke algebra of say tame level $N$.  If I specialize ${\mathbb T}$ to a classical weight $k \geq 2$, then it is proven by Hida that the result is the Hecke algebra of classical $p$-ordinary forms of weight $k$ and level $pN$. (Here by specialize I mean mod out by the prime ideal $p_k \subseteq \Lambda = {\mathbb Z}_p[[\Gamma]]$ generated by $[\gamma] - \gamma^{k-2}$ where $\gamma$ is a topogical generator of $\Gamma$.)  I guess this is Hida's control theorem.
My question is the following: if one takes $\kappa$ to be some $p$-adic weight (but not necessarily classical), how can one describe ${\mathbb T}/ p_{\kappa} {\mathbb T}$?  
I assume the answer is that the result is the Hecke algebra of the space of overconvergent modular forms of weight $\kappa$ and slope 0 -- but I don't have a proof of this nor a reference.  Either would be greatly appreciated!

REPLY [5 votes]: I don't know if this is always true, and I don't think this is known in all cases, even after inverting $p$.
What is known and what is more or less formal to prove (it depends of your starting point, and in particular, of how you define the "$\Lambda$-adic Hecke algebra of tame level $N$) is
that there is a natural surjection, whose kernel is nilpotent, from ${\mathbb T}/p_\kappa {\mathbb T}$ to the  Hecke algebra of the space of overconvergent modular forms of weight $\kappa$ and slope 0. But then, the question of whether the kernel is trivial is difficult to settle without knowing more on the local structure of $\mathbb T$ near the points above $\kappa$.
Edit to answer HP's comment below : Let us consider a "toy model": imagine your
base ring is $\mathbb{Z}_p$, your module of ordinary p-adic form is 
$M=\mathbb{Z}_p^2$, and there is only one Hecke operator around, namely $T$ acting 
on the canonical basis of $M$ by $T(e_1)=p e_1$, $T(e_2)=0$. Then $\mathbb{T}=\mathbb{Z}_p[T]/T(T-p)$ is free of rank $2$ over $\mathbb{Z}_p$,
but on $M/pM$, $T$ acts like $0$ et the Hecke algebra on $M/pM$ is just $\mathbb{F}_p$ which is not $\mathbb{T}/p$.
Now what I say is not that this kind of examples actually happen on the eigencurve, but that
without supplementary hypothesis (like level $1$, artfully restricting to new forms, etc.)
no one, as far as I know, can rule this out.<|endoftext|>
TITLE: (phi, Gamma) module of ordinary elliptic curve
QUESTION [7 upvotes]: Suppose $E$ is an elliptic curve over $\mathbb{Q}_p$ with good ordinary reduction.  Can someone please tell me how to compute the associated $(\phi,\Gamma)$-module of the Tate module of $E$, or give me a reference to where it is computed?

REPLY [4 votes]: As Laurent has already pointed out, the representation is reducible and hence so is the phi-Gamma module, and writing down the two composition factors is easy; describing the extension class is harder. But it can be done: the relevant ext group is one-dimensional, so up to isomophism there is only one non-split extension, and you just need to write down any old non-split extension and you're done. Sarah Zerbes and I describe a way of doing this in our paper "Wach modules and critical slope p-adic L-functions"; see section 4 of http://arxiv.org/abs/1012.0175. There is another approach in one of Pierre Colmez's papers ("La serie principale unitaire" if I remember rightly).<|endoftext|>
TITLE: Valuations on tensor products
QUESTION [11 upvotes]: Let $A$ be a commutative ring, $B$ (resp. $C$) be a commutative $A$-algebra endowed with a valuation $v$ (resp. $w$), not necessarily of rank 1. Assume that $v$ and $w$ induce equivalent valuations on $A$. How to construct a valuation $u$ on $B\otimes_A C$ extending $v$ and $w$?
Without loss of generality, we may assume $A$, $B$, $C$ to be fields. If $B$ is an algebraic extension of $A$, the existence of $u$ follows from the fact that extensions of a valuation to a normal extension field are conjugate to each other [Bourbaki, AC VI 8 Prop. 7]. Thus the only case left to check is when both $B$ and $C$ are purely transcendental over $A$.
Huber lists the existence of $u$ as a "simple property" of valuations [Etale cohomology of Rigid Analytic Varieties and Adic Spaces, 1.1.14 f].  No proof is given there. Are there other references for this?
Added on Aug. 5: Let us denote the value groups of $A$, $B$, $C$ by $\Gamma_A$, $\Gamma_B$, $\Gamma_C$, respectively. The value group of $u$ is an extension of $\Gamma_B$ and $\Gamma_C$ over $\Gamma_A$. How to construct such an extension of linearly ordered Abelian groups? We could put the lexicographic order on $\Gamma_B\times \Gamma_C$, but then we cannot quotient out by the diagonal image of $\Gamma_A$ as the image is not convex.

REPLY [2 votes]: In terms of valuation rings the question is equivalent to the following: given valuation rings $A, B, C$ and injective local ring homomorphisms $A \to B$ and $A \to C$ there exists a ring map $B \otimes_A C \to D$ where $D$ is a valuation ring such that $B \to D$ and $C \to D$ are injective local ring homomorphisms.
To prove this, it suffices to find a specialization $x' \leadsto x$ of points of $\text{Spec}(B \otimes_A C)$ such that $x'$ maps to the generic points of $\text{Spec}(B)$ and $\text{Spec}(C)$ and such that $x$ maps to the closed points of $\text{Spec}(B)$ and $\text{Spec}(C)$. Namely, then we can apply Tag 01J8 to find $D$.
Denote $\kappa_A$ the residue field of $A$ and similarly for $B$ and $C$. Since $\kappa_B \otimes_{\kappa_A} \kappa_C$ is not the zero ring, there exists a point $x$ of $\text{Spec}(B \otimes_A C)$ mapping to the closed points of $\text{Spec}(B)$ and $\text{Spec}(C)$. Pick any maximal point $x'$ of $\text{Spec}(B \otimes_A C)$ specializing to $x$, in other words, $x'$ corresponds to a minimal prime ideal of $B \otimes_A C$. Since $A \to B$ and $A \to C$ are flat ring maps (as torsion free $A$-modules are flat), the ring maps $B \to B \otimes_A C$ and $C \to B \otimes_A C$ are flat as well. By going down for flat ring maps, we see that $x'$ maps to the generic points of $\text{Spec}(B)$ and $\text{Spec}(C)$. This finishes the proof.<|endoftext|>
TITLE: Circles and rational functions
QUESTION [37 upvotes]: Suppose that $\gamma$ is a Jordan analytic curve on the Riemann sphere,
and there exist two rational functions $f$ and $g$ such that
$f$ maps $\gamma$ into a circle, and $g$ maps a circle into $\gamma$.
(All rational functions considered are of degree at least $2$).
Question: Does this imply that $\gamma$ is a circle?
By a "circle" I mean a circle on the Riemann sphere, that is a circle or a straight
line in the plane.
I have only one example of such situation, where $\gamma$ is not a circle,
but in this example, $\gamma$ is not Jordan,
it has the shape of figure 8. 
If one restrict to polynomials $f$ and $g$, the question can be answered relatively
easily, using Ritt's factorization theory for polynomials, or other tools.
For the origin of this problem, see arXiv:1110.6552.

REPLY [26 votes]: Update: I removed the links to the Sage code of the complicated explicit examples, because very much easier examples exist. See below for an example, and this preprint for more details concerning this answer and the computation of explicit examples.
Answer: The answer is no. In the following I'll describe how to find two rational functions $f,g\in\mathbb C(X)$, both of odd prime degree $\ell$, such that the following holds:

 $f(g(X))\in\mathbb R(X)$.
 $g(\mathbb R\cup\{\infty\})$ is a smooth closed Jordan curve in the complex plane.

Let $E$ be an elliptic curve, defined over the reals. For $p\in E(\mathbb C)$ we let $\bar p$ be the complex conjugate of $p$. Choose $E$ such that the following holds:

 There is a point $w\in E(\mathbb R)$ with no $y\in E(\mathbb R)$ with $w=2y$.
 There is a point $z\in E(\mathbb C)$ of order $\ell$ with $\bar z\notin\langle z\rangle$. (Such a point always exists.)

Then $C=\langle z\rangle$ is a subgroup of order $\ell$ of $E$, and $E'=E/C$ is an elliptic curve over $\mathbb C$. Let $\phi:E\to E'$ be the associated isogeny, and $\phi':E'\to E$ be the dual isogeny. Then $\phi'\circ\phi:E\to E$ is the multiplication by $\ell$ map.
Let $\beta$ be the automorphism of order $2$ sending $p\in E(\mathbb C)$ to $w-p$. Similarly, define the involutory automorphisms $\beta'$ of $E'$ and $\beta''$ of $E$ by $\beta'(p')=\phi(w)-p'$ and $\beta''(p)=\ell w-p=\phi'(\phi(w))-p$.
Let $\psi$ be the degree $2$ covering map $E\to E/\langle\beta\rangle=P^1(\mathbb C)$, and define likewise $\psi':E'\to E'/\langle\beta'\rangle=P^1(\mathbb C)$ and $\psi'':E\to P^1(\mathbb C)$.
Let $f(X)$ and $g(X)$ be the rational functions defined implicitly by $\psi'\circ\phi=g\circ\psi$ and $\psi''\circ\phi'=f\circ\psi'$. Note that $\psi$, and $\psi''$ are defined over $\mathbb R$, while $\psi'$ is not.
As the multiplication by $\ell$ map $\phi'\circ\phi$ is defined over the reals, we obtain $f(g(X))\in\mathbb R(X)$.
We next claim that $g(X)$ is injective on $\mathbb R$. Suppose there are distinct real $u,v$ with $g(u)=g(v)$. Pick $p,q\in E(\mathbb C)$ with $\psi(p)=u$, $\psi(q)=v$. Then $\psi'(\phi(p))=g(u)=g(v)=\psi'(\phi(q))$, so $\phi(p)=\phi(q)$ or $\phi(p)=\phi(w)-\phi(q)$. Upon possibly replacing $q$ with $w-q$ we may assume $\phi(p)=\phi(q)$, so $p-q\in C$.
Next we study the effect of complex conjugation. As $\psi$ is defined over the reals, and $\psi(p)=u$ is real, we have $\psi(\bar p)=\psi(p)$, so $\bar p=p$ or $\bar p=w-p$. Likewise, $\bar q=q$ or $\bar q=w-q$. Recall that $p-q\in C$, and $\bar C\cap C=\{0\}$ by condition 2. So we can't have $(\bar p,\bar q)=(p,q)$, nor $(\bar p,\bar q)=(w-p,w-q)$.
Thus without loss of generality $\bar p=p$, $\bar q=w-q$. As $p-q$ and $\bar p-\bar q=p-w+q$ have order $\ell$, we see that $2p-w=r$ with $\ell r=0$ and $r\in E(\mathbb R)$. So $w=2(p+\frac{\ell-1}{2}r)$, contrary to condition 1. Furthermore, the function $g(X)$ behaves well at infinity by this geometric interpretation.
An explicit example, with $\omega$ a primitive third root of unity, is
\begin{align*}
f(X) &= \frac{X^3 - 6(\omega + 1)X}{3X^2 + 1}\\
g(X) &= \frac{2X^3 + (\omega + 1)X}{X^2 - \omega}\\
f(g(X)) &=  \frac{8X^9 - 24X^5 - 13X^3 - 6X}{12X^8 + 13X^6 + 12X^4
- 1}\in\mathbb Q(X).
\end{align*}
First note that $g(\mathbb R\cup\{\infty\})$ is not contained in a circle, for instance because the points $g(0)=0$, $g(\infty)=\infty$, $g(1/2)=(1+2\omega)/3$, and $g(1)=(5+4\omega)/3$ do not lie on a circle.
Secondly, $g(\mathbb R\cup\{\infty\})$ is a Jordan curve, because $g$ is injective on $\mathbb R$: Suppose that $g(x)=g(x+\delta)$ for real $x,\delta$. A short calculation yields $\delta(8\delta^4 + 14\delta^2 + 49)=0$, so $\delta=0$.<|endoftext|>
TITLE: Recent Fast Multiplication Algorithms for Large Integers
QUESTION [8 upvotes]: The STOC 2008 paper "Fast Integer Multiplication using Modular Arithmetic" by De et al
http://arxiv.org/abs/0801.1416 shows how to use $p$-adic numbers instead of $\mathbb C$ used in Furer's multiplication algorithm. Given that Furer's algorithm is not practical for less than a few thousand bits (is there a good estimate of its breakeven point vs Toom-Cook and other algorithms?), I wonder if the De et al algorithm has a lower breakeven point, is easier to implement, and is more practical in general.  
Pointers to follow-up work and/or attempts at implementing the STOC 2008 algorithm would be appreciated.
Igor

REPLY [3 votes]: For new results, in integer multiplication, check the breakthrough paper by:  David Harvey, and Joris Van Der Hoeven, et al . Integer multiplicaion in $O(n*(log$ $n))$. This proves Schonhage Strassens' conjecture from the 1970s that integer multiplication is really possible in $O(n*(log$ $n))$. From, straightforward (school) integer multiplication which is $O( n^{2}) $, to karatsubas' algorithm which is $O(n^{1.58})$, to $O(n*(log$ $n))$, by above authors. The authors use the property of specific multivariate polynomial rings that admit efficient multiplication. 
The authors show that integer multiplication (which is one dimensional) could be represented in a setting of a specific multivariate polynomial ring. Starting with a binary representation of integers, begin with the fixed point coordinate vectors(to a precision), and then go on to utilize them in coefficient rings for that polynomial representation. One could select parameters, and reduce the integer multiplication problem to one of convolution over a ring with a specific structure, reaching the bound. 
The bound is a significant improvement over earlier algorithms, and with many digits the efficiencies are apparent, for a large number of digits billions, and larger scales as author(s) claim its unknown, but good performance is possible.<|endoftext|>
TITLE: When is the degree of this number 3?
QUESTION [10 upvotes]: I am helping a friend of mine, that works in history of mathematics. She is studying the story of the solution of the cubic equation by Cardano. Sometimes she asks me some mathematical questions, that are very hard to motivate from a modern point of view, but that were interesting to Cardano. So please do not ask for motivations. The question is the following. Let $a$, $b$ be rational numbers, with $b$ not a square. Consider the number
$$
t=\sqrt[3]{a+\sqrt{b}}+\sqrt[3]{a-\sqrt{b}}-\sqrt{a^2-b}
$$
Under what conditions on $a$ and $b$ is the degree (over $\mathbb{Q}$) of $t$ equal to $3$? A sufficient condition can be found as follows. Let $P(x) = x^3+\alpha_2 x^2 + \alpha_1 x + \alpha_0$ be a rational polynomial. The general expression of the roots of $P$ is
$$
\sqrt[3]{- \frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} + \sqrt[3]{- \frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} - \frac{\alpha_2}{3},
$$
where
$$
q = \frac{2\alpha_2^3 - 9\alpha_2\alpha_1 + 27\alpha_0}{27}
$$
and
$$
p = \frac{3\alpha_1 - \alpha_2^2}{3},
$$
see here. So we can take $a = -\frac{q}{2}$ and $b = \frac{q^2}{4} + \frac{p^3}{27}$ and we need to force $\sqrt{a^2 -b} = \frac{\alpha_2}{3}$. This boils down to $\alpha_1 = \frac{\alpha_2^2 - 3 \sqrt[3]{3\alpha_2^2}}{3}$. We find that one of the solutions of
$$
x^3 + \alpha_2 x^2+ \frac{\alpha_2^2 - 3 \sqrt[3]{3\alpha_2^2}}{3}x+\alpha_0=0
$$
has the required form (of course we need to assume that $\alpha_2$ is such that $\sqrt[3]{3 \alpha_2^2}$ is rational). In this case $a$ and $b$ are given by the above expressions.
I suspect that if $t$ has degree $3$, then its minimal polynomial must be of this form and that $a$ and $b$ are as above, but I am not able to prove it. Note that the condition that $t$ has degree $3$ implies that it can be written as the sum of two cubic root and a rational number (because of the formula), but it is not completely clear that this way of writing $t$ is unique.

REPLY [7 votes]: Edit: The takeaway is that there is another, exclusive way to generate $a$ and $b$ that given an equation of degree $3$. First, chose parameters $a$ and $s$ such that $2a/s(s^2-3)$ is not a perfect cube and $1-4/s^2(s^2-3)^2$ is not a perfect square. This is most such values of $s$. Then choose $b$ according to the formula:
$b=a^2\left(1-\frac{4}{s^2(s^2-3)^2}\right)$
Then the number asked in the question is this degree $3$ number:
$t=s\sqrt[3]{\frac{2a}{s(s^2-3)}}+ \frac{2a}{s(s^2-3)}$
and can not be written in the way described in the question.
We get that from this analysis:
This is equivalent to asking whether $a^2-b$ is the sixth power of a rational number, since $\sqrt[3]{3\alpha_2^2}=\sqrt[3]{27(a^2-b)}=3\sqrt[3]{a^2-b}$ is rational, and $\sqrt{a^2-b}=\alpha_2/3$ is also rational. Vice versa, if we know that $a^2-b=x^6$ we set:
$\alpha_0=-2a+x^9-3x^5-(2/3)x^6$
$\alpha_1=3x^6-3x^2$
$\alpha_2=3x^3$
We can gain additional insight using Galois theory. Consider the extension $\mathbb Q(\sqrt{a^2-b},\sqrt[3]{a+\sqrt{b}},\sqrt[3]{a-\sqrt{b}},\sqrt{-3})/\mathbb Q(a,b)$. This is a Galois extension of degree $72$, given by adjoining three square roots and then two cube roots. It is easy to check that the Galois group is $S_3 \times S_3 \times S_2$, with the first factor permuting the conjugates of $\sqrt[3]{a^2-b}$, the second permuting the conjugates of $\sqrt[3]{\frac{a+\sqrt{b}}{a-\sqrt{b}}}+\sqrt[3]{\frac{a-\sqrt{b}}{a+\sqrt{b}}}$, and the third permuting the conjugates of $\sqrt{a^2-b}$.
The stabilizer of the formula is Klein four group, generated by a transposition in each $S_3$, the first switching $\sqrt{-3}$ and $-\sqrt{-3}$ and the second switching $\sqrt{-3b}$ and $-\sqrt{-3b}$. Thus, the generic degree is $18$, as Igor Rivin's numerical evidence suggested.
For a specific value of $a$ and $b$, the Galois group is a subgroup of this. To be degree $3$, the Galois group, mod its intersection with the Klein four group, must have order $3$, so the Galois group must have order $6$ or $12$. Whatever subgroup it is, the elements of the field fixed by it are rational numbers. For instance, since every possible subgroup fixes $\sqrt{a^2-b}$, it is always rational.
Assume that $\sqrt{-3b}$ is irrational. Then the subgroup must have order $12$, since $\sqrt{b}$, $\sqrt{-3}$, and $\sqrt{-3b}$ are all irrational, the action on them implies the subgroup must have order a multiple of $4$, thus exactly $12$, so it contains the either Klein four group. There are two subgroups of order $12$ with this property, the one containing the entire first $S_3$ and the one containing the entire second $S_3$.
For the one containing the second $S_3$, the invariants are generated by $\sqrt{a^2-b}$ and $\sqrt[3]{a^2-b}$. This is the case discussed in the question. We are asked to eliminate the other cases.
For the one containing the first $S_3$, the invariants are $\sqrt{a^2-b}$ and $\sqrt[3]{\frac{a+\sqrt{b}}{a-\sqrt{b}}}+\sqrt[3]{\frac{a-\sqrt{b}}{a+\sqrt{b}}}$. Calling them $x$ and $y$, we have the formulas $y^3-3y=\frac{a+\sqrt{b}}{a-\sqrt{b}}+\frac{a-\sqrt{b}}{a+\sqrt{b}}=2\frac{a^2+b}{a^2-b}=4\frac{a^2}{a^2-b}-2$, so we have the nodal cubic curve:
$y^3-3y+2=\left(\frac{2a}{x}\right)^2=z^2$
setting $z=2a/x$, we can parametrize the curve $(y-1)^2(y+2)=z^2$, so $y=s^2-2$, $z=s(s^2-3)$ provides a rational parameterization. (Thanks to Ricky for pointing out it is singular.) At this point we can just set $b=a^2-4a^2/z^2=a^2\left(1-\frac{4}{s^2(s^2-3)^2}\right)$. This gives the formula above.<|endoftext|>
TITLE: How well do we know relative commutants in $L(\mathbb{F}_\infty)$?
QUESTION [5 upvotes]: Let $H=K_1\oplus K_2$ be infinite dimensional Hilbert spaces. Voiculescu's free Gaussian functor gives us free group factors $L(H)$, $L(K_1)$, $L(K_2)$ acting on the full Fock space $\Gamma(H)$ and, moreover, $L(H)=L(K_1)*L(K_2)$.
But what is known about the relative commutant $L(K_1)'\cap L(H)$? 
Certainly $L(K_1)$ is non-amenable, so by Ozawa's result http://arxiv.org/abs/math/0608451 the relative commutant must be atomic. But I don't know any more than that. In particular is it even nontrivial?

REPLY [10 votes]: Given any two tracial von Neumann algebras $(N_1, \tau_1)$ and $(N_2, \tau_2)$ the $L^2$ space of the free product $(N_1 * N_2, \tau)$ canonically decomposes as 
$$
L^2(N_1 * N_2, \tau) = \mathbb C \oplus_{n \in \mathbb N} \bigoplus_{i_1 \not= i_2, i_2 \not= i_3, \ldots, i_{n - 1} \not= i_n } \overline {\otimes}_{k = 1}^n L^2_0(N_{i_k}, \tau_{i_k}),
$$
where $L^2_0(N_i, \tau_i)$ is the orthogonal complement of the scalars.
From this decomposition it's not hard to see that as an $N_1$ bimodule, $L^2(N_1 * N_2, \tau)$ decomposes as a direct sum of one copy of the trivial bimodule $L^2(N_1, \tau_1)$ and copies of the coarse bimodule $L^2(N_1 \overline \otimes N_1, \tau_1 \otimes \tau_1)$.  Specifically, if $\xi \in L^2(N_1 * N_2, \tau)$ is a unit vector which is an ``elementary word'' starting and ending with vectors in $L^2_0(N_2, \tau_2)$ then $x \xi y \mapsto x \otimes y$ extends to an isomorphism from the $N_1$ bimodule generated by $\xi$ and $L^2(N_1 \overline \otimes N_1, \tau_1 \otimes \tau_1)$.
Thus $N_1' \cap (N_1 * N_2) \not= \mathcal Z(N_1)$ if and only if the coarse $N_1$ bimodule has non-zero central vectors, which, by identifying $L^2(N_1 \overline \otimes N_1, \tau_1 \otimes \tau_1)$ with Hilbert-Schmidt operators and then taking a spectral projection, is if and only if $L^2(N_1, \tau_1)$ has a finite dimensional $N_1$-sub-bimodule, which is if and only if $N_1$ has a non-trivial finite dimensional direct summand.
In the case you're looking at $N_1$ is a ${\rm II}_1$ factor and so $N_1' \cap (N_1 * N_2) = \mathcal Z(N_1) = \mathbb C$.<|endoftext|>
TITLE: Is there a categorification of topological K-theory?
QUESTION [8 upvotes]: For a compact Hausdorff topological space $X$, its K-theory $K^0(X)$ is defined to be the Grothendieck group of the isomorphism classes of finite dimensional vector spaces on $X$. For example $K^0(\text{pt})=\mathbb{Z}$.
My question is: is there a categorification of K-theory, just as we can categofify natural numbers to vector spaces?

REPLY [10 votes]: Maybe I am missing something, but is there a reason no one has mentioned the work on 2-vector spaces. One place to start is http://hopf.math.purdue.edu//Baas-Dundas-Rognes/segal60.pdf . They state a conjecture that the algebraic K-theory of the category of 2-vector spaces (over $\mathbb{C}$) is the algebraic K-theory of $ku$. The conjecture is proved in http://arxiv.org/pdf/0706.0531.pdf by Baas, Dundas, Richter and Rognes.

REPLY [8 votes]: One answer these days is to think of $K$-theory as represented by a commutative ring spectrum
(alias $E_{\infty}$-ring spectrum) $K$.  Then there is a perfectly good theory of $K$-module
spectra, to which we can apply Waldhausen's approach to algebraic $K$-theory to construct the
$K$-theory of $K$-theory.  This first appeared in EKMM (Elmendorf-Kriz-Mandell-May. Rings, modules,
and algebras in stable homotopy theory.  AMS 1997) and has been much studied since, for example by Blumberg and Mandell.  The localization sequence for the algebraic $K$-theory of topological $K$-theory.<|endoftext|>
TITLE: When is the hull of a space curve composed of developable patches?
QUESTION [12 upvotes]: Let $C$ be a smooth curve in $\mathbb{R}^3$ that lies entirely on its convex hull,
$\cal{H}(C)$.

Under what conditions on $C$ is $\cal{H}(C)$ the union of developable surface patches?

I believe the patches are all ruled, but perhaps not developable?
Example 1. $C_1$ is the concatenation of four semicircles as shown below.  The surface patches
are subsets of cylinders, or subsets of planes: developable.
          

Example 2. $C_2$ is the curve studied by Ranestad and Sturmfels in
"On the convex hull of a space curve," (arXiv link), and earlier studied by Sottile (who
made the image below):
$$x = \cos(\theta)\;,\; y = \sin(2\theta)\;,\;z = \cos(3\theta) \;.$$
          

Status (to me) unclear, despite its superficial similarity to the "tennisball" curve $C_1$ above.

REPLY [14 votes]: For any point $P$ of the surface there is a point $S$ on the curve, so that the segment $PS$ is on the surface${}^1$. Hence, the surface of the convex hull is a union of ruled surfaces. Let's choose one of them.
Any ruled surface has non-positive Gaussian curvature at each of its points (see e.g. this link).
But for any convex surface, the Gaussian curvature cannot be negative.
It follows that the surface has zero Gaussian curvature. Therefore, it is developable, by definition.
Hence, the answer is "always".


There is  a plane $\alpha$ through $P$, so that $\mathcal H(\mathcal C)$ is completely on one side of the plane. Then the plane $\alpha$ touches $\mathcal C$, say at a point $S$, otherwise $P\notin \mathcal H(\mathcal C)$. If $S\in \alpha\cap \delta \mathcal H(\mathcal C)$, $PS\subset \delta \mathcal H(\mathcal C)$, where $\delta \mathcal H(\mathcal C)$ is the boundary of $\mathcal H(\mathcal C)$.<|endoftext|>
TITLE: Groups with finitely generated center
QUESTION [17 upvotes]: Does every group with a finite classifying space have finitely generated center? 
Remarks:

If $G$ is a finitely generated group with infinitely generated center $Z(G)$,
then the quotient $G/Z(G)$ is not finitely presented (as follows from a result of B.H Newmann).
Finite classifying space means that the group
is the fundamental group of a finite aspherical cell complex.
I suspect the above question is a well-known open problem, but cannot find it stated in the literature, so a reference would be appreciated.
Alperin-Shalen (Inventiones, 1982) showed that the answer is yes
for every subgroup of $GL_n(K)$ where $n>0$ and $K$ is a field of characteristic zero.
The answer is also yes for elementary amenable groups. (I know a proof, but have no reference).

REPLY [3 votes]: The answer to another question implies that there is a finitely presented group which has $\mathbb{Q}$ as its center (or any recursively presentable abelian group). However, you'd have to go through the paper to see if Houcine's proof could produce a group with a finite complex (seems unlikely).<|endoftext|>
TITLE: "Small" maps from sphere to sphere
QUESTION [14 upvotes]: Start with a continuous map $f:S^{n+k} \rightarrow S^n$ (round unit spheres).  The graph of $f$ lives in $S^{n+k}\times S^n$ and suppose it has a surface area (as a subspace of co-dimension $n$).  Now vary $f$ within its homotopy class with the aim of reducing the surface area.
1) Is the lower bound on the surface area always obtained?
2) Are there examples of particular elements in a particular $\pi_{n+k}(S^n)$ with known non-trivial bounds on these surface areas?  Non-trivial constructions of maps in a homotopy class that realize the lower bound?
3) One can also assign a surface area to a homotopy between two elements of a particular homotopy class.  Individually minimizing the surface areas of each element in $\pi_{n+k}(S^n)$ very likely has a cost when it comes to the geometry of the homotopies that arise in the multiplication table of $\pi_{n+k}(S^n)$.  As an alternative to the scenario above, one could pick representatives and homotopies to lower the total surface areas of all the homotopies associated to entries of the multiplication table of $\pi_{n+k}(S^n)$.   Any non-trivial lower bounds available for this story (perhaps when the group has order 2)?

REPLY [11 votes]: Here's an example to show that the infimum is not always attained:
Consider the standard Hopf map $\pi:S^3\to S^2$, which is not null-homotopic, of course, so it follows that the area of the graph in $S^3\times S^2$ of any differentiable map $f:S^3\to S^2$ that is homotopic to $\pi$ is strictly greater than the area of the graph of a constant map, i.e., of $S^3\times\{c\}\subset S^3\times S^2$, where $c\in S^2$ is fixed, namely the volume of $S^3$.
Now, computation shows that, if $\phi_t:S^3\to S^3$ for $0<t\le 1$ is the $1$-parameter family of conformal maps that fix a point $p_0\in S^3$ and its antipode $-p_0\in S^3$, that is the identity for $t=1$ and expands at $p_0$ by a factor of $1/t$ for $t\in (0,1]$, then the area of the graph of $f_t = \pi\circ\phi_t$ in $S^3\times S^2$ converges to the area of $S^3\times\{\pi(p_0)\}\subset S^3\times S^2$.  Indeed, the graph of $f_t$ itself converges to $S^3\times\{\pi(p_0)\}$ outside of an open neighborhood of $\{-p_0\}\times S^2$.
Consequently, the infimum of the areas of the graphs of maps $f:S^3\to S^2$ in the homotopy class of $\pi$ is the volume of $S^3$, but this infimum cannot be obtained by anything in the homotopy class.
Note:  The graph of $\pi:S^3\to S^2$ in $S^3\times S^2$ is a minimal submanifold, of course, but it is not minimizing, i.e., it is unstable.
Addendum:  Actually, it turns out that this computation is just a special case of a much more general phenomenon.  Let $\phi_t:S^{n+k}\to S^{n+k}$ for $0<t\le 1$ be the conformal dilation contracting to a fixed point $p_0\in S^{n+k}$ whose differential at $p_0$ is $t$ times the identity on $T_{p_0}S^{n+k}$.  (In particular, $\phi_1$ is the identity map.)  One then has the following result:
Proposition:  If $f:S^{n+k}\to S^n$ is any $C^1$-map and $k>0$, then the areas of the graphs of the homotopic family $f_t = f\circ\phi_t$ in $S^{n+k}\times S^n$ converge to the volume of $S^{n+k}$ as $t$ goes to $0$.
In particular, the graph area infimum in any homotopy class in $\pi_{n+k}(S^n)$ is the same when $k>0$, namely the volume of $S^{n+k}$, and this can actually be attained only for the trivial homotopy class.<|endoftext|>
TITLE: Is there an explicit example of a coefficient sheaf for which hard Lefschetz fails?
QUESTION [10 upvotes]: Recall that if $X$ is a projective algebraic complex manifold, $L$ is a semisimple $\mathbb C$-local system on $X$ of geometric origin (roughly speaking, this means that $L$ is a cohomology sheaf $R^if_*\mathbb C$ for some algebraic morphism $f:Y\to X;$ see BBD for the precise definition), and $\eta\in H^2(X,\mathbb C)$ is an ample class, then 
$$
\eta^i\cup-:H^{\dim X-i}(X,L)\to H^{\dim X+i}(X,L)
$$
is an isomorphism. This also holds when the projective variety $X$ is allowed to have singularities and $L$ is a perverse sheaf (again semisimple of geometric origin), appropriately shifted.
I'd like to know an example for which this fails; of course, $L$ is no longer of geometric origin. Over finite field, as long as $L$ is assumed semisimple, hard Lefschetz always holds, and conjecturally $L$ is of geometric origin.

REPLY [16 votes]: The good news (or is it bad news?) is that there is no counterexample in the semisimple case. Simpson proved that hard Lefschetz holds for any semisimple local system on a smooth complex projective variety. To be more precise, apply theorem 1 and lemma 2.6 of his paper on Higgs bundles and local systems. More recently, Sabbah and Mochizuki have extended this to some (or perhaps all, I can't quite remember) semisimple perverse sheaves in accordance with a conjecture of Kashiwara.

Addendum  Here is  a counterexample for nonsemisimple local systems (which was something that I had wondered about  myself). Let $X$ be a smooth projective curve with genus $g>1$. Let $\pi=\pi_1(X)$.
Let $\mathbb{Q}_\rho$ denote a nontrivial rank one $\pi$-module with character $\rho$,
and $\mathbb{Q}$ the trivial module. We can see, using Euler characteristics, that
$$Ext^1_\pi(\mathbb{Q},\mathbb{Q}_\rho)\cong H^1(X,\mathbb{Q}_\rho)\not=0$$
Thus we can form a nonsplit extension
$$0\to \mathbb{Q}_\rho \to L\to \mathbb{Q}\to 0$$
We necessarily have $H^0(X,L) = H^0(\pi, L)=0$. On the other hand, by Poincaré duality
$$H^2(X,L) = H^0(X,L^*)^*= H^0(\pi,L^*)^*\not=0$$
So $H^0(X,L)\not= H^2(X,L)$, i.e. hard Lefschetz fails.<|endoftext|>
TITLE: What is a Choice Principle, really?
QUESTION [20 upvotes]: This question is quite soft, and I apologize in advance if it borderline off-topic.
When working in theories between ZF and ZFC the term "choice principle" is heard quite often. For example:
$\quad$ The Axiom of Countable Choice: Every countable family of non-empty sets has a choice function.
We can say, if so, that a choice principle is an assertion about choosing. On the other hand, the above axiom is equivalent [1] to the following assertion:
$\quad$ Every $\sigma$-compact space is Lindelöf.
However this statement does not talk about choosing anything. The first alternative is to define a choice principle as something equivalent to a choice-referring statement, but what about the ordering principle? 
$\quad$ Ordering Principle: Every set can be linearly ordered.
Commonly thought of as a choice principle, the order principle is not really equivalent to any choice principle directly. It does however prove that every family of non-empty and finite sets has a choice function.
Maybe we can define a choice principle as something which proves some sort of a choice statement. Alas this too can be troublesome, consider the Small Violations of Choice principle:
$\quad$ SVC: There exists $S$ such that for every $x$ there is an ordinal $\alpha$ such that $x\leq\mathcal P(\alpha)\times S$.
It is consistent with the existence of an amorphous set AND a countable set of pairs without a choice function that SVC holds, so it doesn't even imply the axiom of choice for pairs.
So we may wish to define a choice principle as follows:

Let $\varphi$ be a sentence in the language of set theory, $\varphi$ is a choice principle if ZF does not prove $\varphi$, but ZFC does.

This definition catches the wanted properties of a choice principle, it is simply an assertion in between the two theories. However in a recent comment on math.SE Carl Mummert suggested that this definition is also problematic, since the assertion: $$\lnot\rm AC\rightarrow\text{Con}(ZF)$$
Is not provable from ZF, but vacuously true in ZFC. 
This leads to the question:

Question: What would be a good definition for Choice Principle (in ZF), which encapsulates the notion of a statement "between" ZF and ZFC, but still avoids vacuous statements as above?


Bibliography:

Brunner, N. σ-kompakte Räume. (German. English summary) [σ-compact spaces]
Manuscripta Math. 38 (1982), no. 3, 375–379.

REPLY [14 votes]: When we speak of a "choice principle" we are discussing the meaning of a statement, not its truth value, i.e., the usual statement AC got its name because of what it means and not because of its truth status. However, logical equivalence preserves only truth values but not meanings. Thus the statements "all unicorns have two horns" and "1 + 1 = 2" are both true, hence equivalent, but they have different meanings. Likewise, the axiom of choice and the well-ordering principle have different meanings, even though they are equivalent. If you wonder about AC being equivalent to a well-ordering principle, you should also wonder about many other equivalences in mathematics that relate statements with different meanings.
As for your question "what is a choice principle, really?" I would say that choice principles are a certain kind of reversal of quantifiers. The axiom of choice can be stated as
$$(\forall x \in A . \exists y \in B . \phi(x,y)) \implies \exists f \in B^A . \forall x \in A . \phi(x, f(x))$$
where $\phi$ is a relation between the sets $A$ and $B$. This form of the axiom of choice does not require any set theory, just a bit of simple type theory and first-order logic (if you read schematically in $\phi$).
Exercise: convince yourself that the above statement is equivalent to AC. Hint: given a family of sets $C_i$ indexed by $i \in I$ let $A = I$, $B = \bigcup_{i \in I} C_i$ and $\phi(i, x) \iff x \in C_i$. Conversely, given $A$, $B$ and $\phi$, let $I = A$ and $C_i = \lbrace y \in B \mid \phi(i,y)\rbrace$.
A category theorist might say that choice is about splitting epis.
Indeed, given a family $C_i$ indexed by $i \in I$, consider the map $e : \coprod_{i \in I} C_i \to I$ defined by $e (i,x) = i$. Then $(C_i)_{i \in I}$ is a family of non-empty sets if, and only if, $e$ is surjective (epi), and it has a choice map if, and only if, $e$ has a right inverse (is split).  Conversely, to split an epi $e : A \to B$ is the same as to give a choice function for the family of sets $C_i = \lbrace x \in A \mid e(x) = i\rbrace$ indexed by $i \in B$. 
Supplement: in answer to Trevor, here is how one might phrase dependent choice categorically. I do not know whether there is a slicker formulation. Given $1 \to A$ and $e, p: B \to A$ with $e$ epi, there is $f: \mathbb{N} \to A$ such that there is a factorization $h: \mathbb{N} \to B$ of the span $f, f \circ \mathrm{succ} : \mathbb{N} \to A$ through the span $e, p: B \to A$. This looks nicer as a commutative diagram, how do I draw one of those? Anyhow, I do not see a particular advantage over the usual formulation. Perhaps someone can improve this.<|endoftext|>
TITLE: Saying things rapidly about integer factorisations
QUESTION [12 upvotes]: Let $N$ be a positive integer. Thanks to the Miller-Rabin test and the work of Agrawal, Kayal and Saxena, these days people have much much faster algorithms for testing whether $N$ is prime or composite than they do for explicitly finding its factorisation.
If you stand at a sufficient distance, the key idea behind both Miller-Rabin and AKS (and every other primality test I've seen) is to work with the ring of residues modulo $N$, and use clever tricks to see whether it has the algebraic properties that it would have if $N$ were prime. Miller-Rabin looks to see if the roots of $1$ behave how Fermat says they should in that ring; AKS tries to verify a certain polynomial identity in it.
Are there any other properties of $N$'s factorisation that can be checked in less time than it takes to actually find that factorisation, using similar ideas? For example, is it reasonable to ask for an algorithm which takes a composite $N$ and says something useful about how many prime factors it has? This question is related to the number of square roots of $1$ in the residue ring, so might conceivably be effectively detectable by some devilishly clever trick. Similarly, the distribution of residue classes of the prime factors modulo small integers is closely related to the number of roots of unity of small orders.
Or is it simply the case that, once you know that the ring is not a field, you just have so little control over it that no good trickery is known?
I read on a webpage about squarefree integers that no such rapid algorithm is known which tests for repeated prime factors, but that doesn't seem to render the general question invalid (though, if nobody can suggest any positive results in this direction, it would lead us to guess that it's hard, perhaps).

REPLY [7 votes]: While the method used is not related to the work you mention in the question, there is recent work of Andrew Booker, Ghaith Hiary and Jon Keating where they develop an algorithm for determining a non trival lower bound (assuming GRH) for the square free part of the number $N$, i.e. if
$$ N=\Delta m^2, $$
where $\Delta$ is square free they obtain a lower bound for $\log \Delta$, by using the Weil explicit formula for certain $L$-functions associated to $N$. I do not find the relevant paper (in progress?), but there is a slide on Hiary's homepage http://www.maths.bristol.ac.uk/~maxgh/ from a talk he has given on the topic (that I attended). In particular they manage to show that assuming the Riemann hypothesis for the relevant Dirichlet L-functions, RSA-210 is not square-full (there exists at least one prime factor in the factorization of RSA-210 that occurs just once), a number that has yet to be factored, so for example it can not be of the form $p^2q^3$. Of course it is likely to be a product of just two prime factors, but we do not yet have a proof of that.<|endoftext|>
TITLE: A remark by Gromov on 4-manifolds
QUESTION [8 upvotes]: Gromov remarks in a a survey on manifolds (p.12) that "it is hard to imagine that there are infinitely many non-diffeomorphic, but mutually homeomorphic, quotients of the hyperbolic 4-space by discrete isometry groups". What is the background of that?

REPLY [10 votes]: I'm not exactly sure what he has in mind. If the smooth Poincare conjecture is false in 4 dimensions, then one could imagine taking a connect sum with a fake 4-sphere (homeomorphic but not diffeomorphic to $S^4$), and getting a manifold which is homeomorphic but not diffeomorphic to a hyperbolic 4-manifold.
However, if the smooth Poincare conjecture is true, there still could be an exotic hyperbolic 4-manifold. The most powerful method for detecting non-diffeomorphic but homeomorphic 4-manifolds is the Donaldson or Seiberg-Witten invariants (it's conjectured that these are equivalent). It is also conjectured that these invariants are trivial on hyperbolic 4-manifolds. If this were true, then one would need a different technique for detecting the exotic smooth structure. 
Another technique for detecting exotic structures is the Rochlin invariant. There is a homotopy $S^3\times \mathbb{R}$ which contains an embedded smooth Poincare sphere, and therefore cannot be smoothly standard since the Poincare sphere cannot bound a contractible manifold since it has odd Rochlin invariant. So one could try to find a manifold homeomorphic to a hyperbolic manifold which contains a smoothly embedded Poincare sphere. There are two difficulties in finding such an example: creating the example, and proving that it is homeomorphic to a hyperbolic 4-manifold. Freedman-Quinn's surgery theorem is not available for groups of exponential growth, so it's not clear what technique one would use for proving homemorphism, maybe some surgery which doesn't affect the homeomorphism type.<|endoftext|>
TITLE: The topology of Arithmetic Progressions of primes
QUESTION [78 upvotes]: The primary motivation for this question is the following: I would like to extract some topological statistics which capture how arithmetic progressions of prime numbers "fit together" in a manner that will be made precise below. 
Setup
Consider a nested family of simplicial complexes $K(p)$ indexed by prime $p \in \mathbb N$ defined as follows: 

the vertices are all odd primes less than or equal to $p$, and
insert a $d$-simplex ($d \geq 2$) spanning $d+1$ vertices if and only if they constitute an arithmetic progression. Of course, one must also insert all faces, and faces-of-faces etc. so that the defining property of a simplicial complex is preserved.

For instance, $K(7)$ has the vertices $3,5,7$ and a single $2$-simplex $(3,5,7)$ along with all its faces. $K(11)$ has all this, plus the vertex $11$ and the simplex $(3,7,11)$. The edge $(3,7)$ already exists so only the other two need to be added. Thus, the fact that $(3,7)$ occurs in two arithmetic progressions bounded by $11$ is encoded by placing the corresponding edge in the boundary of two 2-simplices.
Question
Has someone already defined and studied this complex? What I am mostly interested in is

How does the homology of $K(p)$ change with $p$?

If it helps, here are -- according to home-brew computations -- the statistics for the first few primes (Betti 0 and 1 over $\mathbb{Z}_2$). I've already confirmed that the sequence of Betti-1's is not in Sloane's online encyclopedia of integer sequences. If an intermediate K[p] is missing in the list, that means that the homology is the same as that for the previous prime.
K [3]: 1 0
 K [5]: 2 0
 K [7]: 1 0
 K [13]: 2 0
 K [17]: 2 1
 K [19]: 1 2
 K [23]: 1 4
 K [31]: 1 6
 K [37]: 2 6
 K [43]: 1 7
 K [53]: 1 8
 K [59]: 1 9
 K [61]: 1 10
 K [67]: 1 12
 K [71]: 1 17
 K [73]: 1 20
 K [79]: 1 23
 K [83]: 1 26
 K [89]: 1 31
 K [97]: 1 32
 K [101]: 1 35
 K [103]: 1 41
 K [107]: 1 43
 K [109]: 1 47
 K [113]: 1 53
 K [127]: 1 58
 K [131]: 1 62
 K [137]: 1 67
 K [139]: 1 73
 K [149]: 1 78  
Here's a more concrete question:

Is it true that the $d$-dimensional homology groups of $K(p)$ for $d > 1$ are always trivial?

REPLY [31 votes]: No. 
In fact, for $p = 435052917615787$, this will absolutely be false, as the second homology group will not vanish.
Note that a 2-dimensional "hole" in your complex is a 3-simplex all of whose faces are 2-simplices in $K(p)$, i.e. 4 primes such that each 3 of them lie in some arithmetic progression. 
Of course, you would suspect that such a thing is plausible, and with some effort construct an example.
Here's why this number works:
Consider $p = 398936189798617$. Then, if $d = 2124513401010$, one sees that $p+d$ is not a prime, while $p+2d,p+3d,\ldots,p+15d$ are all primes. Here $p+15d = 435052917615787$ is the prime for which we consider the complex.
In fact, the numbers were taken from the AP-k records at http://primerecords.dk/aprecords.htm#minimal, and $p+2d$ is just a beginning of an AP-23. 
Now, one sees immediately that $p, p+6d, p+10d$ form a simplex, as it is a subsequence of the arithmetic progression of primes $p, p+2d, p+4d, p+6d, p+10d$.
Similarly, $p, p+6d, p+15d$ form a simplex, as do $p, p+10d, p+15d$ and $p+6d, p+10d, p+15d$. These come from sequences with differences $3d$, $5d$, and $d$ respectively. 
However, $p,p+6d,p+10d,p+15d$ can only form a simplex if they all lie in some arithmetic progression of primes. But then its difference must divide $d$, contradicting the fact that $p+d$ is not prime.<|endoftext|>
TITLE: computer algebra system for polynomial algebras over finite fields
QUESTION [8 upvotes]: Is there a computer algebra system that can do arithmetic over polynomial algebras over finite fields where I can specify the extension?
Exempli gratia, if $f(x), g(x) \in \mathbb{F}_p[\mu]/(m(\mu))[x]$, I'd like the CAS to be able to compute things like $f(x + \mu) + g(x)$ where I specify the polynomial $m(\mu)$.
Thanks

REPLY [2 votes]: Here's one way to do it in Macaulay2
restart
R = GF(5)[mu]/(mu^5-mu+1)[x]
f = x^7-1
g = x^2+x+1
sub(f, x => x+mu) + g

returning
      7        6     2 5       2       2                 3     2
o4 = x  + 2mu*x  + mu x  + mu*x  + (2mu  - 2mu + 1)x + mu  - mu

o4 : R

REPLY [2 votes]: In Singular:
> ring R = (5,mu),(x),dp; minpoly=mu^5-mu+1;
> poly f = x^7 - 1;
> poly g = x^2 + x + 1;
> subst(f,x,x+mu)+g;
x^7+(2*mu)*x^6+(mu^2)*x^5+(mu)*x^2+(2*mu^2-2*mu+1)*x+(mu^3-mu^2)<|endoftext|>
TITLE: Proving finite generation by tensoring with $\mathbb{R}$
QUESTION [12 upvotes]: In Chapter III, Theorem 7.4 of The Arithmetic of Elliptic Curves (first edition), Silverman gives the following lemma and proof:

Lemma: Let $M \subset Hom(E_1, E_2)$ be a finitely generated subgroup, and let $M^{div} = \{ \phi \in Hom(E_1, E_2) : [m] \circ \phi \in M$ for some integer $m \geq 1\}$. Then $M^{div}$ is also finitely generated.
Proof: Extend the degree mapping to the finite dimensional real vector space $M \otimes \mathbb{R}$, which we equip with the natural topology inherited from $\mathbb{R}$. Then the degree mapping is clearly continuous, so the set $U = \{\phi \in M \otimes \mathbb{R} : deg(\phi) < 1 \}$ is an open neighborhood of $0$. Further, since $Hom(E_1, E_2)$ is a torsion-free $\mathbb{Z}$-module, there is a natural inclusion $$M^{div} \subset M \otimes \mathbb{R};$$ and clearly $$M^{div} \cap U = \{ 0 \},$$
since every non-zero isogeny has degree at least 1. Hence $M^{div}$ is a discrete subgroup of the finite dimensional vector space $M \otimes \mathbb{R}$, so it is finitely generated.

When I first saw it, this proof felt like absolute voodoo to me. Tensor an endomorphism ring with the reals? What, so you can take $\pi$ or $e$ times an endomorphism?
The point is clearer to me now -- real vector spaces are a nice place to argue that things are finitely generated -- but I'm not sure I would have seen to do this if the proof were left as an exercise.
Does this idea have any natural context? Are proofs like this commonplace in some other area of math?
Thank you!

REPLY [20 votes]: How do you prove that the ring of integers in a number field is finitely generated? One usually embeds them in $\mathbb{R}^{r_1}\times\mathbb{C}^{r_2}$. How do you prove that the units in a number field are finitely generated? One generally embeds them in a hyperplane in $\mathbb{R}^{r_1+r_2-1}$. Then one shows that the group sits as a discrete subgroup, hence is finitely generated. Further, by looking at the co-volume of the resulting lattice, one obtains important arithmetic invariants, namely the discriminant and the regulator. Anyway, one can call almost any technique a trick, but the idea of embedding a group into a real or complex vector space and using volume estimates to prove discreteness is quite a well-established technique. And if one simply has the group and a positive definnite quadratic form, as in the case of End$(E)$ or the canonical height on $E(\mathbb{Q})$, then it is very natural to tensor with $\mathbb{R}$ and extend the quadratic form to put a Euclidean structure on the resulting vector space. 
Of course, as Will Sawin says, one can alternatively view the embedding as being into the dual. But for example, using the height pairing on $E(\mathbb{Q})/tors$, the values of $\hat h$ are real, so even though $E(\mathbb{Q})/tors$ is a torsion-free $\mathbb{Z}$-module, the dual space that one uses has to be Hom$(E(\mathbb{Q})/tors,\mathbb{R})$.
I don't know that this really answers the question, but I hope it gives some indication of why it's not so unusual to think of embedding $M$ into a real vector space as the first step in proving that it is finitely generated. 
Finally, I should mention that I first saw the argument for proving that End$(E)$ is finitely generated in Mumford's Abelian Varieties, but I don't know the origins of the idea.<|endoftext|>
TITLE: Upper bound for largest eigenvalue of 0-1 matrix
QUESTION [5 upvotes]: I have large $(0-1)$-matrices with the additional property that the first row and the first column are almost all $1$'s. My question is

Is there a (good) algorithm to obtain an upper bound for the largest eigenvalue of such a matrix ?

This question actually didn't come up in my own research but some collaborator of mine asked this question at dinner these days. I thus don't know more about the background of the question, but my impression would be that if anything is known, this is the right place to ask.
Thanks, Christian

REPLY [7 votes]: EDIT: this answer was originally written in response to an older version of the question, which merely asked for upper bounds on the largest eigenvalue rather than an algorithm.

Take your matrix to have all entries equal to $1$ to get a matrix which has $n$ as an eigenvalue.
Let $A$ be an $n\times n$ matrix with $0$-$1$ entries. Let $x=(x_1,\dots, x_n)^\top$ and calculate
$$ \begin{aligned}
\sum_{i=1}^n \left\vert \sum_{j=1}^n A_{ij} x_j\right\vert^2
& \leq
n \left( \sum_{j=1}^n \vert x_j\vert \right)^2 \quad\hbox{(triangle inequality)} \\
& \leq
n^2 \sum_{j=1}^n \vert x_j\vert^2 \quad\hbox{(Cauchy-Schwarz)} \\
\end{aligned}
$$
to see that $\Vert Ax \Vert^2 \leq n^2 \Vert x\Vert^2$. So the norm (= largest singular value), and hence the largest eigenvalue, of $A$ is at most $n$. The example I gave at the start shows this is sharp.
The moral is that unless we have more information about this 0-1 matrix, very little informative can be said regarding upper bounds on the largest eigenvalue.<|endoftext|>
TITLE: Given a vector field all of whose integral curves are closed, is the period a smooth function?
QUESTION [6 upvotes]: Disclaimer: The original question consisted of two parts. The first one
  has been answered negatively (see
  below the answers of Sam Lisi and
  Alejandro). It remains the second one.

Background
I am reading about the energy-period relation for Hamiltonian Systems.
In Weinstein's formulation (cf. Abraham, Marsden, Foundations of Mechanics 2nd Ed, page 198) this relation amounts to:

$(\ast)$ Given an Hamiltonian system $(M,\omega, X_H)$, let be $\Phi$ the flow of $X_H$ and $\text{per}:=\{(t,x)\in\mathbb R\times M\mid\Phi(t,x)=x\}.$
  If $N$ is a smooth submanifold contained in $\text{per},$ then $\left.dt\wedge dH\right|_N=0,$ i.e. $t=t(H)$ on $N,$ (the period depends only on the energy.)

Question
In Guillemin, Stenberg, Geometric Asymptotics, between pages 170-171, I have additionally found that, when all integral curves of $X_H$ are periodic, we can take $N=\text{per}$ in $(\ast),$ which should mean that in such a case $\text{per}$ is a smooth submanifold of $\mathbb R\times M.$  
In order to justify this last point I was wondering myself:


If $X$ is a non singular vector field on $M,$ all of whose integral curves are periodic, and $\tau(p)$ denotes the period of the orbit through $p,$ then $\tau:M\to\mathbb R$ is smooth? 
otherwise, how to prove that in such a case $\text{per}$ is a submanifold?


What I have tried about point 2
Probably I am missing something because my guess is that if there were a principal bundle structure $(M,p,X,\mathbb S^1)$ such that the $\mathbb S^1$-orbits are the trajectories of $X$ then the period $\tau:M\to\mathbb R$ should be smooth because of the relation $\zeta=\tau X_H,$ where $\zeta$ is the infinitesimal generator of the action.
But I don't know how to proceed without this additional hypothesis.  
Edit1 (After Sebastian's answer about point 1): As illustration of my difficulties with point 1, I imagine that $M$ is the Moebius band $[0,1]\times\mathbb R/\sim$ and $X=\frac{\partial}{\partial x}$ then the period is $$\tau([(x,y)]_{\sim})=\begin{cases}1&\text{if }y=0\\\2&\text{if }y\neq 0\end{cases}$$

REPLY [7 votes]: Sam is completely right. In general, the period function $\tau\colon M\to\mathbb{R}$ is not even continuous. 
A very nice reference for a (counter-)example to Giuseppe's question is the paper A counterexample to the periodic orbit conjecture, by Dennis Sullivan. In the paper, Sullivan constructs a singularity-free flow on a compact 5-manifold such that all its orbits are periodic and function $\tau$ is unbounded!<|endoftext|>
TITLE: Evaluating a limit similar to the Euler constant
QUESTION [5 upvotes]: In the course of studying a certain complex-valued functional equation, I have had a need to evaluate the following limit:
$$\gamma_\mathcal{T}=\lim_{n\to\infty}\left(-\frac{i}{2}\sum_{k=1}^n \frac1{ik+k^{3/2}}-\log\left(1+\frac{i}{\sqrt n}\right)\right)$$
which is structurally similar to the usual limit definition for the Euler constant $\gamma$.
So far as I can tell, there seems to be no elementary closed form for this limit, so I set about trying for numerical estimation.
The problem is that the convergence of this limit looks to be excruciatingly slow. Even with the help of a sequence extrapolation method, I only managed to produce a few good digits:
$$\gamma_\mathcal{T}\approx-0.5-0.9300125396i$$
I am wondering if there are more efficient, alternative methods for numerically evaluating this limit. Thanks in advance!

REPLY [2 votes]: As already mentioned by Fedor and Igor, you can ignore the logarithmic term since it zeroes out at $\infty$, and you can just concentrate on the series
$$-\frac{i}{2}\sum_{k=1}^\infty \frac1{ik+k^{3/2}}$$
Using Laplace transform techniques, your sum can be transformed into the integral
$$-\frac12-\frac{i}{\sqrt\pi}\int_0^\infty \frac{F(\sqrt{u})}{\exp\,u-1}\mathrm du$$
where $F(z)$ is Dawson's integral.
I don't know of a closed form for this integral, but Mathematica easily evaluates this numerically:
-1/2 - I NIntegrate[DawsonF[Sqrt[u]]/(E^u - 1), {u, 0, Infinity}, 
    Method -> "DoubleExponential", WorkingPrecision -> 50]/Sqrt[Pi]
-1/2 - 0.93001253961059515359034795785857166233326206076173 I<|endoftext|>
TITLE: generalizing the ultrapower
QUESTION [6 upvotes]: Given an 2-valued measure $\mu$ on a set $I$, and structures $M_i \ (i \in I)$, one can construct the ultrapower $\prod_{i\in I}M_i / U$ (where $U$ is the ultrafilter associated with $\mu$.) One can then prove nice theorems about this structure - e,g, Łoś's theorem.
But suppose now that rather than beginning with a 2-valued measure, we begin with a real valued measure on the set $I$, and structures $M_i \ (i \in I)$. Is there any sort of analagous structure that we can build and prove useful results about?

REPLY [5 votes]: The reduced power and reduced product constructions are generalizations of the ultrapower construction that use filters instead of ultrafilters, and you can construct a filter from a measure (and hence a reduced power) simply by taking the collection of all sets whose complement is measure 0. See the book "A Course in Universal Algebra" at http://www.math.uwaterloo.ca/~snburris/htdocs/UALG/univ-algebra.pdf for the definition of a reduced power. 
There is a weak version of Łoś's theorem that holds for reduced powers known as the Feferman-Vaught theorem, but it is a bit messier for reduced powers than for ultrapowers. This result basically gives a method of determining the truth value of a sentence in a reduced power from the truth value of different sentences in the factors of the reduced power, and this result also holds for generalizations of the reduced power construction such as limit reduced powers, reduced products, and Boolean ultraproducts. The Feferman-Vaught theorem can be found in standard texts on model theory such as Hodges and the text by Chang and Keisler. Let me now state the Feferman-Vaught theorem.
Given any first order sentence $\phi(x_{1},...,x_{n})$ in some first order language $\mathcal{L}$, the algorithm finds a sequence $(\sigma,\theta_{1},...,\theta_{m})$ of formulas such that $\sigma(z_{1},...,z_{m})$ is a formula in the language of Boolean algebras, and each formula $\theta_{i}$ has at most the variables $x_{1},...,x_{n}$ free and where we have the following:
Assume $I$ is an index set, $Z$ is a filter on the set $I$, and $\mathcal{A}_{i}$ is an $\mathcal{L}$-structure for $i\in I$. Assume also that $f_{1},...,f_{n}\in\prod_{i\in I}A_{i}$, and given $1\leq j\leq m$, let $R_{j}=\{i\in I|\mathcal{A}_{i}\models\theta_{j}(f_{1}(i),...,f_{n}(i))\}$. Then we have $$\prod_{i\in I}\mathcal{A}_{i}/Z\models\phi(f_{1}/Z,...,f_{n}/Z)$$ if and only if
$$P(I)/Z\models\sigma(R_{1}/Z,...,R_{m}/Z).$$
I personally found the above result to be very useful since I was able to determine the sentence algebras and the elementary classes of a certain variety using a version of the above result. Of course, the above result is not as powerful as Łoś's theorem since it does not automatically give you a non-standard model of a structure like the real numbers.<|endoftext|>
TITLE: Semisimplicity of Frobenius operation on etale cohomology?
QUESTION [12 upvotes]: Let $X_0$ be a variety defined over a finite field of characteristic $p \neq l$. 
Is it true, that the action of the frobenius on the l-adic cohomology $H_l^*(X)$ is semisimple (say for smooth $X_0$)? If not, what would be a counter-example?

REPLY [21 votes]: Let $X$ be a smooth projective variety over a finite field 
$\mathbb{F}_{q}$ of caracteristic $p$ and let $l$ be a prime number different from p.
 We consider the following statement :
(A) The action of the Frobenius on the etale cohomology $H^{i}_{et}( X_{\overline{\mathbb{F}}_{q}}, \mathbb{Q}_{l})$ is semisimple.
How to suppress the projective hypothesis is the subject of the mathoverflow question 
link text
(A) is true in the following cases :
1) $X=A$ an abelian variety (and so for $X$ a curve via the jacobian).
As mentionned in comment by Emerton, it is a consequence of the Weil's
work on the Riemann hypothesis in this case.
Fix a polarization on $A$. For $x$ an endomorphism of $X$ which gives an endomorphim on $H^{1}_{et}$, we can define an endomorphism $x'$ (' : "Rosati involution") by $x' = *x^{T}* $
where in the middle we have the transposition with respect to the intersection product
and * comes from the duality theory of abelian varieties ( the polarisation gives a identification between $H^{1}_{et}(A)$ and $H^{1}_{et}(\check{A})$).
Weil proved that $Tr(xx')>0$ if $x$ is non-zero. Let $F$ be the (geometric) Frobenius. For
$x = q^{-1/2}F$, we have x'=$x^{-1}$. So $Tr(aa')$ is a definite positive bilinear form 
on the $\mathbb{Q}$ algebra generated by x and is preserved by multiplication by $x$ : so
multiplication by $x$ is unitary which shows that $x$ is semi-simple (and eigenvalues 
of modulus one gives the Riemann hypothesis).
2) $X$ a K3 surface. As mentionned in comment by shenghao, it is a consequence of the work of Deligne : link text
The result is deduced from the case of abelian varieties 
via the Kuga-Satake construction (of course there is a non-trivial thing to do because
Kuga-Satake construction is a priori of transcendental nature but Deligne did it).
For $X$ general, (A) is conjectured. It is a consequence of standard conjectures. More precisely, things should work as in the case of abelian varieties. We can still define 
$x \mapsto x'$ at the cohomological level but $Tr(xx')>0$ is conjectural : a standard conjecture 
of Lefschetz type implies $x'$ algebraic if $x$ is, which permits to use a trace formula 
expressing $Tr(xx')$ as an intersection product. The positivity should then be a consequence of a standard conjecture of Hodge type.
For more details, as mentionned in comment by Damian Rössler, see Kleiman "The standard conjectures" (whose some details depend on Kleiman, "Algebraic cycles and the Weil conjectures").<|endoftext|>
TITLE: How do you see that higher genus surfaces are not homogeneous?
QUESTION [9 upvotes]: I am trying to get some intuition about why the torus and the sphere are the only surfaces which can be realised as homogeneous spaces.  On the one hand, I know this is true because there is the result that homogeneous spaces must have non-negative Euler characteristic:
A Structure Theorem for Homogeneous Spaces, Mostow, G, Geometriae Dedicata, (114) 2005, 87-102
However, on the other hand, a higher genus surface can be realised as a quotient of hyperbolic space by a group of isometries.  The latter would seem (in my head) to give rise to a hyperbolic surface where the points look the same; all of the points have the same curvature, for instance.
My question is then: How do you distinguish the points of such a hyperbolic surface?

REPLY [3 votes]: One way to distinguish points of a hyperbolic surface $S$ is to show that the local geometry of the Voronoi graph of a point $p$ is changed when $p$ is perturbed. The Voronoi graph is a connected 1-complex consisting of the locus of points on $S$ for which there are two or more shortest paths to $p$; its complement is an open 2-cell which is the locus of points for which there is a unique shortest path to $p$. Equivalently, it is the projection to $S$ of the Voronoi diagram for the full pre-image of $p$ under a locally isometric universal covering map $\mathbb{H}^2 \mapsto S$. You need a nontrivial fundamental group to get a nonempty Voronoi graph, and a noncyclic fundamental group to get a Voronoi graph more complicated than a bi-infinite geodesic and therefore possessing at least one 0-cell (point of valence three or more).
Suppose that $x$ is a certain 0-cell of the Voronoi graph, a point at which there are three or more shortest paths to $p$. If $x$ has valence three, which is the "generic" situation, then $p$ can be perturbed so that the three angles of the 1-cells incident to $x$ are changed; this can be verified by some simple hyperbolic trigonometry calculations. If $x$ has valence four or more, then $p$ can be perturbed to change the local topology of the Voronoi graph near $x$, breaking $x$ into two or more vertices of smaller valence. 
By the way, with regard to the two topological types of closed Euclidean surfaces, while the torus is always homogeneous in any Euclidean metric, the Klein bottle is never homogeneous, because an orientation reversing simple closed geodesic is unique in its homotopy class.

Further remarks: I addressed Igor's first question in my comment below. To address his second question, the answer I am offering to the OP's question is that points $p \in S$ can be locally distinguished by the local geometry of their Voronoi graph $V(p)$. By this I mean two things: (1) for each $p,q \in S$, there exists an isometry of $S$ taking $p$ to $q$ if and only if $V(p)$ is locally isometric to $V(q)$; (2) for $q$ near $p$ there does not exist a local isometry between $V(p)$ and $V(q)$. 
To fill this in somewhat: the edges of $V(p)$ are geodesic arcs, and the local geometry of $V(p)$ is determined by the topological type of its embedding into its regular neighborhood $Nhd(V(p))$, by the lengths of its edges, and by the angles around its vertices. To prove the ``if'' direction of (1), once the local isometry type of $V(p)$ is determined, there is a unique way to fill in $Nhd(V(p)) - V(p)$ by an open hyperbolic disc. The cheapest proof of (2) is to use (1) together with the fact that the isometry group of $S$ acts discretely on $S$, which of course requires already knowing the latter fact, using a proof that depends for example on facts about geodesics explained in the other answers.
As alluded to, there is a much more expensive proof of (2) which uses hyperbolic trigonometry, which under special assumptions goes like this. Assume first that vertices of $V(p)$ have valence 3. Cut $S$ open along $V(p)$ to get a convex hyperbolic polygon containing $p$. Assume further that there is a perpendicular geodesic from $p$ to each side of this polygon (not true in general). Use those perpendiculars, and the geodesics from $p$ to the vertices, to cut the polygon into hyperbolic right triangles. Write down equations relating the shapes of these right triangles to the local geometry of the Voronoi graph. Use these equations to prove (with much agony) that as $p$ is perturbed the local geometry of the Voronoi graph changes.<|endoftext|>
TITLE: Can the friendship graph be determind by its adjacency spectrum?
QUESTION [5 upvotes]: Let $n\geq 1$ be an integer. The Friendship Graph (or Dutch windmill graph or $n$-Fan) $F_n$ is a graph that can be constructed by coalescence $n$ copies of the cycle graph $C_3$ with a common vertex. By construction, the friendship graph $F_n$ is isomorphic to the windmill graph $Wd\left(3,n\right)$.
Can $F_n$ be determined by its adjacency spectrum?
By the adjacency spectrum of a graph, we mean the multiset of the eigenvalues of the adjacency matrix of the graph. For a graph $G$, we denote by $Spec(G)$ its adjacency spectrum.
A graph $G$ is said to be determined by its adjacency spectrum, if $Spec(G)=Spec(H)$ for some graph $H$, then $G\cong H$. 
It is known that the friendship graph can be determined by the signless Laplacian spectrum.
See [Discrete Math. 310, No. 21, 2858-2866 (2010).]

REPLY [4 votes]: This is answered in the following recent paper:
The graphs with all but two eigenvalues equal to ±1
By Sebastian M. Cioabă, Willem H. Haemers, Jason Vermette, Wiseley Wong
One may download it from
http://arxiv.org/abs/1310.6529
The answer is no! But there is only one exception that is $F_{16}$ and all other friendship graphs can be determined by spectrum.<|endoftext|>
TITLE: Companion of the pointclass of inductive sets
QUESTION [5 upvotes]: This question is about the notion of a companion for a Spector class, as defined in Moschovakis's book Elementary Induction on Abstract Structures.
I am interested in Spector classes on $\mathbb{R}$, which are just a type of boldface pointclass.  The smallest one is IND, the class of (boldface) inductive sets, which I will consider as a typical example.
The companion of a Spector class $\bf \Gamma$ on $\mathbb{R}$ is a structure $(M,\in,R)$ with certain properties (listed in the book) such that $\bf \Gamma$ is the class of pointsets that are $\Sigma_1$-definable  in $M$ with real parameters from the relation $R$.
The companion of $\bf \Gamma$ is not unique, but its underlying set $M$ is unique and also the class of relations on $M$ that are $\Sigma_1$-definable from $R$ in $M$ with real parameters is unique.
The pointclass IND can also be described as the class of pointsets that are $\Sigma_1$-definable over $M$ from parameters in $\mathbb{R} \cup \lbrace\mathbb{R} \rbrace$ where $M = L_\kappa(\mathbb{R})$ is the least admissible level of $L(\mathbb{R})$.  We must allow $\lbrace\mathbb{R} \rbrace$ itself as a parameter here or we would just get the ${\bf \Sigma}^1_2$ sets.
For any companion $(M,\in,R)$ of IND we must have $M = L_\kappa(\mathbb{R})$.
Question: Is there a companion $(M,\in,R)$ of IND where the relation $R$ has a simple definition over $M = L_\kappa(\mathbb{R})$ (simpler than in Moschovakis's general construction of a companion?)  Maybe something that is already studied in the fine structure of $L(\mathbb{R})$?

REPLY [4 votes]: The relation $R$ can be empty; i.e., if $\kappa$ is the least ordinal such that $L_\kappa(\mathbb{R})$ is admissible, then the structure $(L_\kappa; \in, \emptyset)$ is a companion of the pointclass $\mathrm{IND}$ of inductive sets.  We can prove a somewhat more general statement.
Assume that $\kappa$ is an ordinal such that

$J_\kappa(\mathbb{R})$ is admissible (in which case $J_\kappa(\mathbb{R}) = L_\kappa(\mathbb{R})$, but there may be points later in the argument where it is necessary to use the Jensen hierarchy) and
$\kappa$ begins a $\Sigma_1$-gap in $L(\mathbb{R})$, meaning that $J_\alpha(\mathbb{R})$ is not a $\Sigma_1(\mathbb{R} \cup \{\mathbb{R}\})$-elementary substructure of $J_\kappa(\mathbb{R})$ for any ordinal $\alpha <\kappa$. 

Then the structure $(L_\kappa; \in, \emptyset)$ is a companion for the pointclass $\mathbf{\Sigma}_1^{J_\kappa(\mathbb{R})}$. (So letting $J_\kappa(\mathbb{R})$ be the least admissible level of $L(\mathbb{R})$ we obtain the desired result for the pointclass $\mathrm{IND}$.)
The criterion for "companionship" that was not clear to me when posting the question was projectability:  the existence of a $\mathbf{\Delta}_1^{J_\kappa(\mathbb{R})}$ partial surjection $\mathbb{R} \dashrightarrow J_\kappa(\mathbb{R})$.
The existence of a $\Sigma_1^{J_\kappa(\mathbb{R})}$ partial surjection $\mathbb{R} \dashrightarrow J_\kappa(\mathbb{R})$ is well-known.  Moreover, from this we can easily get a $\Delta_1^{J_\kappa(\mathbb{R})}$ partial surjection as follows.
(I have edited my answer below to replace the more convoluted argument that I wrote before.)
Let $F: \mathbb{R} \dashrightarrow J_\kappa(\mathbb{R})$ be a $\Sigma_1^{J_\kappa(\mathbb{R})}$ partial surjection and let $\theta$ be a $\Sigma_1$ formula defining it over $J_\kappa(\mathbb{R})$.
Then define $G: \mathbb{R} \dashrightarrow J_\kappa(\mathbb{R})$ by letting $(x,a) \in G$ if and only if there is an $\alpha < \kappa$ such that
$$J_\alpha(\mathbb{R}) \models \theta(x,a) \quad\And\quad \forall \xi < \alpha\, J_\xi(\mathbb{R}) \models \forall y \in \mathbb{R}\, \neg\theta(y,a).$$
Then $F \subset G$ and it is easy to check that $G$ is $\Delta_1^{J_\kappa(\mathbb{R})}$ and that the range of $G$ is equal to the range of $F$, which is all of $J_\kappa(\mathbb{R})$.<|endoftext|>
TITLE: How do facts about the homotopy type of cell complexes shed light on analytic number theory?
QUESTION [13 upvotes]: I just saw this link text interesting MO question, with a link to this paper, which uses facts from the topology of cell complexes to derive facts of an analytic number theory flavor.
From the perspective of an analytic number theorist, what insight does the topology offer? This approach is capable of proving results whose statements don't involve any topology, for example Theorem 4.1 of the link. Presumably these proofs (and those in the papers cited upon which they depend) could be translated out of the language of algebraic topology and into pure combinatorics and number theory; how big of a mess would this make out of the proofs?
Björner constructs cell complexes for which the Euler characteristic gives the summatory function of the Möbius function -- which is natural, as this is still elementary combinatorics in both cases. However, by the end of the paper he is quoting what appear to be distinctly nontrivial theorems in topology. Is it easy to summarize what these theorems are capable of saying from the number-theoretic point of view?
The proofs of his theorems rely on results from analytic number theory (e.g. Theorem 2.3). To what extent might one hope for results to flow in the other direction?

REPLY [5 votes]: There are other places where topology interacts with number theory. The beautiful book  by Hirzebruch and Zagier  The Atiyah-Singer index theorem and elementary number theory is a good place to see more examples based on rather deep topological results. 
The theory of Hilbert modular surfaces is another subject  where topology/geometry  meets number theory. According to Atiyah, the investigations of these  surfaces lead to  a remarkable result  called the Atiyah-Patodi-Singer index theorem which  when applied to Hilbert modular surfaces reveals a connection between the signatures of these  varieties and the $L$-functions  of the fields of algebraic integers naturally associated to these varieties. Hirzebruch's memoir on this subject is a joy to read. 
More than  two decades ago it was observed  that lattice point counts for certain rational polytopes are intimately related to intersection theory on certain toric varieties.
These are some examples that immediately jumped to my mind.<|endoftext|>
TITLE: Failure of a basic fact from Representation Theory
QUESTION [7 upvotes]: Recently I have been working with a certain subgroup of $GL_{10}(\mathbb{F}_2)$ and for various reasons was fairly sure it contained a normal subgroup isomorphic to $A_5$. Today I was able to affirmatively show that this presumed copy of $A_5$ does indeed exist in the bigger group. Now the matrices composing this copy of $A_5$ are themselves $2\times 2$ matrices over a certain commutative 5-dimensional algebra $\mathcal{S}$ (necessarily with zero divisors) over $\mathbb{F}_2$; this gives rise to a 2-dimensional representation of $A_5$ over $\mathcal{S}$ which I am fairly sure is irreducible (as opposed to just indecomposeable).
Now it is a fairly standard exercise early in learning representation theory to show that a simple group cannot have an irreducible representation of dimension 2. As the proof of this simple fact relies only on considerations involving the characters of elements of order 2, does the existence of the above representation rely only on the fact that I am working in characteristic 2, or is it related to the fact that I am working with rings with zero divisors as opposed to an algebraically closed field, or something else more subtle? I am primarily curious as to whether it is solely related to the characteristic, as I am also working with some related groups defined over commutative rings in other characteristics which I suspect also contain subgroups isomorphic to $A_5$. Hopefully someone with more background in modular representation theory than I can shed some helpful light on this situation.

REPLY [22 votes]: Well, the simple groups ${\rm SL}(2,2^{n})$ ( $n >1$) certainly have $2$-dimensional irreducible representations in characteristic $2$, and in fact $A_{5} \cong {\rm SL}(2,4)$
(to see this, just note that ${\rm SL}(2,4)$ has order $(4+1)(4^{2}-4) = 60 ).$
The standard argument to show that a non-Abelian finite simple group can't have a $2$-dimensional irreducible representation does not work in characteristic $2$, since it relies on the fact that an involution (which must be non-central using simplicity) must have determinant $-1$, since one of its eigenvalues must be $1$ and the other $-1.$ But again since $G$ is non-Abelian simple, every representation has trivial determinant, a contradiction. But this line  argument does not work over a field of characteristic $2$.
Note for historical interest, related to comments in Qiaochu's answer: Since a finite group of odd order is solvable, and since the dimension of absolutely irreducible modules in any characteristic divide the group order when the group is solvable, it follows that a finite group of odd order can't have a $2$-dimensional absolutely irreducible module in any characteristic. However, this requires the rather heavyweight Feit-Thompson theorem, so may be regarded as a little unsatisfactory in the context of this discussion. So I outline a different argument. This still requires some relatively difficult group theory, namely the Hall-Higman theorem, or its descendants, a Theorem of E. Shult and the theory of $p$-stability. The fact that we need from this theory is that if $p$ and $q$ are odd primes, and a $p$-group $P$ normalizes a $q$-group $Q$ and the group $PQ$ has a faithful linear representation in characteristic other than $q$,  then every element of $p$ which acts with minimum polynomial of degree less than $p$ centralizes $Q.$
      Hence if a finite group $G$ of odd order has a faithful $2$-dimensional representation in some finite characteristic $p$, then it follows that for each prime divisor $q$ of $|G|$ other than $p$, we have that $N_{G}(Q)/C_{G}(Q)$ is a $q$-group for each $q$-subgroup $Q$ of $G.$ By a Theorem of Frobenius, $G$ has a normal $q$-complement. It follows that $G/O_{p}(G)$ is nilpotent, where $O_{p}(G)$ is the largest normal $p$-subgroup of $G.$
Hence (without using the classification of finite simple groups, or the odd order theorem),
we see that a non-Abelian simple group of odd order can't have a $2$-dimensional irreducible representation in any characteristic.

REPLY [13 votes]: To reinforce Geoff's answer, I'd emphasize that the "fairly standard exercise" mentioned in the question is only standard when you consider characteristic 0 irreducible representations as is usually done in a first course (then it's also usual to start with a big field like $\mathbb{C}$ to avoid extra complications).
When dealing with modular representations, there are lots of further issues.
Qiaochu attempts an outline of the "standard exercise" which shows clearly where problems arise.   In his first step, you need characteristic 0 (and a splitting field) to be confident that the degree of an irreducible representation divides the group order.   This is definitely false in general.   And then of course characteristic 2 is especially dangerous for dealing with the negative of the $2 \times 2$ identity matrix.   
The bottom line is that you have to avoid this "standard exercise" and focus just on the special situation you are studying, in order to avoid confusion.
[ADDED] 1) Concerning the "standard exercise" (say over $\mathbb{C}$), I recall that decades ago it appeared as a Math Monthly problem (which I wrote an answer for) and later became an exercise in at least one advanced algebra text.   Given an absolutely irreducible representation of degree 2 for a (necessarily nonabelian) finite simple group, classical character theory implies that the group order is even, so you can apply Cauchy etc.   Of course, Feit-Thompson gets you there as well, but that's overkill.   However, in odd characteristic that does help to get through the other steps of the proof, but in characteristic 2 you'd still face $I = -I$.  
2) While the degrees of irreducible representations of simple groups of Lie type are far from being known explicitly in general, over a splitting field of the defining characteristic you can see easy examples where degrees fail to divide the group order.   For instance, take $\mathrm{PSL}_2(\mathbb{F}_7)$ in characteristic 7.   Here you have an irreducible of degree 5 whereas 5 doesn't divide the group order 168.<|endoftext|>
TITLE: Knot diagrams, sets of moves and equivalence relations
QUESTION [11 upvotes]: Short version:  Does anyone study equivalence classes generated by a given set of "moves" (in the sense of, but not limited to, Reidemeister moves) on the set of knot diagrams?
Yes, I understand that the concept of knot has a natural geometrical significance, that one usually views knot diagrams as a tool to study underlying knots, that the value of Reidemeister moves lies in how they preserve and generate the equivalence relation of isotopy.  So, yes, I see why a knot theorist might reasonably have little interest, say, in looking at local moves not preserving isotopy.  
So I'm asking this question in the spirit of abstraction for its own sake.  But there is a precedent.  A "symmetry theorist" studies groups because they capture the set of symmetries of important objects.  But combinatorial group theory studies equivalence classes of strings under moves...and symmetry, if it enters the story at all, does so as a tool.  
That said, it would be interesting if one could add an extra move to the Reidemeister moves that produced a coarser but computationally tractable classification.
No need to retread the ground here -- http://en.wikipedia.org/wiki/Reidemeister_move -- so, for example, I already understand that knot theorists know what happens with only Reidemeister type II and III moves.  I am interested in stories like this, where one gets a finer equivalence relation that isotopy, but equally interested in sets of local moves that don't preserve isotopy and thus generate equivalence relations either coarser to, or simply incomparable with, isotopy.

REPLY [8 votes]: The Delta move generates the equivalence relation of linking numbers for links,
as proved by Matveev and Murakami Nakanishi. 

There's the double Delta move which Naik and Stanford showed generate S-equivalence. 

A student of Freedman studied "slide equivalence" of knot diagrams. Unfortunately
it wasn't published though.<|endoftext|>
TITLE: Order-increasing bijection from arbitrary groups to cyclic groups
QUESTION [35 upvotes]: In his answer to this previous MO question, Gjergji Zaimi referred to the statement that for every finite group $G$ of order $n$, there is a bijection $\sigma \colon G \to \mathbb{Z}/n\mathbb{Z}$ satisfying
$$o(\sigma(g))\geq o(g)$$
for all $g\in G$.
However, as Marty Isaacs pointed out, the "proof" of this fact that appeared in the Amer. Math. Monthly is flawed (or at least incomplete).


Does anyone know of a valid proof (or counterexample) of this fact?


Added:
Since this question has been without answer for about 2 weeks now, I'm adding some more details that might help (perhaps not everyone has access to the paper in Amer. Math. Monthly.)
The "proof" in loc. cit. goes as follows:
Let $G_k = \{\,g \in G: g^k = 1\,\}$.
For $d \mid n$, we inductively define a set $S_d$ consisting of $\phi(d)$ elements in $G_d$, where $\phi$ is Euler's totient function. These sets will be pairwise disjoint; since
$\sum_{d \mid n} \phi(d) = n$, they thus partition $G$.
Let $S_1 = \{1\}$. Suppose that $k$ divides $n$ and that $S_d$ has been constructed for all $d < k$.
By a theorem of Frobenius (see M. Hall, The Theory of Groups, Macmillan, 1959, p. 137), $|G_k|$ is a multiple of $k$. Also $G_k$ contains $S_d$ for each $d$ dividing $k$. Since $|G_k| > k$ and $\sum_{d \mid k} \phi(d) = k$, there remain at least $\phi(k)$ other elements in $G_k$, and we take $\phi(k)$ of them to form $S_k$.
The cyclic group $Z_n$ of order $n$ has exactly $\phi(d)$ elements of order $d$ for each $d$ that properly divides $n$. We construct a bijection $\sigma\colon G \to Z_n$ that maps $S_d$ into the elements of order $d$ in $Z_n$. Thus $o(\sigma(g)) > o(g)$ for all $g \in G$.  $\quad\square$

The mistake in the proof is in the last sentence of the second paragraph, where the authors overlooked the fact that some of the sets $S_d$ where $d$ does not divide $k$ (but still $\gcd(d,k) \neq 1$) might have used some of the elements of $G_k$, leaving less than $\phi(k)$ elements left in $G_k$.
As F. Ladisch (I think, I hope I'm giving proper credit) pointed out in a comment to an earlier answer by G. Zaimi that was deleted, the proof is fundamentally incorrect, in the sense that only using the combinatorial data implied by Frobenius's Theorem cannot be sufficient to prove the result. For example, let $n=12$, and let
$$|G_1| = 1, |G_2| =  4, |G_3| = 3, |G_4| = 4, |G_6| = 6, |G_{12}| = 12. $$
(This would correspond to a non-existing group with $1$ element of order $1$, $3$ elements of order $2$, $2$ elements of order $3$, and $6$ elements of order $12$.)
Then the sets $G_k$ fulfill the requirements given by Frobenius's Theorem, but still there is no bijection to $Z_n$ satisfying the required statement.

Any ideas are welcome!

REPLY [10 votes]: This is only a partial answer:  
The assertion is true for solvable groups. In fact, when $N$ is a normal elementary abelian $p$-subgroup of $G$ and $\sigma\colon G/N \to C_{|G/N|}$ an "order multiplying bijection" (that is, $o(\bar{g})$ divides $o(\sigma(\bar{g}))$ for all $\bar{g}\in G/N$), then one can lift $\sigma$ to an order multiplying bijection $\hat{\sigma}\colon G\to C_{|G|}$. This means that for every $g\in G$ we find an order multiplying bijection 
between cosets $gN \to cC_{|N|}$, where $\bar{c}= \sigma(\bar{g})$ (I use overbars to denote the canonical epi's $G\to G/N$ and $C_{|G|}\to C_{|G/N|}$).
To see this, first an easy observation:  

$(*)\qquad $ For any $g\in G$, the order $o(g)$ divides $po(\bar{g})$.  

Indeed,  $\langle g \rangle \cap N$ is cyclic, so has order $p$ or $1$.
It follows that the orders of elements in the coset $gN$ are $o(\bar{g})$ or $po(\bar{g})$.
For $c\in C_{|G|}$, the situation is this: if $p$ does not divide $o(\bar{c})$, then in $cC_{|N|}$ the orders $o(\bar{c})p^k$ occur, where $p^k$ divides $ |N|$, and $o(\bar{c})$ occurs exactly once. If $p$ divides $o(\bar{c})$, then all elements in $cC_{|N|}$ have order $o(\bar{c}) |N|$.  
It follows from these observations that whenever $o(\bar{g}) \mid o(\bar{c})$, we find an order multiplying bijection $gN \to c C_{|N|}$. When $p \mid o(\bar{c})$, in fact any bijection does, and if $p$ does not divide $o(\bar{c})$, then $p$ does not divide $o(\bar{g})$, and there is at least one pre-image $g$ of $\bar{g}$ such that $p$ does not divide $o(g)$ (Schur-Zassenhaus or more elementary). We map such a $g$ to the unique element in $cC_{|N|}$ having order $o(\bar{c})$, and then map the rest of $gN$ onto the rest of $cC_{|N|}$.  
Remarks

It follows that a minimal counterexample to the assertion of the question has trivial Fitting subgroup. One could try to reduce to simple groups, but I don't see how, since we don't have something like $(*)$ for arbitrary normal subgroups.  
Write $H\preceq G$ if there is an order multiplying bijection from $H$ to $G$. This defines a prae-order on the class of finite groups of fixed order, and we want to show that $G\preceq C_{|G|}$ for all $G$. Now at least examples of small size suggest that nonsolvable groups are minimal or close to minimal with respect to this prae-order. For example, $S_5$ and $A_5\times C_2$ are minimal with respect to $\preceq $ among groups of order 120, and $SL(2,5)$ covers both with respect to $\preceq$. Moreover, there are quite a few groups 'between' these groups and $C_{120}$. This supports the intuition that there are "more degrees of freedom" in choosing an order multiplying bijection when the group is non-solvable, since such groups have in general few elements of (arithmetically) large order and many of small order.<|endoftext|>
TITLE: homomorphism into reductive groups
QUESTION [7 upvotes]: Let $k$ be an algebraically closed field with char($k$)$= p > 0$. 
Let $P$ be a finite $p$-group. For any homomorphism
$\rho : P \rightarrow GL(n,k)$ we know that the image $im(\rho)$ can be
put inside the group of upper triangular matrices(standard borel) 
after conjugation. The question I have is what can we say about homomorphisms
of $P$ into a general reductive or even semi-simple algebraic groups.
(The usual argument for $GL(n,k)$ case can be adapted to $SL(n,k)$ as well.
)

REPLY [4 votes]: This is a (rather) expanded version of Yves' answer, because for the specific focus of the question one can understand the proof without getting into many other technicalities of the Borel--Tits paper (which has bigger fish to fry, related to fields of definition and much more). We will explain the proof of the following result (which is given over perfect fields rather than just algebraically closed fields so that we include finite fields):
${\bf Theorem}$ (Borel--Tits): In  a connected linear algebraic group $G$ over a perfect field $k$, every subgroup $H \subset G(k)$ consisting of unipotent elements is contained in a "canonical" connected unipotent smooth closed $k$-subgroup (i.e., one whose formation is compatible with respect to $k$-isomorphisms in the pair $(G,H)$ and perfect extension on $k$).
This will use a beautiful construction in section 2 of the Borel--Tits paper. The following is an exposition of some arguments in that paper, with an eye towards the proof of the preceding Theorem (which thereby answers the question posed); by all means one should look at the original paper too (all I have done is provide some more explanation about various aspects of the argument).
Proof: Due to the canonicity claim applied in the setting of Galois descent relative to $\overline{k}$ over $k$, we may and do assume $k$ is algebraically closed (and the compatibility with respect to any further perfect extension of the ground field will be clear from the construction). Since the Zariski closure of $H$ is (smooth and) unipotent (but perhaps disconnected), we can replace $H$ with its Zariski closure and instead work with unipotent smooth closed subgroups $U$ in $G$ (e.g., finite constant ones when in positive characteristic). We induct on the dimension of $G$ (dimension 0 being trivial).  We can also replace $U$ with $U \cdot R_u(G)$ so that $U$ contains $R_u(G)$, and if $R_u(G) \ne 1$ then we can pass to $G/R_u(G)$ to drop the dimension and conclude, so we can assume that $G$ is reductive. We can assume $U \ne 1$, so by nilpotence of unipotent algebraic groups there is a nontrivial central element $u$ in $U(k)$. 
We will show that $u$ lies in the unipotent radical of a parabolic subgroup $P$ that is "canonical" in terms of $u$ (i.e., one stable under all automorphisms of $G$ that preserve $u$, or even just preserve the cyclic subgroup generated by $u$). Once this is proved, such a $P$ is stable under $U$-conjugation on $G$ (as that leaves the central $u$ in $U(k)$ invariant), so by the self-normalizing property of parabolic subgroups it would follow that $U \subset P$.  But $P \ne G$ (since $R_u(P)$ contains $u \ne 1$ whereas we arranged that $G$ is reductive), so we can use dimension induction to conclude that $U$ lies in a connected unipotent subgroup of $P$, but without canonicity in $(G,U)$. However, we gain the property that $U$ is "embeddable" (in the terminology of Borel--Tits), meaning that it does occur inside some connected unipotent subgroup (equivalently, in the unipotent radical of a Borel subgroup). 
Observe that $u$ lies in the unipotent radical of some parabolic subgroup, or equivalently in the unipotent radical of some Borel subgroup: this is immediate from Grothendieck's important theorem that $G(k)$ is covered by the subgroups $B(k)$ for Borel subgroups $B$ of $G$ (a result that is proved near the end of section 14 of Borel's textbook but possibly not stated as an official result there; it can be extracted nonetheless). Thus, the Zariski closure of the cyclic subgroup generated by $u$ is "embeddable".  Moreover, we saw above via the dimension induction that once this cyclic subgroup is realized inside the unipotent radical of a "canonically associated" parabolic subgroup of $G$, then $U$ itself is "embeddable".  Hence, by two applications of the following result (which is to be regarded as a finer version of Grothendieck's theorem, but whose applicability in our inductive strategy ultimately rests on Grothendieck's theorem) we would be done:
${\bf Refined}$ ${\bf Theorem}$ (Borel--Tits). Any embeddable unipotent smooth
closed subgroup $U$ of a connected linear algebraic group $G$ over a perfect field $k$ is contained in the unipotent radical of a "canonically associated" parabolic $k$-subgroup $P$. (As before, "canonical" means that the formation of $P$ is compatible with respect to $k$-isomorphisms in the pair $(G,U)$ and perfect extension on $k$.)
[Note that this result is interesting even when $U$ is connected, and it is interesting in characteristic 0. Also, the embeddable hypothesis is automatic once this is all over, so we see that the unipotent radicals of the minimal parabolic $k$-subgroups are precisely the maximal unipotent smooth closed $k$-subgroups; in particular, if $G$ has no proper parabolic $k$-subgroups then $G(k)$ contains no nontrivial unipotent smooth closed $k$-subgroups. The proof gives a "canonical" $P$, but it is not uniquely determined by the stated conditions in general, as we see by considering $G = \prod G_i$ for several pairwise non-isomorphic $k$-simple semisimple $G_i$ with $U \subset G_1$. I don't know offhand whether one can do better when $G$ is $k$-simple or absolutely simple.]
Proof:  By Galois descent from $\overline{k}$ we may assume $k$ is algebraically closed (and it will be clear from the construction that we have compatibility with any further perfect extension of the ground field).  To construct $P$, we will work with the normalizer $N(U)$, which might be non-unipotent and even more disconnected. Consider the unipotent radical of $N(U)$ (i.e., maximal unipotent connected normal subgroup) anyway.  Note that $R_u(N(U))$ trivially contains $U^0$, but it isn't clear if it contains $U$ or not.  So following Borel--Tits (see 2.4 in their paper), let's put $U$ back in: define the closed subgroup
$$L(U) = U \cdot R_u(N(U))$$
that is unipotent and does contain $U$. By the "canonicity" of $L(U)$ in terms of $U$, any automorphism of $G$ that carries $U$ back to itself must do the same for $L(U)$. Also, just as we assumed that $U$ is embeddable (i.e., lies in the unipotent radical of a Borel), we claim the same holds for $L(U)$.  This amounts to checking that $L(U)$ has a fixed point on the projective variety of Borel subgroups of $G$.  By hypothesis $U$ has a fixed point, so the closed (!) locus of Borels containing $U$ is a non-empty projective variety (perhaps disconnected).  Certainly $N(U)$ acts on that locus, so the connected solvable $R_u(N(U))$ does too.  By the Borel fixed point theorem, $R_u(N(U))$ has a fixed point in that locus, so we have found a Borel subgroup of $G$ that contains $L(U)$.  Hence, we lose nothing by replacing $U$ with $L(U)$.
We iterate a few times, and eventually reach the situation that the dimension stops growing.  That is, we may assume $L(U)$ has the same dimension as $U$. Thus, the connected normal subgroup $R_u(N(U))$ in $L(U)$ must be contained in $U^0$, so $L(U) = U$.  In other words, we  $R_u(N(U)) \subset U$, so the containment $U^0 \subset R_u(N(U)) = R_u(N(U)^0)$ is an equality. Thus, the normal unipotent subgroup $U$ in $N(U)$ has $R_u(N(U))$ as its identity component.  At this stage, we shall prove that $U$ is already connected (but we really need to go further and get the canonical parabolic, or else the plan of the argument would collapse; in fact, we will show that $N(U)$ is parabolic).
By our arranged "embeddable" hypothesis on $U$, we can pick a Borel subgroup $B$ of $G$ that contains $U$, so $L := B \cap N(U)$ is a solvable (possibly disconnected) closed subgroup of $N(U)$ that contains $U$. The connected solvable $L^0$ must lie in a Borel subgroup $B'$ of $N(U)^0$, and if $B''$ denotes the opposite Borel subgroup then $R_u(B' \cap B'') = R_u(N(U)^0) = U^0$.  We have $B'' = gB'g^{-1}$ for some $g \in N(U)^0$, so $R_u(L^0 \cap gL^0g^{-1}) \subset U^0$.  Equivalently, $R_u(L \cap gLg^{-1}) \subset U^0$. That is, 
$$R_u(N(U) \cap B \cap gBg^{-1}) \subset U^0$$
for some $g \in N(U)^0$ and some Borel $B$ of $G$ containing $U$.  Clearly $U$ is contained in $gBg^{-1}$ too, so 
$$U \subset B \cap gBg^{-1}.$$
But any intersection of two Borel subgroups is always connected (even contains a maximal torus), and unipotent radicals are functorial for connected solvable groups, so 
$$U \subset R := R_u(B \cap gBg^{-1}).$$ 
We now show that this inclusion is an equality (so $U$ is connected).
Suppose to the contrary that it is a strict inclusion. Consideration of the descending central series of the connected unipotent $R$ (whose steps must eventually emerge from $U$, maybe at the first step or maybe later) yields a closed connected  subgroup $U'$ in $R$ that normalizes $U$ but is not contained in $U$. This connected unipotent $U'$ then lies inside the identity component of the solvable $N(U) \cap B \cap gBg^{-1}$, so it lies in its unipotent radical, which we have seen is contained in $U^0$. The resulting inclusion $U' \subset U^0$ contradicts that $U'$ is not contained in $U$, so we have shown that $U = R$.  Thus, now $U$ is connected, so we conclude that $U = U^0 = R_u(N(U))$.  
Now it suffices to prove that $N(U)$ is a parabolic in $G$; i.e., we just have to produce a Borel of $G$ inside $N(U)$, where $U = R_u(B \cap B')$ for Borel subgroups $B$ and $B'$ of $G$. We'll show that $B$ is contained in $N(U)$ (by exploiting that also $U = R_u(N(U))$).  Since $U$ contains $R_u(G)$, it is harmless to pass to $G/R_u(G)$ so that $G$ is reductive (if it wasn't already).  We will soon make serious use of the links between reductive groups and root systems (and the uniqueness characterizations of root groups).
There is a maximal torus $T$ of $G$ contained in the smooth connected $B \cap B'$, so $T$ normalizes $R_u(B \cap B') = U$. Thus, $N(U)$ contains $T$. Since $R_u(B)$ is generated by the simple positive root groups for $\Phi(B,T)$ (this is a delicate fact in small positive characteristics), it suffices to show that each of those normalizes $U$.  The $T$-stable connected subgroup $U \subset R_u(B)$ must be generated by the root groups for the nontrivial $T$-weights that occur on its Lie algebra, and we just have to consider simple positive $a \in \Phi(B,T)$ for which the root group $U_a$ is not contained in $U = R_u(B \cap B')$, so $U_a$ is not contained in $B'$.  Hence, $-a \in \Phi(B',T)$ (so $U_{-a} \subset B'$).  
Applying to $B$ the "reflection" associated to the simple $a \in \Phi(B,T)$ gives a Borel subgroup $B_a$ containing $T$ and having the same $T$-weights as $B$ except that $a$ is replaced with $-a$.  Thus, $R_u(B \cap B_a)$ is directly spanned by the root groups associated to $\Phi(B,T) - (a) = \Phi(B_a,T) - (-a)$, so it is normalized by $U_{-a}$ (as $-a$ is simple for $\Phi(B_a,T)$). We conclude that $U_{-a}$ normalizes $R_u(B \cap B_a) \cap B'$, so it normalizes its unipotent radical, which is directly spanned by the roots in 
$$(\Phi(B,T) - (a)) \cap \Phi(B',T) = \Phi(B,T) \cap \Phi(B',T) = \Phi(R_u(B \cap B'),T)
= \Phi(U,T).$$
In other words, $R_u(B \cap B_a) \cap B'$ has unipotent radical $U$, so $U_{-a}$ normalizes $U$ yet $U_{-a}$ is not contained in $U^0$ (since $-a$ is not even a $T$-weight for $B$).  Now recall that $U^0 = R_u(N(U)) = R_u(N(U)^0)$, so the $T$-weight $-a$ occurring on $N(U)^0$ does not arise on the unipotent radical, and hence (by reductivity of the quotient  $N(U)^0/R_u(N(U)^0)$ that contains $T$) the $T$-weight $-(-a) = a$ occurs on $N(U)^0$.  This forces $U_a \subset N(U)^0$, so $U_a$ normalizes $U$.  To summarize, for each simple $a \in \Phi(B,T)$, either $U_a \subset U$ or else $U_a$ normalizes $U$, so either way always
$U_a \subset N(U)$.  This forces $B \subset N(U)$, so we are done. QED<|endoftext|>
TITLE: Conventional names for finite categories
QUESTION [7 upvotes]: I'm looking for, or hoping to inspire the creation of, a list of conventional names for categories that come up often.
For example, we have the terminal category $\fbox{$\bullet$}$, a nice name. I've heard this category $\fbox{$\bullet\to\bullet$}$ called the free arrow category. That's fine by me. What about the category with one object $x$ and an one arrow $p^n\colon x\to x$ for each natural number $n$ (and the obvious composition law)? It's hard to draw it without a package; here's my best attempt: $\fbox{$\bullet\circlearrowleft$}$. I'd like to call this the free loop category. But is that standard? 
The categories $[n]$ for $n\in\mathbb{N}$ might be drawn $\fbox{$\bullet^0\to\bullet^1\to\cdots\bullet^n$}$. I might call this the length-$n$ chain category. What about $\fbox{$\bullet\rightrightarrows\bullet$}$? I might call this the parallel arrows category. Would one know what I meant by the two equalized arrows category or the two coequalized arrows category? Hint: they each have three objects and four non-identity morphisms. But what if I didn't want a certain commutative diagram to hold there, i.e. I wanted to name the related five-morphism categories?
The point I hope is clear. Does anyone know of a definitive list of names for important diagram categories? 
Thanks!

REPLY [3 votes]: I notice that the categories considered for naming here are all the domains, or shapes, of basic diagrams; an object, an arrow, an endomorphism (n.b., my instinct was just to call that $\mathbb{N}$), a composable sequence, parallel arrows, equalized arrows...  Not that diagrams in these categories aren't also interesting (a composable sequence in a composable sequence category, e.g., is well-worth half-an-hour's thought), but as diagram domains is where they all come up first for most of us; so why not call them what they are?

$\fbox{$\phantom{X}$}$, the trivial diagram domain/the shape of the empty diagram
$\fbox{$\bullet$}$, the object diagram domain/the shape of an object
$\fbox{$\overset\bullet\circlearrowleft$}$, the endomorphism diagram domain/the shape of an endomorphism...

Of course, to establish a convention, one must write a famous textbook. Good luck with that!<|endoftext|>
TITLE: Adjoining a new isolated point without changing the space
QUESTION [5 upvotes]: Suppose $X$ is a $T_1$ space with an infinite set of isolated points. Show that if $X^\sharp = X \cup \lbrace \infty \rbrace$ is obtained by adding a single new isolated point, then $X$ and $X^\sharp$ are homeomorphic.
I am almost embarrased to raise this, which seems obvious.  The proof must be simple, but it eludes me for now.  Maybe it is an exercise in some textbook.  You can clearly establish a 1-1 equivalence between the isolated points of $X$ and those of $X^\sharp$.  But it is not clear how this equivalence would extend to the closure of the isolated points.
The theorem is easy when $X$ is compact $T_2$ and $cl(D) = \beta(D)$, where $D$ is the set of isolated points.

REPLY [9 votes]: In my answer to https://mathoverflow.net/questions/26414 I descibed a somewhat simpler-looking example than Nik's, but proving that it works may be harder.  Take two copies of $\beta\mathbb N$ and glue each non-isolated point of one copy to the corresponding point of the other copy.  Any way of "absorbing" a new isolated point into the two copies of $\mathbb N$ forces a relative shift of those two copies, which forces corresponding shifts of the non-isolated points, which in turn conflicts with the gluing.  The perhaps surprising thing about this example is that, if you add two isolated points, the result is (easily) homeomorphic to the original.<|endoftext|>
TITLE: triviality of determinant sheaf
QUESTION [6 upvotes]: On a smooth algebraic variety X, every coherent sheaf F has a finite resolution by
locally free sheaves. Using such resolution, we can define the determinant
of F, det F, which is a line bundle on X.
My question is :
why if the support of F is of codimension greater or equal to 2
is the determinant of F trivial ?
It is mentionned without proof on the book "The geometry of moduli spaces of
sheaves", D. Huybrechts, M. Lehn.
I have verified this result on some explicit examples for which I know some
explicit locally free resolutions but I don't see how to do the general case.

REPLY [16 votes]: Outside the support of $F$, the resolution is an exact sequence, so the alternating tensor product of the determinants is trivial. On a smooth scheme, a line bundle trivial outside a codimension $2$ subset is trivial.<|endoftext|>
TITLE: Characterizing $\omega_1$-like dense linear orderings
QUESTION [5 upvotes]: I recently came upon the following theorem which was attributed to J. Conway:
For each $A\subset \omega_1$, let $\Phi(A)$ be a linear ordering of type $\sum_{\alpha<\omega_1} \tau_\alpha$, where $\tau_\alpha$ is $\eta$ (the order-type of the rationals) whenever $\alpha\notin A$, and $1+\eta$ whenever $\alpha\in A$. The theorem asserts that for every $\omega_1$-like dense linear ordering is isomorphic to some $\Phi(A)$, $A\subset\omega_1$.
My questions:
(1) Does the theorem generalize to higher cardinalities?
(2) The reference given was J. Conway's Ph.D. Thesis, but I did not manage to find anything on the web. Any help?
Thank you in advance.

REPLY [7 votes]: Concerning question 2, the theorem that you mention in the case of
$\omega_1$ is not actually difficult. I had briefly sketched a
proof of it at the conclusion of my answer to the math.SE question
Linearly ordered sets somewhat
similar to $\mathbb{Q}$, which is concerned with these types of orders---what I called $\mathcal{Q}_A$ is the same as your $\Phi(A)$. Here is a complete argument:
Theorem. Every $\omega_1$-like dense linear ordering is
isomorphic to $\Phi(A)$ for some $A\subset\omega_1$.
Proof. Suppose that $L$ is an $\omega_1$-like dense linear order.
Let $\langle x_\alpha\mid
\alpha\lt\omega_1\rangle$ be any increasing cofinal $\omega_1$-sequence in
$L$, containing none of its limit points (i.e., scattered). Let
$\tau_\alpha$ be the interval of points above 
$\cup_{\beta\lt\alpha}\tau_\beta$ and below the point $x_\alpha$. This
is either $\eta$ or $1+\eta$, depending on whether
$\cup_{\beta\lt\alpha}\tau_\beta$ has a supremum in $L$ or not. These
intervals therefore realize $L$ as
$\Sigma_{\alpha\lt\omega_1}\tau_\alpha$, as desired. Thus, $L$ is
$\Phi(A)$, where $A$ is the set of $\alpha$ where that supremum
exists. QED
In that previous answer, I proved that the $\Phi(A)$ orders are
determined up to isomorphism essentially by the equivalence of $A$
modulo the club filter.
Theorem. $\Phi(A)$ is isomorphic to $\Phi(B)$ if and only
if $A$ and $B$ agree on having $0$ and also agree modulo the club
filter, meaning that there is a closed unbounded set
$C\subset\omega_1$ such that $A\cap C=B\cap C$. In other words,
this is if and only if $A$ and $B$ agree on $0$ and are equivalent
to $B$ in $P(\omega_1)/\text{NS}$, as subsets modulo the
nonstationary ideal.
It would seem to be an interesting question to inquire in your
style whether one may extend this beyond $\omega_1$ to higher
cardinals.
The anwer, unfortunately, is negative. Suppose $\kappa\gt\omega_1$ is any cardinal. Let $A$ be the empty set, and let $B$ be any nonstationary set containing some ordinals with uncountable cofinality; for example, consider the singleton set $B=\{\ \omega_1\ \}$. These two sets agree on a club, since both omit a club. But meanwhile, $\Phi(A)$ and $\Phi(B)$ are not
isomorphic, because the former has all points having cofinality
$\omega$, but the latter has points of uncountable
cofinality.<|endoftext|>
TITLE: Rank two vector bundles on a curve of genus two
QUESTION [12 upvotes]: I recently learned of an interesting result of Narasimhan and Ramanan from 1969, which says that moduli space of rank two vector bundles with trivial determinant on a curve $X$ of genus two is naturally isomorphic to $\mathbb PH^0(\operatorname{Pic}^1X,\mathcal L_\Theta^{\otimes 2})$ (this vector space has dimension four).  I'd like to understand this isomorphism better in the context of the Narasimhan--Seshadri theorem.
First, fix a closed topological surface $X$ of genus two.  Let:
$$M_{X,SU(2)}=\operatorname{Hom}(\pi_1(X),SU(2))//SU(2)$$
denote the $SU(2)$ character variety of $X$.  For a complex structure $\sigma$ on $X$, let $M_{X,\sigma,\operatorname{rk}2}$ denote the moduli space of rank two vector bundles on $X$ with trivial determinant (actually, there is a technical stability/equivalence relation which should be included, but I will ignore this).  According to the Narasimhan--Seshadri theorem, $M_{X,SU(2)}$ and $M_{X,\sigma,\operatorname{rk}2}$ are naturally diffeomorphic (again, there are some qualifications to this which I am ignoring; in particular both spaces are usually singular).
Now I want to recall the isomorphism $M_{X,\sigma,\operatorname{rk}2}\to\mathbb PH^0(\operatorname{Pic}^1X,\mathcal L_\Theta^{\otimes 2})$.  To any rank two vector bundle $E$, we consider the subvariety $C_E$ of $\operatorname{Pic}^1X$ consisting of bundles $\xi$ for which there is an exact sequence:
$$0\to\xi\to E\to\xi^{-1}\to 0$$
(that is, $E$ is an extension of $\xi^{-1}$ and $\xi$).  Then Narasimhan and Ramanan prove that $C_E$ is a divisor on $\operatorname{Pic}^1X$ and is linearly equivalent to $2\Theta$.  This gives a map $M_{X,\sigma,\operatorname{rk}2}\to\mathbb PH^0(\operatorname{Pic}^1X,\mathcal L_\Theta^{\otimes 2})$, which Narasimhan and Ramanan go on to show is an isomorphism.  (That was only a rough outline).
OK, now let's reinterpret the map $M_{X,\sigma,\operatorname{rk}2}\to\mathbb PH^0(\operatorname{Pic}^1X,\mathcal L_\Theta^{\otimes 2})$ in terms of the Narasimhan-Seshadri theorem.  Remember that $\operatorname{Jac}X$ is $\operatorname{Hom}(\pi_1(X),U(1))$.  Thus for a homomorphism $\rho:\pi_1(X)\to SU(2)$ (corresponding to a vector bundle of rank two), we can define the subvariety $C_E$ as those $U(1)$ representations $\alpha:\pi_1(X)\to U(1)$ for which $\rho$ can be conjugated to have the form:
$$\left(\begin{matrix}\alpha&\beta\cr 0&\alpha^{-1}\end{matrix}\right)$$
This is a subset of $\operatorname{Hom}(\pi_1(X),U(1))=\operatorname{Jac}X$.  Now according to Narasimhan and Ramanan, it should be a subvariety of $\operatorname{Jac}X$ for any complex structure on $X$.  This seems a bit unlikely to me, because there is a large moduli of complex structures on $X$.  Also, somehow I've constructed naturally $C_E\subseteq\operatorname{Jac}X$, but according to the construction in Narasimhan and Ramanan I should be getting $C_E\subseteq\operatorname{Pic}^1X$, which is really not the same thing canonically.
I suppose I'm getting confused in applying the Narasimhan-Seshadri theorem.  Any assistance is appreciated!

REPLY [7 votes]: Dear unknown, by NR, there exists always a family of degree -1 line sub bundles for every holomorphic vector bundle of rank two with trivial determinant on a genus 2 surface. As they have degree -1, they do not admit flat connections. This means that when your representation $\rho$ has the upper triangular form you wrote down, the representation $\alpha$ will correspond to a degree $0$ subbundle $L$ and not to a degree -1 subbundle. Of course, this implies that your bundle $E$ is only semi-stable and not stable, or equivalently, $\rho$ is reducible, which is not the generic case. (In that case, $C_E$ can be explicitly computed as $L\Theta\cup L^*\Theta$ which clearly depends on the Riemann surface structure.)<|endoftext|>
TITLE: Normal subgroup of classical groups
QUESTION [8 upvotes]: For $r\leq n$, consider the following reduction homomorphism 
$$
\pi_{n,r}: {\rm SL}_2(\mathbb{Z}/(p^n\mathbb{Z}))\to {\rm SL}_2(\mathbb{Z}/(p^r\mathbb{Z})).
$$
Bourgain and Gamburd in their paper "Expansion and random walks in ${\rm SL}_d(\Bbb Z/(p^n
\Bbb Z))$" mentioned that the set 
$$\{\ker\pi_{n,r}\},$$
gives all normal subgroups of ${\rm SL}_2(\Bbb Z/(p^n\Bbb Z))$. Probably this is standard, but I am not able to prove it.
Can any one please give me a hint or a reference for this fact? 
I remember that I read somewhere that a modification of  this fact also valid for 
$${\rm SL}_k(\mathbb{Z}/(p^n\mathbb{Z})),$$ 
and the symplectic group 
$${\rm Sp}_{2k}(\mathbb{Z}/(p^n\mathbb{Z}).$$
But I can not find where I saw these. I would be most thankful if anyone can help me with these.

REPLY [7 votes]: Let me answer the question for $p>3$ and $SL_2$. I imagine that a similar method will work for the other cases but I haven't checked.
Some notation: $G=SL_2(\mathbb{Z}/p^n\mathbb{Z})$, $Z=\{I, -I\}$ and, for $i=1,\dots, n$,
$$K_i := \ker \pi_{n,i}.$$
Proposition: The proper normal subgroups of $G$ are $K_i$ and $K_i\times Z$ for $i=1,\dots, n$.
Sketch of basic steps of proof:

If $n=1$, $G$ is quasisimple and the result is immediate. Assume from here on that $n>1$.
Observe that $K_1$ is a normal
$p$-subgroup of $G$, that
$|K_1|=p^{3(n-1)}$ and $G/K_1\cong
   SL_2(\mathbb{Z}/p\mathbb{Z})$, a
quasisimple group since $p>3$.
Observe, next that, for $i=1,\dots,
   n$, |$K_i| = p^{3(n-i)}$ and,
moreover, $$\{1\}=K_n \lhd K_{n-1}
   \lhd \cdots \lhd K_1$$ is a chain of
normal subgroups. In fact this is an
upper central series for the group
$K_1$, i.e $K_{i-1}/K_i = Z(K_1/K_i)$
and $K_{i-1}/K_i$ is elementary
abelian of order $p^3$.
Now $G$ acts naturally on the group
$K_1$ by conjugation. The upper
central series structure just
described implies that this action
induces an action of $G/K_1 =
   SL_2(\mathbb{Z}/p\mathbb{Z})$ on the
groups $K_{i-1}/K_i$. Thus the group
$K_{i-1}/K_i$ becomes a 3-dimensional
module for the group
$SL_2(\mathbb{Z}/p\mathbb{Z})$. It is
easy to see that for $i=2,\dots, n$,
these modules are isomorphic. Fact to
check: This module is irreducible.
Now let $N$ be a normal subgroup of
$G$. Suppose first that $N\cap K_1$
is trivial. Then $N$ is isomorphic to
a normal subgroup of
$SL_2(\mathbb{Z}/p\mathbb{Z})$, i.e.
$N$ is trivial, equal to $Z$, or
isomorphic to
$SL_2(\mathbb{Z}/p\mathbb{Z})$. In
the latter case we would have
$G=K_1\times N$ and it is easy to
check that this does not happen. Thus
$N$ is trivial or equal to $Z$.
Assume next that $N\cap K_1$ is
non-trivial. In particular $N\cap K_1$ is
a non-trivial normal subgroup of
$K_1$. We use the following easy
fact: A non-trivial normal subgroup
of a $p$-group intersects the center
of that $p$-group non-trivially. Thus
$N\cap K_{n-1}$ is non-trivial and is
a normal subgroup of $G$. But, since
$SL_2(\mathbb{Z}/p\mathbb{Z})$ acts
irreducibly on the module $N\cap
   K_{n-1}$, $N$ must contain $K_{n-1}$.
If $N\cap K_1 = K_{n-1}$, then there
are three possibilities for $N$, namely $N=
   K_{n-1}$, $N=K_{n-1} \times Z$ or $N=K_{n-1}.
   SL_2(\mathbb{Z}/p\mathbb{Z})$. If $n=2$, the last possibility corresponds to $G$ and we are done. If $n>2$, then the
last possibility is impossible just
as before.
Now the proof is completed by
observing that $G/K_{n-1} \cong
   SL_2(\mathbb{Z}/p^{n-1}\mathbb{Z})$ and
appealing to induction.

Final remark: I've read some of Bourgain & Gamburd's work dealing with $SL_2$. They tend to (implicitly) consider the center as a trivial normal subgroup as their work deals with asymptotics on $p$ which are unaffected by $Z$. This explains the apparent inaccuracy of their assertion that the $K_i$ are all of the normal subgroups of $SL_2(\mathbb{Z}/p^n\mathbb{Z})$.<|endoftext|>
TITLE: Modular representation theory: central idempotents in $\mathbb{Z}_p[G]$
QUESTION [8 upvotes]: Let $G$ be a finite group and let $p$ be a prime dividing the order of $G$. Let $\chi$ be a $\mathbb{C}_p$-valued irreducible character of $G$ and let $e_{\chi} = |G|^{-1}\chi(1)\sum_{g \in G} \chi(g^{-1})g$ be the associated primitive central idempotent in $\mathbb{C}_p[G]$. Let $\mathbb{Q}_{p}(\chi)=\mathbb{Q}_p(\chi(g) : g \in G)$ be the character field. Let $H=\mathrm{Gal}(\mathbb {Q}_{p}(\chi)/\mathbb{Q}_p)$ and let $e=\sum_{h \in H} e_{\chi^h}$ ($H$ acts on characters in the usual way.) Then $e$ is a central primitive idempotent of $\mathbb{Q}_p[G]$. Let $v_p$ denote the usual $p$-adic valuation.
Claim: $v_p(|G|)=v_p(\chi(1))$ if and only if
$e \in \mathbb{Z}_p[G]$ and $e\mathbb{Z}_p[G]$ is a maximal $\mathbb{Z}_p$-order.
If $v_p(|G|)=v_p(\chi(1))$ then it is clear that $e \in \mathbb{Z}_p[G]$. That $e\mathbb{Z}_p[G]$ is a maximal $\mathbb{Z}_p$-order follows from Jacobinski's formula for the central conductor of $\mathbb{Z}_p[G]$ in a maximal order (see Curtis-Reiner, Methods of representation theory, vol 1 section 27).
For the converse, I can prove the claim for $p$ odd again using Jacobinski's formula and some calculations of the different of the extension $\mathbb{Q}_p(\chi)/\mathbb{Q}_p$.
Question: can anyone provide a proof of counterexample for the missing part for $p=2$?
Here is a related claim that would prove the claim and make everything much simpler if true:
If $e \in \mathbb{Z}_p[G]$ then $\mathbb{Q}_p(\chi)/\mathbb{Q}_p$ is unramified (i.e. $\mathbb{Q}_p(\chi) \subseteq \mathbb{Q}_p(\zeta_n)$ for some $n$ relatively prime to $p$).
Also, maybe I can drop the maximal order part of the claim altogether?
I have a reasonable knowledge of ordinary representation theory but have only really started to look at modular representation theory in the past few days. I know that this is related to "blocks of defect zero", but in the books I have looked at (Serre, Curtis-Reiner) it is assumed that the ground field is "sufficiently large", which doesn't really help me. But I suspect this is an easy problem for someone who knows the subject well.

REPLY [7 votes]: EDIT: There is a MUCH simpler proof than my first one, which I found after looking up the proof of (90.4) in Curtis-Reiner (Rep. Th. of finite groups...) mentioned by Florian Eisele in the comments:
The central idempotent $e\in \mathbb{Z}_p[G]$ is supported on $p$-regular elements. (By the way, a quite elementary proof of this fact can be given using the ideas in a paper of M. Leitz (Proc. Amer. Math. Soc. 128 (2000), no. 11, 3149–3152, MR1676316), see also Külshammers paper cited at the end.)
Since 
$$ e = \frac{\chi(1)}{|G|}\sum_{h\in H}\sum_{g\in G} \chi(g^{-1})^h g, $$
it follows that the character $\beta= \sum_{h\in H} \chi^h$ vanishes on $p$-singular elements. Let $P$ be a Sylow $p$-subgroup of $G$. Then $\beta$ vanishes on $P\setminus 1 $, so
the multiplicity of the trivial character of $P$ in the restriction $\beta_P$ is 
$$ (\beta_P, 1_P) = \frac{\beta(1)}{|P|}= \frac{|H|\chi(1)}{|P|}. $$
On the other hand, we have
$$ (\beta_P,1_P) = \sum_{h\in H} (\chi^h_P,1_P)= |H|(\chi_P,1_P). $$
Thus $|P|=|G|_p$ divides $\chi(1)= |P|(\chi_P,1_P)$, q.e.d.<|endoftext|>
TITLE: Proving univariate polynomials (defined by sums, binomial coeffs, etc.) are nonnegative: is it 'routine'?
QUESTION [18 upvotes]: My colleagues and I are working on a project related to an old paper of C. Borell and we have boiled it down to the following problem: 
Show, for all integers $1 \leq i \leq k$, that the univariate real polynomial $P(x) = \frac12 \binom{2k}{2i} (1+x)^{2k-2i} + \frac12 Q(x)$ is everywhere nonnegative, where $Q(x) = \sum_{j=i}^k \binom{2k}{2j}(1-x)^{2k-2j}\binom{j}{i}(-4x)^{j-i}$.  
(For what it's worth, Maple recognizes $Q(x)$ as $\binom{2k}{2i} (1-x)^{2k-2i} \text{hypergeom}([-(k-i), -(k-i)+\frac12], [i+\frac12], -4x/(1-x)^2)$.
I'm not asking MO to prove this (although I suppose if someone saw how to do so immediately I wouldn't turn down the answer).  Instead, I'm asking "Is this 'routine'?" in the sense of the word used in the Petkovsek-Wilf-Zeilberger A=B book?  In other words, would Doron Zeilberger say, "Oh yes, just type the following into Maple and it will produce a proof of the claim"?  In other other words, does this question fall into a class of problems known to be decidable?
Of course, for any fixed $i$ and $k$ the problem is 'routine'.  E.g., for $k = 4$, $i = 2$, we have $P(x) = 70x^4-168x^3+804x^2-168x+70$ and furthermore I can coax my computer into proving that's nonnegative.  (Say, by obtaining a sum-of-squares representation like $P(x) = 42(x-1)^4 + 28x^4 + 552x^2+28$.)
I don't know the  "A=B technology" very well: my question is whether it, or any other techniques, can be used to automatically prove these inequalities.  I would also be happy to accept an answer explaining why proving these inequalities is not obviously 'routine' and will require some ingenuity.  
UPDATE:  Doron Zeilberger wrote me an email describing why at least 'half' of this problem is routine: namely, that for $i$ symbolic and $k-i$ numeric the nonnegativity can be proven by computer.  He preferred not to post himself but said I could post his ideas here; I will do so once I get a chance to think about them.

REPLY [16 votes]: In good old times the word "routine" meant "easily done by a trained person". In this sense, the problem is, indeed, routine. Change the notation to $s=j-i$, $p=k-i$. Observe that negative $x$ are not a problem. For positive $x$, denote $y^2=\frac{4x}{(1-x)^2}$ and rewrite the inequality as 
$$
{2(p+i)\choose 2i}(1+y^2)^p+\sum_{s=0}^p (-1)^s{2(p+i) \choose 2(s+i)}{s+i\choose i}y^{2s}\ge 0.
$$
Now start with $i=0$. Then the ugly junk goes away and we get $(1+y^2)^p+\Re[(1+iy)^{2p}]\ge 0$, which is a no-brainer ($|z|^{2p}+\Re[z^{2p}]\ge 0$ for all complex $z$). Now just define $F_0(y)=\Re[(1+iy)^{2p}$, $F_m(y)=y\int_0^yF_{m-1}(t)dt$ and $G_m(y)=\frac{1}{(2m-1)!!}y^{2m}(1+y^2)^p$. The general inequality is equivalent to $G_m+F_m\ge 0$ (my $m$ is your $i$ but, since I used $i$ for the imaginary unit, to use it for the index now would be quite unfortunate). However $y\int_0^y G_{m-1}(t)dt\le y\int_0^y\frac{1}{(2m-3)!!}t^{2m-2}(1+y^2)^p dt=G_m(y)$ ($(-1)!!=1$, of course). 
But, as you said, today "routine" means something completely different and I just retreat in shame from the brave new world with my outdated language and ideas...<|endoftext|>
TITLE: Does this notion of pseudoprime relative to a matrix appear in the literature?
QUESTION [9 upvotes]: Let $M$ be a square matrix with integer entries. Then Fermat's little theorem for matrices holds:
$$\text{tr}(M^p) \equiv \text{tr}(M) \bmod p.$$
This follows by an examination of the action of the Frobenius map on the roots of the characteristic polynomial of $M$ over a splitting field; there is also a combinatorial proof. (Incidentally, I have no idea who to credit for this result.)
This suggests the following definition generalizing the notion of Fermat pseudoprime, the notion of Fibonacci pseudoprime (it feels to me like the Wikipedia article has the names backwards), and the notion of Perrin pseudoprime: a pseudoprime to base $M$ is a positive integer $n$ such that
$$\text{tr}(M^n) \equiv \text{tr}(M) \bmod n.$$
This definition appears to be related to Grantham's notion of Frobenius pseudoprime, although it might be weaker. For concreteness, let $M$ be the companion matrix of your favorite monic irreducible polynomial over $\mathbb{Z}$. 
Anyway, this looks like a reasonable primality test to me (choose random matrices, exponentiate them, etc.), but I don't really know anything about the subject. Does this notion appear anywhere in the literature? How does it compare to other primality tests?

REPLY [8 votes]: I think this is weaker than the notion of Frobenius pseudoprime. The trace of a matrix is the sum of the roots of its characteristic polynomial, so you're testing whether the sum of the roots remains invariant upon raising to the pth power. The Frobenius test is stronger than that.<|endoftext|>
TITLE: String diagrams for (weak) monoidal categories
QUESTION [11 upvotes]: In a strict monoidal category, where the associator, left and right unitor are identity morphisms we have the following relations between (string) diagrams:

(source)
where $i_x$ and $e_x$ are the unit and counit isomorphisms.
For the first relation for example, the LHS starts by $1\otimes x$ and goes to $x\otimes 1$, but the two are really the same object $x$ since we're in a strict monidal category, so we go from $x$ to $x$, which is exactly what happens in the RHS.
What I don't understand, is why such equalities are still valid in a weakened monoidal category, where now $x\otimes 1$ and $1\otimes x$ are merely isomorphic.

REPLY [17 votes]: The validity of string diagram equalities should be viewed as a form of coherence. What does an equality of two string diagrams tell us? Well, given such an equality, we can fix an arbitrary parenthesization and unitization of the input and the output. The associators and unitors are suppressed in a string diagram with the understanding that any two valid ways of adding them in give the same morphism. So in your example of the first zig-zag diagram, we can take both the input and output to be $x$, in which case the right-hand side is just the identity, and the left-hand side is given by the composite
$$x \stackrel{\lambda_x^{-1}}{\to} 1 \otimes x \stackrel{i_x \otimes x}{\to} (x \otimes x^{\vee}) \otimes x \stackrel{\alpha_{x, x^{\vee}, x}}{\to} x \otimes (x^{\vee} \otimes x) \stackrel{x \otimes e_x}{\to} x \otimes 1 \stackrel{\rho_x}{\to} x.$$
Here I've added associators and unitors where needed to get domains and codomains of morphisms to match up; I can do this with the confidence that had I chosen another way of adding associators and unitors, the composite morphism would be the same. This is what coherence tells us.
On the other hand, we could take the input to be $1 \otimes x$ and the output to be $x \otimes 1$. In this case, the right-hand side would represent the morphism
$$1 \otimes x \stackrel{\lambda_x}{\to} x \stackrel{\rho_x^{-1}}{\to} x \otimes 1,$$
while the left-hand side would be given by
$$1 \otimes x \stackrel{i_x \otimes x}{\to} (x \otimes x^{\vee}) \otimes x \stackrel{\alpha_{x, x^{\vee}, x}}{\to} x \otimes (x^{\vee} \otimes x) \stackrel{x \otimes e_x}{\to} x \otimes 1.$$
Of course, there are infinitely many possible choices of input and output, since I can just tensor 1 arbitrarily many times on the right and left and parenthesize this however I want. But once I fix a choice of input and output, each string diagram defines an unambiguous morphism.<|endoftext|>
TITLE: Ext of cyclic module
QUESTION [10 upvotes]: Hello,
I asked this question already on StackExchange with no answer, maybe it will be better suited here. It is well known that $\operatorname{Ext}^1_\mathbb{Z}(\mathbb{Z}/p\mathbb{Z},A) \simeq A/pA$. Is there a generalization of this formula outside PIDs? I mean $\operatorname{Ext}^1_R(R/I,M) \simeq M/IM$ for maximal or prime ideals? Does it hold for Dedekind domains? The classic proof I know for abelian groups seem to depend on principality of the ideal and I was unable to either think of different way make it work or to find an counterexample. Thanks for any input.

REPLY [8 votes]: Interesting question! Here is a generalization which includes the Dedekind domain case:

Proposition: Let $R$ be a Noetherian regular domain with $n=\dim R$  and $I\subset R$ an ideal such that $R/I$ is artinian and Gorenstein. Then for a finitely generated $R$-module $M$, we have $Ext^n(R/I,M)\cong M/IM$.

(when $R$ is Dedekind $R/I$ is locally a quotient of a DVR, so it is locally a hypersurface, thus Gorenstein).
Proof. It is enough to prove for $M=R$ because of the following 

Claim: if $N$ is an artinian $R$-module then the natural map $Ext^n_R(N,R)\otimes M \to Ext^n_R(N,M)$  is an isomorphism. 

Here's why: We have $pd_RN = n$. Let $P_{\bullet}: 0\to P_n \to \cdots \to P_0$ be a projective resolution of $N$. Then one compute the LHS by taking the cohomology at the end of $Hom(P_{\bullet},R)$ and tensor with $M$, and the RHS by taking cohomology at the end of $Hom(P_{\bullet},M)$. But tensor product is right-exact, proving that we get the same answer. 
Assuming we can prove for $M=R$, then apply the claim with $N=R/I$ completes the proof.
Now we prove it for $M=R$. Since $R/I$ is artinian, we only need to prove it after localizing at each maximal ideal containing $I$. Thus we can assume $(R,m)$ is local. But by Local Duality $Ext^n(R/I,R)$ is isomorphic to the  Matlis dual of $H^0_m(R/I) =R/I$. But since $R/I$ is Gorenstein this dual is actually isomorphic to  $R/I$. QED

REPLY [6 votes]: In general $Ext^1_R(R/I,M) \not\cong M/IM$. 
As an example take a finite abelian group $G$ and set $R := \mathbb{Z}G$ and let $I:= I_G = \ker(\mathbb{Z}G \to \mathbb{Z},\; g \mapsto 1)$ be the augmentation ideal. $I_G$ is a prime ideal since $\mathbb{Z}G/I_G \cong \mathbb{Z}$. Then, with trivial coefficients 
$$Ext_R^1(R/I,\mathbb{Z})=Ext_{ZG}^1(\mathbb{Z},\mathbb{Z})=H^1(G;\mathbb{Z})=Hom(G,\mathbb{Z})=0$$
while $\mathbb{Z}/I_G\mathbb{Z}=\mathbb{Z}$. The latter holds because $I_G$ is generated by $g-1$ $(g \in G)$ and $(g-1) \cdot 1 = 0$. 

In general, the following holds:
$$Ext^1_R(R/I,M)=\dfrac{Hom_R(I,M)}{i(M)}$$
where $i: M \to Hom_R(I,M),\; m \mapsto (x \mapsto xm)$. 
If $I$ is finitely generated by $a_1,...,a_n$ we can choose a presentation $R^n \to I \to 0$ that induces an embedding $0 \to Hom_R(I,M) \to Hom_R(R^n,M) \cong M^n$. Thus 
$$Ext_R^1(R/I,M) \le \dfrac{M^n}{\lbrace (a_1m,...,a_nm) \mid m \in M \rbrace}.$$

REPLY [5 votes]: Let $(R,\mathfrak m, k)$ be an arbitrary (say Noetherian) local ring of dimension at least $2$ and $M$ an arbitrary non-zero (say finitely generated) $R$-module of $\mathrm {depth}_R M\geq 2$. Then $\mathrm{Ext}^1_R(R/\mathfrak m,M)=0$, but $M/\mathfrak m M\neq 0$ by Nakayama.<|endoftext|>
TITLE: "Nice" sigma-algebra on set of measurable functions
QUESTION [17 upvotes]: In topology, given topological spaces $X$ and $Y$, the compact-open topology is considered, under the relatively mild requirement that $X$ be locally compact Hausdorff, to be the most "natural" topology on the set $\mathcal{C}(X,Y)$ of continuous functions $X\to Y$. (I'm not going to even attempt to define "natural" here---take it to mean whatever seems most appropriate).
Is there something analogous, in the sense of being somehow "natural", for the set of measurable functions between two measure spaces?

REPLY [16 votes]: There is an impossibility theorem: If you let $\mathcal{L}$ be the the space of Borel-measurable functions $f:[0,1]\to[0,1]$, and $e:\mathcal{L}\times [0,1]\to[0,1]$ the evaluation given by $e(f,x)\mapsto f(x)$, then there is no $\sigma$-algebra on $\mathcal{L}$ such that the evaluation is jointly measurable. The result is a consequence of the rather complicated classification result in  R. Aumann, Borel Structures for Function Spaces, Illinois Journal of Mathematics 5 (1961), pp. 614-630. Easier proofs of the main results can be found in the paper "Borel Structures for Function Spaces" (yes, same title) by B.V. Rao, Colloquium Mathematicum, 1971.
A $\sigma$-algebra on measurable functions I have actually seen used is the following: If $(S,\mathcal{S})$ and $(T,\mathcal{T})$ are measurable spaces, we endow the family of measurable functions between them with the $\sigma$-algebra generated by sets of the form $\{f:f(s)\in B\}$ with $s\in S$ and $B\in\mathcal{T}$. The author used this $\sigma$-algebra to show that to each Markov kernel from $S$ to $T$, there corresponds a certain probability measure on this $\sigma$-algebra. The paper is H. v. Weizsäcker Zur Gleichwertigkeit zweier Arten der Randomisierung, Manuscripta Mathematica 11 (1974).<|endoftext|>
TITLE: Do the following set of Dehn twists generate the mapping class group?
QUESTION [7 upvotes]: If $S$ is the surface illustrated below, do the Dehn twists about the red curves generate the mapping class group $\operatorname{MCG}(S,\partial S)$?

http://math.stanford.edu/~pardon/img/dehntwists.png

REPLY [10 votes]: Humphries arguments are in 
Humphries, Stephen P.
Generators for the mapping class group. Topology of low-dimensional manifolds (Proc. Second Sussex Conf., Chelwood Gate, 1977), pp. 44–47,
Lecture Notes in Math., 722, Springer, Berlin, 1979. 
The key argument for the $\ne 2g$ case is that the symplectic group $Sp(2g, \mathbf F_2)$  is not generated by $2g$ transvections (which maybe was well known). Steve and I joined up to present this argument and more in our paper 
``Orbits under symplectic
transvections II:  the case  $K=\bf F_2$'', Proc. London
Math. Soc. (3) 52 (1986) 532-556.
and Steve followed this up with much more work in this area.<|endoftext|>
TITLE: Homology-Cohomology Pairing
QUESTION [8 upvotes]: The Steenrod algebra $\mathcal{A}^*$ is the algebra of cohomology operations for the spectrum $\mathrm{H}\mathbb{F}_2$, i.e. $\mathcal{A}^*\simeq (\mathrm{H}\mathbb{F}_2)^*(\mathrm{H}\mathbb{F}_2)$.
The dual Steenrod algebra $\mathcal{A}^\vee _*$ can be interpreted as the coalgebra of homology cooperations, i.e. $\mathcal{A}^\vee _\ast\simeq (\mathrm{H}\mathbb{F}_2) _*(\mathrm{H}\mathbb{F}_2)$ and it coacts on mod 2 homology groups of cell complexes. This is a consequence of the perfect pairing between mod 2 homology and cohomology. Unfortunately, this does happens because $\mathbb{F}_2$ is a field, which means that we may in trouble if change our coefficients to something else. The situation may get even worse with generalized cohomology theories.
Now let us say we are given a generalized cohomology theory $E$. The ring $E^\ast E$ acts on cohomology groups of any space $X$: we have map $E^*E\otimes_{E^\ast} E^\ast(X)\rightarrow E^\ast(X)$, which is natural in $X$. The coaction may be a little tricky: it is not true that $E_\ast E$ coacts on $E_*(X)$. However, upon requiring that $E_\ast E$ is a flat $E_*$-module, we get a coaction as well. So let us say that we are in this situation. In addition to this we have a pairing $E^*E\otimes_{E_*}E_\ast E\rightarrow E_\ast$.
My questions is: do these pairings work well together or there are some tricky things that has to account for? In particular, $E_\ast E$ has distinct left and right $E_*$-module structures, and I fear this may come into play.

REPLY [5 votes]: Propositions 17.10 and 17.11 of Switzer's book seem to be examples of things working nicely.
For those who don't have the book to hand: let $E$ be a commutative ring spectrum such that $E_\ast(E)$ is flat as a right module over $E_\ast=E_\ast(*)$, and let $X$ be any spectrum.
Let $\psi_X\colon E_\ast(X)\to E_\ast(E)\otimes_{E_\ast} E_\ast(X)$ be the coaction map, and let $c\colon E_\ast(E)\to E_\ast(E)$ be the conjugation map (which switches the left and right counits). 
Then for any elements $a\in E^\ast(E)$, $y \in E^\ast(X)$ and $u\in E_\ast (X)$ with $\psi_X(u) = \sum_i e_i\otimes u_i$, we have
$$ 
a\cdot u = \sum_i \langle a,ce_i\rangle u_i
$$
and
$$
\langle ay,u\rangle = \sum_i (-1)^{|y||e_i|}\langle a,e_i\langle y,u_i\rangle \rangle,
$$
where  $\langle \cdot , \cdot \rangle\colon E^\ast(X)\otimes_{E_\ast} E_\ast(X)\to E_\ast$ is the Kronecker pairing.<|endoftext|>
TITLE: Equivariant handle decompositions
QUESTION [6 upvotes]: Suppose I have some smooth closed high-dimensional manifold $M$ acted on smoothly by a finite group $G$.  By a metric averaging procedure, we can equip $M$ with a smooth Riemannian metric so that $G$ acts by isometries.  I can't necessarily pick a $G$-invariant morse function $f:M\to\mathbb R$, but nevertheless, I can certainly pick a smooth function $f:M\to\mathbb R$ which, though perhaps not Morse, still has only isolated "nice" critical points in some precise sense.  We therefore conclude:

There is a "handle" decomposition of $M$ (where I haven't said what I mean by "handle") which is preserved by $G$.  Thus $G$ just permutes (and/or acts on individually) the handles.

I am interested in knowing to what extent this can be generalized to the case of an "action up to homotopy".  More specifically, suppose we have $G\to\operatorname{Homeo}(M)/\text{homotopy}$.  To what extent can we "decompose" M into simple pieces in a $G$-invariant way?  If it helps, then it is OK to assume that the action of $G$ is "close to the identity" in a vague coarse sense.
(I am essentially just interested on the case of high-dimensional $M$, but of course the question makes sense in any dimension).

REPLY [3 votes]: This foundational paper: Arthur G. Wasserman, Equivariant Differential Topology, Topology, 8(1967), 127-150, has section 4 dealing with equivariant Morse theory for manifolds with a smooth action of a compact Lie group, including equivariant handle attaching, equivariant Morse Lemma, etc. See MathSciNet Review. The last result of the paper concludes that a compact G-manifold is a union of "handle bundles over orbits".<|endoftext|>
TITLE: Efficient computation of integer representation as a sum of three squares
QUESTION [17 upvotes]: Recently I've been studying the problem of integer representation as sum of three squares. Most of the articles that I've found study the function $r_m(n)$ which counts the number of representations of $n$ as the sum of $m$ squares. However, this is not what I am interested in. What I'm looking for is an efficient way (for some given $n$) to find $x$, $y$ and $z$ such that $n = x^2 + y^2 + z^2$. I need to find at least one such representation. Can you recommend me some articles that study this problem?
P.S. I believe that Emil Grosswald's book "Representation of integers as Sums of Squares" contains the answer. However, I could not find this book on my university's web-site.

REPLY [24 votes]: This problem is discussed in my paper with Rabin, Randomized algorithms in number theory,
Commun. Pure Appl. Math. 39, 1985, S239 - S256.  We give an algorithm that, assuming a couple of reasonable conjectures, will produce a representation as a sum of three squares in polynomial time.<|endoftext|>
TITLE: Motivation and unsolved problems of TQFT
QUESTION [13 upvotes]: I have been studying topological quantum field theory by mainly reading the Turaev's book.
I'd like to know if there are unsolved problems that motivate mathematicians to study TQFT, like Riemann's hypothesis for number theory.
I also would like to know if there is a paper or book that list big or small unsolved problems of TQFT. If not, could you suggest some problems here? I have been learning TQFT but I don't know what to do by myself as a graduate student.
Thank you.

REPLY [10 votes]: There are various open classification problems: classify modular tensor categories (the input for Reshetikhin-Turaev type theories), classify semisimple pivotal 2-categories (the input for Turaev-Viro type theories).  There is a vague conjecture, popular among physicists, that all examples of modular tensor categories are obtainable in some way from the standard $Rep(U_q(\mathfrak g))$ examples.<|endoftext|>
TITLE: Which concept of dimension of a ring of functions on a manifold, gives the dimension of the manifold?
QUESTION [9 upvotes]: Let $R$ be a ring of (smooth?) functions on a (connected?) manifold of dimension $n$. What concept of dimension (of the ring $R$) gives the dimension of the manifold? To what class of rings does this concept apply?

REPLY [3 votes]: There is an approach to this question through a smooth version of the Hochschild-Kostant-Rosenberg (HKR) Theorem, which is related to Robert Bryant's answer.  The smooth version is due to Alain Connes in its original form.
First, the original HKR theorem says that for a regular affine $k$-algebra $R$ (think of the algebra of polynomial functions on a smooth affine variety), there are isomorphisms
$$
\Lambda^\bullet (\Omega_{R/k}) \cong \mathrm{Tor}_\bullet^{R^e}(R,R) \overset{\mathrm{def}}{=} HH_\bullet(R)
$$
and
$$
\Lambda^\bullet (\mathrm{Der}(R)) \cong \mathrm{Ext}_{R^e}^\bullet (R,R) \overset{\mathrm{def}}{=} HH^\bullet(R).
$$
Here $\Omega_{R/k}$ is the $R$-module of Kahler differentials and $\mathrm{Der}(R)$ is the $R$-module of derivations of $R$, which are algebraic analogues of 1-forms and vector fields on $R$, respectively.  As Donu Arapura points out in his comment, Kahler differentials are the predual to derivations, rather than the other way around, as one normally defines forms to be dual to vector fields in the differential-geometric setting.
Also $R^e = R \otimes R$, and $HH_\bullet(R)$ and $HH^\bullet(R)$ are the Hochschild homology and cohomology of $R$, respectively.
The upshot is that when $R$ is the coordinate ring of a smooth affine variety, we can find the dimension of that variety by looking at the highest degree in which the Hochschild homology or cohomology does not vanish.
Moving to the smooth case one needs to be quite careful, however.  For a smooth manifold $M$, the Kahler differentials of $C^\infty(M)$ are not the same as the module of smooth 1-forms, as can be seen in this question.
The extra structure that $C^\infty(M)$ possesses is that of a Frechet algebra, where the seminorms are given by sup-norms of partial derivatives on compact subsets of $M$.
Hochschild homology and cohomology can be adapted to the setting of certain topological algebras, and then there is an analogue of the HKR theorem.  One formulation of it says that for $A = C^\infty(M)$, there are isomorphisms
$$
_cHH_\bullet(A,A) \cong \Omega^\bullet(M)
$$
and 
$$
_cHH^\bullet(A,A) \cong \Gamma^\infty(\Lambda^\bullet(TM)),
$$
where $\Omega^\bullet(M)$ is the algebra of differential forms, and $\Gamma^\infty(\Lambda^\bullet(TM))$ is the space of smooth polyvector fields on $M$, and the subscript $c$ indicates the continuous Hochschild homology and cohomology.
So again, the dimension of $M$ can be determined as the top degree in which the continuous Hochschild homology and cohomology do not vanish.
This was proved by Connes (see Chapter 3, Section 2 of his book Noncommutative Geometry, plus references therein) for compact manifolds.  For noncompact manifolds it is written up concisely in the paper On Continuous Hochschild Homology and Cohomology Groups, by Markus Pflaum.  Connes' version uses the language of de Rham currents, which are dual to differential forms.
Another source to look at is Chapter 8 of the book Elements of Noncommutative Geometry by Gracia-Bondia, Varilly, and Figueroa.<|endoftext|>
TITLE: Equivalent definitions of invertible modules
QUESTION [26 upvotes]: Let $R$ be commutative unital ring, and $M$ an $R$-module. $M$ is called invertible (a.k.a. projective module of rank one), if it is finitely generated, and $M_{\mathfrak{p}} \cong R_{\mathfrak{p}}$ for every $\mathfrak{p} \in \operatorname{Spec} R$. Equivalently, there are $a_1, \dotsc, a_n \in R$ such that $(a_1, \dotsc, a_n) = R$ and $M_{a_i} \cong R_{a_i}$ for every $i$. There is a third equivalent definition: $M$ is finitely generated, and there is an $R$-module $N$ such that $M \otimes_R N \cong R$.
See e.g. Bourbaki: Commutative Algebra, II, 5.4, Theorem 3, or Proposition 19.8 in Pete Clark's note on Commutative Algebra. Both of these references assume that $M$ is finitely generated in the third definition.
My question: is the finitely generatedness really necessary in the last definition? It seems to me that if $M \otimes_R N \cong R$, then $M$ and $N$ are automatically finitely generated. This would make the last definition really simple.
Remark: I have a proof in mind, so as a second question: is the following argument correct?
If $\varphi \colon M \otimes N \to R$ is an isomorphism, then $\varphi^{-1}(1) = \sum_{i=1}^s x_i \otimes y_i$ for some $x_i \in M$ and $y_i \in N$. Then let $M'$ be the submodule of $M$ generated by $x_1, \dotsc, x_s$. The composition $M' \otimes N \xrightarrow{\sigma} M \otimes N \xrightarrow{\varphi} R$ is surjective (because $\sum_{i=1}^r x_i \otimes y_i \in M' \otimes N$ goes to $1 \in R$), and $\varphi$ is an isomorphism, so $\sigma$ is also surjective. Let $M'' = M/M'$, so $0 \to M' \to M \to M'' \to 0$ is exact. Then $M' \otimes N \xrightarrow{\sigma} M \otimes N \to M'' \otimes N \to 0$ is also exact. However $\sigma$ is surjective, thus $M'' \otimes N = 0$. But then
$$
0 = (M'' \otimes N) \otimes M \cong M'' \otimes (N \otimes M) \cong M'' \otimes R \cong M'',
$$
therefore $M = M'$. So $M$ is indeed finitely generated.

REPLY [9 votes]: Here are two categorical comments.
The first is that $M$ is an invertible module in the sense of having an inverse with respect to the tensor product (and why would you pick any other definition? "Invertible" is right there in the name!) iff the functor 
$$\text{Mod}(R) \ni N \mapsto N \otimes_R M \in \text{Mod}(R)$$ 
is an autoequivalence. This functor sends the unit module $R$ to $M$, from which it follows that invertible modules inherit any categorical properties that the unit possesses. Finite presentation turns out to be such a categorical property (namely, it is the categorical property of compactness), as is projectivity, and since both properties hold of $R$ they necessarily hold of any invertible module. Finite presentation and projectivity together are equivalent to another categorical condition called tininess. 
This is a completely general observation valid in any monoidal category: invertible objects always inherit any categorical properties the unit possesses.
The second is that we can do just as well with a natural weakening of invertibility, namely dualizability. It's straightforward to show, with a proof very similar to the proof in the OP (that doesn't pass through compactness or tininess), that the dualizable modules are precisely the finitely presented projective modules; see here for details.<|endoftext|>
TITLE: Rational solutions to x^3 + y^3 + z^3 - 3xyz = 1
QUESTION [5 upvotes]: I can show that there infinitely many solutions to this equation. Is it possible that the set
of rational solutions is dense?

REPLY [6 votes]: The answer is yes, the rational points on your surface lie dense in the real topology.
Let's consider the projective surface $S$ over $\mathbb{Q}$ given by $X^3+Y^3+Z^3-3XYZ-W^3=0$. It contains your surface as an open subset, so to answer your question we might as well show that $S(\mathbb{Q})$ is dense in $S(\mathbb{R})$. Observe that $S$ has a singular rational point $P = (1:1:1:0)$. Since $P$ is singular, the intersection of $S$ with a plane $V$ that contains $P$ is a singular cubic $C_V$, with the rational point $P$ on it. It is a well-known and easy fact that on such curves, the rational points are dense (except if $C_V$ consists of three lines conjugate over $\mathbb{Q}$, but there are only finitely many $V$ for which this happens). 
Well, in order to approximate any real point $R \in S(\mathbb{R})$ to within distance $\epsilon$, we just pick a plane $V_0$ defined over $\mathbb{Q}$, which we may choose such as to be within distance $\epsilon$ to $R$. But then also the rational points on $C_{V_0}$, which lie dense in its real locus, lie within $\epsilon$ of $R$.<|endoftext|>
TITLE: An infinite set of identities using Stirling numbers 1st kind - are they all zero?
QUESTION [8 upvotes]: I have the following set of series involving the Stirling numbers 1'st kind and binomials, which can be understood as a set of dot-products of row- and column-vectors of two infinite matrices (where R and C indicate rows and columns, beginning at zero):
$$ w_{R,C} =\sum_{k=\max(R,C)}^\infty (-1)^k {s_1(1+k,1+k-R)\over k!} \cdot (-1)^C (1+C)^k \cdot \binom {1+k}{1+C} $$
I've tested this heuristically for several R and C and always approximated zero; also wolfram-alpha can evaluate this explicitely to zero if feeded with      

sum (-1)^k * StirlingS1(k+1,1+k-R)/k! * (1+C)^k * binomial(1+k,1+C), for k=max(C,R) to infty      

where we replace $C$, $R$ and $\max(C,R)$ with actual values.     
However, I've no option to let wolfram-alpha answer this in general.      
I've proved this for $C=0,1,2$ and the first few $R$ using exponential generating functions, but again, a general proof is out of reach for me (possibly I'm overlooking something trivial like telescoping...), so I ask for help here.     

The convention for Stirling numbers first kind as in Math'ica, indexes beginning at zero:       
$ \small \qquad \qquad \begin{array} {rrrrr}
 1 & . & . & . & . & . \\\
 0 & 1 & . & . & . & . \\\
 0 & -1 & 1 & . & . & . \\\
 0 & 2 & -3 & 1 & . & . \\\
 0 & -6 & 11 & -6 & 1 & . \\\
 0 & 24 & -50 & 35 & -10 & 1
 \end{array} $

If some background is of interest: here are the questions on MSE
https://math.stackexchange.com/questions/16228    // question of some user which motivated me to look at an example
https://math.stackexchange.com/questions/89853    // my follow-up question dealing with the current problem
and a more worked out treatize on this in a pdf-file 
 http://go.helms-net.de/math/divers/InverseNullmatrix.pdf 

[update] Hmm, after 1 1/2 years I've looked at the question again and still do not have an idea how to construct a proof for the whole set of identities. To possibly stimulate helpful answers here I'll insert pictures of the matrices - perhaps it helps to get an immediate idea when the patterns are more visible/obvious than in the bare formula above. 
This is (the top-left-segment of) the matrix $M$ in question.           
 
This are the L and D factors of the L D U-decomposition. Because it seems convenient to recognize familiar numbers I've documented the product LD = L D 
 
This is the U factor:                    
 

This is the reciprocal of U (call it UI):             
 
This is the reciprocal of LD (call it LDI):             
 
and in the limit for infinite size of the UI and LDI, the product UI * LDI = MI = 0  by hypothese. 
 
Here are the matrices UI and LDI in a near-symbolic display, the coefficients $s1[r,c]$ are the Stirling numbers first kind.
 

 
Reformulating the dotproducts using their exponential generating functions it is not difficult to prove the identities for a couple of examples.
But what is missing is the proof for the full set of dotproducts.
[/update]

REPLY [7 votes]: Cancelling a few terms, you see that we want to show that
$$
\sum_{k=C}^\infty (-1-C)^k {s_1(1+k,1+k-R)(1+k)\over (k-C)!}=0.
$$
Let us further simplify this equation via the coordinate transformations $C\leftarrow -C-1$ and $k\leftarrow k+1$. We then want to prove that
$$
I_{R,C}:=\sum_{k=-C}^\infty \frac{C^k}{(k+C)!}\cdot ks_1(k,k-R)=0
$$
for all $C\leq-1$ and $R\geq0$.
Recall the recurrence relation
$$ks_1(k,k-R)=s_1(k,k-R-1)-s_1(k+1,k-R).$$
Using summation by parts, we obtain
$$
\begin{aligned}
I_{R,C}
&=\sum_{k=-C}^\infty \frac{C^k}{(k+C)!}\cdot (s_1(k,k-R-1)-s_1(k+1,k-R))\\
&=B+\sum_{k=-C}^\infty \left(\frac{C^{k+1}}{(k+1+C)!}-\frac{C^k}{(k+C)!}\right)\cdot s_1(k+1,k-R)\\
&=B+\sum_{k=-C}^\infty \frac{C^{k+1}-C^k(k+1+C)}{(k+1+C)!}\cdot s_1(k+1,k-R)\\
&=B-\sum_{k=-C}^\infty \frac{C^k}{(k+1+C)!}\cdot (k+1)s_1(k+1,k-R)\\
&=B-\sum_{k=-C+1}^\infty \frac{C^{k-1}}{(k+C)!}\cdot ks_1(k,k-R-1)\\
&=-\frac{I_{R+1,C}}C
\end{aligned}
$$
with boundary term
$$
\begin{aligned}
B&=\lim_{k\rightarrow\infty}\frac{C^k}{(k+C)!}s_1(k,k-R-1)+\frac{C^{-C}}{(-C+C)!}s_1(-C,-C-R-1)\\
&=\frac{C^{-C-1}}{(-C+C)!}\cdot(-C)s_1(-C,-C-R-1).
\end{aligned}
$$
We can therefore use induction and only have to show $I_{R,C}=0$ for $R=0$. But as Gottfried Helms pointed out, this is easy.<|endoftext|>
TITLE: Why are cup-i products and Steenrod Squares often (always?) unary?
QUESTION [5 upvotes]: One way to define the Steenrod Operations is to use the cup-i product, as in Mosher and Tangora's book.  It basically says, given the chain complex from mod-2 homology $C_\ast$, define
$D_0 : C_\ast\to C_\ast\otimes C_\ast$, 
so that the cup product is given (on cocycles)
$(u\cup v)(\sigma) = (u\otimes v)(D_0\sigma)$
and then for higher $i$, define $D_i$ so that 
$D_{i-1}+\rho D_{i-1} = D_i\partial + \partial D_i$
where $\rho$ is the flipping map.  Then the $\cup_i$ product is just
$(u\cup_i u)(\sigma) = (u\otimes u)(D_i\sigma)$
And then define for $[u]\in H^n$
$Sq^{2n-i}([u]) = [u\cup_{i}u]$
This definition seems perfectly well-defined as a binary operation, and yet wherever I've seen it done it has only even been used as a unary operation.  
Is there a reason why this is the case, why either the product is undefined as a binary product or not useful as a binary product or just too hard to use?
Is this a dumb question?
Thanks,
-Joseph

REPLY [8 votes]: Under suitable hypotheses, Gugenheim and I use the binary $\cup_1$ and especially the fact that it is a graded derivation (Hirsch formula) as the key to giving a calculation of $H^*(G/H)$ as the torsion product over $H^*(BG)$ of $R$ and $H^*(BH)$, where coefficients are taken in a suitable commutative ring $R$.  This even works for suitable $H$-spaces.   See
http://www.math.uchicago.edu/~may/BOOKS/GugMay.pdf and 
http://www.math.uchicago.edu/~may/PAPERS/MNApril20.pdf<|endoftext|>
TITLE: Dimensions of unitary representations of group extensions
QUESTION [8 upvotes]: Is the property of having a bound on the dimensions of irreducible representations preserved by an extension? 
For example $G_1=\mathbb{Z}$ and $G_2=\mathbb{Z}/2\mathbb{Z}$ are discrete abelian groups each of which only have 1 dimensional irreducible representations, while the group
$$\mathbb{Z}\rtimes\mathbb{Z}/2\mathbb{Z}$$
has only irreducible representations of dimensions one and two.
One of the purposes of this question for me is to get a feel for how difficult problems of this kind are as I really do not know very much about unitary representations of groups.
The case when the groups considered are finitely generated discrete (elementary) amenable would be very helpful to me.

REPLY [13 votes]: A discrete group has the property you describe if and only if it is virtually abelian. I must admit to never learning the proof, but some foraging on MathSciNet indicates the result is due to Isaacs and Passman:

Isaacs, I. M.; Passman, D. S.
  Groups with representations of bounded degree.
  Canad. J. Math. 16 1964 299--309 |
  MathReview

(I seem to remember hearing that there are proofs which get better estimates on the degree of the abelian subgroup, but I may be misremembering.)
The class of virtually abelian groups has good hereditary properties. However, it is not stable under semi-direct products: indeed there are metabelian groups, arising as crossed products, which are not virtually abelian, e.g. integer Heisenberg, or the ax+b group over the rationals. Indeed, in the former case once can write down  infinite-dimensional unitary irreps. So I think you need to restrict the kinds of extension you are considering if you want this property to be preserved.
If you are interested in the problem beyond the discrete setting, then I think one place to start would be with the paper of C. C. Moore

Calvin C. Moore, 
  Groups with finite dimensional irreducible representations.
Trans. Amer. Math. Soc. 166 (1972), 401--410. |
  MathReview

(A locally compact group $G$ is called a Moore group if all its (continuous) unitary irreps are finite-dimensional. It is a hard result of Thoma that if a discrete group is Moore then there is a uniform bound on the degrees of the irreps, which is the property described in your question. Of course non-discrete compact groups show that the two concepts are in general distinct.)

REPLY [10 votes]: This property is not preserved under extension. The discrete Heisenberg group $H_3(\mathbb{Z})$, which consists of integer matrices of the form
$$\left[ \begin{array}{ccc} 1 & a & b \\\
 0 & 1 & c \\\
 0 & 0 & 1 \end{array} \right],$$
is a central extension of $\mathbb{Z} \times \mathbb{Z}$ by $\mathbb{Z}$. It has finite quotients $H_3(\mathbb{F}_p)$ with irreducible representations of dimension $p$ (this follows from the fact that $H_3(\mathbb{F}_p)$ is non-abelian and has order $p^3$).<|endoftext|>
TITLE: When is an HNN-extension finitely presented?
QUESTION [11 upvotes]: Let $G=\langle H, t; K^t=K^{\prime}\rangle$ be an HNN-extension of $H$, with $t$ inducing the isomorphism $\phi: K\rightarrow K^{\prime}$. I was wondering if the following question can be answered, and if so what is the answer,

When is $G$ finitely presented?

It seems "obvious" that the answer should be "when $H$ is finitely presentable and $K$ is finitely generated". However, Grigorchuk's group is not finitely presentable but does have a finitely-presentable HNN-extension (Lysenok’s extension).
EDIT: Now, I am not expecting the question to be answerable. What I would really like is an incomplete answer, along the lines of,

If you put these restrictions on $H$ and $K$ (and perhaps on the isomorphism $\phi$) then you can say something.

REPLY [8 votes]: Here are my  comments combined into an answer.
For ascending HNN extensions, i.e. $H=K$, $\phi\colon H\to K'$ an injective endomorphism (as in Baumslag-Remeslennikov case (see above), in the Grigorchuk case, and many others) one needs, as Ben Steinberg ponted out that $H$ has a finite L-presentation (named after Igor Lysenok, who proved that the Grigorchuk group $G_1$ has such a presentation) with respect to the endomorphism $\phi:H\to K′$. That is there are finite number of relations $r_1=1,...,r_k=1$ so that the set of relations $\{\phi^m(r_j)=1\mid m\ge 0,1\le j\le k\}$ defines $H$. Here we consider $\phi$ as a substitution $x\mapsto u_x$ where $u_x$ is any word representing $\phi(x)$ in $H$, $x$ a generator of $H$. Indeed, in this case $G$ (generated by the finite generating set $X$ of $H$ and the free letter $t$) has finite presentation consisting of relations $r_1,...,r_k$ and the HNN relations $x^t=\phi(x), x\in X$. I think that the converse statement should also be true: if the HNN extension is finitely presented then $H$ has a finite L-presentation with respect to $\phi$. 
More examples can be found in Sapir, Mark, Wise, Daniel T., Ascending HNN extensions of residually finite groups can be non-Hopfian and can have very few finite quotients. J. Pure Appl. Algebra 166 (2002), no. 1-2, 191–202 and in Olʹshanskii, Alexander Yu.; Sapir, Mark V. Non-amenable finitely presented torsion-by-cyclic groups. Publ. Math. Inst. Hautes Études Sci. No. 96 (2002), 43–169. In both cases it was crucial that the "extended" group $H$ has an L-presentation, in fact it was constructed as such. 
For arbitrary HNN extensions the situation is more difficult but not hopeless, I think that necessary and sufficient conditions can be found in many more cases.<|endoftext|>
TITLE: Classifying spaces of topological groups that are not well-pointed
QUESTION [7 upvotes]: Let $G$ be a topological group.
The geometric bar construction $BG = B_{\bullet}(pt, G, pt)$ together with $EG = B_{\bullet}(pt,G,G)$ and the map $EG \to BG$ yields the universal principal $G$-bundle at least, when the identity $e \in G$ is a closed cofibration (a condition that is often called well-pointed). 

It is claimed in the book by Rudyak "On Thom spectra, Orientability and Cobordism" in theorem 1.65 (iii) that this still holds true, if $G$ is not well-pointed. Is this an error in the book or did I miss something? 

I only had the chance to look up one of the references that Rudyak gives. In "Classifying spaces and fibrations" by May, it is stated in theorem 8.2 that $EG \to BG$ is a principal $G$-bundle in case $e \in G$ is non-degenerate basepoint.
I guess my question is:

How bad is the fibration $EG \to BG$ in the case that $G$ is not well-pointed. Does $BG$ still have the right homotopy type / weak homotopy type?

REPLY [3 votes]: I don't have a real answer but possibly relevant observations and speculations. I imagine $BG$ can be pretty horrible if $G$ is not well-pointed, unless of course one uses the fat realization, when I imagine the fat construction of $EG$ does not give a bundle.  One remark is that $G$ is homotopy equivalent to a well-pointed monoid (not group) $G'$, as noted in Remarks 9.3 of Classifying spaces and fibrations.  One might try to compare the categories of principal $G$-fibrations and principal $G'$-fibrations.  In another direction, for a space to be well-pointed means that it is cofibrant in the Hurewicz (or Strom) model structure on based topological spaces.  As will probably be apparent, I haven't really thought about this, but one might ask for a Hurewicz model structure on topological groups for which cofibrant approximation gives a homotopy equivalent well-pointed group.<|endoftext|>
TITLE: Tail bound for Poisson random variable
QUESTION [11 upvotes]: Is the following fact about Poisson random variables true?

For any $\lambda \in (0,1)$ and integer $k > 0$, if $X$ is a Poisson random variable with mean $k \lambda$, then $\Pr(X < k) \geq e^{-\lambda}$.

It clearly holds for $k=1$, and for any fixed $\lambda$ it's easy to see that it holds for all sufficiently large $k$, but for intermediate values of $k$ it's not obvious to me.

REPLY [12 votes]: A classical inequality of Teicher (1955) asserts

Proposition (Teicher). Let $X \sim \mathrm{Pois}(\lambda)$. Then, $\mathbb P(X \leq [\lambda]) > e^{-1}$.

A modification of his argument will allow us to prove the following.
Proposition. For $\lambda \in (0,1)$, let $X_{n\lambda} \sim \mathrm{Pois}(n\lambda)$. Then the sequence $b_n := b_{n,n\lambda} = \mathbb P(X_{n\lambda} < n)$ is monotonically increasing. In particular, $b_n \geq e^{-\lambda}$ for all $n$.
Proof. First, note that, by considering a Poisson process with rate 1, we have, 
$$
b_{n+1,\mu} = \sum_{x=0}^n \frac{e^{-\mu} \mu^x}{x!} = \int_\mu^\infty \frac{x^n e^{-x}}{n!} \,\mathrm{d}x \,,
$$
for all $n$ and $\mu$. Now,
$$
b_{n+1,n\lambda} - b_{n+1,(n+1)\lambda} = \int_{n\lambda}^{(n+1)\lambda} \frac{x^n e^{-x}}{n!} \,\mathrm{d}x = \int_0^1 (\lambda(y+n))^n \frac{e^{-\lambda(y+n)}}{n!} \lambda \,\mathrm{d}y \,,
$$
where the last equality follows from the substitution $y = (x-n\lambda)/\lambda$.
We can rewrite the last integral as
$$
\frac{e^{-\lambda n}(\lambda n)^n}{n!} \lambda \int_0^1 (1+y/n)^n e^{-\lambda y} \,\mathrm{dy} <  \frac{e^{-\lambda n}(\lambda n)^n}{n!} = b_{n+1,n\lambda} - b_{n,n\lambda} \,
$$
where the inequality follows from facts that $(1+y/n)^n < e^y$ and (upon integrating) $e^{1-\lambda} < \lambda^{-1}$, true for any $\lambda \in (0,1)$.
But, then $b_{n,n\lambda} < b_{n+1,(n+1)\lambda}$ which is what was to be shown. Since $b_{1,\lambda} = e^{-\lambda}$, the second part of the proposition statement holds.<|endoftext|>
TITLE: cohomology and $j_!$
QUESTION [6 upvotes]: I have a projective variety $X$ and an open immersion $j : U \to X$.
Say I have a sheaf, locally free in my case of interest, $\mathcal{S}$ on $U$. Is there any reasonable relationship between $H^i(X,j_! \mathcal{S})$ and $H^i(U,\mathcal{S})$? What if I add that I know that $H^i(U,\mathcal{S}) = 0$ for $i>0$. I'm hopeful the latter can imply that $H^i(X,j_! \mathcal{S}) = 0 $ for $i>0$.

REPLY [7 votes]: Consider $X = \mathbb{P}^{1}$, $U =\mathbb{A}^{1}$ and j the inclusion of a affine chart in the projective line. The complement of the affine chart is a point P, let i be the inclusion of this point in the projective line.
Consider the sheaf $\mathcal{F} = \mathcal{O}(-2)$ on the projective line. One has $dim H^{1}( \mathbb{P}^{1}, \mathcal{F} ) =1$.
One has an exact sequence :
$0 \to j_{!} ( \mathcal{F}_{U} ) \to \mathcal{F} \to i_{*} ( \mathcal{F}_{P} ) \to 0$
$i_{*} ( \mathcal{F}|_{P} ) $ is a skyscraper sheaf over P so its $H^{1}$ is 0.
$H^{1}$ of $\mathcal{F}$ is not 0. By long exact sequence in cohomology, we have 
$H^{1}$ of $j_{!}( \mathcal{F}|_{U} )$ is not 0.
But on the other hand  $\mathcal{F}|_{U}$ is a trivial sheaf over a affine scheme so its 
$H^{1}$ is 0.
This example seems to show that the expected relation is not true.

REPLY [4 votes]: If $X$ is smooth and we consider sheaves of $k$-modules, $k$ a commutative ring, then $H^*(X, j_!\mathcal{S})\cong H^*_c(U,\mathcal{S})$; the latter  is equipped with a non-degenerate pairing $$H^*_c(U,\mathcal{S})\otimes H^{2d-*}(U,\mathcal{S}^{\vee})\to k,$$
where $\mathcal{S}^\vee$ is the Verdier dual local system (i.e., the local system that is constructed from the representation of $\pi_1(U)$ dual to the one that gives $\mathcal{S}$) and $d=\dim X$. So if $k$ is a field, we do get a statement relating the vanishing of $H^*(X, j_!\mathcal{S})$ and $H^{*}(U,\mathcal{S}^{\vee})$.<|endoftext|>
TITLE: Circle-arc number of a knot
QUESTION [10 upvotes]: I would like to build knots in $\mathbb{R}^3$ from arcs
of unit-radius (planar) circles, joined together at points where
the tangents match.  Thus the knot will have
curvature $1$ at all but the joints.
Here is an example of how two arcs might join:

 
 
 
 
 
 
 
 


Define the circle-arc number $C(K)$ of a knot $K$ as the fewest
number of such arcs from which one can build a nonselfinterecting
curve in space representing $K$.
This number is analogous to the
stick number of a knot,
except that the pieces are arcs,
and there is a tangent-joining condition.
I would be interested to learn of bounds on $C(K)$ in
terms of other knot quantities, for example,
the stick number, or the
crossing number cr$(K)$.
Here is an example of what I have in mind.
It appears that one might be able to build a trefoil from six
arcs, something like this:

 
 
 
 
 
 
 
 


However, the above picture is actually planar, and I have not
verified carefully that this is achievable in $\mathbb{R}^3$!
Has this concept been studied before?  If so, pointers would
be welcomed. Thanks!
Addendum. The trefoil can be realized with six arcs:
      
(The black triangle vertices indicate the circle centers on the plane before their arcs are twisted into 3D.)

REPLY [8 votes]: You can bound this number from below by the crossing number. The projection of the arcs of two unit circles can cross in at most two points, so $cr(K)\leq C(K)(C(K)-1)$. Also, you ought to be able to bound it quadratically from above by the grid number. If you have a knot presented by a grid diagram, you can represent the knot by a linear number of segments of linear length. Each one of these can be made into a linear number of arcs of unit circles by putting in wiggles. Since grid number is bounded above linearly by crossing number, one obtains inequalities of the form $$ \sqrt{cr(K)}\leq C(K)\leq A\cdot cr(K)^2$$ for some constant $A$. Notice that the grid number can sometimes be like $O(\sqrt{cr(K)})$, so I don't expect the upper bound to be sharp. For example, for certain torus knots you'll have $C(K)=O(cr(K))$.<|endoftext|>
TITLE: Eigenvalues, singular values, and the angles between eigenvectors
QUESTION [10 upvotes]: Suppose the $n \times n$ matrix $A$ has eigenvalues $\lambda_1, \ldots, \lambda_n$ and singular values $\sigma_1, \ldots, \sigma_n$. It seems plausible that by comparing the singular values and eigenvalues we gets some sort of information about eigenvectors. Consider:
a. The singular values are equal to the absolute values of eigenvalues if and only if the matrix is normal, i.e., the eigenvectors are orthogonal (see http://en.wikipedia.org/wiki/Normal_matrix , item 11 of the "Equivalent definitions" section ).
b. Suppose we have two distinct eigenvalues $\lambda_1, \lambda_2$ with eigenvectors $v_1, v_2$. Suppose, hypothetically, we let $v_1$ approach $v_2$, while keeping all the other eigenvalues and eigenvectors the same. Then the largest singular value approaches infinity. This follows since $\sigma_{\rm max} = ||A||_2$ and $A$ maps the vector $v_1 - v_2$, which approaches $0$, to $\lambda_1 v_1 - \lambda_2 v_2$, which does not approach $0$.
It seems reasonable to guess that the ``more equal'' $|\lambda_1|, \ldots, |\lambda_n|$ and $\sigma_1, \ldots, \sigma_n$ are, the more the eigenvectors look like an orthogonal collection.  So naturally my question is whether there is a formal statement to this effect.
Note: I asked this question on math.SE about a week ago.

REPLY [11 votes]: Suprprisingly, no, or at least not using any obvious measure of eigenvalue equality. Let $M$ be a matrix that is very non-normal, and consider the matrix:
$\left(\begin{array}{cc} I+\epsilon M & 0 \\ 0 & 2I + \epsilon M\end{array}\right)$
This matrix has a bunch of eigenvalues very close to $1$ and a bunch of eigenvalues very close to $2$, and similarly for the singular values. Thus, the difference between the eigenvalues and singular values, properly normalized to account for the total variance of eigenvalues, is very very small.
But the eigenvectors of this matrix are the same as the eigenvectors of $M$, repeated twice, and thus are very non-orthogonal. (Except, I guess, that half the eigenvectors are orthogonal to the other half .)<|endoftext|>
TITLE: State of knowledge on the Commutative W-spaces which appear in "Model Categories of Diagram Spectra"
QUESTION [6 upvotes]: This is a follow-up question to another question I asked last month. In MMSS's "Model Categories of Diagram Spectra," the authors consider many different models for spectra and prove monoidal Quillen equivalences between them. One of the models is $W$-spaces, i.e. diagrams of shape $W$ in $Top$ where $W$ is the category of based spaces which are homeomorphic to finite CW complexes, and $Top$ means compactly generated spaces.

Does the category of commutative monoids in $W$-spaces inherit a model structure?

By commutative here I mean strictly commutative, not $E_\infty$. My interest is for the background section of a paper I'm working on now about when commutative monoids form a model category. If it's not known one way or the other I can try my theorem on the category of $W$-spaces, but I suspect that category won't satisfy my hypotheses. If it's known to be true, maybe I can recover it as a special case of my theorem. If it's known to be false, that would be good to know too.
Note that this question comes up on page 5 of MMSS and the authors say they don't know. On page 47 they point out the obstacle which prevented them from proving CommMon($W$-spaces) inherits a model structure, and it's the same obstacle Peter May pointed out in his answer to my previous question. They couldn't prove that every relative $Sym(K^+)$-cell complex was a stable equivalence (where $K^+$ are the positive stable generating trivial cofibrations) because they couldn't prove $W$-spaces have the property that for every cell $S$-module $M$, $(E\Sigma_j)_+ \wedge_{\Sigma_j} M^j \to M^j/\Sigma_j$ is a weak equivalence. If someone had proven this in the past decade I'm sure Peter May would know. What I'm wondering is if someone found a different way to prove CommMon($W$-spaces) forms a model category or if someone found evidence that it can't, similar to the evidence in this answer that CDGA can't be a model category, or on page 13 of this paper of Schwede's that CommMon($\Gamma$-spaces) can't be a model category. Even if no hard evidence exists, I'd like to hear some soft evidence, i.e. why one would or would not believe this to be true.
The reason one might hope the answer is yes is that Theorems 0.1 and 0.4 prove that $W$-spaces are Quillen equivalent to Orthogonal Spectra and Symmetric Spectra, and that this Quillen equivalence carries over when we pass to the model categories of monoids (the fact that such model structures exist follows from the fact that all these categories satisfy the pushout product and monoid axioms). Both Symmetric Spectra and Orthogonal Spectra have a notion of positive model structures, which makes the category of commutative monoids inherit a model structure. It's natural to wonder if $W$-spaces has something like that too.
MMSS also raise another natural question (on page 6): 

Is $Ho(E_\infty$ $W$-spaces$) \cong Ho($Commutative $W$-spaces$)$?

It seems like with all the work that has been done on operads in the last decade this should be known. Can someone confirm that and give a reference?

REPLY [5 votes]: There is a very illuminating paper I did not know about when I last answered a similar question:
Tyler Lawson. ``Commutative $\Gamma$ rings do not model all commutative ring spectra". Homology,
Homotopy and Applications Vol 11, 2009, 189-194.  The title says it all.  Together with the monoidal functor from $\Gamma$-spaces to $W$-spaces of MMSS, Tyler's result makes clear that commutative $W$-rings cannot give a homotopy category equivalent to the categories of orthogonal, symmetric, or EKMM commutative ring spectra.  Incidentally, the question has a minor slip.  David writes: "By commutative here I mean strictly commutatative, not $E_{\infty}$".  As explained in EKMM, strictly commutative EKMM ring spectra are Quillen equivalent to $E_{\infty}$ ring spectra as defined way back in the 1970's. Those still have the nicest connection to spaces, namely $E_{\infty}$ ring spaces.  See
my paper: What precisely are $E_{\infty}$ ring spaces and $E_{\infty}$ ring spectra.
Geometry & Topology Monographs 16(2009), 215--282 (and on my web page) for a modern overview.<|endoftext|>
TITLE: Where do nonstandard elliptic curve angles come from?
QUESTION [6 upvotes]: This is a question which has bounced around my head over the past few years.  At the same time, I am answering https://mathoverflow.net/questions/104421/riemann-hypothesis-for-zeta-function-of-algebraic-curves-over-fields-of-infinite with another question.
Let $E$ be an elliptic curve over $Q$.  Let $u$ be an nonprincipal ultrafilter on the set of prime numbers.  
For each prime $p$ (at which $E$ has good reduction, let's say), let $\pm \theta_p$ be the elliptic curve angle at $p$.  In other words, $a_p = 2 \sqrt{p} \cdot \cos(\theta_p)$.  Then, by the compactness of the interval $[-1,1]$, there is a nonstandard elliptic curve angle $\theta_u$ naturally associated to the set $(\theta_p)$ and $u$.
I've been wondering if there's any other way to produce these nonstandard angles.  For example, let $\sigma$ be a "generic" field automorphism of the complex numbers $C$, in the sense that $(C,\sigma)$ is a model of $ACFA$.  Can one produce an elliptic curve angle $\theta$ directly from the data $(E, C, \sigma)$?
What's so difficult here is that, in the transfer from characteristic $p$ to characteristic $0$, it is so difficult to figure out how to handle things like $\sqrt{p}$.  The only hope, that I can see, would be to think of $a_p$ as a $p$-adic number (use $p$-adic cohomology), and then transfer the result to a Laurent series field (so $a_u$ might belong to $C((\varpi))$ and $a_u / \sqrt{\varpi}$ would be well-behaved).  But this is all "pie in the sky" for now.
Any ideas?  Anyone thought about RH in models of ACFA?

REPLY [2 votes]: It is a consequence of the generalized Sato-Tate conjecture, that given a non-CM elliptic curve over $\mathbb Q$, any element of $Gal(\bar{\mathbb Q}/\mathbb Q)$, and a real number $\in[−1,1]$, one can construct an ultrafilter on the primes such that Frobenius converges to that element and the angle of Frobenius converges to that real number.
Thus any attempt to answer this question must somehow make use of the transcendentals. I have no idea how one might do that.
EDIT: By ACL's answer to my question, the nonstandard angle of Frobenius is totally independent from all first-order statements about the nonstandard elliptic curve.<|endoftext|>
TITLE: Conjugcy classes in GL(F_2) ? GL(F_q)
QUESTION [6 upvotes]: How to deduce a formula (see below) for number of conjugacy classes in GL_n(F_2) ? (More generally F_q) ?
Is there some description of conjugacy classes or we just know how many of them but do not know how to describe them ? 
Can someone send me a paper, please ?:
"Pairs of commuting matrices over a finite field"
Walter Feit and N. J. Fine
http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.dmj/1077468920
FORMULA from OEIS
The number a(n) of conjugacy classes in the group GL(n, q) is the coefficient of t^n in the infinite product: product k=1, 2, ... (1-t^k)/(1-qt^k) - Noam Katz (noamkj(AT)hotmail.com), Mar 30 2001.
Simple observations
Clearly if characteristic polynom is different then matrices are not conjugated.
For each char. pol. it is easy to give a matrix with such char. pol. so we get (q-1)q^{n-1}
possibilities at least.
But if char. pol has a roots with multiplicities - then we may have several Jordan cells
and several conjugacy classes with same characteristic polynomial.
So it is not clear for me how to count them.
Diagonal elements of corresponding Jordan cells may not live in F_q but in some alg. extentsion
of it, these makes me completely puzzled - is it possible to control this ? Hardly... 
Related and not so much but nice  questions :)
Number of conjugacy classes in generic finite group?
Sizes of twisted conjugacy classes of $PSL(n,q)$
How many conjugacy classes of subgroups does GL(2,p) have?
Decomposition of GL(2,p) into irreducible representations
Centralizers in GL(n,p)
Products of Conjugacy Classes in S_n
The number of conjugacy classes and the order of the group
Arithmetic relations between degrees of irrchar and cardinals of conjugacy classes
Can we bound degrees of complex irreps in terms of the average conjugacy class size?
Symmetric group irreps in tensor products of exterior products of the standard representation

REPLY [5 votes]: Edit: Unfortunately, the first version of this answer was wrong, so I had to rewrite it almost entirely.
So the formula you want to prove is:
$$
 \sum_{n=1}^\infty C_{n,q} t^n =\prod_{k=1}^\infty \frac{1-t^k}{1-qt^k}
$$
where $C_{n,q}$ denotes the number of conjugacy classes in $\textrm{GL}_n(q)$.
Note that
$$
   \frac{1-t^k}{1-qt^k} =1+\sum_{i=1}^\infty (q^i-q^{i-1}) t^{ki}
$$
As Paul Garrett notes in his answer, $q^i-q^{i-1}$ is the number of monic polynomials in $\mathbb F_q[x]$ of degree $i$ with non-vanishing constant term. One usual way of counting conjugagy classes of matrices in $\textrm{GL}_n(q)$ (or $\textrm{Mat}_n(\mathbb F_q)$, if one admits the eigenvalue $0$) is counting isomorphism classes of $\mathbb F_q[x]$-modules of dimension $n$, which by the structure theorem on f.g. modules over PID's are given by
$$
   \mathbb F_{q}[x]/(p_k(x)) \oplus \mathbb F_{q}[x]/(p_k(x)\cdot p_{k-1}(x)) \ldots \oplus  F_{q}[x]/(p_k(x)\cdot p_{k-1}(x)\cdots p_{1}(x))
$$
where $p_1,\ldots,p_k$ are monic polynomials with constant term $\neq 0$ (uniquely determined by the isomorphism type of the module) such that $1\cdot \textrm{deg}(p_1)+2\cdot \textrm{deg}(p_2) \ \ldots + k \cdot \textrm{deg}(p_k) = n \textrm{ (the dimension of the module)}$. Note that some of the $p_i$ may very well be equal to $1$.
From the above considerations it is clear that we can parametrize isomorphism classes of $\mathbb F_q[x]$-modules by giving a sequence of monic polynomials which eventually becomes constant $=1$, by letting the module given above correspond to the sequence
$$
(f_1,f_2,f_3,\ldots) =(p_k, p_{p-1},\ldots,p_1,1,1,\ldots)
$$
Note that in this sequence, $f_m$ constributes $m\cdot \textrm{deg}(f_m)$ to the dimension of the module. It follows
$$
\begin{array}{rcl}
\sum_{n=1}^\infty C_{n,q} t^n &=& \prod_{m=1}^{\infty}\left(\sum_{i=0}^\infty \#\textrm{(choices for $f_m$ that contribute $mi$ to the dimension)}\cdot t^{mi}\right)\newline &=&\prod_{m=1}^{\infty} \left(1+\sum_{i=1}^\infty (q^i-q^{i-1}) t^{mi}\right) =\prod_{m=1}^\infty \frac{1-t^m}{1-qt^m}
\end{array}
$$<|endoftext|>
TITLE: Road to Solovay's Land.
QUESTION [8 upvotes]: In the first semester of 2012 I took a course in General Topology and Set Theory, at undergraduate level. For topology, I was instructed to use Engelking's General Topology; albeit I had a great difficult to approach it, I got used to the text and did (and I'm still doing) some exercises (but none of the problems until now). For the Set Theory course we used Jech & Hrbacek's Introduction to Set Theory, which I think was suitable for my level back there. In these courses, I heard about Boolean Algebras, Forcing, Independence Proofs, Models, Topological Games, Kunen's book (which I just bought a copy), and others interesting things that caused me to change my favorite mathematical area (in fact I was a physics undergrad student when this year began).
In this semester, I enrolled myself in Measure and Integration course, also at undergraduate level, where I discovered about Solovay's model, which completely drove me to madness. 
I'm looking for advice about my background and the path that I have to follow to reach these mentioned topics; is too early to begin? do my background is sufficiently enough to start? And where to begin with ? what books do I have to read?
P.S.: I had no background in mathematical logic, the only thing I can do is some proofs with truth-tables.

REPLY [10 votes]: Let me give an alternative ending to Noah's road map. The splitting point is at symmetric models.
After you've understood the basics of forcing well, you can switch to Kanamori's The Higher Infinite. In the chapter about the real numbers and forcing he again reviews forcing (and if you're new to this - such review is always good) and constructs Solovay's model in a very clear approach.
He avoids [1] talking about symmetric models (which can be a rather complicated tool) by using the "external" construction: we add some sort of generic set to $V$ then we consider an inner model of $V[G]$ which is $HOD(\mathbb R)$ or $L(\mathbb R)$, the latter being thrown around a lot in discussions about models of set theory without choice.
In Kanamori you can find a good introduction to large cardinals (if you haven't run into them in previous steps) which also play a role in Solovay's construction, although that appears in another chapter of the book.
I want to add that studying the construction of symmetric extensions is a good idea. This is an extremely illuminating construction which sheds a lot of light on how set theory works, at least this is how I felt in the past year. However for this particular case I think that using the approach of relative constructibility is better.

Footnotes:

This is not entirely true that Kanamori avoids the symmetric models because as it turns out all symmetric models are $HOD(A)$ (whatever that means) of some generic set $A$. In the case of Solovay's model it is just much simpler to use this sort of construction rather going through the complication of symmetric forcing.<|endoftext|>
TITLE: Irreducible homology 3-spheres that bound smooth contractible manifolds
QUESTION [14 upvotes]: Some examples of irreducible homology 3-spheres that bound  smooth contractible 4-manifolds are listed in the comment to problem 4.2 in  Kirby's problem list, and all of them happen to occur among the 
Brieskorn spheres $\Sigma(p,q,r)$ modelled on $\widetilde{SL}_2(\mathbb R)$,
i.e. such that $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}<1$ (except for the standard $S^3$, of course). 
 Question. Are there any known homology 3-spheres that bound smooth contractible 4-manifolds and are modelled on other geometries, e.g. NIL or the hyperbolic space?
EDIT: Glazner in the paper "Uncountably many contractible 4-manifolds" constructed some other examples but I cannot recognize the geometry. (Glazner's six page paper
is easily googlable by title, and it gives an explicit representation for the fundamental group, denoted $G_n$ on page 40).

REPLY [10 votes]: This is a variation of Agol's answer. Take any contractible special polyhedron $X$, that is a compact polyhedron such that every point has a neighborhood of one of these three types:

(source: unipi.it) 
and such that the natural stratification is a cellularization, i.e. all the points as in the center (right) form disjoint open 1-cells (2-cells).
A famous example is Bing's house: 
Now thicken 4-dimensionally the polyhedron $X$. You do this first by thickening the 1-skeleton, and this can be done in a unique way (we only construct orientable 4-manifolds). The rest can be thickened in various ways, but for simplicity we restrict ourselves only to locally flat thickenings, i.e. thickenings that locally lie in a (smooth) 3-dimensional slice (a very reasonable assumption). As shown by Turaev, the locally flat thickenings are parametrized by assigning a (half-)integer at each 2-cell of the polyhedron (these numbers are called gleams), which measures the "Euler number" of the thickening. This is very similar to the description of the 4-manifold via a Kirby diagram. For instance in the Bing's house there are three 2-cells and each needs to be coloured by an integer.
The boundary 3-manifolds of all these thickenings are all obtained by Dehn surgery on a fixed link $L$ in a connected sum $\sharp_h(S^2\times S^1)$ of some $h$ copies of $S^2\times S^1$. Different gleams yield different surgeries on the same link $L$. The nice thing here is the following: Costantino and Thurston have shown that the link $L$ is always hyperbolic; the complement $\sharp_h(S^2\times S^1)\setminus L$ can be constructed by gluing some regular ideal octahedra. You need two regular octahedra for each vertex of $X$, so that the volume of the hyperbolic manifold is precisely twice the volume of the regular ideal hyperbolic octahedron times the number of vertices of $X$.
Now as Agol said you only need to use Thurston Dehn filling theorem that ensures you that by avoiding a finite set of gleams on each face the resulting 3-manifold is hyperbolic. That is, the boundary of the resulting thickened 4-manifold is hyperbolic. Moreover the 6 theorem tells you that if all the gleams are bigger than 6 you are certainly done (because we know very well how the cusps are). Hence you can decorate Bing's house with a triple of integers bigger than 6 and you get plenty of hyperbolic 3-manifolds bounding contractible 4-manifolds.<|endoftext|>
TITLE: Vector fields on path spaces
QUESTION [10 upvotes]: I've been reading Chen's original works on iterated integrals and in order to consider differential forms on the path space $PM$ of a smooth manifold $M$ he gives $PM$ the following "differentiable space" structure:
Let $N$ be a smooth manifold. A continuous map $\alpha: N \to PM$ (where $PM$ has the compact open topology) is said to be smooth if the adjoint map $\tilde{\alpha}: N \times I \to M$ defined by $(n,t)\mapsto \alpha(n)(t)$ is smooth in the usual sense. The smooth map $\alpha$ is a plot if $N$ is an open convex subset of $\mathbb{R}^n$ for some $n$. Thus, we are modeling $PM$ locally with plots of varying dimension.
Chen defines a differential n-form $\omega$ on $PM$ as a rule which assigns to every plot $\alpha: U \to PM$ a differential form $\omega_{\alpha} \in \Omega^n(U)$. We define $(d\omega)_{\alpha}=d\omega_{\alpha}$. It turns out that for Chen's purposes one does not need to develop more calculus tools on $PM$. He shows a De Rham type result: the cohomology of the complex $\Omega^*(LM)$ (where $LM$ is the free loop space) is isomorphic to the real singular cohomology of $LM$.
However, for other purposes it is useful to consider forms as alternating tensors and to do this in this context we need a notion of vector fields on $PM$. I've always thought of a tangent vector at $\gamma \in PM$ as a vector field $T_\gamma$ along $\gamma$ on $M$, so a vector field on $PM$ assings each point $\gamma \in PM$ a vector field along $\gamma$. However, following Chen, the natural way to define vector fields to make it compatible with his notion of differential forms is as follows: a vector field $T$ on $PM$ is a rule which assings to each plot $\alpha: U \to PM$ a vector field $T_{\alpha}$ on $U$. 
How do we reconcile these two notions of vector fields on $PM$? Are they equivalent?

REPLY [5 votes]: Good question!
There is a fairly well developed theory of diffeological spaces, with many recent references. The only difference between diffeological spaces and Chen's smooth spaces is that the plots of a diffeological space have open subsets $U \subset \mathbb{R}^n$ as their domains, while the ones of a smooth space have convex subsets. For questions concerning differential forms and tangent vectors, this seems to be irrelevant, so I'll continue with diffeological spaces.
As Dimitri pointed out in his comments, a differential form on a diffeological space $X$ is not just a form $\psi_c$ on the domain $U_c$ of each plot $c: U_c \to X$. It must be required for two plots $c_1: U_1 \to X$ and $c_2:U_2 \to X$ such that one has a smooth map $f: U_1 \to U_2$ with $c_2 \circ f = c_1$, the forms satisfy 
$$
\psi_{c_1} = f^{*}\psi_{c_2}.
$$ 
As also remarked by Dimitri, vector fields can neither be pulled back nor be pushed forward. Hence there is no analogous definition of vector fields. 
Yet the concept of a tangent space to a general diffeological space $X$ can be defined. A good reference is the paper "Diffeological spaces" by Martin Laubinger (Proyecciones Journal of Mathematics, Vol. 25, No2, pp. 151-178). Roughly, the tangent space at a point $x\in X$ is the sum of all tangent spaces $T_0U_c$ over all plots $c:U_c \to X$ centered at $x$, i.e. $0 \in U_c$ and $c(0)=x$. Then there is an equivalence relation that enforces the chain rule for maps.
Smooth manifolds form a full subcategory of diffeological spaces, by regarding a smooth manifold $M$ as a diffeological space with plots all smooth maps $c:U \to M$, for all open subset $U \subset \mathbb{R}^n$ and all $n\in \mathbb{N}$. Thus, for a smooth manifold $M$ we have now the classical tangent space, and the diffeological tangent space. Laubinger proves that both tangent spaces are isomorphic. 
Apparently less well-known is a theorem of M. Losik showing that Fréchet manifolds also form a full subcategory of diffeological spaces (with the plots defined in the very same way as for smooth manifolds). Thus, for a Fréchet manifold we also have two concepts of a tangent space, the classical one and the diffeological one. I don't know a reference for the statement that both concepts coincide, as in the smooth case, but I have looked at Laubinger's proof and I don't see why the same proof should not work for Fréchet manifolds. 
Now let's look at the path space $PM$ or the loop space $LM$ or at any other mapping space of a compact smooth manifold into a smooth manifold $M$. I continue with $LM$ for simplicity. There are, a priori, two diffeologies on the loop space $LM$. The first one is the one from the theorem of Losik, i.e. the one where the plots are the Fréchet-smooth maps from open sets $U$ to $LM$. The second one is the one described in the question, it is called the functional diffeology, with plots those maps $c: U \to LM$ whose adjoint $U \times S^1 \to M$ is smooth (demanding continuity for $c$ is superfluous). Luckily, the two diffeologies coincide (see Lemma A.1.7 in my paper Transgression to Loop Spaces and its Inverse, Part I).
And here is the result: classical Fréchet theory says that the tangent space of $LM$ at a loop $\tau$ consists of vector fields along $\tau$. The earlier claimed generalization of Laubinger's theorem shows that it coincides with the diffeological tangent space, and the coincidence of the diffeologies on $LM$ shows that it can be defined via plots $c$ whose adjoints $U \times S^1 \to M$ are smooth maps. This is, I think, the best one could hope for.<|endoftext|>
TITLE: How can I tell if a variety is normal?
QUESTION [13 upvotes]: Suppose $R$ is a subalgebra of ${\mathbb C}[x_1,...,x_N]$ generated by polynomials $p_1,...,p_k.$ I know that ${\mathbb C}[x_1,...,x_N]$ is the integral closure of $R$.
Is there an algorithm to determine if $R={\mathbb C}[x_1,...,x_N]$ which would work for large $N$? (Say $N=20$). 
Clearly, it is a question about the normality of the variety $X=Spec(R).$
The problem is that I don't know relations between the generators of $R$ (and these may be difficult to compute for large $N$)-- otherwise I could check if $X$ is smooth and if the map ${\mathbb C}^n\to X$ is an immersion -- conditions which imply that $R={\mathbb C}[x_1,...,x_N]$.

REPLY [4 votes]: Consider the ideal $(p_1(x_1,\dots,x_n)-p_1(y_1,\dots,y_n),\dots,(p_k(x_1,\dots,x_n)-p_k(y_1,\dots,y_n))$. If the polynomials generate the ring, this is the ideal $(x_1-y_1,\dots,x_k-y_k)$. This is clear, but the converse is also true, since a map is a closed immersion if it is proper and separates points and tangent vectors, which we can check using the ideal.
Proof: To check that  two points $P$ and $Q$ are separated, choose a $p_i$ that does not vanish at $P,Q$. To check that two tangent vectors $t_1$ and $t_2$ at $P$ are separated, choose a $p_i$ such that the derivative of $p_i$ at $P$ in the direction $(t_1,t_2)$ is nonzero, which there must be since $t_1$ and $t_2$ differ along some coordinate $x_j$, so the derivative of $x_j-y_j$ is nonzero.
The map is proper since it is a finitely-generated integral closure, thus finite, thus proper.
A closed immersion of affine schemes is a quotient by an ideal, but this map has no kernel, so it is an isomorphism as long as $(p_1(x_1,\dots,x_n)-p_1(y_1,\dots,y_n),\dots,(p_k(x_1,\dots,x_n)-p_k(y_1,\dots,y_n))=(x_1-y_1,\dots,x_k-y_k)$<|endoftext|>
TITLE: Equivalence of forcing automorphisms
QUESTION [5 upvotes]: Suppose that $P$ is a forcing poset in $V$. If $\pi$ is an automorphism of $P$ then $\pi$ extends to automorphisms of the names by induction: $$\pi\dot x = \lbrace(\pi p,\pi\dot y)\mid (p,\dot y)\in\dot x\rbrace$$
I've been stuck on the following proposition for quite some time now and I don't see an argument for, nor an obvious counterexample again:

Let $\mathscr G$ be a group of automorphisms of $P$. Suppose that $\dot x$ is a $P$-name and $G$ is $P$-generic over $V$ then the equivalence relation over $\mathscr G$ defined as: $$\pi\sim_G\sigma\iff (\pi\dot x)^G=(\sigma\dot x)^G$$ is multiplicative? Namely, is the set $\lbrace\pi\in\mathscr G\mid\pi\sim_G\mathrm{id}_P\rbrace$ is a normal subgroup of $\mathscr G$?

I am particularly interested in the case where $P$ is a Cohen forcing, in case it isn't true in general.
Edit: To restrict the question even more (after Joel's answer), what if we assume that $\dot x$ is hereditarily symmetric with respect to a normal filter of subgroups over $\mathscr G$?

REPLY [4 votes]: I think Joel's example can be easily modified to use a hereditarily symmetric name.  Use Cohen's original model for the negation of choice, i.e., use Cohen forcing to add an $\omega$-sequence of generic reals, and let the filter be the finite-support filter on the group $\mathcal G$ of permutations of the positions in the $\omega$-sequence.  Then use Joel's example working on  the first real in the $\omega$-sequence.  (If you don't like the fact that Joel's $\pi$ and $\tau$ aren't in $\mathcal G$, then adjoin them and extend the filter accordingly.)<|endoftext|>
TITLE: product operation: name and notation
QUESTION [6 upvotes]: For every product in a category, there exists an operation $\langle\cdot, \cdot\rangle$ that turns morphisms $f : C \to A$ and $g : C \to B$ into morphisms $\langle f, g\rangle : C \to A \times B$. How are this operation and its results usually called? Calling $\langle f, g\rangle$ the product of $f$ and $g$ does not make sense to me, since I would expect $f \times g$ to be the product of the two morphisms.
Furthermore there is the generalisation of binary products in the form of arbitrary products $\prod_{i \in I} A_i$ for families $\{A_i\}_{i \in I}$ of objects. These products have, of course, variants of the $\langle\cdot, \cdot\rangle$-operator. What is the usual notation for these variants? So far, I wrote $\langle\{f_i\}_{i \in I}\rangle : C \to \prod_{i \in I} A_i$ for families of morphisms $f_i : C \to A_i$. Is this standard?

REPLY [7 votes]: Probably $\langle f, g\rangle : C \to A \times B$ is most often just called the arrow to the product.  You are right it should not be called a product arrow.  People who want a specific name for the operation have called it the "pairing arrow."
I would write $\langle f_i\rangle_{i \in I} : C \to \prod_{i \in I} A_i$ without curly brackets in the angle brackets and i would call it the arrow to the product.  A more specific name for the operation could be "tupling arrow."<|endoftext|>
TITLE: A map inducing isomorphisms on homology but not on homotopy
QUESTION [10 upvotes]: As a consequence of the Whitehead theorem, Spanier's Algebraic Topology book has on 7.6.25 the following theorem:

A weak homotopy equivalence induces isomorphisms of the corresponding integral singular homology. Conversely, a map between simply connected spaces which induces isomorphisms of the corresponding integral singular homology groups is a weak homotopy equivalence.

Is it possible that a map between (non-simply connected) topological spaces induces an isomorphism on all homology groups, and yet is not a weak homotopy equivalence? If so, I would be glad to have an example.

REPLY [3 votes]: Several decades ago this question led to the concept of nilpotent space, nilpotent CW complex.  You can google that, along with "Emmanuel Dror", who later changed his name to Emmanuel Dror Farjoun.<|endoftext|>
TITLE: Around Continuum Hypothesis
QUESTION [10 upvotes]: It's now well known that Continuum Hypothesis (CH) is independent from standard axioms of set theory: one can assume that $c=\aleph_1$ or assume that $c \neq \aleph_1$. Let us assume the second case-then the natural queation rise:
What 'values' $c$ can take? I know that it's impossible that $c=\aleph_{\omega}$? Is it possible to give simple criterium for a cardinal $\alpha$ to have a property: $c=\alpha$ in 
some model of set theory? In particular, is there a cardinal $\alpha$ defined not using the term $c$ for which we have $c<\alpha$? And the last question, is it possible for $c$ to be weakly inaccessible?

REPLY [11 votes]: Any cardinal of uncountable cofinality can consistently be the cardinal of the continuum.  In particular, it is consistent that the cardinal of the continuum is weakly inaccessible (assuming, of course, that inaccessible cardinals are consistent.)  The relevant paper is Solovay's "$2^{\aleph_0}$ can be anything it ought to be" but nowadays the result is also in standard textbooks.<|endoftext|>
TITLE: (Co)localization of the derived category
QUESTION [6 upvotes]: Let me start saying that a similar question can be stated for general locally Noetherian Grothendieck categories but I state it for categories of modules as it is simpler. So we fix a right Noetherian ring (non-commutative) and we let $Mod(R)$ be the category of right $R$-modules.
Recall the following theorem due to H. Krause (for its proof Brown's representability is used).
Theorem 1.
Let $\mathcal T$ be a triangulated category with small coproducts which is well generated. Let $H : \mathcal T \to \mathcal A$ be a cohomological functor into an abelian category $\mathcal A$ which has small coproducts and exact $\alpha$-filtered colimits for some regular cardinal $\alpha$. Suppose also that $H$ preserves small coproducts. Then there exists an exact localization functor $L:\mathcal T \to\mathcal T$ such that for each object $X$ we have $LX=0$ if and only if $H(X[n])=0$ for all $n \in \mathbb Z$.
Fix now a hereditary torsion theory $\tau$ on $Mod(R)$. We go on defining in three steps an exact localization functor of the derived category $L_\tau:{ D}(R)\to { D}(R)$.
(1) Denote by 
$$H^n:{\bf D}(R)\to Mod (R)$$
 the usual $n$-th cohomology, for every $n\in\mathbb Z$. It is clear that each $H^n(-)$ is cohomological and preserves coproducts. 
(2) Fix a hereditary torsion theory $\tau$ on $Mod(R)$. The $\tau$-localization functor 
 $$Q_\tau:Mod(R)\to \mathcal A_\tau=Mod(R)/\mathcal T_{\tau} ,$$
where $\mathcal T_{\tau}$ is the hereditary torsion class associated to $\tau$, is exact  and preserves coproducts. Furthermore, $\mathcal A_\tau$ is a Grothendieck category and so it has small coproducts and exact colimits. 
(3) For every $n\in\mathbb Z$ denote by 
$$H_\tau^n:{\bf D}(R)\to \mathcal A_\tau$$
the composition of the above two functors, that is $H_\tau^n(-)=Q_\tau H^n(-)$. By (1) and (2) one can easily derive that $H_\tau^n(-)$ is cohomological and preserves coproducts. 
Now we have all the instruments to construct the localization functor $L_\tau(-)$:
Corollary.
Let $\tau$ be a hereditary torsion theory on $Mod(R)$. Then there exists an exact localization functor $L_\tau:{\bf D}(R)\to {\bf D}(R)$ such that $L_\tau(X)=0$ if and only if the $n$-th cohomology of $X$ is $\tau$-torsion for every $n\in\mathbb Z$.
Proof.
Consider the functor $\prod_{n\in\mathbb Z}H^n_\tau:{\bf D}(R)\to \prod_{n\in\mathbb Z}\mathcal A_\tau$. By the above discussion, this functor is a cohomological functor preserving coproducts from ${\bf D}(R)$ to a bicomplete abelian category with exact colimits. Let $X\in {\bf D}(R)$, then $\prod_{n\in\mathbb Z}H^n_\tau(X)=0$ if and only if $H_\tau^n(X)=0$ for every $n\in\mathbb Z$, if and only if $Q_\tau H^n(X)=0$ for every $n\in\mathbb Z$. This is equivalent to say that all the cohomologies of $X$ are in the kernel of $Q_\tau(-)$ that is, they are $\tau$-torsion. Now, to prove the existence of $L_\tau(-)$ it is enough to apply Theorem 1.

Question. In the above notation, is it possible to prove that for a given object $X\in{\bf D}(R)$, $L_\tau(X)$ belongs to the smallest localizing subcategory of ${\bf D}(R)$ containing $X$?

More specifically, consider an indecomposable injective module $E$ and let $\tau$ be the hereditary torsion theory cogenerated by $E$. In the commutative case, there is a unique prime ideal associated to $E$ and localizing with respect to $\tau$ is the same as localizing at that prime ideal. In particular, localized modules can be constructed as a direct limit of $R$-modules and this construction can be "lifted" to the derived category to answer positively to the above question.
In the non-commutative case, there is no prime ideal associate in general but we can suppose it if we suppose our ring to be right FBN (in the general case $E=E(C)$ with $C$ a cocritical module which can be chosen of the form $R/I$ with $I$ an irreducible ideal). But even in the case when the localization at $\tau$ is Ore, the localized modules are constructed as direct limits in the category of Abelian groups and after that they are induced with the structure of modules. This makes no sense (to me) in ${\bf D}(R)$.

REPLY [2 votes]: I think the answer is no in the following very simple example.
Let $R$ be the path algebra of the quiver $\bullet\rightarrow\bullet$ over a field $k$. Then since $R$ is hereditary and of finite representation type, the objects of ${\bf D}(R)$ are just coproducts of shifts of copies of the three indecomposable representations $[k\rightarrow0]$, $[k\rightarrow k]$ and $[0\rightarrow k]$.
Let $E$ be the injective representation $[k\rightarrow 0]$, so the $\tau$-torsion objects for the hereditary torsion theory $\tau$ cogenerated by $E$ are the representations of the form $[0\rightarrow V]$, and the localization functor $L_{\tau}$ just takes $[U\rightarrow V]$ to $[U\rightarrow 0]$.
In particular, if $X=[k\rightarrow k]$. then $L_{\tau}(X)=[k\rightarrow0]$, but this is not in the localizing subcategory generated by $X$, which contains only coproducts of shifts of copies of $X$.<|endoftext|>
TITLE: Does any identity holding in all finite-dimensional Lie algebras hold in all Lie algebras?
QUESTION [9 upvotes]: Equivalently, is the free Lie algebra on finitely many generators over a fixed field $k$ (say of characteristic not equal to $2$) residually finite-dimensional in the sense that any nonzero element remains nonzero in some finite-dimensional quotient? Some quick Googling on my part was not successful here. 
Motivation: If this is false, an identity holding in all finite-dimensional Lie algebras which doesn't hold in all Lie algebras isolates a potentially interesting class of infinite-dimensional Lie algebras (namely the ones satisfying the identity).

REPLY [9 votes]: Maybe I am missing something but let me try. The free Lie algebra is a subalgebra of the free associative algebra made Lie via the bracket. The free associative algebra is residually finite dimensional by truncating polynomials. Hence its Lie algebra is residually finite dimensional. Here I assume finitely generated but that is enough. 
clarification. I wrote this answer before going to bed so let me say it better. There is a functor from associative algebras to Lie algebras which sends A to the vector space A with the commutator bracket $[a,b]=ab-ba]$. It is then clear that this functor preserves residual finite dimensionality. 
It is well known that the free Lie algebra is obtained by applying this Functor to the free associative algebra and generating a sub-Lie algebra by the letters. Since the free associative algebra is residually finite by truncating polynomials we are done.

REPLY [6 votes]: I believe this is entirely equivalent to Benjamin Steinberg's answer, but one way to see this is that the quotient of a free Lie algebra on finitely many generators by the $n$th term $\mathfrak{F}_n$ of the lower central series (i.e., the ideal generated by $n$-fold bracketings of generators) is a finite dimensional Lie algebra, namely, the free Lie algebra of nilpotence class $n$ on the same generating set. Each element of the free Lie algebra lies outside of $\mathfrak{F}_{d + 1}$, where $d$ is the maximal bracket-nesting depth of the element.<|endoftext|>
TITLE: Projective dimension of simple module
QUESTION [5 upvotes]: Let $R$ be a (not necessarily commutative) ring and $M$ a simple right $R$-module. Then $\mathfrak{m}=Ann(M)$ is a maximal ideal of $R$. It is seems known that
$$
pdim_{R}(M)=pdim_{R_{\mathfrak{m}}}(M_{\mathfrak{m}})
$$
where $R_{\mathfrak{m}}$ is the localization of $R$ with respect to $\mathfrak{m}$. "$\ge$"-part is obvious as localization is an exact functor mod-$R$ to mod-$R_{\mathfrak{m}}$. Could anyone give me a proof or a reference of this equality? 
Thanks.

REPLY [8 votes]: In general, it is not true that $Ann_R(M)$ is a maximal two-sided ideal of $R$ if $M$ is a simple $R$-module. For example, let $k$ be a field of characteristic zero, let $\mathfrak{g} = \mathfrak{sl}_2(k)$ and let $A = U(\mathfrak{g}) / \langle C \rangle$ where $C$ is the Casimir element. Then the Verma module $V$ of highest weight $-2$ for $\mathfrak{g}$ is a simple faithful $A$-module (so $Ann_A(V) = 0$), but the zero ideal in $A$ isn't maximal since it's contained in the proper two-sided ideal $\mathfrak{g}\cdot A$.
Annihilators of simple modules are called primitive ideals; whilst any maximal two-sided ideal $I$ in a ring $R$ is necessarily primitive (simply pick a maximal left ideal $J$ containing $I$ and consider the simple module $R/J$), the converse is not true as the above example shows.
As Ralph notes, in the general setting of non-commutative rings, the localisation $R_{\mathfrak{m}}$ does not exist: one needs certain Ore conditions to be satisfied by the set $S$ of elements in $R$ that are non-zero-divisors modulo $\mathfrak{m}$.
Even if the Ore localisation $R_\mathfrak{m}$ exists, it could happen that the Ore localisation $M_{\mathfrak{m}}$ is zero. This is actually the case for the $A$-module $V$ considered above: since $A$ is a Noetherian domain, the set $S$ of all non-zero elements of $A$ is an Ore set, but $V$ is a torsion $A$-module so $V_S = 0$. It's known that $pd_A(V) = 1$ in this case, so the inequality $pd_R(M) \geq pd_{R_{\mathfrak{m}}}(M_{\mathfrak{m}})$ is strict in general.
In the positive direction, however, it is known that flat dimension behaves well under Ore localisation. More precisely: suppose that $S$ is a two-sided denominator set in a ring $R$ and let $N$ be an $R_S$-module. Then the flat dimension of $N$ as an $R_S$-module is equal to the flat dimension of $N$ as an $R$-module; this is Proposition 7.4.2(iii) of the book Noncommutative Noetherian Rings by McConnell and Robson.
For a finitely generated module $M$ over a Noetherian ring $R$, flat dimension coincides with projective dimension (see section 7.1.5 of the book cited above). This is enough to give a positive answer to your question under the following extra hypotheses:

the annihilator $\mathfrak{m}$ of your simple module $M$ is localisable on both sides 
the ring $R$ is Noetherian
the set $S$ of regular elements mod $\mathfrak{m}$ acts invertibly on $M$.

If this is the case then the Ore localisation $M_{\mathfrak{m}} = M_S$ coincides with $M$, so we can deduce that 
$pd_R(M) = fd_R(M) = fd_R(M_{\mathfrak{m}}) = fd_{R_{\mathfrak{m}}}(M_{\mathfrak{m}}) = pd_{R_{\mathfrak{m}}}(M_{\mathfrak{m}})$
using the results cited above.
The third condition is known to hold for example whenever $R / \mathfrak{m}$ is an Artinian ring, since in this case the image of $S$ in $R /\mathfrak{m}$ is precisely the set of units of $R / \mathfrak{m}$ and therefore acts invertibly on $M$.<|endoftext|>
TITLE: What can be the cohomology ring of a manifold?
QUESTION [17 upvotes]: Is there something well-known about which ring can be a cohomology ring of a manifold? More concretely, I would be interested in the following question: does there exists an $m$, s.t. for each $r,s\in\mathbb{N}$ and $\beta_{i,j}^k\in\mathbb{Z}$, there exists a connected compact $m$-manifold $M$ s.t. $H^2(M)\simeq \mathbb{Z}^r$, $H^4(M)\simeq \mathbb{Z}^{s}$ and $x_i\smile x_j=\sum \beta_{i,j}^k y_k$ for the cohomology generators? I would be grateful for some reference..

REPLY [29 votes]: You probably meant $\beta_{ij}^k=\beta_{ji}^k$ (the cup-product is commutative). With that condition, there is always a simply-connected compact $7$-manifold $M$ whose only non-trivial cohomology groups are $H^0(M)\cong\mathbb{Z}$, $H^2(M)\cong\mathbb{Z}^r$, and $H^4(M)\cong\mathbb{Z}^s$, and such that the cup-product is as you indicate. 
The manifold $M$ can be constructed from a CW-complex $X$ defined as the mapping cone of a certain map $\varphi\colon\vee_s S^3\rightarrow\vee_rS^2$. There is a subpolyhedron $K\subset\mathbb{R}^7$ of the same (even simple) homotopy type of $X$, see http://math.berkeley.edu/~stall/embkloz.pdf. Take $M$ to be the closure of a nice regular neighbourhood of $K\subset\mathbb{R}^7$. This $M$ is a compact simply-connected $7$-manifold (with boundary). The manifold $M$ is (simply) homotopy equivalent to $X$.
Let me now indicate how to define $\varphi$ to get the desired cohomology ring, see Baues' book on 4-dimensional complexes.
Recall that $\pi_3(\vee_rS^2)\cong\Gamma(\mathbb{Z}^r)$. Here $\Gamma$ is Whitehead's functor, defined for any abelian group $A$ by the existence of a universal quadratic map $\gamma\colon A\rightarrow\Gamma(A)$ (not a homomorphism), i.e. a map such that $\gamma(a)=\gamma(-a)$ and the cross effect map $A\times A\rightarrow \Gamma(A)\colon (a,b)\mapsto (a|b)=\gamma(a+b)-\gamma(a)-\gamma(b)$ is bilinear. The isomorphism $\pi_3(\vee_rS^2)\cong\Gamma(\mathbb{Z}^r)$ is defined by the map $\mathbb{Z}^r\cong \pi_2(\vee_rS^2)\rightarrow \pi_3(\vee_rS^2)\colon f\mapsto f\eta$ given by pre-composition with the Hopf map $\eta\colon S^3\rightarrow S^2$, which is universal with those properties. The group $\Gamma(\mathbb{Z}^r)$ is free abelian of rank $\binom{r+1}{2}$. If $\{e_1,\dots,e_r\}\subset \mathbb{Z}^r$ is a basis, then $\{\gamma(e_1),\dots,\gamma(e_r)\}\cup\{(e_i|e_j) \,;\, 1\leq i < j \leq r\}\subset \Gamma(\mathbb{Z}^r)$ is a basis.
There is a natural homomorphism $\tau\colon \Gamma(A)\rightarrow A\otimes A$ defined by the quadratic map $A\rightarrow A\otimes A\colon a\mapsto a\otimes a$. If $A=\mathbb{Z}^r$ then $\tau(\gamma(e_i))=e_i\otimes e_i$ and $\tau(e_i|e_j)=e_i\otimes e_j+e_j\otimes e_i$. 
A homotopy class $\varphi\colon\vee_s S^3\rightarrow\vee_rS^2$ is essentially the same as a homomorphism $f=\pi_3(\varphi)\colon \mathbb{Z}^s\rightarrow \Gamma(\mathbb{Z}^r)$. The cohomology of the mapping cone $X$ is obviously $H^0(X)\cong\mathbb{Z}$, $H^2(X)\cong\mathbb{Z}^r$, $H^4(X)\cong\mathbb{Z}^s$, and zero otherwise. The cup-product $H^2(X)\otimes H^2(X)\rightarrow H^4(X)$ turns out to be the $\mathbb{Z}$-linear dual of $\tau f\colon \mathbb{Z}^s\rightarrow \Gamma(\mathbb{Z}^r)\rightarrow \mathbb{Z}^r\otimes \mathbb{Z}^r$. Hence, if $\{e_1,\dots,e_r\}\subset \mathbb{Z}^r$ and $\{\bar e_1,\dots, \bar e_s\}\subset\mathbb{Z}^s$ are the canonical bases, given by the inclusions of the factors of the wedges, it is enough to define $f$, and hence $\varphi$, as $f(\bar e_k)=\sum_{i=1}^r  \beta_{ii}^k\gamma(e_i)+\sum_{1\leq i<k \leq r}\beta_{ij}^k(e_i|e_j)$.<|endoftext|>
TITLE:  comparing Hodge structures on cohomology of conjugate varieties
QUESTION [5 upvotes]: What can one say about the relation between the Hodge decompositions 
of $H^\*(X,C)$ and $H^{*}(X_\sigma,C)$
for a complex algebraic smooth projective variety $X$ and $\sigma$ an automorphism
of the field of complex numbers ? I am also interested in conjectural results, e.g.
whether Standard Conjectures imply something about the relationship of
the Hodge structure of conjugate varieties $X$ and $X_\sigma$.
For example, it appears that the Hodge numbers $h^{ab}=dim_C H^a(X,\Omega_X^b)$
and $h_\sigma ^{ab}=dim_C H^a(X_\sigma,\Omega_{X_\sigma}^b)$ are necessarily
the same (being defined via de Rham cohomology); is this right (cf. 
discussion in the comments)? 
what other invariants/subspaces etc are the same or known to be possibly different ? 
Can one define a related Galois (or $Aut(C/Q)$) action on the Hodge structures,
something like where an automorphism $\sigma\in Aut(C/Q)$ takes the Hodge structure of $X$ 
into that of $X_\sigma$'s ? 
I know of several references that show that conjugate varieties
may be rather different topologically. F.Charles, 
Conjugate varieties with distinct real cohomology algebras
lists most references I know of and well, proves what's in the title.
Serre constructed two two non-homotopic conjugated varieties with different fundamental groups;
his proof exploits the fact that for a CM elliptic curve, its fundamental group has different
$EndE$-module structures for different embeddings. Abelson constructs two non-homotopic conjugated varieties with the same finite fundamental groups.

REPLY [8 votes]: As Jason & Will  have already commented, the Hodge numbers are the same for
conjugate pairs because of GAGA and the fact algebraic coherent cohomology behaves well with respect to field extensions. In fact, there would be an isomorphism of filtered vector spaces
$$(\mathbb{H}^i(X,\Omega^\bullet), F^\bullet)\otimes_\mathbb{C} \mathbb{C}_\sigma
\cong (\mathbb{H}^i(X_\sigma,\Omega^\bullet), F^\bullet)$$
Nevertheless,  the Hodge structures need not be the same because this may not be compatible with integral structures. 
For example, if $X$ is an elliptic curve with transcendental $j$-invariant, we
can find an automorphism with $j(X)\not= j(X_\sigma)$ so that $H^1(X)\not\cong H^1(X_\sigma)$ as Hodge structures. (So assuming I've understood your last question, that would be a no.)<|endoftext|>
TITLE: Open problems in sub-Riemannian geometry
QUESTION [7 upvotes]: What are some open problems in sub-Riemannian geometry?
I am interested especially in problems concerning connections and curvature, but any contribution is welcomed.

REPLY [5 votes]: Andrei Agrachev recently posted an article on the arXiv:  http://arxiv.org/abs/1304.2590 entitled "Some open problems" that you might find interesting.  He discusses several  open problems in control theory and sub-Riemannian geometry that are of considerable interest.<|endoftext|>
TITLE: Under exactly what (extra) conditions (if any) is a connected Hausdorff manifold with a Riemannian metric a metric space?
QUESTION [20 upvotes]: The setting is that manifolds are Banach manifolds, not necessarily finite dimensional. No other assumption is made about the topology of the manifold. In particular, it is not assumed to be regular or normal. Of course, this means that the manifold is not assumed to be paracompact. There is no assumption made about the manifold admitting partitions of unity. In this setting, both Lang (Fundamentals of Differential Geometry, 1999, Springer-Verlag) and Abraham, Marsden, and Ratiu (Manifolds, Tensor Analysis, and Applications, 1988, Springer-Verlag) prove that a connected Hausdorff manifold with a Riemannian metric is a metric space. Both proofs, it seems to me, suffer from the same flaw. The proofs are similar enough that I'll just refer to Lang's. We start with a connected Hausdorff manifold $X$ and a Riemannian metric, $g$ on $X$. No special assumptions are made about the Hilbert space, $E$, on which the manifold is modeled. For example, $E$ may or may not be separable. Start by defining a length function, $L_g$ which assigns a real number $L_g(\gamma)$ to each piecewise $C^1$ path, $\gamma$, from $J=[a,b]$ into $X$, based on the metric, $g$. The distance function, $d_g:X\times X\to\mathbb{R}$ is then defined by $d_g(x,y)=\inf\{L_g(\gamma)\}$ over all piecewise $C^1$ paths, $\gamma$, defined on $J$ with $\gamma(a)=x$ and $\gamma(b)=y$.
Without any difficulty, $d_g$ is a pseudo-metric. The first main point of the proof is to show that $d_g$ is actually a metric. So we start with distinct points
$x$ and $y$ of $X$ and set out to show $d_g(x,y) > 0$. We have a chart $(U, \phi)$
at $x$ for $X$ with $\phi(U)$ open in $E$, and we can arrange $U$ to be small enough that $y$ is not in $U$,
since the manifold is assumed to be Hausdorff. Working in $\phi(U)$ we find an
$r>0$ such that the closed ball $D(\phi(x),r)$ is contained in $\phi(U)$ and such
that certain other properties hold. Let $S(\phi(x),r)$ be the boundary of
$D(\phi(x),r)$. Then we define $D(x,r)=\phi^{-1}(D(\phi(x),r))$ and
$S(x,r)=\phi^{-1}(S(x,r))$, both subsets of $U$.
Since $\phi$ is a homeomorphism, $D(x,r)$ and $S(x,r)$ are closed in $U$ (not
necessarily closed in $X$). To me, this is a key stumbling point, as I'll
explain. We next let $\gamma:J \to X$ be any piecewise $C^1$ path in $X$ from
$x$ to $y$. Both proofs make the following assumption: since $x$ is in $D(x,r)$
and since $y$ is not in $U$, the path $\gamma$ must cross $S(x,r)$. Neither author
explicitly proves this assumption (and AMR doesn't even state it).
When I set out to prove this, using the continuity of $\gamma$ and the
connectedness of $J$, I quickly run into the need to show that $D(x,r)$
is closed in $X$, not just in $U$. If $X$ were known to be regular, it would
not be a problem to take $r$ small enough that $D(x,r)$ was closed in $X$.
But as I mentioned at the beginning, I don't know that $X$ is regular.
If I could show that the pseudo-metric topology for $X$ induced by $d_g$
was the same as the original manifold topology, I would also get that
$X$ was regular. But I don't see how to do that without first completing
the first part of the proof.
The whole question seems to be, can I make $r$ small enough that $D(x,r)$ stays
away from the topological (in the original manifold topology of $X$) boundary
of $U$? But this does not seem to be a local issue, since it depends on what is
closed in $X$ which in turn, depends on what is open everywhere in $X$ including
outside of $U$.
So, the question is, are other assumptions necessary, or is it possible to fix the proof so that no other assumptions need be made?

REPLY [11 votes]: There are infinite dimensional weak Riemannian manifolds with vanishing geodesic distance
(in the sense as defined in the question). These are modeled on nuclear Frechet spaces, but the results extend to Sobolev completions of high enough order ($>\dim/2 +2$). They are still weak Riemannian manifolds (i.e., the Riemann metric does not generate the topology on the tangent spaces).
The first example was the $L^2$ metric on $\text{Emb}(S^1,\mathbb R^2)/\text{Diff}(S^1)$, as shown in the first paper below. Then it turned out that the right invariant $L^2$-metric
on each full diffeomorphism group also has this property, also Sobolev metrics for 
Sobolev order $<1/2$ ($\le 1/2$ on $\text{Diff}(S^1)$).
In particular, Burgers' equation and KdV are nonlinear PDE's corresponding to geodesic equations for metrics with vanishing geodesic distance.
All the papers are in the arXiv or on my homepage.  

Peter W. Michor; David Mumford: Riemannian geometries on spaces of plane curves. J. Eur. Math. Soc. (JEMS) 8 (2006), 1-48.
Peter W. Michor; David Mumford. Vanishing geodesic distance on spaces of submanifolds and diffeomorphisms. Documenta Math. 10 (2005), 217--245 (written later)  
Martin Bauer, Martins Bruveris, Philipp Harms, Peter W. Michor: Vanishing geodesic distance for the Riemannian metric with geodesic equation the KdV-equation. Ann. Glob. Anal. Geom. 41, 4 (2012) 461-472.
Martin Bauer, Martins Bruveris, Philipp Harms, Peter W. Michor: Geodesic distance for right invariant Sobolev metrics of fractional order on the diffeomorphism group. Ann. Glob. Anal. Geom. 44, 1 (2013), 5-21. 
Martin Bauer, Martins Bruveris, Peter W. Michor: Geodesic distance for right invariant Sobolev metrics of fractional order on the diffeomorphism group. II. 7 pages. To appear in: Ann. Glob. Anal. Geom.


Edit.
  Some more remarks: 

If you have a strong Riemannian metric (such that $g_x$ induces the topology on $T_xM$ for each $x$), then you have a Hilbert manifold, the Riemannian exponential mapping is a local diffeomorphism, and by the Gauss lemma geodesic distance describes the manifold topology.
If the Riemannian metric is weak (so $g_x: T_xM\to T_x^*M$ is injective only), Then:
 
(1) one has to prove that the connection exists and is smooth.

(2) Even for a Banach manifold where the Riemannanian exponential mapping is automatically a local diffeomorphism, the Gauss lemma is not true in general, since the exponential mapping is a diffeomorphism on a neighborhood of $0\in T_xM$, but this need not be a $\|\cdot\|_{g_x}$-ball.
 
This is what happens all the examples described in the papers above. 
The first paper has an example (concentric spheres, towards the end) of an incomplete geodesic where the conjugate points are dense.
We believe, that vanishing geodesic distance is tied to the fact, that sectional curvature is locally positive unbounded: behind mountain you always find a shorter geodesic.
For weak Riemannian metrics of low Sobolev order we do not know whether the geodesic equation is well posed. Of course KdV and Burgers are, but Burgers' close relative, the $L^2$-metric on $\text{Imm}(S^1,\mathbb R^2)$ or $\text{Imm}(S^1,\mathbb R^2)/\text{Diff}(S^1)$, are not known to have well posed geodesic equation. 

In the paper

arXiv:1202.5122 Right-invariant Sobolev metrics ${H}^{s}$ on the diffeomorphisms group of the circle. Joachim Escher (IFAM), Boris Kolev (LATP),

the geodesic equation for the Sobolev $1/2$-metric on $\text{Diff}(S^1)$ is shown to be well posed, but the third paper above shows that it has vanishing geodesic distance. 
The OP question stats from a (Banach)-manifolds and asks for Riemannian metric on it.
But even if you find one, there are many more obstacles until one ends up with a metric space described by geodesic distance. 
In my view, the metric (or even the geodesic equation) is more important than the manifold, which you can adapt to the metric somewhat.<|endoftext|>
TITLE: Exponentiating 4 by 4 matrix analytically
QUESTION [6 upvotes]: Does there exist an analytical method by which i can exponentiate a 4 by 4 matrix, in the same way as the general 2 by 2 matrix case in pauli matrix basis. I have dirac matrices (which are composed of direct products of pauli matrices) as my basis for 4 by 4 matrices. 
I need an analytical way !
Any reply is appreciated.
regards

REPLY [8 votes]: Perhaps I misunderstand the question.  When you say you have Dirac matrices, does that mean that you are computing the exponential of a liner combination of Dirac matrices?  If so, then there is a very simple analytical formula in any dimension: just use the Clifford relations in the exponential series.
More concretely, suppose that you would like to compute the exponential of a matrix $X := \sum_i x^i \Gamma_i$, where the Dirac matrices $\Gamma_i$ obey the Clifford relation
$$
\Gamma_i \Gamma_j + \Gamma_j \Gamma_i = - 2 g_{ij} I~,
$$
with $I$ the identity matrix.  Then it follows from this relation that
$$
X^2 = - x^2 I~,
$$
where I have introduced the (indefinite, if $g_{ij}$ has indefinite signature) "squared norm"
$$
x^2 = \sum_{i,j} x^i x^j g_{ij}~.
$$
If $x^2$ = 0, then
$$
\exp X = I + X
$$
and if $x^2 \neq 0$, then letting $x = \sqrt{x^2}$ (which could be imaginary),
$$
\exp X = \cos x I + \frac{\sin x}{x} X~.
$$
Added (for the "heathens")
Quiaochu's comment is correct.  Here are some more details.
Let $V$ be a finite-dimensional real vector space with a non-degenerate inner product $\left<-,-\right>$.  Let $Cl(V)$ be the corresponding Clifford algebra.  Let $\rho: Cl(V) \to \operatorname{End}(M)$ be an irreducible representation of $Cl(V)$.  Let $(e_i)$ be a basis for $V$.  Then $\Gamma_i := \rho(e_i)$ are called Dirac matrices of $CL(V)$ in the representation $M$.<|endoftext|>
TITLE: Vanishing of $\mathrm{Ext}^i_R(N, R)$
QUESTION [12 upvotes]: $\DeclareMathOperator\Ext{Ext}$This question is related to Are the supports of $Ext^i(M,N)$ eventually periodic? .
Let $(R, \mathfrak{m}, k)$ be a Noetherian local ring of dimension $d$. It well known that
Theorem $R$ is Gorenstein iff $\Ext^i_R(k, R) = 0$ for some $i > d$, and iff $\Ext^i_R(k, R) = 0$ for all $i > d$.
By above Theorem we have: $\Ext^i_R(k, R) = 0$ for some $i > d$ iff $\Ext^i_R(k, R) = 0$ for all $i > d$.
The following question consider in a generalization of this situation for modules of finite length.
Question: Let $(R, \mathfrak{m}, k)$ be a Noetherian local ring of dimension $d$, $N$ an $R$-module of finite length. Is it true that: $\Ext^i_R(N, R) = 0$ for some $i > d$ iff $\Ext^i_R(N, R) = 0$ for all $i > d$?

REPLY [5 votes]: No.
Let's work with an Artinian local ring $R$.  Let $X$ be a module satisfying $\mathrm{Ext}_R^i(X,R)\neq 0$ for all $i$.  (These are plentiful; for example, assume $R$ is non-Gorenstein and let $X$ be the residue field.)  Define  a module $Q$ by "Serre's trick": take generators $\chi_1, \dots, \chi_r$ for $\mathrm{Ext}_R^1(X,R)$ and consider the short exact sequence corresponding to $(\chi_1, \dots, \chi_r) \in \mathrm{Ext}_R^1(X,R^r)$:
$$0 \to R^r \to Q \to X \to 0$$
Then $\mathrm{Ext}_R^1(Q,R) =0$, since the long exact sequence of $\mathrm{Ext}$ looks like
$$\cdots \to \mathrm{Hom}_R(R^r,R) \to \mathrm{Ext}_R^1(X,R) \to \mathrm{Ext}_R^1(Q,R) \to \mathrm{Ext}_R^1(R^r,R)=0$$
and the map $\mathrm{Hom}_R(R^r,R) \to \mathrm{Ext}_R^1(X,R)$ is cooked up precisely to be surjective.  On the other hand, $\mathrm{Ext}_R^i(Q,R) = \mathrm{Ext}_R^{i}(X,R) \neq 0$ for all $i\geq 2$.
One can jazz this up a bit, using a result of Jorgensen–Şega (journal link): There exists a local Artinian ring $R$ and a family $\{M_s\}_{s\geq 1}$ of reflexive $R$-modules such that (among other things) $\mathrm{Ext}_R^i(M_s,R)\neq 0$ if and only if $1 \leq i \leq s-1$.  They give a completely explicit construction of $R$ and the modules $M_s$.  Taking one of the $M_s$ in place of $X$ above, one obtains modules $Q_{a,b}$ for which $\mathrm{Ext}_R^i(Q_{a,b},R)$ vanishes up to $i=a-1$, is nonzero for $a \leq i \leq b$, and vanishes again for $i \geq b+1$.  Then taking direct sums gives essentially arbitrary behavior of vanishing and non-vanishing.<|endoftext|>
TITLE: What is known about the polynomial factorization of power series?
QUESTION [12 upvotes]: Some power series factorize;  $1+\sum_{n=1}^\infty x^n=\prod_{n=1}^\infty (1+x^{2^n})$ and $1+\sum_{n=1}^\infty x^{2n}/(2n+1)!=\prod_{x=1}^\infty (1+x^2/n^2\pi^2)$ for example; while others do not----in particular, $1+\sum_{n=1}^\infty x^n/n!$. Specifically the question is: what is known about necessary and sufficient conditions on the coefficients of the power series $1+\sum_{n=1}^\infty a_nx^n$ (assuming a positive radius of convergence) for it to factorize as $\prod_{n=1}^\infty (1+ b_nx+c_nx^2)$, where the coefficients $a_n$, $b_n$, and $c_n$ are real constants?
Remarks  The Weierstrass factorization theorem doesn't directly answer the question, since here we allow Taylor series of non-entire functions (e.g. the first example above) and exclude non-polynomial factors. The roots of the question are algebraic: loosely stated, under what conditions can the fundamental theorem of algebra be pushed to infinity? A positive radius of convergence is supposed because formal power series with no functional meaning make me uncomfortable, and admitting them might complicate the answer. I previously posted an unanswered version of this question on Math.StackExchange.
Edit  Thanks to juan and Douglas Zare for showing my error in the preamble of the question by citing the interesting factorization of the exponential series by Gingold et al. As Aaron Meyerowitz indicates, restriction to linear and quadratic factors leads respectively to quite different situations. Alexandre Eremenko answered the linear case. Allowing polynomial factors of unrestricted degree opens a wide vista. The rather simple power series $1$ includes among its polynomial product representations, valid for $|x| \lt r_0$, such forms as
$$\left(1-\dfrac{p(x)}{r} \right) \prod_{n=0}^\infty \left[1+\left(\dfrac{p(x)}{r} \right)^{2^n} \right],$$
where $r_0$ is an arbitrary positive constant, $p(x)$ is an arbitrary polynomial, and $r$ is any real constant strictly exceeding $|p(x)|$ whenever $|x|\leqslant r_0$. Myriad expansions like these may be woven into any polynomial product expansion of any power series. I find this vista daunting and so will stick to the specified quadratic case, which remains unanswered so far.

REPLY [3 votes]: Here is an observation which shows the delicacy of factorization: Your second product $$\prod_{n=1}^\infty (1+x^2/n^2\pi^2)=\sum_{n=0}^\infty x^{2n}/(2n+1)!=\frac{\sinh{x}}{x}$$ is absolutely convergent but is only conditionally convergent if it is written as $$\prod_{n=1}^\infty (1+ix/n\pi)(1-ix/n\pi).$$ So an order dependent factorization does not sound like the fundamental theorem of algebra.
We can leave complex numbers out entirely by shifting to a famous product of Euler (which is not an Euler product.) 
$$\prod_{n=1}^\infty (1-x^2/n^2\pi^2)=\sin{x}/x=\sum_{n=0}^\infty (-x^2)^{n}/(2n+1)!$$ Of course this was not rigorously justified at the time and I think Euler was perhaps content to leave it as $$\prod_{n=1}^\infty (1+x/n\pi)(1-x/n\pi).$$ The heuristic justification (one treatment is page 3 of this article) was that the product gives a function $f(x)$ with $f(0)=1$ and a single root of $f(n\pi)=0$ for non-zero integers $n$. Hence it should be $\sin{x}/x$ for which we do know the power series. 
In volume I of Polya's Mathematics and Plausible Reasoning it is related that the motivation was to find the exact (and unknown) value of $\sum 1/n^2$ which does converge to about $1.644934.$ Of course comparing the coefficient of $x^2$ gives $\sum 1/n^2=\pi^2/6.$ The conclusion is "(extremely) daring", says Polya,   but it does agree with the known value and the method yields other similarly valid looking results, so it seems acceptable.<|endoftext|>
TITLE: seeking an integer parameterization for A^2+B^2=C^2+D^2+1
QUESTION [6 upvotes]: I'm looking for a complete [integer] parameterization of all integer solutions to the Diophantine equation
$A^2+B^2=C^2+D^2+1$,
analogous to the classical parameterization of the Pythagorean equation, i.e.
$A^2+B^2=C^2 \implies t,m,n \text{ such that } (A,B,C)=t(m^2-n^2,2mn,m^2+n^2)$.
Dickson's History contains many references and examples, but most appear to be inadequate, incomplete, or simply incorrect. Barnett and Bradley independently reached almost the same parameterization of the more general equation
$A^2+B^2+C^2=D^2+E^2+F^2$,
but I have so far been unable to reduce their parameterization(s) to one which solves the first equation I posted.
Any help or further references would be greatly appreciated.
Thanks!
Kieren.

REPLY [5 votes]: Such a parametrization is not possible.
Proof. Suppose A,B,C,D are polynomials with integer coefficients (in any number of variables) and A^2+B^2 = C^2+D^2+1. 
Then we have a parametrization for the congruence subgroup  H of SL(2,Z) reducing to
the permutation matrices modulo 2 (see Noam D. Elkies Aug 13 '12 at 15:49 above).
This H appears in Example 14 in my Annals paper. The first rows in H are (A+C, B+D).
Modulo 2, (A+C)(B+D) = 0, hence either  A+C or C+D is always even. But those rows contain both (1,0) and (0,1). This is a contradiction. QED.<|endoftext|>
TITLE: Example of non-projective variety with non-semisimple Frobenius action on etale cohomology?
QUESTION [25 upvotes]: This question was motivated by a more general question raised by Jan Weidner here.  In general one starts with a variety $X$ (say smooth) over an algebraic closure of a finite field $\mathbb{F}_q$ of characteristic $p$.   Here there is a natural action of a Frobenius morphism $F$ relative to $q$.   Given a distinct prime $\ell$, there is an induced operation of $F$ on etale cohomology groups (with compact support) $H^i_c(X, \overline{\mathbb{Q}_p})$.   When $X$ is projective, this action is conjectured to be semisimple on each of the finite dimensional vector spaces involved.   But it seems that semisimplicity can fail when $X$ isn't projective.   My basic question is:

Is there an elementary example where the Frobenius action fails to be semisimple?  (References?)

Of course, etale cohomology developed in response to the Weil conjectures and related matters in number theory.   Here there is a lot of deep literature which 
I'm unfamiliar with, but I'd like to get some insight into the narrow question of what does or doesn't force semisimplicity for non-projective varieties.
My interest lies mainly in Deligne-Lusztig varieties and their role in studying characters of finite groups of Lie type.   Such varieties $X_w$ are indexed by Weyl group elements and are locally closed smooth subvarieties of the flag variety for a reductive group $G$, with all irreducible components of equal dimension.   Here the finite subgroup $G^F$ acts on the etale cohomology, commuting with $F$, and the resulting virtual characters (alternating sums of characters on cohomology spaces) are the D-L characters.   
Characters of finite tori also come into play here, but I'm thinking first about the trivial characters of tori which lead to "unipotent" characters.   These include essential but mysterious "cuspidal" unipotent characters which can't
be extracted from the usual induced characters obtained by parabolic induction.   
For example, the Chevalley group $G_2(\mathbb{F}_q)$ typically has 10 unipotent characters (at the extremes the trivial and the Steinberg characters), with four being cuspidal.    Those four appear in etale cohomology groups of a variety $X_w$ with $w$ a Coxeter element: the variety has dimension 2, with four characters (three cuspidal, the other Steinberg) in degree 2, one (cuspidal) in degree 3, and one (the trivial character) in degree 4.  Miraculously, it always happens in the Coxeter case that $F$ acts semisimply (here with 6 distinct eigenvalues: the Coxeter number) and its eigenspaces afford distinct irreducible characters.   In the year after he and Deligne finished their fundamental paper (Annals, 1976), Lusztig worked out the Coxeter case in a deep technical paper here.   This was followed by a more complete determination of cuspidal unipotent characters, and then much more.   The Coxeter case seems to be unusually well-behaved in this program.
P.S. As I suspected, there's more going on under the surface of my basic question about semisimplicity than meets the eye.   As an outsider to much of the algebraic geometry framework I can appreciate the outline of Dustin's answer though not yet the details.   My question came from wondering whether there are ways to shortcut some of the older steps taken by Lusztig, but the wider questions here are obviously important.   I'll have to see how far my motivation (in a manner of speaking) takes me.
And Wilberd: thanks for the proofreading, which is not one of my favorite things to do.  (Though I somehow got "just bonce" into a book that was supposedly proofread.)

REPLY [27 votes]: Actually, the semisimplicity should hold with no hypotheses on X, so no example should exist.  In fact it is generally expected that, with char. 0 coefficients and over a finite field (both hypotheses being necessary), every mixed motive is a direct sum of pure motives -- so the question for arbitrary varieties reduces to that for smooth projective ones.
The reason is as follows:  the different weight-pieces have no frobenius eigenvalues in common (by the Weil conjectures), so the weight filtration can be split by a simple matter of linear algebra.  (And the splitting will even be motivic since frobenius is a map of varieties.)
Edit: In response to Jim's comment, let me try to provide a clearer argument (2nd edit: no longer using the Tate conjecture).  I claim that if we assume the existence of a motivic t-structure over F_q w.r.t. the l-adic realization in the sense of Beilinson's article http://arxiv.org/pdf/1006.1116v2.pdf, then provided that H^i_c(X-bar) is Frobenius-semisimple for smooth projective X, it is in fact so for aribtrary X.
Indeed, given a motivic t-structure, its heart is an artinian abelian category where every irreducible object is a summand of a Tate-twist of an H^i(X) for X smooth an projective, and furthermore there are no extensions between such irreducibles of the same weight (this is all in Beilinson's article).
That's all true over a general field.  But now let's argue that, in the case of a finite field, there also can't be extensions between such irreducibles of different weights; then in the motivic category all of our H^i_c(X-bar) of interest will be direct sums of summands of H^i(X)'s, and we'll have successfully made the reduction to the smooth projective case.
So suppose M and N are irreducible motives of distinct weights over F_q, and say E is an extension of M by N.  Consider the characteristic polynomials p_M and p_N of Frobenius acting on the l-adic cohomologies of M and N.  By Deligne, they have rational coefficients and distinct eigenvalues, so we can solve q * p_N == 1 (mod p_M) for a rational-coefficient polynomial q.  But then (q*p_N)(frobenius) acting on E splits the extension (recall from Beilinson's article that the l-adic realization is faithful under our hypothesis), and we're done.

Later commentary: apparently, when I wrote this I was a little too excited about the perspectives offered by motives.  I should emphasize the point essentially made by Minhyong Kim, that the reduction from the general case to the proper smooth case likely doesn't require any motivic technology, and should even be independent of any conjectures.  One just needs to know that there's a weight filtration on l-adic cohomology of the standard type where the pure pieces are direct sums of direct summands of appropriate cohomology of smooth projective varieties.  As Minhyong says, this probably follows from Deligne's original pure --> mixed argument, via use of compactifications and de Jong alterations.  Or at least that's what it seems to me without having gone into the details.  I'm sure someone else knows better.<|endoftext|>
TITLE: Normal form for trace-free real cubic forms in 3 variables under SO(3)-action?
QUESTION [5 upvotes]: I'm looking at irreducible, real representations of $SO(3)$.  The 5-dimensional irrep is isomorphic to the space of trace-free quadratic forms on $\mathbb{R}^3$, and we all know that any such quadratic form can be diagonalized by the standard $SO(3)$ action on $\mathbb{R}^3$.  My question is: is there any analogous "normal form" result for the 7-dimensional irrep, regarded as the space of trace-free cubic forms on $\mathbb{R}^3$, under the standard action of $SO(3)$?
More generally, I'd like to better understand the geometry of the orbits of these actions.  Can anyone recommend a reference?
-Jeanne

REPLY [6 votes]: One way to do this is to reduce this to understanding sets of points on the $2$-sphere up to a certain equivalence.  If you let $\mathcal{H}_k$ denote the homogeneous polynomials $p$ of degree $k$ on $\mathbb{R}^3$ that are harmonic, i.e., that satisfy $\Delta p = 0$, then $\mathcal{H}_k$ is an irreducible $\mathrm{SO}(3)$-representation of dimension $2k{+}1$.  The space $\mathcal{P}_k$ of all homogeneous polynomials of degree $k$ can be written as the direct sum
$$
\mathcal{P}_k = \mathcal{H}_k \oplus (R\cdot\mathcal{H}_{k-2})
\oplus (R^2\cdot\mathcal{H}_{k-4}) \oplus \cdots 
$$
where $R = {x_1}^2+{x_2}^2+{x_3}^2$.  In particular, there is a canonical 'harmonic part' $H(p)\in \mathcal{H}_k$ for any $p\in \mathcal{P}_k$, and the map $H:\mathcal{P}_k\to \mathcal{H}_k$ is linear.  Then one has the following `factorization result':
Proposition:  Any $p\in \mathcal{H}_k$ can be 'factored' in the form
$p = H(\ell_1\ell_2\cdots\ell_k)$,
where the 'factors' $\ell_i\in \mathcal{H}_1$ satisfy $|\ell_1| = |\ell_2| = \cdots = |\ell_k|$ and the elements $\ell_i$ are unique up to permutation and replacement of an even number of them by their negatives.
This 'harmonic factorization' is $\mathrm{SO}(3)$-equivariant, so finding normal forms for harmonic polynomials of a fixed degree $k$ reduces to normalizing sets of $k$ points on the $2$-sphere up to permutation and replacing an even number of them by their negatives, which is a combinatorial problem.
The proof of the above proposition goes through the standard representation theory of $\mathrm{SU}(2)$, which is the double cover of $\mathrm{SO}(3)$.<|endoftext|>
TITLE: Computing determinants of matrices of linear forms
QUESTION [8 upvotes]: Suppose we have three $n \times n$ matrices $A$, $B$, $C$ with floating point entries. We would like to compute the polynomial $\det (xA+yB+zC)$. At least in Mathematica, and I think in all computer algebra systems, this will take $n!$ steps; Mathematica chokes around $n=15$. 
What's the smart way to do this? I have some ideas, but they are all complicated enough that I don't want to implement them before hearing from others.
This question is on the border between MO, cstheory and Mathematica, but I suspect that this is the right place to start.
ADDED IN RESPONSE TO QUESTIONS BELOW The trouble with Gaussian elimination is that you have to divide by polynomials in $(x,y,z)$, and the expressions soon get huge. Interpolation, probably at roots of unity so that the interpolation matrix will be unitary, is my best idea, but I wanted to see if there was a better one before I tried it.

REPLY [3 votes]: To just add to the answers already here, it is possible to extract the individual coefficients of this polynomial exactly using the following method. Let
\begin{equation}
h(x) = \sum_{j = 0}^d c_j x^j
\end{equation}
be a polynomial of degree $d$, then
\begin{equation}
c_j = \frac{1}{d + 1} \sum_{k = 0}^d e^{-\frac{2 \pi i j k}{d + 1}} h(e^{\frac{2 \pi i k}{d + 1}})
\end{equation}
since
\begin{align}
\frac{1}{d + 1} \sum_{k = 0}^d e^{-\frac{2 \pi i j k}{d + 1}} h(e^{\frac{2 \pi i k}{d + 1}}) &= \frac{1}{d + 1} \sum_{k = 0}^d e^{-\frac{2 \pi i j k}{d + 1}} \left(\sum_{l = 0}^d c_l e^{\frac{2 \pi i k l}{d + 1}} \right) \\
&= \sum_{l = 0}^d \left(\frac{1}{d + 1} \sum_{k = 0}^d e^{\frac{2 \pi i (l - j) k}{d + 1}} \right) c_l \\
&= \sum_{l = 0}^d \delta_{jl} c_l \\
\frac{1}{d + 1} \sum_{k = 0}^d e^{-\frac{2 \pi i j k}{d + 1}} h(e^{\frac{2 \pi i k}{d + 1}}) &= c_j
\end{align}
where $\delta_{jl}$ is the Kronecker delta and $i$ is the imaginary unit. Letting 
\begin{equation}
f(x, y, z) = \det(x A + y B + z C)
\end{equation}
we can expand $f$ in powers of $x, y, z$ using multilinearity of the rows or columns of the matrix inside the determinant as
\begin{equation}
f(x, y, z) = \sum_{j = 0}^n x^j g_j(y, z) = \sum_{j = 0}^n x^j \left(\sum_{k = 0}^{n - j} a_{j, k} y^k z^{n - j - k}\right)
\end{equation}
Note that, since one choice of $x, y, z$ must be made for every row or column in the expansion, the powers of these variables must sum to $n$ in every term. This polynomial is completely specified by the coefficients $\{a_{j, k}\}$, and to extract them, it is sufficient to only evaluate $f$ at $z = 1$
\begin{equation}
f(x, y, 1) = \sum_{j = 0}^n x^j g_j(y, 1) = \sum_{j = 0}^n \sum_{k = 0}^{n - j} a_{j, k} x^j y^k
\end{equation}
One query of the polynomial 
\begin{equation}
g_j(y, 1) = \sum_{k = 0}^{n - j} a_{j, k} y^k
\end{equation}
is given by the above formula
\begin{equation}
g_j(y, 1) = \frac{1}{n + 1} \sum_{p = 0}^n e^{-\frac{2 \pi i j p}{n + 1}} f(e^{\frac{2 \pi i p}{n + 1}}, y, 1)
\end{equation}
Similarly,
\begin{equation}
a_{j,k} = \frac{1}{n - j + 1} \sum_{q = 0}^{n - j} e^{-\frac{2 \pi i k q}{n - j + 1}} g_j(e^{\frac{2 \pi i q}{n - j + 1}}, 1)
\end{equation}
Therefore
\begin{equation}
\det(x A + y B + z C) = \sum_{j = 0}^n \sum_{k = 0}^{n - j} a_{j,k} x^j y^k z^{n - j - k}
\end{equation}
where
\begin{equation}\
a_{j, k} = \frac{1}{(n + 1)(n - j + 1)}\sum_{p = 0}^{n} \sum_{q = 0}^{n - j} e^{-2 \pi i \left(\frac{jp}{n + 1} + \frac{kq}{n - j + 1} \right)} \det(e^{\frac{2 \pi i p}{n + 1}} A + e^{\frac{2 \pi i q}{n - j + 1}} B + C)
\end{equation}
requiring the evaluation of only $O(n^4)$ determinants to completely characterize the polynomial.<|endoftext|>
TITLE: Floer homology and Invariants for Einstein Field Equations?
QUESTION [10 upvotes]: Motivation: There have been the instanton (anti-self dual connection) solutions to the Yang-Mills equation $d_A^\ast F_A=0$ which extremize the YM energy $\int_M|F_A|^2$, leading to the Donaldson invariants and even a Floer homology. There have been the monopole (connection + spinor) solutions to the Seiberg-Witten equations $D_A\psi=0$ and $F_A^+=\psi\otimes\psi^\ast-\frac{1}{2}|\psi|^2$ which extremize the Chern-Simons-Dirac functional, leading to the SW invariants and a nice Floer homology. These utilize the fundamental particles in the Standard-Model of physics... but not of General Relativity, where the gravitons arise.
So I would be interested in a Floer homology and/or invariants arising from gravitational instantons (Riemannian metrics), i.e. solutions to the Einstein Field Equations $R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R-g_{\mu\nu}\Lambda=0$ in vacuum (no stress-energy term $T$). Here $\Lambda$ is the "cosmological constant", which we may or may not want to assume is zero. Surely these have been studied extensively. (This term 'gravitational instanton' is used first (I think) in Stephen Hawking's seminal 1977 paper "Gravitational Instantons", and basic examples are the Schwarzschild and Taub-NUT metrics.)
Should I expect something to arise? Are there immediate obstacles? Otherwise this would have been done by now, right?
Downfall?: Perhaps the moduli space is too big, or boring, or unknown.
Progress?: Witten has even shown that (2+1)-dimensional gravity (with no cosmological constant) on $M=\Sigma\times \mathbb{R}$ (compact surface $\Sigma$) is an ISO(2,1) Chern-Simons theory, i.e. the equations of motion of the CS-action are precisely the field equations. And if there is a cosmological constant $\Lambda\ne 0$, then the same result holds when we replace the gauge group ISO(2,1) by SO(3,1) or SO(2,2) depending on the sign of $\Lambda$.
More: There is something to be said from Witten's recent paper "Analytic Continuation Of Chern-Simons Theory", but I am not ready to understand it.

REPLY [11 votes]: Let me clarify a couple of issues from the previous answers/comments:
1) The linearization of $Rc-\tfrac{1}{2}Rg$ has mixed signs, and for this reason the equation $\partial_t g=-(Rc-\tfrac{1}{2}Rg)$ is bad, a kind of coupled backwards/forwards heat equation, i.e. no short time existence.
2) The linearization of $Rc$ is elliptic however (after fixing the gauge), and the Ricci flow $\partial_tg=-2Rc$ indeed is a good equation. In terms of functionals, it's better to consider the Perelman functional (whose gradient is $Rc$ up to gauge) instead of the Einstein-Hilbert 
functional (whose gradient is $Rc-\tfrac{1}{2}Rg$).
3) There was the issue of 1st order vs. 2nd order: While the YM-equation ($D^*F=0$) is second order in $A$, there is also the 1st order equation $F^+=0$ (antiselfdual connections). Solutions of this 1st order equation are special solutions of the YM-equation (as follows immediately from the Bianchi identity). There is a similar story for Ricci-flat metrics: Here the "special" solutions are the ones with special holonomy. Having holonomy $SU,Sp,G_2$ or $Spin_7$ implies that the metric is Ricci-flat. E.g. by solving the first order system $d\psi=0,d*\psi=0$ for a 3-form $\psi$ on a 7-manifold you get Ricci-flat metrics with holonomy $G_2$. Indeed, there is a proposal by Simon Donaldson for a higher order gauge theory based on $G_2$ (though, instead of just counting $G_2$-structures, the idea there is actually to "count" the number of associative submanifolds...)<|endoftext|>
TITLE: Topological spaces determined by generalized metric spaces
QUESTION [12 upvotes]: At http://en.wikibooks.org/wiki/Real_Analysis/Metric_Spaces you can find the standard definition of a metric space: a set $X$ given with a function $d:X\times X\to\mathbb{R}$ that satisfies properties 1 through 4. Later on the page it defines open ball and open set, and proves that arbitrary unions and finite intersections of open sets are open. (The page has a few mistakes; in particular, in the second paragraph of the proof, the unions should be intersections.) In other words, the open sets form a topology.
What I find remarkable is that none of properties 1 through 4 are needed for this proof. So consider a set $X$ and an arbitrary function $d:X\times X\to\mathbb{R}$. We can define open balls and open sets using the same definitions, producing a topology on $X$. Such topologies can be very different from those that arise from true metrics; for example, if $d$ is identically 0 we get the indiscrete topology.
Can we prove anything interesting about which topologies can arise this way? In particular, does every finite topology arise this way?

REPLY [12 votes]: Some partial answers to OP´s original question:
1) Any first-countable space (and in particular any finite topology) can be obtained in this way.
Let $X$ be first-countable, and for each $x \in X$ fix a local neighborhood base $\{B_n^x:n \in \omega \}$ such that $X=B_0^x$ and $B_{n+1}^x \subseteq B_n^x$. We can now define the (not necessarily symmetric) function $$d(x,y)=\inf \left \{\frac{1}{n+1}:y \in B_n^x \right \}.$$
It is not hard to see that the topology defined by this function (defining balls centered at $x$ as $B(x,\varepsilon):=\{y : d(x,y) < \varepsilon \}$) is the same as the original topology on $X$.
2) One can get non-first-countable topologies.
If we let $X=\{a_{n,m} : n,m \in \omega \} \cup \{b_n :n \in \omega\} \cup \{c\}$ and define $d(b_n,a_{n,m})=d(c,b_n)=\frac{1}{n+1}$, $d(x,x)=0$ and $d(x,y)=1$ otherwise, we obtain Arens´ space which is first-countable at every point except at $c$.
3) Not every topology can be obtained in this way.
If $X=D \cup \{\infty\}$ is the one-point compactification of an uncountable discrete space, then the topology of $X$ cannot come from a "generalized metric". Just note that in this situation the balls $B(\infty,1/n)$ have to be open and hence cofinite; thus their intersection is cocountable and therefore the topology induced by the generilized metric wouldn´t be Hausdorff (but the original topology is).
Edit: here is another example for 3), prompted by Marcos´ comment, which is interesting because it shows that the class of spaces that we are looking at is not closed under subspaces: 
The Arens-Fort space, which is just the subspace $\{a_{n,m} : n,m \in \omega \} \cup \{c\}$ of the Arens´ space defined in 2), can´t be obtained by this procedure. The reason is that the balls $B(x,\varepsilon)$ would have to be open (this is not hard to see) but then $\{B(c,1/n):n \in \omega\}$ whould be a countable local base at $c$.<|endoftext|>
TITLE: When are incompressible surfaces isotopic into a two-skeleton?
QUESTION [5 upvotes]: Haken proved that an incompressible surface in a triangulated irreducible 3-manifold is isotopic to a surface which is normal with respect to the triangulation (Theorie der Normalflächen. Acta Math. 105 1961 245–375).  
While normal surfaces are tremendously useful, I want my surfaces to be isotopic into the two-skeleton and I am unsure when I can conclude that they indeed are isotopic into the two-skeleton.
I suspect that there are triangulated 3-manifolds out there containing incompressible surfaces which are not isotopic into the two-skeleton, but I am hoping that if the 3-manifold has a metric and the tetrahedra are "small enough" compared to the injectivity radius of the 3-manifold, then incompressible surfaces can be isotoped into the two-skeleton.
Can someone point me to a reference or explain whether or not there are reasonable properties of a triangulation which imply that incompressible surfaces are isotopic into the two-skeleton?

REPLY [10 votes]: Unfortunately, you can't always expect every incompressible surface to be isotoped into the $2$-skeleton. There are only finitely many subsets of the $2$-skeleton, but there can be infinitely many nonparallel incompressible surfaces. This is simplest in the $3$-torus, but there are many other examples. Jaco's stair construction gives incompressible surfaces of arbitrarily high genus in a surface (of positive genus) times a circle.<|endoftext|>
TITLE: A conjecturally easier version of matrix Descartes rule of signs.
QUESTION [5 upvotes]: This is (quite obviously) inspired by this question. Let $C_i$ be symmetric positive definite matrices. Then is it true that there is exactly one symmetric positive definite $X$ such that 
$F(X) = X^n - \sum_{i=0}^n C_i \circ X^i = 0$, where $\circ$ denotes the Schur (component-wise) product (and exponentiation is with respect to that same product) Notice that unlike in the inspiring question, the Schur product is commutative.

REPLY [7 votes]: I suspect you mean "is it true that there is exactly one symmetric $X$ such that $F(X) = X^n - \sum_{i=0}^{n-1} C_i \circ X^i = 0$", which is more in line with the inspiration question. If you meant something else, I'll just delete this later today.
In that case, the answer is no. Take 
$$
C_2 = 
\begin{bmatrix}
1 & 2 \\\
2 & 5
\end{bmatrix}
\quad
C_1 = 
\begin{bmatrix}
1 & 1 \\\
1 & 2
\end{bmatrix}
\quad
C_0 = 
\begin{bmatrix}
2 & -2 \\\
-2 & 24
\end{bmatrix}
$$
Then we have four polynomials defining the entries in the matrix $X$ (really three by symmetry). The two diagonal entries are unique by Descartes rule of signs, being the real solutions to $z^3 -z^2 -z - 2 =0$, and $z^3 -5z^2 -2z - 24 =0$, so $2$ and $6$ - but the off diagonal entries are the solutions to $z^3 - 2z^2 -z +2 = 0$, so they can be $2, 1$ or $-1$. Each of these gives a positive definite matrix.<|endoftext|>
TITLE: Convergence of Markov chains in terms of relative entropy
QUESTION [6 upvotes]: Consider a finite state, irreducible Markov chain with a rate matrix $Q$ and a stationary distribution $\pi$. Suppose the chain starts with the initial distribution $p$ at time $0$, then at time $t$ the distribution is given by $$p_t = p*e^{Qt}$$
The relative entropy of $p_t$ with respect to the $\pi$ (also known as the Kullback-Leibler divergence $D(p_t || \pi)$) is given by
$$D(p_t || \pi) = \sum_i p_t(i) \log \left(\frac{p_t(i)}{\pi(i)}\right)$$
where $i$ ranges over all the states.
It seems well known that $D(p_t || \pi)$ monotonically decreases as $t$ increases. (For instance, section $2.9$ of Cover and Thomas's book here https://web.cse.msu.edu/~cse842/Papers/CoverThomas-Ch2.pdf).
My question is:

Is $D(p_t || \pi)$ a convex function of $t$?

I simulated for many random matrices using Matlab and did not find a counter example. Any references would be appreciated!

REPLY [4 votes]: Turns out $D(p_t||\pi)$ need not always be convex. The following paper demonstrates a counter-example in section $4.2$- http://arxiv.org/abs/0712.2578<|endoftext|>
TITLE: Series representation for Euler-Mascheroni constant
QUESTION [9 upvotes]: I found this formula for the Euler-Mascheroni constant $\gamma$. 

Just wondering whether such a formula already exists in literature?
  Also, wanted to know whether there are formulas that converge faster than this?

$$\gamma = \sum_{k = 1}^{\infty} \frac{1}{2^k k} - \sum_{k = 1}^{\infty}
   \frac{\zeta \left( 2 k + 1 \right)}{2^{2 k} \left( 2 k + 1 \right)} $$
UPDATE:
Thanks for your reply quid. I just came across this while doing some calculations with the zeta function. The calculations are a bit too long to be posted, but in short it derives from 
$$\zeta(s) = \frac{s+1}{2(s-1)} + \frac{s}{8} - \frac{s(s+1)}{2\pi^2}\int_1^\infty \frac{(\tan^{-1}\cot(\pi x))^2}{x^{s+2}}dx$$.

REPLY [18 votes]: In his 1887 paper Table des valeurs des sommes $S_k = \sum_{1}^\infty n^{-k}$ (Acta Mathematica 10 (1887), 299-302; volume available online), Stieltjes used almost exactly this formula to compute Euler's constant to 33 decimal places.  Of course as quid points out you need to know the zeta values to do this, but the main point of this paper was to compute those values, so he was just getting Euler's constant as a corollary.  He uses a slight variant of the formula, with $\zeta(2k+1)-1$ in place of $\zeta(2k+1)$ for faster convergence (and a corresponding adjustment in the other term, which becomes $1+\log 2 - \log 3$).  He derives the formula by taking the Taylor series expansion of $\log \Gamma(1+x)$ and using it to compute $\log \Gamma(1+1/2) - \log \Gamma(1-1/2)$.<|endoftext|>
TITLE: Orbits of independent sets of the hypercube
QUESTION [5 upvotes]: How does one enumerate the distinct orbit classes of independent sets of the hypercube modulo symmetries of the hypercubes? 
The counting of the number of independent sets in an $n$-dimensional hypercube modulo symmetries of the hypercube has been done up to $n=5$ by D.Eppstein as seen in the OEIS. We do have a bound on the number of independent sets of a regular graph as found by Y.Zhao. But I'm not aware of other results to this old problem of mine.
Does anyone know of a resource on how to go about this enumeration? I've already coded a brute-force program to do this listing and had reproduced Eppstein's results up to $n=5$. The $n=6$ took too much time and memory space on my personal computer that it always crashed. But this was years ago. I was just reminded of this recently, so here's my first post on MO.

REPLY [6 votes]: This is an easy problem for an orderly algorithm, which can generate the independent sets without isomorphs with no direct isomorphism checks between different independent sets.
It takes a couple of minutes on a recent computer and concludes that there are there are 519195 independent sets in the 6-cube up to isomorphism.
The numbers of each size (from 0 to 32) are:
1, 1, 5, 11, 56, 182, 742, 2323, 6714, 15233, 29200, 44870, 58711, 64916, 64902, 58982, 51064, 40835, 30884, 21145, 13511, 7631, 4058, 1859, 839, 316, 133, 42, 19, 5, 3, 1, 1<|endoftext|>
TITLE: Are there "unsociable" irreps? (Definition inside)
QUESTION [9 upvotes]: If, with a bit abuse of notation, $U\notin R\bigotimes{R}\bigotimes{R}\bigotimes{...}$ (i.e., regardless how long you clebsch up $R$, $U$ won't appear in the expansion), I call $U$ unsociable with respect to $R$ (in the group $G$). Clearly everything else is unsociable with the $1$ irrep, since (duh) $1\bigotimes{1}\bigotimes...=1$. For finite groups there can be nontrivial (ahem) cases: For $G=C_{3v}$ (the symmetry group), $A_2\bigotimes{A_2}=A_1$ and thus $E$ is unsociable w.r.t. $A_2$.
But what is the deal with Lie groups? E.g. is the defining irrep of $E_7$ unsociable w.r.t. the adjoint one?

REPLY [10 votes]: This is meant as an extended comment (sometimes correction) to things said in various answers and comments.   As a public service I'll try to fill in some of the history and references.  I recall some of this coming up previously on MO (exercise: find it).
In the framework of representation (or character) theory of a finite group $G$ over $\mathbb{C}$, Burnside led the way in his 1911 treatise (Chap. XV, Thm. IV): given a faithful representation of $G$, all irreducible representations (or characters) occur in some tensor power.    A proof of this is given in the 1962 Curtis-Reiner book (32.9); a later addendum to that book refers also to the slightly later work of Brauer and his student Steinberg.   Note that faithfulness is an obvious necessary condition here, not emphasized in most sources.   
Steinberg, Complete sets of representations of algebras, Proc. Amer. Math. Soc. 13 (1962), 746-747, offered a simpler proof of Burnside's theorem.  Then Brauer, A note on theorems of Burnside and Blichfeldt, Proc. Amer. Math. Soc. 15 (1964), 31-34, provided his own version of Burnside's theorem with a number of nice refinements: limiting how many tensor powers are needed, limiting the size of the characteristic 0 field, commenting on what remains true in prime characteristic.   Both of these papers are available online at the AMS website. 
Meanwhile, in his 1946 book Theory of Lie Groups (Chapter VI), Chevalley treated compact Lie groups and essentially provided a variant of Burnside's theorem.   But here it's essential to use both the faithful representation and its contragredient (dual).   The proof is in the context of the ring of representative functions, Tannaka duality, Peter-Weyl theorem.   
In a later short note, G.I. Lehrer provided a much more algebraic proof of Chevalley's theorem (again for compact Lie groups):  Produits tensoriels de grepresentations de groupes de Lie, C.R. Acad. Sci. Paris 275 (1972).   He also refers to Steinberg's paper.
From compact Lie groups it's not so far to the finite dimensional representation theory of a semisimple Lie algebra or algebraic group in characteristic 0.  But Burnside's theorem and its variants take their sharpest form for groups, whereas a single Lie algebra may belong to simple Lie or algebraic groups or the same Lie type but different isogeny type.    This conceals the requirement that one start with a faithful representation.<|endoftext|>
TITLE: Basis of a group
QUESTION [6 upvotes]: Let $G$ be a finite group. I will say that a set of a subgroups $H_1,\ldots ,H_k$ defines a basis for a group $G$ if any subgroup $H$ of $G$  there exists $S\subset [k]$ such that $H=\cap_{i\in S}H_i$.
My question is does it possible to give any upper bounds on the size of the minimal basis for $G$. 
For example does it possible to prove that for any $G$ there exists a basis of size at most $|G|^{10}$?
For example for an Abalian group it is easy to show that there is always exists a basis of size $|G|$. For example in case $A=Z_p^n$ it will be all maximal subgroups.

REPLY [6 votes]: This is an extended comment. An element of a lattice is called meet-irreducible if it cannot be expressed as a meet of two elements containing it. For a finite lattice, the meet irreducible elements are the unique minimal generating set for the lattice under meet. Your bases are exactly meet generating subsets of the subgroup lattice and the unique minimal basis is the set of meet irreducible subgroups (sometimes called primitive subgroups in the literature). It is known that the minimal degree of a faithful permutation representation of G is a lower bound for the size of this set. I am not sure what upper bounds are known. 
You might want to look at Johnson, D. L. Minimal permutation representations of finite groups. Amer. J. Math. 93 (1971), 857–866.<|endoftext|>
TITLE: Complemented Subspaces and Riesz-Thorin interpolation
QUESTION [5 upvotes]: Riesz-Thorin interpolation may sometimes be applied to subspaces (of $\ell^p$ or $L^p$) when these are complemented and the spaces in the complementation comes from a common dense subspace. To be a bit more formal, if there exists $A$ and $B$ such that if $A_p = \overline{A}^{\ell^p}$ and $B_p = \overline{B}^{\ell^p}$, and if $\forall q \in [q_1,q_2]$, $A_q \oplus B_q = \ell^q$, then the interpolation may be done between values of $q$ in this interval. 
I was wondering what techniques are available to show this situation holds...? To make the question more precise, here is a case (which may be constructed):
$\textbf{Question}$:  fix some $K>1$ and assume that two subspaces $A_p$ and $B_p$ of $\ell^p(\mathbb{N})$ are given by the closure of the span of $\lbrace a_i \rbrace_{i \in \mathbb{N}}$ and $\lbrace b_i \rbrace_{i \in \mathbb{N}}$ respectively. These spanning elements are combinatorially not too messy: the $a_i$ and $b_i$ may be written as the a difference of characteristic functions $\chi_S - \chi_T$ where $S$ and $T$ are finite sets in $\mathbb{N}$ of cardinality less than $K$. Furthermore, for a given $n \in \mathbb{N}$, the number of $i$ such that $a_i$ or $b_i$ is non zero at $n$ is less than $K$. Finally, $A_2$ is actually the (orthogonal) complement of $B_2$ in $\ell^2(\mathbb{N})$. Are there $p \in [1,\infty]$ (and $p \neq 2$), for which one may conclude that $A_p + B_p = \ell^p(\mathbb{N})$? 
(For $p>2$ one has obviously that $\overline{A_p + B_p} = \ell^p(\mathbb{N})$.) 
The description of the span of the spaces avoid the examples known to me of uncomplemented subspaces ($1 \leq p<\infty$), but it may very well not exclude all possibilities. Apologies in advance if this makes an easy negative answer. 
EDIT: I realized the answer may not be positive for $p$ "too small" (where "too small" depends perhaps on $K$) as one would obtain contradictions with some known results.

REPLY [2 votes]: You do get complementation in the case you mention.  Two key facts that you did not state explicitly but which follow easily from your hypotheses are that every unit vector $e_n$ is in the linear span of at most $K$ of the $a_i$ and $b_i$, and the span of $K$ of the $a_i$ and $b_i$ is contained in the span of at most $N=2K^2$ unit vectors.  Using these facts, it is not hard to verify that the orthogonal projection $P$ onto $A$ is bounded in the $\ell_1$, which by interpolation and duality gives you what you want. To check boundedness in $\ell_1$, you just have to give a bound on $\|Pe_n\|_1$ that is independent of $n$.  But on the span of $N$ unit vectors, the $\ell_1$ norm is dominate by $N^{1/2}$ times the $\ell_2$ norm.
EDIT:  As Antoine points outs, my answer is wrong.<|endoftext|>
TITLE: The set of Godel numbers of true sentences.
QUESTION [15 upvotes]: Tarski's Theorem on the undefinability of truth gives me a bit of a headache, and as a beginner I am still trying to grapple with its consequences. Here's a question.
Let $T$ be the set of Godel numbers of sentences $\sigma$ such that $V \models \sigma$. 
Now, I know that by Tarski's theorem, there is no formula that defines $T$ - that is to say, there is no first order formula $\tau (x)$ such that for each $n$, $n \in T \leftrightarrow V \models \tau (n)$. 
I can live with that, because obviously some sets are not definable. (Indeed, most are not definable.) But must $T$ exist somewhere in $V$, even if it not definable? The argument is not so obvious to me. Or will this depend on background assumptions? (Whether $V=L$, or whether this or that large cardinal exists, and so on ...)

REPLY [7 votes]: Allow me to complement the other answers by mentioning that Kelly Morse set theory KM proves the existence of a satisfaction predicate for first-order set-theoretic truth, and consequently in Kelly-Morse set theory you can prove the existence of your truth sets as desired.
A satisfaction predicate is a class $T$ of pairs $\langle \varphi,a\rangle$, which obeys the following Tarskian conditions:

(atomic truth) $T$ contains $\langle \varphi,\vec a\rangle$ for atomic $\varphi$ just in case $\varphi(\vec a)$ holds.
(Boolean combinations) $T$ contains $\langle \varphi\wedge\psi,\vec a\rangle$ just in case it also contains $\langle \varphi,\vec a\rangle$ and $\langle \psi,\vec a\rangle$, and similarly, it contains $\langle \neg\varphi,\vec a\rangle$ just in case it doesn't contain $\langle \varphi,\vec a\rangle$. 
(quantifiers) $T$ contains $\langle \exists x\varphi,\vec a\rangle$ if and only if there is some $b$ such that $T$ contains $\langle\varphi,(b,\vec a)\rangle$, where we place $b$ in the right position to be interpreted by variable $x$. 

A partial satisfaction predicate, in contrast, satisfies these requirements for all formulas up to a specified syntactic complexity. It is not difficult to prove that if $T$ is a partial satisfaction class for $\Sigma_n$ formulas, then one may define from it a partial satisfaction class for $\Sigma_{n+1}$ formulas, simply by applying one more step of the inductive definition of truth. Furthermore, one may prove by induction that if there is a partial satisfaction class for $\Sigma_n$ formulas, then it is unique. 
It follows from those observations that one may prove in KM by induction on the natural numbers that for every natural number $n$ there is a partial satisfaction class for $\Sigma_n$ formulas, and furthermore, that these partial satisfaction classes agree with each other, cohering into a full satisfaction class $T$ that works with all formulas.
In particular, in Kelly-Morse, you have your desired truth sets about first-order truth.
(Note that the induction that partial satisfaction classes exist is a second order claim, and although it can be proved by induction in KM, this part of the argument cannot be carried out in ZFC.)<|endoftext|>
TITLE: what exactly is the moduli functor for classifying elliptic curves with (full) level N structure?
QUESTION [12 upvotes]: So, when people say, "the moduli problem of classifying elliptic curves over $\mathbb{C}$ with level $N$ structure", there are usually two associated functors I've seen:

$P_N : \textbf{Ell}\rightarrow\textbf{Sets}$, where $\textbf{Ell}$ is the category of elliptic curves $E\rightarrow S$ over $S$ and morphisms are cartesian squares, and
$P_N(E/S) = \text{set of isomorphisms } \alpha : E[N]\rightarrow (\mathbb{Z}/N\mathbb{Z})^2 \text{ of determinant 1}$

$F_N : \textbf{Sch}\rightarrow\textbf{Sets}$, where
$F_N(S) = \text{set of isomorphism classes of pairs } (E/S,\alpha) \text{ with } \alpha\in P_N(E/S) $


I apologize for the length of this post, but this has been terribly confusing for me.
Alright, so I know that for $N\ge 3$, both functors are representable by the modular scheme $Y(N) := \Gamma(N)\backslash\mathcal{H}$ which are fine moduli schemes, and that's got something to do with the fact that there are no automorphisms of elliptic curves $E/S$ fixing any $\alpha\in P_N(E/S)$.
However, in the case $N = 1,2$, the modular curve $Y(2) := \Gamma(2)\backslash\mathcal{H}$ only gives you a coarse moduli scheme.
How should I think of the relation between the two above functors? In a way, a representing object $E/S$ for $P_N$ gives you both the universal elliptic curve $E$ and the base moduli scheme $S$ in one fell swoop. However, the functor $P_N$ doesn't seem like a naturally phrased moduli problem, since being able to represent $P_N$ just says:
"there is an elliptic curve $E/S$ such that for any other elliptic curve $E'/S'$, a level structure on $E'/S'$ is equivalent to a morphism $S'\rightarrow S$ such that $E'\cong E\times_S S'$."
(after thinking about it for a bit, it seems you can show that $P_N$ representable $\Longrightarrow$ $F_N$ representable)
On the other hand, the functor $F_N$ is much more natural, in that a representing object for $F_N$ much more clearly parametrizes elliptic curves with level structure. However, Peter Bruin's article (http://user.math.uzh.ch/bruin/moduli.pdf) and Katz/Mazur's book (specifically thm's 3.6 and 4.7) both seem to imply that if $P_N$ is not rigid (eg, $N = 1,2$), then even if $F_N$ is representable by an object $M$, the object $M$ might not carry a universal family. I am further confused by the fact that wikipedia (in the section on Fine Moduli Spaces) says that the universal family exists and must correspond to $\text{id}_M\in\text{Hom}(M,M)$.
I'm assuming wikipedia is wrong.
If wikipedia is wrong, then in the case $N = 2$, is the functor $F_2$ representable? If it is, is $\Gamma(2)\backslash\mathcal{H}$ the representing object?
When people talk of the moduli problem of classifying elliptic curves with full level $N$ structure, which functor are they referring to?
...onto stacks...
For $N = 1,2$, there is no fine moduli scheme, and hence at least $P_N$ is not representable. However, is there a fine moduli stack? (does that mean anything?)
In general, would I be correct in saying that a stack for a moduli problem is basically just the moduli functor itself? Can you make this more precise? (though I guess you'd have to replace "isomorphism classes of..." with the objects themselves)
Are there meaningful moduli problems that aren't stacks?
Thanks for bearing with me

will

REPLY [13 votes]: Your $F_N$ is the functor people would usually mean when they talk about the functor classifying elliptic curves with full level N structure (though it's a bit nicer if you replace $(Z/nZ )^2$ with $\mu_n \times Z/nZ$, so that the determinant takes values in $\mu_n$ on both sides.)
Wikipedia is not wrong.  (Wikipedia is surprisingly seldom wrong!)  Your F_2 is indeed not representable by a scheme, which is to say that the scheme known as Y(2) is not a fine moduli space. To say it was a fine moduli space would be precisely to say it represents the functor in question.  That's why Y(2) doesn't have to have a universal family over it.
Yes -- the stack is the same thing as the functor; but we're keeping track of certain facts about that functor when we say it's a stack.  For that matter, in the case where the functor is representable by a scheme, the scheme is also the same thing as a functor; but in that case we don't think of it as or refer to it as a functor, because that would make us seem very pretentious.<|endoftext|>
TITLE: Unusual decomposition of 3x3 real symmetric matrices - is this possible?
QUESTION [9 upvotes]: If $M$ is a 3x3, real symmetric matrix, then I know there are a few ways to decompose $M$ as
$M = A^T D A$,
where $D$ is a real diagonal matrix: e.g., this can always be done for some $A \in SO(3)$, or for some lower triangular $A$.  Can it always be done for some $A \in SO(2,1)$?
-Jeanne

REPLY [12 votes]: Unfortunately, the answer is 'no'.  You are basically asking whether you can simultaneously diagonalize two quadratic forms in three variables, and the answer is that, 'generically' you can (and you always can if some linear combination of the two is definite), but there are special pairs that cannot be simultaneously diagonalized.  
This happens already in dimension $2$.  You can't simultaneously diagonalize $x^2$ and $xy$, for example.  I think you cannot simultaneously diagonalize $-x^2 + y^2 + z^2$ and $2xy + z^2$ (if I remember the example correctly).  
Generally, if the (real) null cones of the two indefinite quadratic forms are tangent, then they can't be simultaneously diagonalized.<|endoftext|>
TITLE: Torsion in $H^2(X,\mathbb{Z})$ induced by torsion in $H_1(X,\mathbb{Z})$
QUESTION [6 upvotes]: Let $X$ be a topological space. The universal coefficient theorem says that there is a short exact sequence
$$
0\rightarrow Ext(H_{k-1}(X,\mathbb{Z}),\mathbb{Z})\rightarrow H^{k}(X,\mathbb{Z})\rightarrow Hom(H_{k}(X,\mathbb{Z}),\mathbb{Z})\rightarrow 0
$$
for $1\le k$. In particular a torsion element in $H_{1}(X,\mathbb{Z})$ induces a torsion element in $H^2(X,\mathbb{Z})$. 
I would like to understand the statement above in the following situation. Consider a $n$-torus $(S^1)^n$-fibration $f:X\rightarrow Y$. Assume that $H_{1}(Y,\mathbb{Z})=\pi_{1}(Y)=\mathbb{Z}/2\mathbb{Z}$ and monodromy representation is given by $(-id,\dots,-id)$ on $(S^1)^n$. Then the Spectral sequence associated with $f$ shows
$$
H_{1}(X,\mathbb{Z})\cong H_{1}(Y,\mathbb{Z})=\mathbb{Z}/2\mathbb{Z}.
$$ 
Is it possible to "see" the torsion in $H^2(X,\mathbb{Z})$ induced by $H_{1}(X,\mathbb{Z})$? 
One may want to use Poincare duality (to realize the torsion as a submanifold etc), so please take your favorite $Y$ (satisfying the condition above or something similar) and $n\in \mathbb{N}$.  
Thank you for your assistance.

REPLY [5 votes]: Having the cycles and boundaries of degree one, it's possible to construct a cohomology class in $H^2(X)$ that corresponds to your torsion element. This is explained in the following: 
1. Homological algebra
Let $(C,d)$ be a chain complex of free abelian groups, $G$ an abelian group and let 
$$\varphi: Ext(H_{q-1}(C),G) \to H^q(C,G)$$ be the monomorphism from UCT. An inspection of 
the proof in [Spanier: Algebraic Topology, Theorem 5.5.3] reveals: Set $Z_q = \ker(d_g)$, $B_{q-1}=\text{im}(d_q)$. Then there is an exact sequence 
$$\text{Hom}(Z_{q-1},G) \to \text{Hom}(B_{q-1},G) \to Ext(H_{q-1}(C),G) \to 0.\hspace{50pt}(\ast)$$

If $e \in Ext(H_{q-1}(C),G)$ is represented by $f: B_{q-1} \to G\;$ then 
  $\varphi(e) \in H^q(C,G)$ is represented by the cycle $f \circ d_q: C_q \to G$. 

2. The finitely generated case
For simplicity let's assume each $C_q$ has finite rank. Let $\alpha \in H_{q-1}(C)=Z_{q-1}/B_{q-1}$ 
be a generator of a direct summand of order $n$. Choose a representative $x \in Z_{q-1}$ 
of $\alpha$. We can find decompositions $Z_{q-1} = Z' \bigoplus \mathbb{Z}x,\;B_{q-1}=B' \oplus \mathbb{Z}nx$ with $B' \le Z'$. Let $f: B_{q-1} \to \mathbb{Z},\;f|B'=0,\;f(nx)=1$. Then $f$ represents $e \in Ext(H_{q-1},\mathbb{Z})$ via $(\ast)$ and $e$ corresponds to $\alpha$ under the non-canonical isomorphism $Ext(H_{q-1}(C),\mathbb{Z}) \cong H_{q-1}(C)_{tor}$. By 1., $h=f \circ d_q: C_q \to \mathbb{Z}$ is a cocycle that represents $\varphi(e)$. 
3. An example
Let $X$ be the CW-complex in exercise 4 with $\Pi_1(X)=Q_8$ (quaternion group) and let $C=C(X)$ be the cellular chain complex. $X$ has 0-cells $x, y$, 1-cells $a,b,c,d$, 2-cells $p, q, r$ and one 3-cell. The boundary operator is given by: 
$$d_1(a)=d_1(b)=d_1(c)=d_1(d)=x-y\hspace{130pt}$$
$$d_2(p)=a-b+c-d,\;\;d_2(q)=c-b+d-a,\;\;d_2(r)=d-b+a-c$$
Set $A = a-b,\; B = a-c,\;C=a-d$. Then $Z_1 = \langle A,B,C\rangle$  and $B_1 = \langle A-B+C,2B,2C \rangle$. 
Hence $B$ represents a homology class in $H_1(X)$ of order 2. Define $f: B_1 \to \mathbb{Z}$ by $$f(A-B+C)=0,\; f(2B)=1,\; f(2C)=0.$$ Now by 2., the cocycle $h=f \circ d_2$: 
$$h(p)=0,\;h(q)=0,\;h(r)=1$$
represents a cohomology class in $H^2(X)$ of order 2. 
Remark: Note that we didn't need to consider $d_3: C_3 \to C_2$ in order to construct our torsion class in $H^2(X)$. This, in contrast, would be necessary if we computed $H^2(X)$ directly.<|endoftext|>
TITLE: Instances where an existence result precedes the constructive version
QUESTION [22 upvotes]: The basic motivation here is to encourage and inspire - via examples - the pursuit of alternate proofs of existing results that might be more accessible and intuitive by cataloging success stories. Here's the question:

Are there good examples of instances in mathematics research where an existence-only result preceded - by some number of years - the corresponding constructive result?

I am aware of one nice example from my own field, which I hope is illustrative of the type of answer that would be nice to collect here: Tucker's Lemma, which is a discrete version of the famous Borsuk-Ulam Theorem, was first proved in the following paper by contradiction:

A. W. Tucker. Some topological properties of disk and sphere. In Proc. First Canadian Math. Congress, Montreal, 1945, pages 285–309. University of Toronto Press, Toronto, 1946.

The first constructive proof (which is starkly different from the original), did not appear in the literature until

R. M. Freund and M. J. Todd. A constructive proof of Tucker’s combinatorial lemma. J. Combin. Theory Ser. A, 30(3):321–325, 1981.

I did try just searching google for "A constructive proof of" and similar search strings.
This does provide some examples, but the results are not filtered by importance of the result in question, or the extent of difference between the old existence result and the newer constructive one.
Only one example per answer, please!

REPLY [6 votes]: Hironaka's original desingularization algorithm had constructive aspects but as I understand, also had aspects / computations which are not constructive and could not be done by hand or with a computer.  
Resolution algorithms now typically are based upon blowing up the strata of some invariant (where heuristically the singularities are at their worst) and so they are constructive if you can compute the invariant.  But these invariants might be for all intents and purposes completely incomputable, for example based on infinite amounts of data.
New algorithms have simplified many aspects on this, and indeed, algorithms have been implemented in computers, and it isn't so hard to write down all the various invariants as you follows modern algorithms. 
You can see 
Hironaka desingularisation theorem -- new proofs in literature?
for more discussion.<|endoftext|>
TITLE: Why does the de Rham double complex compute the algebraic de Rham cohomology ?
QUESTION [6 upvotes]: If $\mathcal F^\bullet$ is a chain complex of sheaves, to compute the hypercohomology, you take the cohomology of an injective resolution of $\mathcal F^\bullet$, i.e., a chain complex of injectives $\mathcal I^\bullet$ which is quasi-isomorphic to $\mathcal F^\bullet$.
It seems to be a well-known fact that for the algebraic de Rham complex $\Omega^\bullet_{X/S}$ you can compute this hypercohomolgy in practice by taking the total cohomology of the Cech/de Rham double complex $(\check{C}^i(X,\Omega^j), d, \check{d})$.
Does anyone know of a reference for this (somewhere in EGA?), or is there an easy proof ?

REPLY [5 votes]: This has nothing to do with the details of the problem; it is a generalization of Leray's theorem to hypercohomology. If $\mathcal{F}_{\bullet}$ is a complex of sheaves, and $U_i$ is an open cover of $X$ such that $H^q(U_{i_1} \cap \cdots \cap U_{i_r}, \mathcal{F}_j)=0$ for $q \geq 1$ and all $(i_1, \ldots, i_r)$ and $j$, then the hypercohomology of $\mathcal{F}_{\bullet}$ is computed by the cohomology of the Cech-$\mathcal{F}$ double complex.
I don't have my books on hand to find you a reference, but it isn't hard.<|endoftext|>
TITLE: Exponentiable objects in a category, valued in a larger, containing category
QUESTION [9 upvotes]: Recall that when dealing with topological spaces one usually likes dealing with a subcategory of $Top$ which is convenient, one facet of which is that it is cartesian closed. However to get to a similar point with smooth manifolds one needs to consider things like diffeological spaces. Not that there is anything wrong with that. But we have a partial solution if we are just looking for exponentiable objects, and willing to consider infinite-dimensional smooth manifolds (usually Frechet manifolds). 
More formally, an object $A$ of category $C$ with binary products is exponentiable if the functor $-\times A\colon C\to C$ has a right adjoint. The classification of which topological spaces are exponentiable is well known, and cartesian closed categories are defined by the fact that every object is exponentiable.
But in the category of (Hausdorff, finite-dimensional) smooth manifolds the only exponentiable objects are the compact manifolds of dimension at most zero. But we can still sensibly talk about smooth mapping spaces between a general compact manifold and an arbitrary manifold, where the mapping space is an object in the category of Frechet manifolds $Frech$, in which the category $Diff$ of finite-dimensional smooth manifolds sits as a full subcategory.
There are clear analogies with, say, finite CW-complexes, where the 'internal' hom is a topological space of a rather more infinite nature. Similarly, we can consider the mapping presheaf $X \mapsto C(X\times A,Y)$ on C.
What I would like to know is if there is a name for this sort of phenomenon, that we have a category $C$, and full embedding $C\hookrightarrow D$ and $D$-valued mapping objects for certain objects of $C$: these are objects of $C$ which are exponentiable as objects of $D$.
It seems to fall into some gap between cartesian closedness and enrichment, but I don't have a way of making that precise.

REPLY [4 votes]: Fernando is right that it has something to do with Kan extensions. However, it is not about Kan extensions along an inclusion, but more about extension of an inclusion.
I thought about similar issues a few years ago (and recently --- yesterday) , but in a slightly different context --- after the excellent answer by Todd Trimble to my question Completion of a category, I wondered if there was a general 2-categorical setting that could explain such constructions (I was mainly interested in carrying to a 2-categorical setting the highly related concept of Day convolution). 
Now I try to slowly reproduce some of these ideas.
I shall introduce the concept of a Yoneda triangle (perhaps I should call it the right Yoneda triangle, because there are obvious dual concepts).
Let $\mathbb{W}$ be a 2-category. A Yoneda triangle in $\mathbb{W}$ consists of 1-morphisms $y \colon A \rightarrow \overline{A}$, $f \colon A \rightarrow B$, $g \colon B \rightarrow \overline{A}$ together with a 2-morphism $\eta \colon y \rightarrow g \circ f$ which exhibits $g$ as a pointwise left Kan extension of $y$ along $f$ and exhibits $f$ as an absolute left Kan lifting of $y$ along $g$. (BTW: these data are exactly what led Mark Weber to strengthen the definition of a Yoneda structure introduced by Street and Walters).
The idea of a Yoneda triangle is that, we have a morphism $y \colon A \rightarrow \overline{A}$ which plays the role of a "defective identity" and for a given morphism $f \colon A \rightarrow B$ we try to characterise its right adjoint up to the "defective identity" $y$.
Example: [Yoneda triangles in $\mathbf{Cat}$]
If we take $\mathbb{W}$ to be the 2-category $\mathbf{Cat}$ of locally small categories, functors and natural transformations, then the condition that $G$ is a pointwise left Kan extension of $Y$ along $F$ reduces to:
$$G(-) = \int^{A\in\mathbb{A}} \hom(F(A), -) \times Y(A)$$
(where the coend has to be interpreted as the colimit of $Y$ weighted by $\hom(F(=), -)$ in case the category is not tensored over $\mathbf{Set}$). 
And the condition that $F$ is an absolute left Kan lifting of $Y$ along $G$ reduces to:
$$\hom(Y(-), G(=)) \approx \hom(F(-), =)$$
Particularly, if $Y$ is dense, than $G$ is canonically a pointwise Kan extension --- from density we have:
$$G(-) \approx \int^{A\in\mathbb{A}} \hom(Y(A), G(-)) \times Y(A)$$
and using the formula for an absolute lifting: 
$$G(-) \approx \int^{A\in\mathbb{A}} \hom(F(A), -) \times Y(A)$$
Example: [Adjunction as Yoneda triangle]
It is folklore that an adjunction $f \dashv g$ in a 2-category $\mathbb{W}$ may be equally characterised in the following way: $f$ is an absolute left lifting of the identity along $g$. In such a case $g$ is automatically a pointwise left extension of the identity along $f$ and $\mathit{id}, f, g$ together with the unit of the adjunction form a Yoneda triangle.
Example: [Yoneda triangle as a relative adjunction]
There is an old concept of so called "relative adjunction", which is defined in the same way as the Yoneda triangle, but without the requirement that $g$ is a left Kan extension. Note however, that in such a case $g$ need not be uniquely determined by $f$.
Let me move to the more specific example that you asked about.
Example: [Yoneda triangle along Yoneda embedding]
Let $F \colon \mathbb{A} \rightarrow \mathbb{B}$ be a functor between locally small categories (or more generally, a locally small functor). There is also an inclusion $y_\mathbb{A} \colon \mathbb{A} \rightarrow \mathbf{Set}^{\mathbb{A}^{op}}$. One may easily verify that these data may be always extended to the Yoneda triangle with $G(-) = \hom(F(=), -) \colon \mathbb{B} \rightarrow \mathbf{Set}^{\mathbb{A}^{op}}$ --- which reassembles the fact that every functor always has a "distributional" right adjoint. The same is true for internal categories and for categories enriched in a complete and cocomplete symmetric monoidal closed category, and generally (almost by definition) for any 2-category equipped with a Yoneda structure. 
The essence of the above example is that because the Yoneda functor $y_\mathbb{B} \colon \mathbb{B}\rightarrow \mathbf{Set}^{\mathbb{B}^{op}}$ is a full and faithful embedding, functors $F\colon\mathbb{A} \rightarrow \mathbb{B}$ may be thought as of distributors 
$$y_\mathbb{B} \circ F = \hom(=, F(-))$$
Every distributor arisen in this way has a right adjoint distributor $\hom(F(=), -)$ in the bicategory of distributors. The distributor $\hom(F(=), -)$ has actually the type $\mathbb{B} \rightarrow \mathbf{Set}^{\mathbb{A}^{op}}$, which is the only think that may prevent $F$ of having the ordinary (functorial) right adjoint $G \colon \mathbb{B} \rightarrow \mathbb{A}$ --- just recall, that we say that $F$ has a right adjoint, if there exists $G$ such that:
$$y_\mathbb{A} \circ G \approx \hom(F(=), -)$$
which means:
$$\hom(=, G(-)) \approx \hom(F(=), -)$$
Unfortunately, as a non-mathematician I will not help you with your other examples involving highly mathematical and completely non-understandable terms like a topological space or a manifold, so perhaps you have to calculate the other examples yourself :-)
However, I will give you another example that actually led me to the above considerations. One may similarly define the concept of a Yoneda bi-triangle and a Yoneda monoidal bi-triangle.
Example: [2-powers from Yoneda triangle]
The motivating example is to start with a 2-functor $J \colon \mathbb{W} \rightarrow \mathbb{D}$ equipping a 2-category $\mathbb{W}$ with proarrows, and an extension $Y \colon \mathbb{W} \rightarrow \overline{\mathbb{W}}$ embedding "small objects" into "locally small" (or large) objects in $\overline{\mathbb{W}}$. Then to extend these data to the Yoneda triangle, we have to find a functor $P \colon \mathbb{D} \rightarrow \overline{\mathbb{W}}$ representing a proarrow $A \nrightarrow B$ as a morphism $A \rightarrow P(B)$ in $\overline{\mathbb{W}}$, and a natural transformation $\eta \colon Y \rightarrow P\circ J$ playing the role of a familly of Yoneda morphisms $\eta_A \colon A \rightarrow P(A)$.  
The archetypical situation is when we take $\mathbb{W} = \mathbf{cat}$, $\overline{\mathbb{W}} = \mathbf{Cat}$, $\mathbb{D} = \mathbf{Dist}$, where  $\mathbf{cat}$ is the 2-category of small categories, $\mathbf{Cat}$ is the 2-category of locally small categories, and $\mathbf{Dist}$ is the bicategory of distributors between small categories. Then $J \colon \mathbf{cat} \rightarrow \mathbf{Dist}$, $Y \colon \mathbf{cat} \rightarrow \mathbf{Cat}$ are the usual embeddings, $P \colon \mathbf{Dist} \rightarrow \mathbf{Cat}$ is the covarinat 2-power pseudofunctor $\mathbf{Set}^{(-)^{op}}$ defined on distributors via left Kan extensions, and $\eta_\mathbb{A} \colon \mathbb{A} \rightarrow \mathbf{Set}^{\mathbb{A}^{op}}$ is the Yoneda embedding of a small category $\mathbb{A}$.
We know that there are isomorphisms of categories:
$$\hom_{\mathbf{Dist}}(\mathbb{A}, \mathbb{B}) \approx \hom_{\mathbf{Cat}}(\mathbb{A}, \mathbf{Set}^{\mathbb{B}^{op}})$$
where $\mathbb{A}$ and $\mathbb{B}$ are small. Therefore, to show that $P$ is a (bi)pointwise left Kan extension it suffices to show that $Y$ is 2-dense. However, $Y$ is obviously 2-dense, because the the terminal category is a 2-dense subcategory of $\mathbf{Cat}$ and $Y$ is fully faithful.
The point is that in most situations $\mathbb{D}$ is a monoidal (bi)category, where the monoidal structure is inherited from the closed structure on $\mathbb{W}$. Moreover, functors and the natural transformation constituting the Yoneda triangle are (lax)monoidal. This means that monoids in $\mathbb{W}$ are mapped to the (pro)monoids in $\mathbb{D}$ which are mapped to monoids in $\overline{\mathbb{W}}$. If I am not mistaken this observation leads to an abstract characterisation of the concept of the Day convolution (and in a similar manner one may try to define a Dedekind-MacNeille completion of an object).
In our archetypical situation, a category $\mathbb{A} \times \mathbb{B}$ is mapped by $P$ to $\mathbf{Set}^{\mathbb{A}^{op} \times \mathbb{B}^{op}}$ and the missing morphisms making the unit of the triangle lax monoidal:
$$\mathbf{Set}^{\mathbb{A}^{op}} \times \mathbf{Set}^{\mathbb{B}^{op}} \rightarrow \mathbf{Set}^{\mathbb{A}^{op} \times \mathbb{B}^{op}}$$
is given by the convolution of the distributional identity $\mathbb{A} \times \mathbb{B} \nrightarrow \mathbb{A} \times \mathbb{B}$:
$$\langle F, G \rangle \mapsto \int^{A \in \mathbb{A}, B \in \mathbb{B}} F(A) \times G(B) \times \hom(-, A) \times \hom(=, B) = F(-) \times G(=)$$
Now, a promonoidal category $M \colon \mathbb{A} \times \mathbb{A} \nrightarrow \mathbb{A}$ is mapped by $P$ to:
$$H \mapsto \int^{\langle A, B \rangle \in \mathbb{A}\times \mathbb{A}} H(A, B) \times M(-, A, B)$$
and by composing it with the above map:
$$\langle F, G \rangle \mapsto  \int^{\langle A, B \rangle \in \mathbb{A}\times \mathbb{A}} F(A) \times G(B) \times M(-, A, B)$$
we obtain the well-known formula for convolution.
One may also go in the other direction --- starting from the composition $P \circ J$ satisfying monoidal-like laws and try to find a right or left resolution in the category of (right/left) modules over monoid on $P \circ J$. If I am not mistaken, the left resolution (the Eilenberg-Moore object) of $P \circ J$ in our archetypical situation consists of the category of cocomplete categories and cocontinous functors and the right resolution (the Kleisli object) consists of the bicategory of distributors (i.e. the category of free cocomplete categories and cocontinous functors).
(BTW: this in some sense relates the concept of a proarrow equipment with the concept of a Yoneda structure.)
(BTW: perhaps the concept of a 2-topos should be defined as a Yoneda monoidal bi-triangle induced by the embedding of a 2-category of small objects into a category of bigger objects relatively to a category of "relations" in $\mathbb{W}$, which, for some purposes may be defined as the 2-category of discrete fibred spans, and for another purposes may be defined as the 2-category of codiscrete cofibred cospans).<|endoftext|>
TITLE: About a letter by Richard Palais of 1965.
QUESTION [42 upvotes]: In Cushman and Bates, Global Aspects of Classical Integrable Systems, 1997, I have read 

In a widely circulated but unpublished letter in 1965, Palais explained the symplectic  formulation of Hamiltonian mechanics.

I would like to know if, in the meanwhile, this letter was made available.

REPLY [114 votes]: I haven't thought about that letter for a very long time, but as far as I can recall I didn't ever make it publicly available, and I don't think any of the friends to whom I sent it did either. However, I am a bit of a pack-rat, so after Ryan Budney alerted me that this question had appeared on MO I did some searching in my piles of old papers, and I found and scanned what I am pretty sure was the mathematical content of the letter, and here is a link to the scan. 
http://vmm.math.uci.edu/PalaisLetterOnSymplectic.pdf
(I'm not sure if I just sent out copies of this with a cover letter or rather wrote a letter in which I copied the contents of the above.)
Note, that the first page of the above is a scan of page 159 of "Foundations of Mechanics" by Ralph Abraham and Jerry Marsden. It says that the letter was from around 1962, which is what I recall. Ralph was one of the recipients.  Dick Palais

REPLY [15 votes]: I would like to know if, in the meanwhile, this letter was made available

Yes! (See here).<|endoftext|>
TITLE: Invariants of symmetric matrices
QUESTION [10 upvotes]: $\newcommand{\eS}{\mathscr{S}}$ $\DeclareMathOperator{\SO}{SO}$ $\newcommand{\eP}{\mathscr{P}}$ $\newcommand{\bR}{\mathbb{R}}$ $\DeclareMathOperator{\tr}{tr}$ Let $m>1$ be an integer and denote by $\eS_m$ the vector space of symmetric $m\times m$ real matrices. The group $\SO(m)$ acts by conjugation on $\eS_m$.  The space of $\SO(m)$-invariant quadratic polynomials on $\eS_m$ is spanned by the two polynomials
$$ A \mapsto \tr A^2,\;\;A\mapsto (\tr A)^2. $$
Let $\SO(m-1)$ be the subgroup of $\SO(m)$ consisting of orthogonal transformations of $\bR^m$ that fix a unit vector $\eta$.
What is the space of $\SO(m-1)$-invariant on $\eS_m$. More precisely, can one explicitly write a basis of this space?
Clearly $\tr A^2$ and $(\tr A)^2$ are such polynomials, and so are
$$ A\mapsto (A^2\eta,\eta),\;\;A\mapsto (A\eta,\eta)^2, $$
where $(-,-)$ denotes the natural inner product on $\bR^m$. Are there any more $\SO(m-1)$ invariant quadratic polynomials? (I am inclined to believe that the above is the complete list.) Update After Robert Bryant's answer I lost my initial inclination.

REPLY [12 votes]: This is a trick question, right?  It's not true when $m=2$ because, then $\mathrm{SO}(m{-}1)=\mathrm{SO}(1)$ is trivial, so that all polynomials on $2$-by-$2$ symmetric matrices are invariants, and the four quadratics you mention clearly don't span the the six-dimensional space of all quadratics.  
Moreover, for $m>2$, you are missing $\mathrm{tr}(A)(A\eta,\eta)$.  When $m>2$, these five do span the space of $\mathrm{SO}(m{-}1)$-invariant quadratic polynomials, as you can see by realizing that, when $m>2$, the space $\mathcal{S}_m$ of symmetric $m$-by-$m$ matrices splits under $\mathrm{SO}(m{-}1)$ into a direct sum
$$
\mathcal{S}_m = \mathbb{R}\oplus\mathbb{R}\oplus\mathbb{R}^{m-1}\oplus \mathcal{S}^0_{m-1}
$$
of $\mathrm{SO}(m{-}1)$-irreducible modules, where the only equivalences are between the first two (trivial) summands, and where $\mathcal{S}^0_{m-1}$ means the trace-zero symmetric $(m{-}1)$-by-$(m{-}1)$ matrices.  (These two projections to $\mathbb{R}$ are given by the invariant linear forms $\mathrm{tr}(A)$ and $(A\eta,\eta)$.)  [This splitting is valid for $m=2$, of course, but then the third summand is another copy of $\mathbb{R}$ and the fourth summand has dimension $0$.]
However, you switched from quadratic polynomials at the beginning to all polynomials.  Was that intentional, or did you just want quadratics?<|endoftext|>
TITLE: Action of k* on a variety induces grading?
QUESTION [8 upvotes]: Let $V$ be a $\Bbbk$-variety such that $\Bbbk^\times$ (as an algebraic group) acts algebraically on $V$. Given any $f\in\Bbbk[V]$, let us call $f$ homogeneous of degree $d$ if for all $v\in V$ and all $\lambda\in\Bbbk^\times$, we have $f(\lambda.v)=\lambda^d f(v)$. 
My question is: Does this define a grading on $\Bbbk[V]$? 
I was convinced that it is true, but I am running into difficulties. Let us first assume $\Bbbk=\mathbb{C}$, the ground field should not be an obstruction. The linear span of $\Bbbk^\times f$ decomposes since $\Bbbk^\times$ is reductive, but I don't see how to turn this into a grading on all of $\Bbbk[V]$. 
If it is true, I would really like to see a proof - it should use as little machinery as possible.

REPLY [16 votes]: Turning the action map of varieties into a map of rings, we get a ring map $\phi$ from $k[V]$ to $k[V][t,t^{-1}] $, the coordinate ring with an extra invertible variable (the coordinate on $k^*$) adjoined.  Now, for any function $\phi(f)=\sum_{i\in \mathbb{Z}}f_it^i$ for some $f_i$'s, almost all of which are 0.  Note that $f=\sum f_i$, which we obtain by restricting the function to $t=1$. Using associativity, applying $\phi$ again to the $f_i$'s is the same as applying pull-back by the multiplication map to t.  Thus, as functions on $V\times k^*\times k^*$ (letting $t,u$ be the two coordinates)
$$\sum_{i\in \mathbb{Z}}\phi(f_i)u^i=\sum_{i\in \mathbb{Z}} f_i t^i u^i$$ 
since the pull-back of the coordinate by multiplication is just the product of the coordinates .  Thus, $\phi(f_i)=f_it^i$.  
We can define the grading by letting $f$ be homogeneous of degree $i$ if $\phi(f)=ft^i$.  We have already seen that every element can be written uniquely as a sum of such elements (the $f_i$'s), and this is multiplicative since $\phi$ is a ring homomorphism.
Alternatively, we can note that we have proven that the span of the $f_i$'s is an finite-dimensional invariant subspace containing $f$, so we can apply your argument.  In general, essentially the same argument shows that the action of any affine algebraic group on the coordinate ring of any affine variety by pull-back is a locally finite action: any function is contained in a finite-dimensional invariant subspace.<|endoftext|>
TITLE: Homotopy theory of topological stacks/orbifolds
QUESTION [17 upvotes]: Motivation $\newcommand{\T}{\mathscr{T}}$
I have many times found myself saying some variant of the following. Let $\T_g$ be the Teichmüller space of a surface of genus $g$, and $\Gamma_g$ its mapping class group. The quotient $\T_g/\Gamma_g$ is the moduli space of curves $M_g$. It is a deep fact that $\T_g$ is in fact diffeomorphic a ball, which implies that $M_g$ is a model for the classifying space $B\Gamma_g$ of the mapping class group. In particular, the cohomology of $M_g$ is just the group cohomology of $\Gamma_g$.
...well, almost. Since $\Gamma_g$ does not act freely, $M_g$ is in fact not a $B\Gamma_g$. However, all the stabilizers are finite groups, and this implies via a spectral sequence argument that the rational cohomology of $M_g$ coincides with the rational cohomology of $\Gamma_g$. 
Most algebraic geometers seem to ignore these issues by working instead with stacks or orbifolds. Indeed, the stack quotient $[\T_g/\Gamma_g]$ is the moduli stack of curves $\mathcal M_g$, which is in any case the more fundamental object of study. 
Question
My question is whether the topological arguments in the first two paragraph can be carried out in a more highbrow way using orbifolds or topological stacks. I am vaguely aware that Noohi's work on topological stacks includes setting up a homotopy theory of topological stacks, but I know almost nothing about any of this. So the question should be interpreted as "does there exist a developed homotopy theory of topological stacks where the following question can be asked and answered".

Question 1. Are any two quotients $[E/G]$ and $[E'/G]$, where $E$ and $E'$ are contractible spaces with a not necessarily free group action by $G$, homotopy equivalent as topological stacks?

My second question is more speculative since I know even less about rational homotopy theory.

Question 2. Let $\mathcal X$ be a topological stack, with coarse moduli space $X$. Suppose that all isotropy groups of $\mathcal X$ are torsion. Is $\mathcal X \to X$ a rational homotopy equivalence?

REPLY [19 votes]: Here is a simple way to talk about the homotopy type of a stack.  Let $\mathfrak{X}$ be a stack and $f: U \to \mathfrak{X}$ a representable surjective submersion (an atlas) from a space $U$ (e.g., the map from the Teichmuller space to the moduli stack.)  Now, form the pullback of $f$ along itself:  $U\times_{\mathfrak{X}} U$.  This comes with two maps to $U$ and there is a diagonal map from $U \to U\times_{\mathfrak{X}} U$.  All together, these maps give a topological groupoid.  The nerve of this topological groupoid is a simplicial space and the geometric realization of which is a space that one can regard as representing the homotopy type of the stack.  
Here are some easy/nice properties of the above notion of homotopy type that are easy to check.

Given a space $X$ with a $G$ action, the homotopy type of $[X/G]$ is the Borel construciton, aka homotopy quotient, $EG \times_G X$.  In particular, the answer to your Question 1 is affirmative, and the homotopy type of the moduli stack of curves is exactly $B\Gamma_g$.
One can define singular and de Rham cohomology of a stack and these invariants coincide with the integral and rational cohomology of the homotopy type of the stack.  This is in fact almost a tautology since, for example, the de Rham cohomology can be defined by taking a covering by a manifold, forming iterated pullbacks (to produce a simplicial manifold), taking the de Rham algebra of this simplicial manifold to get a cosimplicial dga, and then taking the totalization to get a dga.
It follows from property 1 above that the answer to your question 2 is also affirmative.
[Edit] This notion of homotopy type is well-defined because one can check that any two atlases determine Morita equivalent topological groupoids which then have weakly equivalent nerves.

If I remember correctly, Noohi uses a slightly more sophisticated notion of homotopy type.  He defines a universal weak equivalence to be a representable morphism from a space $U$ to a stack $\mathfrak{X}$ such that the pullback along any morphism from a space to $\mathfrak{X}$ is a weak equivalence.  $U$ can then be regarded as the homotopy type of the stack.  I think this is more of less equivalent to the naive version I explained above, but it has the advantage of being a bit more functorial and there might be some other technical advantages I can't remember.  David Carchedi will probably be able to give more details.<|endoftext|>
TITLE: Maximal number of maximal subgroups
QUESTION [15 upvotes]: Let $G$ be a finite group. I want to find an upper bound on the number of the maximal subgroups. My questions is does it possible to prove that the number of maximal subgroups of any finite group $G$ is at most $|G|^{100}$?
One can easily find that any subgroup is generated by at most $\log|G|$ elements thus the number of subgroups(in particular maximal subgroups) is at most $|G|^{\log|G|}$. 
Does it possible to improve this upper bound. 
For an abelian group the number of maximal subgroups is at most $|G|$ and in fact I do not know any example where the number of maximal subgroups is more than $|G|$.
I am almost sure that I am not the first who is asking this question I would like to know if the answer to this question is known or either this is a hard question.

REPLY [8 votes]: The document I linked to above is sufficiently striking as to warrant an answer of its own. I hope it complements the community wiki above.
As mentioned above the relevant conjecture in this area is due to Wall:

Conjecture The number of maximal subgroups of a finite group $G$ is less than the order of $G$.

This has been the subject of much study with the landmark work (until recently) being the above-cited work of Liebeck, Pyber and Shalev. In addition to the result mentioned above they show that the conjecture is true if the group $G$ is simple, up to a finite number of exceptions.
Now a quote from the linked document is relevant:

This largely directed attention to
  composite groups, where Wall in his original paper had at least shown the conjecture to be true for finite solvable groups. The key remaining cases were known to be semidirect products of a vector space V with a nearly simple finite group G acting faithfully and irreducibly on it.

It turns out that in this case Wall's conjecture implies some bounds on the cohomology groups $H^1(G,V)$. And, as the document relates, examples have now been found which violate these bounds. In particular, Wall's conjecture does not hold.
In light of this development, the bound $C|G|^{3/2}$ mentioned above, also due to Liebeck, Pyber and Shalev, assumes greater importance. Although, as the linked document mentions, it is likely that the value $3/2$ can be reduced a great deal.
One final interesting quote:

A conjecture of Aschbacker and Guralnick, not made at the conference... would
  now rise to be the main conjecture in maximal subgroup theory. (The conjecture states that it is the number of conjugacy classes of maximal subgroups that is bounded, less than the number of conjugacy classes of elements in the group.)

Anyone interested should definitely read this document. Not only is it interesting mathematically, it's a very engaging account of how this recent breakthrough was achieved.<|endoftext|>
TITLE: What are the algebras for the double dualization monad?
QUESTION [31 upvotes]: Let $k$ be a field, and let $\mathbf{Vect}$ denote the category of vector spaces (possibly infinite-dimensional) over $k$.  Taking duals gives a functor $(\ )^*\colon \mathbf{Vect}^{\mathrm{op}} \to \mathbf{Vect}$.  
This contravariant functor is self-adjoint on the right, since a linear map $X \to Y^*$ amounts to a bilinear map $X \times Y \to k$, which is essentially the same thing as a bilinear map $Y \times X \to k$, which amounts to a linear map $Y \to X^*$.  It therefore induces a monad $(\ )^{**}$ on $\mathbf{Vect}$.
What are the algebras for this monad?
Remarks

I assume this is known (probably since a long time ago).
The first paper I came across when searching for the answer was Anders Kock, On double dualization monads, Math. Scand. 27 (1970), 151-165.  I'm pretty sure it doesn't contain the answer explicitly, but it's possible that it contains some results that would help.
The monad isn't idempotent (that is, the multiplication part of the monad isn't an isomorphism).  Indeed, take any infinite-dimensional vector space $X$.  Write our monad as $(T, \eta, \mu)$.  If $\mu_X$ were an isomorphism then $\eta_{TX}$ would be an isomorphism, since $\mu_X \circ \eta_{TX} = 1$.  But $\eta_{TX}$ is the canonical embedding $TX \to (TX)^{**}$, and this is not surjective since $TX$ is not finite-dimensional.
There's another way in which the answer might be somewhat trivial, and that's if $(\ )^*$ is monadic.  But it doesn't seem obvious to me that $(\ )^*$ even reflects isomorphisms (which it would have to if it were monadic).  
There's a sense in which answering this question amounts to completing the analogy:


sets are to compact Hausdorff spaces as vector spaces are to ?????

Indeed, the codensity monad of the inclusion functor (finite sets) $\hookrightarrow$ (sets) is the ultrafilter monad, whose algebras are the compact Hausdorff spaces.  The codensity monad of the inclusion functor (finite-dimensional vector spaces) $\hookrightarrow$ (vector spaces) is the double dualization monad, whose algebras are... what?  (Maybe this will help someone to guess what the answer is.)

REPLY [23 votes]: Tom, I believe $(-)^\ast: \mathbf{Vect}^{op} \to \mathbf{Vect}$ is monadic, essentially because all objects in $\mathbf{Vect}$, in particular $k$ as a module over $k$ as ground field, are injective. 
For instance, to check that $(-)^\ast$ reflects isomorphisms, suppose $f: V \to W$ is any linear map. We have two short exact sequences 
$$0 \to \ker(f) \to V \to im(f) \to 0$$ 
$$0 \to im(f) \to W \to coker(f) \to 0$$
Because $k$ is injective, the functor $(-)^\ast = \hom(-, k)$ preserves short exact sequences: 
$$0 \to im(f)^\ast \to V^\ast \to \ker(f)^\ast \to 0$$ 
$$0 \to coker(f)^\ast \to W^\ast \to im(f)^\ast \to 0$$ 
and if $f^\ast$, the composite $W^\ast \to im(f)^\ast \to V^\ast$, is an isomorphism, then $W^\ast \to im(f)^\ast$ is injective, which forces $coker(f)^\ast = 0$ and therefore $coker(f) = 0$. By a similar argument, $\ker(f) = 0$. Therefore $f$ is an isomorphism. 
The remaining hypotheses of Beck's theorem (in the form given in Theorem 2, page 179, of Mac Lane-Moerdijk) are similarly easy to check. Obviously $\mathbf{Vect}^{op}$ has coequalizers of reflexive pairs since $\mathbf{Vect}$ has equalizers. And $(-)^\ast: \mathbf{Vect}^{op} \to \mathbf{Vect}$ (which has a left adjoint, as pointed out) preserves coequalizers; this is equivalent to saying that $\hom(-, k)$, as a contravariant functor on $\mathbf{Vect}$, takes equalizers to coequalizers, or takes kernels to cokernels, but that's the same as saying that $k$ is injective, so we're done. 
Oh, incidentally, double dualization is not a commutative or monoidal monad, if I recall correctly. 
Edit: In a comment below, Tom asks for a more concrete description of $\mathbf{Vect}^{op}$ along the lines of topological algebra. I suspect the way to go is to see $\mathbf{Vect}$ as the Ind-completion (or Ind-cocompletion) of the category of finite-dimensional vector spaces, and therefore $\mathbf{Vect}^{op}$ as the Pro-completion of the opposite category, which is again $\mathbf{Vect}_{fd}$. I think I've seen before a result that this is equivalent to the category of topological $k$-modules which arise as projective limits of (cofiltered diagrams of) finite-dimensional spaces with the discrete topology, or something along similar lines, but I'd have to look this up to be sure. There might be pertinent material in Barr's Springer Lecture Notes on $\ast$-autonomous categories, but again I'm not sure. 
Edit 2: Ah, found it. $\mathbf{Vect}^{op}$ is equivalent to the category of linearly compact vector spaces over $k$ and continuous linear maps. See Theorem 3.1 of this paper for example: arxiv.org/pdf/1202.3609. The result is credited to Lefschetz.<|endoftext|>
TITLE: Algebraic K-theory with commutative semirings?
QUESTION [5 upvotes]: My question is basically given in the title: Are there any references for a generalization of algebraic K-theory to the scenario where the domain of the functors consists of commutative semirings (provided this has been done at all)?

REPLY [9 votes]: There are a few papers out there dealing with a slightly different focus - algebraic K-theory over the "field with one element" $\mathbb{F}_1$.  Some of the frameworks for $\mathbb{F}_1$ algebra include semirings, so these papers might contain material that covers what you are interested in as well.
The thesis of Nicolai Durov arXiv:0704.2030 describes a setting for algebraic geometry over a class of objects more general than rings.  These objects are commutative algebraic theories.  Commutative rings, commutative semirings and commutative monoids all form full subcategories of commutative algebraic theories.  Among the many things Durov does in his thesis, he includes some discussion of algebraic K-theory in the final chapter.
The "blueprints" of Oliver Lorscheid (arxiv.org/1103.1745 and its sequels) also contain semirings as a full subcategory.  According to the abstract, he will eventually get to K-theory of blueprints, which will contain K-theory of semirings as a special case.<|endoftext|>
TITLE: Optimizing the condition number
QUESTION [16 upvotes]: Suppose I have a set $S$ of $N$ vectors in $W=\mathbb{R}^m,$ with $N \gg m.$ I want to choose a subset $\{v_1, \dots, v_m\}$ of $S$ in such a way that the condition number of the matrix with columns $v_1, \dotsc, v_m$ is as small as possible -- notice that this is trivial if $S$ does not span $W,$ since the condition number is always $\infty,$ so we can assume that $S$ does span. The question is: is there a better algorithm than the obvious $O(N^{m+c})$ algorithm (where we look at all the $m$-element subsets of $S?$). and how much better? One might guess that there is an algorithm polynomial in the input size, but none jumps to mind.

REPLY [12 votes]: Update: This recent paper on this topic may also be of interest; it's quite short and claims to have a fully constructive approach.

I think the closest to answering your question is the following paper.
"Column subset selection, matrix factorization, and eigenvalue optimization", by J. A. Tropp.
In Proc. 2009 ACM-SIAM Symp. Discrete Algorithms (SODA), pp. 978-986, New York, NY, Jan. 2009. SODA .pdf or a longer arXiv version.
From the abstract:

Most research from the algorithms and numerical linear algebra communities focuses on a variant called rank-revealing QR, which seeks a well-conditioned collection of columns that spans the (numerical) range of the matrix.
....
a celebrated result of Bourgain and Tzafriri demonstrates that each matrix with normalized columns contains a large column submatrix that is exceptionally well conditioned. Unfortunately, standard proofs of this result cannot be regarded as algorithmic. This paper presents a randomized, polynomial-time algorithm that produces the submatrix promised by Bourgain and Tzafriri.<|endoftext|>
TITLE: Classification of long exact sequences
QUESTION [13 upvotes]: Let $\mathcal C$ be the category of long exact sequences of finitely generated abelian groups almost all of whose entries vanish.
The category $\mathcal C$ is naturally additive as a subcategory of complexes of abelian groups.

Question: Can we write down a complete list of isomorphism classes (up to translation) of indecomposable objects of $\mathcal C$?

It is easy to see that the number of such isomorphism classes is countably infinite.
Here are some indecomposable objects:
$$
\cdots\to 0\to\mathbb Z\overset{m}{\to}\mathbb Z\to\mathbb Z_m\to 0\to\cdots,
$$
$$
\cdots\to 0\to\mathbb Z_{(m,n)}\to\mathbb Z_n\overset{m}{\to}\mathbb Z_n\to\mathbb Z _{(m,n)}\to 0\to\cdots,
$$
where $m$ is a natural number, $n$ is a prime power and $(m,n)$ denotes the greatest common divisor.
But there is more; for instance, if $p$ is prime then the indecomposable object
$$
\cdots\to0\to\mathbb Z_p\to\mathbb Z_{p^2}\overset{p}{\to}\mathbb Z_{p^2}\overset{p}{\to}\cdots\overset{p}{\to}\mathbb Z_{p^2}\overset{p}{\to}\mathbb Z_{p^2}\to\mathbb Z_p\to 0\to\cdots
$$
can have any finite "length."
If this classification problem has been solved, a reference would be great. Otherwise I would very much appreciate any idea/hint towards a general solution.
(I've added the noncommutative-algebra tag because chain complexes can be considered as modules over a certain non-commutative ring. The question I am asking is a sub-problem of classifying all finitely generated indecomposables for this ring.)

REPLY [10 votes]: In a series of recent papers, Schmidmeier and Ringel show than the classification of monomorphisms in the category of finitely generated $\mathbb Z/p^n$-modules is a wild problem of representation theory for $n>6$. Hence, your problem is also wild. There's no hope to get what you want.<|endoftext|>
TITLE: A diagonal operation on power series
QUESTION [9 upvotes]: Given a formal power series $f(x,y)=\sum_{n,m\ge 0}f(n,m)\: x^n y^m$ in two variables $x$ and $y$ over some field of characteristic zero, e.g. the field of complex numbers $\mathbb C$, we define a new formal power series in one variable $t$:
$\Delta(f)(t):=\sum_{i\ge 0} f(i,i) \:t^i$.
It is known that if $f(x,y)$ is rational $\Delta(f)(t)$ is in general not rational (I think it is algebraic though).
Example: $f(x,y)=\frac{1}{1-x-y}$ leads to $\Delta(f)(t)=(1-4t)^{-1/2}$.
The question is: Is there a constructive way (i.e. algorithm or explicit formula) to calculate 
$\Delta(f)(t)$ for rational (or algebraic) $f(x,y)$ ?

REPLY [6 votes]: The reference to Stanley's book given by Qiaochu Yuan provides indeed an enlightening exposition on diagonals of bivariate rational functions. Note that the method using a residue mentioned at the end of example 6.3.4 is better suited for computations than the one using Puiseux series (Theorem 6.3.3).
This being said, it is hard (impossible?) to find a reference with a complete algorithm for the computation of algebraic equations for bivariate diagonals in characteristic 0. This is one of the motivations for a recent paper of Alin Bostan, Bruno Salvy and myself on that subject: Algebraic Diagonals and Walks: Algorithms, Bounds, Complexity. 
H.-C. Herbig's request is met there by Algorithm 3 and formula (4).<|endoftext|>
TITLE: Solving the field membership problem using Grobner bases
QUESTION [8 upvotes]: Is there an easy way to determine whether a set of elements in a field generates the whole field or only a subfield?
Specifically, I have a subfield of $k(x,y)$ described in terms of a canonical set of generators, and I have two other elements of this ring that I think should be sufficient to generate the whole thing but I need to prove it.
In particular, Muller-Quade and Steinwandt's paper "Basic Algorithms for Rational Function Fields" (http://www.sciencedirect.com/science/article/pii/S0747717198902462) seems to describe exactly what I'd like to do (check if the canonical generators are inside the field generated by my two elements, and write one in terms of the other) but I'm not familiar enough with the notation they use in their algorithms to really follow them.
Has this sort of algorithm been implemented in a computer algebra program (eg Magma or Maple) that I could use, or is there another way to do it?
If anyone could shed some light on what's happening in the paper above that would also be appreciated, as I'd be interested in generalizing the approach to the q-division ring (i.e. rational functions in two $q$-commuting variables, $q \in k^{\times}$) if possible.

REPLY [3 votes]: Between David's answer and much staring at the paper I linked to above, I think I've mostly figured out what's going on there so I thought I'd post a summary of their method here as well in case anyone else had a similar problem.  I'm focusing here on $how$ to use their algorithm, not $why$ it works: I found it was a lot easier to follow the theory once I knew where I was going!
Their setup is: given a field $k(x_1,\dots, x_n) = k(\mathbf{x})$ and a set of elements $\mathbf{g} = \{g_1, \dots, g_m\}\subset k(\mathbf{x})$, we can form the field $k(\mathbf{g})$ generated by the elements of $\mathbf{g}$ and then ask: given $f \in k(\mathbf{x})$, do we have $f \in k(\mathbf{g})$ and if so, how can we write $f$ in terms of the generators?  The authors' aim is to answer this without resorting to using the lex ordering, which as mentioned above can be very slow.
So let $g_i = n_i/d_i$, and introduce a new variable $T_i$ for each $g_i$.  Collect together all of the prime divisors of each of the denominators $d_i$ into a set $\{p_1, \dots, p_r\}$ (eliminate any repeats) and introduce a new variable $Y_i$ for each $p_i$.
In the polynomial ring $k[Y_1, \dots, Y_r, x_1, \dots, x_n, T_1, \dots, T_m]$ define the ideal $I=\langle n_1-T_1d_1, \dots, n_m-T_md_m, Y_1p_1-1,\dots,Y_rp_r-1\rangle$ and compute a Groebner basis of $I$.  The only restriction on ordering is that you should have $\textbf{Y} >> \textbf{x} >> \textbf{T}$, within each block they can be ordered however you like.
Eliminate all terms involving any $Y_i$ from the Groebner basis; call the resulting set $G$.  Let $H$ be the subset of $G$ involving terms only in the $T_i$: this may be empty. (This step seems to deal with syzygies between the generators, giving you the simplest result at the end and avoiding dividing by zero at any point.)
Let $S=\langle H \rangle$ and define $L = Q(k[\mathbf{T}]/S)$, so if $H$ was empty this is just $k(\mathbf{T})$.  Let $\pi$ be the natural embedding $k[\mathbf{T}][\mathbf{x}] \hookrightarrow L[\mathbf{x}]$ and $G' := \pi(G)$.
If $f=n_f/d_f \in k(\mathbf{x})$ is the element of interest then introduce a new variable $A$ (so we're now in the ring $L[\mathbf{x},A]$) and reduce the polynomial $n_f-Ad_f$ to normal form wrt the Groebner basis $G'$: you should obtain an expression $N-AD$.
Optionally at this point, you can reduce $N$ and $D$ to their simplest form by reducing their numerators wrt $\pi(H)$; recall that $N,D \in L[\mathbf{x}]$ where $L$ is a localization of $k[\mathbf{T}]$ so they could be rational functions in the $T_i$.
If $D = 0$ then $f \not\in k(\mathbf{g})$.  If $D \neq 0$ then solve $N-AD = 0$, i.e. $A=N/D$.  If $N/D \not\in k(\mathbf{T})$ then $f \not\in k(\mathbf{g})$, otherwise $f=N(g_1, \dots, g_m)/D(g_1,\dots,g_m)$.

Whew!  I hope that's fairly clear if you just want to apply the algorithm; if you want to understand why each step works, the paper I linked to in the question has all the technical details.  I suspect it's essentially the same approach as David's answer, with a few tweaks to avoid necessarily using the lex ordering and to deal with relations between the set of generators.<|endoftext|>
TITLE: Brown representability for non-connected spaces
QUESTION [47 upvotes]: In many places (on MO, elsewhere on the Internet, and perhaps even in some textbooks) one finds a statement of the classical Brown representability theorem that looks something like this:

If $F$ is a contravariant functor from the (weak) homotopy category of topological spaces $\mathrm{Ho}(\mathrm{Top}_\ast)$ to the category $\mathrm{Set}_\ast$ of pointed sets which sends coproducts (i.e. wedges) to products and weak/homotopy pushouts (or CW triads) to weak pullbacks, then $F$ is repesentable.

However, I have been unable to find a convincing proof of this statement.  All the proofs I have found fall into one of these classes:

Proofs of an analogous statement where $\mathrm{Ho}(\mathrm{Top}_*)$ is replaced by its subcategory of pointed connected spaces.
Proofs which purport to prove the above statment, but which actually implicitly assume all spaces are connected at some point.  (It's sometimes hard to tell whether a given proof falls into this class or the previous one, since some classical algebraic topologists seem to use "space" to mean "connected pointed space".)
Proofs that all cohomology theories are representable by a spectrum.  Here the suspension axiom implies that the behavior on connected spaces determines the whole theory.

The issue is that the collection of pointed spheres $\{S^n | n\ge 0\}$ is not a strongly generating set for all of $\mathrm{Ho}(\mathrm{Top}_*)$: a weak equivalence of pointed spaces is still required to induce isomorphisms of homotopy groups with all basepoints, whereas mapping out of pointed spheres detects only homotopy groups at the specified basepoint (and hence at any other basepoint in the same component).
Is the above version of the theorem true, without connectedness hypotheses?  If so, where can I find a proof?

REPLY [37 votes]: I was thinking more about this question and found another paper by Heller which offers an answer (unfortunately a negative one). The paper is

Peter Freyd and Alex Heller, Splitting homotopy idempotents. II. J. Pure Appl. Algebra 89 no. 1-2 (1993) pp 93–106, doi:10.1016/0022-4049(93)90088-B.

This paper introduces a notion of conjugacy idempotent. It is a triple $(G, g, b)$ consisting of a group $G$, an endomorphism $g \colon G \to G$ and an element $b \in G$ such that for all $x \in G$ we have $g^2(x) = b^{-1} g(x) b$. The theory of conjugacy idempotents can be axiomatized by equations, so there is an initial conjugacy idempotent $(F, f, a)$. The Main Theorem of the paper says (among other things) that $f$ does not split in the quotient of the category of groups by the conjugacy congruence.
Now $f$ induces an endomorphism $B f \colon B F \to B F$ which is an idempotent in $\mathrm{Ho} \mathrm{Top}$ and it follows (by the Main Lemma of the paper) that it doesn't split. It is then easily concluded that $(B f)_{+} \colon (B F)_{+} \to (B F)_{+}$ doesn't split in $\mathrm{Ho} \mathrm{Top}_{*}$.
The map $(B f)_{+}$ induces an idempotent of the representable functor $[-, (B F)_{+}]_{*}$ which does split since this is a $\mathrm{Set}$ valued functor. Let $H \colon \mathrm{Ho} \mathrm{Top}_{*}^\mathrm{op} \to \mathrm{Set}$ be the resulting retract of $[-, (B F)_{+}]_{*}$. It is half-exact (i.e. satisfies the hypotheses of Brown's Representability) as a retract of a half-exact functor. However, it is not representable since a representation would provide a splitting for $(B f)_{+}$.
The same argument with $B f$ in place of $(B f)_{+}$ shows the failure of Brown's Representability in the unbased case.<|endoftext|>
TITLE: Examples for open disc bundle which is not vector bundle
QUESTION [8 upvotes]: William Browder showed in "Open and closed disc bundles", Ann. of Math. (2) 83 (1966), 218-230 that there are open disc bundles over some complex which cannot be isomorphic to a vector bundle.
My question is: Is there any related examples?
For example: What is the least dimensions for the base complex and the dimension of the fiber?
Or is there any explicit example for this complex?

REPLY [3 votes]: Some remarks which amount in some way to an answer.
(1) Let  $\text{Diff}(D^n)$ be the diffeomorphisms of $D^n$ which restrict to the identity on
the boundary.
When $n\gg k$ is large, $\pi_k(\text{Diff}(D^n))\otimes \Bbb Q$ was computed
by Farrell and Hsiang. The answer is that this is $\cong \Bbb Q$ when $k = 4j-1$ and $n$ is odd. In every other case this group is trivial.
(2) Let $\text{Diff}(D^n,S^{n-1})$ be the group of diffeomorphisms of $D^n$ which aren't necessarily the identity on the boundary of $D^n$. 
Then as in Ryan's answer there's a homotopy fiber
sequence
$$
C(S^{n-1}) \to \text{Diff}(D^n,S^{n-1}) \to O(n)
$$
where $C(S^{n-1})$ is the concordance space of the $(n-1)$-sphere. This is the group of
diffeomorphisms of $S^{n-1} \times [0,1]$ which restrict to the identity on 
$S^{n-1} \times 0$.
As Ryan notes
$$
\pi_k(\text{Diff}(D^n,S^{n-1})) \cong \pi_k(C(S^{n-1})) \times  \pi_k(O(n)) .
$$
Again, when $n\gg k$, we can compute $\pi_k(O(n)) \cong \pi_i(O)$ by Bott periodicity.
Rationally, it's given by $\Bbb Q$ in dimensions congruent to $3 \text{ mod } 4$
and trivial otherwise.
Also $ \pi_k(C(S^{n-1}))\otimes \Bbb Q$ can be computed when $n \gg k$. It's also 
$\cong \Bbb Q$ in dimensions congruent to $3 \text{ mod } 4$ and trivial otherwise.
So we get, for $n \gg k$, that $\pi_k(\text{Diff}(D^n,S^{n-1})) \otimes \Bbb Q$ is
isomorphic to $\Bbb Q \oplus \Bbb Q$ in dimensions congruent to $3 \text{ mod } 4$ and trivial otherwise.<|endoftext|>
TITLE: Regarding the Thurston norm and the ways that a three-manifold can fiber over the circle
QUESTION [8 upvotes]: I'm learning about the Thurston norm and am trying to understand the implications that the existence of fibered faces has for the ways in which a given three-manifold $M$ can fiber over the circle. In particular, I am interested in the following question:

Let $M$ be a compact, oriented three-manifold without boundary, and suppose that $M$ is irreducible and atoroidal, so that the Thurston norm is non-degenerate. Suppose further that $M$ admits a fibering over $S^1$, so that there exists a fibered face for the unit norm ball. Is it then true that $M$ fibers over $S^1$ with fiber $S_g$ the closed surface of genus $g$ for infinitely many $g$?

I believe the answer is yes but I haven't seen a discussion of a result of this type in any of the sources I have consulted (Thurston's paper, this paper of McMullen, and the books by Kapovich, Calegari, and Candel-Conlon mentioned in this question.) The argument I constructed relies on Theorem 6.1 in the aforementioned McMullen paper, which gives conditions under which a class $\phi \in H^1(M, \mathbb{Z})$ is Poincare-dual to a connected norm-minimizing surface $S$. Specifically, for any $\phi$ that is primitive (i.e. maps onto $\mathbb{Z}$) and for which $b_1(M_\phi)$ is finite (where $M_\phi$ is the cyclic covering of $M$ associated to the kernel of $\phi$), a connected, Thurston-norm minimal $S$ dual to $\phi$ exists. In the case where $\phi$ is associated to a fibration $M \to S^1$, it seems to me that the condition on the finiteness of $b_1$ is automatically satisfied: McMullen remarks that in these circumstances we have $b_1(M_\phi) = b_1(F)$ where $F$ is the fiber of the fibration associated to $\phi$.
Now the argument goes as follows: given any primitive $\phi$ in the cone on the fibered face (such $\phi$ exist with arbitrarily large Thurston norm), take the $S$ associated to $\phi$ by the McMullen theorem. By construction, $S$ is dual to $\phi$, which is in turn dual to the fiber $F$ of the associated fibration, so that $S$ and $F$ represent the same homology class. Per a lemma in Calegari's book, a norm-minimizing surface in an irreducible manifold is incompressible. Then, following the remark of Thurston at the beginning of his Section 3, any incompressible surface homologous to a fiber is in fact isotopic to the fiber; in particular they have the same Euler characteristic. Since we can take $|| \phi ||$ to be arbitrarily large, we find that $\chi(S)$, and hence the genus of the fiber, can be arbitrarily large. 
My question then is whether the answer to my above question is yes, and if so, is the argument I give correct, and is there a simpler way to show it? Should this be surprising? Is there a simple example of this phenomenon, preferably one where the various fiberings are easy to "see"? I know of examples when $M$ has boundary, namely link complements in the three-torus, but I would like an example with closed fibers.

REPLY [10 votes]: The answer is: yes if the rank of $H_2(M;\mathbb{Z})$ is $\ge 2$; and no if the rank is $1$ because in that case there is up to isotopy a unique connected surface bundle structure on $M$. The proof uses nothing more than what is in Thurston's original article MR0823443, although there are probably multiple other ways to do it; I'll mention another proof due to Fried.
In the case of rank $\ge 2$, consider a fibered face, and let $C \subset H_2(M;\mathbb{Z})$ be the open cone over that face whose fiber has homology class in $C$. Fix one particular fibration over the circle. Since rank$\ge 2$, $C$ contains primitive elements of $H_2(M;\mathbb{Z})$ arbitrarily far from the origin, i.e. having arbitrarily large norm. So it remains to check that the norm evaluated on the homology class of a fiber equals minus the Euler characteristic of that fiber. To put it another way, you just need to check that a fiber of a fibration over the circle is norm minimizing in its homology class. This is a consequence of the theorem in Thurston's original article saying that any compact leaf of a taut transversely orientable foliation is norm minimizing; in more detail, what he proves is that the absolute value of the Euler class of the foliation evaluated on the compact leaf (which equals $|\chi(S)|$), equals the norm evaluated on that leaf.
Another proof I particularly like is Fried's formula for the restriction of the norm to $C$, from his article here. Consider the fiber $F$, consider its pseudo-Anosov monodromy $\phi : F \to F$, let $S_\phi \subset F$ be the set of singularities of $\phi$, and let $C_\phi \subset M$ be the suspension of $S_\phi$, so $C_\phi$ is a collection of oriented circles in $M$ where $M$ is regarded as the mapping torus of $\phi$. Turn $C_\phi$ into a 1-cycle by assigning to each of its components a coefficient $\frac{p}{2}-1$ where $p$ is the number of prongs of the corresponding singularity in $S_\phi$. The conclusion of Fried's theorem is that the norm is equal to the absolute value of the intersection number with $C_\phi$ (which is obviously equal to $|\chi(S)|$ by the Euler-Poincare formula for $\chi(S)$ expressed in terms of the singularities of $\phi$. Fried proves that intersection number with $C_\phi$ is equal to the norm throughout the entire open cone $C$.<|endoftext|>
TITLE: Fast Algorithms for Distinguishing Squarefrees
QUESTION [6 upvotes]: Given $n$ I need to quickly determine whether $p^2|n$ for any prime $p$. I've already considered an algorithm that strips all primes not exceeding $n^{1/3}$ and then we can easily check that $n$ is not an exact square (except one).
I've heard mentioned that there is an $O(n^{1/6})$ algorithm in fact but no details were given and I can't see how. I expect this would be an optimal algorithm for small enough $n$. Of course it would be possible to use a sub-exponential algorithm to factor $n$ fully so this question is really about what shortcuts are available when we only care whether $n$ is squarefree.

REPLY [7 votes]: The $N^{1/6}$ algorithm is explained in this work by Booker, Hiary, Keating, which is devoted to a MUCH cooler algorithm (conditional on the GRH, and subexponential (though super polynomial))

REPLY [5 votes]: It might seem surprising, but as far as I know there is really not that much to say beyond: factor as fast as you can and check. (And this might be the 'truth' though I do not think there is a proof showing the equivalence of the two problems.)
And, this problem is also well studied as it arises in Computational Algebraic Number Theory; 
for example in Zassenhaus Algorithm do compute integral basis for maximal orders to decompose a discriminant into squarefree and squre part is the main computational bottleneck.
I do not know what this order $n^{1/6}$ algorithm is, but this is (as you say) worse than factoring for large $n$.
Cf. http://mathworld.wolfram.com/Squarefree.html between equation (2) and (3) for a reference for about what I say.

Added: The reference provided by Igor Rivin is very interesting and pertinent to the question, to save rushed MO-readers some time, I extract some info from it.
The '$O(n^{1/6})$ algorithm' runs more or less along the lines sketched out already by OP. 
Since checking whether an integer is a square is easy, it suffices to detect factors up to size $n^{1/3}$. 
Now, this can be done Pollard--Strassen algorithm that allows to detect factors of size at most $B$ in time $O(n^{\varepsilon}B^{1/2})$, which for $B=n^{1/3}$ yields the claimed thing (up to the epsilon and this can improved a bit, but things stay slightly above $n^{1/6}$).
The running time for the squarefree-proving algorithm under GRH is of order $\exp( (\log n )^{c+o(1)} )$, where this $c$ is defined in a certain complicated way, and is conjectured to be $2/3$. 
For comparison, and as mentioned in that reference, there are factoring algorithms with expected running time $\exp( (\log n )^{c'+o(1)} )$ with $c'$ equal to $1/2$ (Elliptic Cuves, Quadratic Sieve) and $1/3$ for Number Field Sieve. 
So for this squarefree-proving algorithm one can prove (under GRH) something on its running time that one cannot prove for factoring algorithms. 
Thus, it seems to me, if one is mainly interested in actually establishing/computing that a number is squarefree or not (as opposed to needing provable information/bounds on the difficulty of the task) it might still be better to go with factoring, as one expects it to be faster; and if ever the expectations should be wrong then, under typical circumstances, one could still do something else.<|endoftext|>
TITLE: $\pi_1$ Sequence of Topological Groups
QUESTION [6 upvotes]: Consider a connected topological group $G$ (not necessarily Lie). You have some maps $G\times G\to G$, such as projection to either summand, or multiplication $(g,h)\mapsto gh$. Now let's look at a slightly more complicated but naturally-occuring map: $(g,h)\mapsto ghg^{-1}h^{-1}$, i.e. $G\times G\to [G,G]\hookrightarrow G$.  What goes on at the fundamental group level?  
In other words, is it true that $\pi_1(G\times G)\to\pi_1([G,G])\to\pi_1(G)$ is exact? 
I have a rather ad hoc reason to believe that the first map is trivial (as $\pi_1$ is abelian here, the commutator $[g,h]$ will unwind itself to the constant loop) and so I would want the second map to be injective.
Update: The comments below take care of this when $G$ is a Lie group!
So what can obstruct $\pi_1$ being injective on $[G,G]\hookrightarrow G$ for non-Lie groups?
Update: It has also been pointed out that this works for finite-dimensional topological groups!
That leaves a possible counterexample for the infinite-dimensional case.

REPLY [4 votes]: This is really more of a comment, but it kind of answers one of the OP's question, so I am indulging myself: It is a result of W. Browder (Annals, 1961) that $\pi_2$ of a finite dimensional $H$-space is trivial, so the result holds in that setting. I learned of this (and also that this is not true without the finite dimensionality assumption) from @Allen Hatcher's answer to this (very relevant) question:
Homotopy groups of Lie groups<|endoftext|>
TITLE: The (Sigma) Algebra of Convex Sets
QUESTION [7 upvotes]: This is a question-by-proxy for a colleague from computer science. I'm sure many people here are already aware that convex decomposition forms an important sub-field of both computational geometry and vision (in the CS sense). The prototypical problem in this field is as follows: 

Consider a nice compact, connected set $B \subset \mathbb{R}^n$ with non-zero volume (this set is typically a polyhedron). Find a minimal collection of closed convex subsets $C_j \subset B$ such that the union of $C_j$'s equals $B$, and all pairwise intersections $C_i \cap C_j$ for $i \neq j$ have zero volume.

A lot of work has gone into this, presumably because convex objects are considered simpler to analyze than arbitrary compact blobs. I'm no expert in the field, but I believe it is quite standard: algorithms have been implemented in CGAL, based on Chazelle's pioneering work.
A lot of the theory breaks down almost immediately if one starts to step away from polyhedra. My question here can be best explained with this picture:

Note that there is no convex decomposition for the square minus disc on the left. So,

Which compact sets with non-zero volume are definable as finite unions and intersections of convex sets and their complements?

And of course, the natural follow-up: if we allow countable rather than finite unions and intersections and restrict our domain to some compact, convex subset of $\mathbb{R}^n$, then the definable sets naturally generate a sigma algebra.

Is this convex sigma algebra the same as the traditional one with open sets as a basis?

On one hand, every open ball is a countable union of closed convex subsets. But what about the other direction?

REPLY [5 votes]: This is tangential to the posed question, not an answer.  But there has been work on a
discrete version of this concept, called in the literature the convex deficiency tree.
For example, here is Fig. 1 from the paper, V. Shapiro,
"A Convex Deficiency Tree Algorithm for Curved Polygons,"
International Journal of Computational Geometry & Applications, Vol. 11, No. 2 (2001) 215-238
(World Scientific link):
         
One emphasis in this paper is on "well-formed Boolean expressions" for representing shapes.<|endoftext|>
TITLE: Loop space: De Rham cohomology
QUESTION [8 upvotes]: How to calculate  the DeRham cohomology of the free loop space $LM= C^\infty(S^1,M)$ as a Frechet manifold?.   
Edit: It  will be enough for me to know: 
When $H^1_{DR}(LM)$ is not $\{0\}$. 
Bounty is ending within 5 hours.

REPLY [3 votes]: Assuming that de Rham cohomology is ordinary cohomology (which shouldn't be too hard by the standard sheaf-theoretic proof?), $H^1 = Hom(\pi_1,\mathbb{R})$. Now the base point fibration gives rise to a short exact sequence
$$
1 \rightarrow \pi_2(M) \rightarrow \pi_1(LM) \rightarrow \pi_1(M) \rightarrow 1
$$
Hence yes, if $H^1(M;\mathbb{R}) \neq 0$, then also $H^1(LM;\mathbb{R}) \neq 0$. Maybe more importantly, in the simply-connected case $H^1(LM) = H^2(M)$, something that one can also see by hand.<|endoftext|>
TITLE: Composition Series
QUESTION [7 upvotes]: Which finite groups have uniqueness for the ordered sequence of composition factors (up to isomorphism)?

REPLY [12 votes]: Here is a characterization. A group $G$ has two different composition series if and only if it has a factor $H/K$ which is a direct product of two non-isomorphic simple subgroups, where $H$ is a subnormal subgroup of $G$, $K$ is a normal subgroup of $H$. Indeed, if such $H/K$ exists, then clearly there are two different composition series. Conversely, suppose that there are two different composition series $A_0=1 < A_1 < A_2 < ... < A_n = G$ and $B_0=1 < B_1 < B_2 < ... < B_n=G$. Let $j$ be the last index with $A_{j-1}\ne B_{j-1}$ (non-isomorphic), $n\ge j\ge 1$. Let $H=A_j=B_j$. Note that $A_{j-1}$ and $B_{j-1}$ are normal in $H$. Hence $A_{j-1}B_{j-1}$ is normal in $H$. Since it is bigger than $A_{j-1}$, we have $H=A_{j-1}B_{j-1}$. Hence $K=A_{j-1}\cap B_{j-1}$ is normal in $H$ and $A_{i-1}/K$ is isomorphic to $H/B_{j-1}$ hence simple. Similarly $B_{j-1}/K$ is simple. Now $H/K$ has two different normal simple subgroups $A_{j-1}/K$ and $B_{j-1}/K$ with trivial intersection, so $A_{j-1}/K$ and $B_{j-1}/K$ commute and form a direct product. Therefore $H/K$ is isomorphic to the direct product of two non-isomorphic simple groups $A_{j-1}/K$ and $B_{j-1}/K$.<|endoftext|>
TITLE: On surjections, idempotence and axiom of choice
QUESTION [10 upvotes]: The following assertion is trivial in ZFC, or even in much weaker theories. Is it also true in ZF?
(I couldn't find it in the Consequences site so far.)

If $A$ is an infinite set such that $A$ can be mapped onto $A\times 2$ then $|A\times 2|=|A|$

The problem is that we cannot necessarily choose from every fiber of $f$, so we cannot construct an injection from $A$ to $f^{-1}(A\times\lbrace 0\rbrace)$, which will prove the assertion.
While I'm on the topic, is it possible for a D-finite set to have such property? It is possible for a D-finite set to be surjected onto a larger set than itself, but what about that large?

REPLY [2 votes]: A simpler version (using the length of a sequence for coding, rather than its entries):  Let $X$ be infinite but Dedekind finite.   Let $S$ be the set of all finite 1-1 sequences from $X$.  Then $S$ is Dedekind finite. 
Now define a surjective map $f$ from $S$ onto $S\times \{0,1\}$ as follows.

If $s\in S$ has length $2k+i$ with $i\in \{0,1\}$, map $s$ to $(s\mathord\restriction k, i)$ where $s\mathord\restriction k$ is the initial segment of $s$ of length $k$.


To get a surjective  map from  $S$ onto $S^{<\omega}$, we  fix a surjective function $$H:\omega\to (\omega^{<\omega})^{<\omega}$$
with the property that all numbers appearing in $H(n)$ will be $< n$. 
Now define $f:S\to S^{<\omega}$ as follows:  Given a sequence $s\in S$ of length 
$n$, write $H(n)$ as $ (h_0,\ldots, h_{k-1})$, and let $F(s) := (s\circ h_0, \ldots, s\circ h_{k-1})$. (If some $h_i$ is not 1-1, let $F(s):=()$; alternatively, modify the function $H$ so that its range is the set of all sequences of 1-1 sequences.)
It is clear that $f$ is onto: given any sequence $t=(t_0,\ldots, t_{k-1})$ of elements of $S$, first find a 1-1 sequence $s$ enumerating  the union of the ranges of the $t_i$.  So there 1-1 sequences $h_i$ such that t $t_i = s\circ h_i$ for all $i$.  Now extend $s$ (in a 1-1 way) to a sequence $r$ whose length codes the family of the $h_i$'s. We will have $f(r)=t$.<|endoftext|>
TITLE: Mathematical computer desk
QUESTION [6 upvotes]: D. Gibb, from the Mathematical Laboratory, University of Edinburgh,
describes a Computer Desk in his book A course in interpolation and numerical integration for the
mathematical laboratory, G. Bell & Sons, Ltd., 1915, available here:
Where computation is performed to any considerable extent, computer's desk will be found useful. Those used in the mathematical laboratory of the University of Edinburgh are 3' 0" wide, 1'9" from front to back, and 2'6  1/2" high. They contain a locker, in wich computing paper can be kept without being folded, and a cupboard for books, papers, drawing-board, arithmometer, or instruments. Each desk is supplied with a copy of Barlow's tables (which gives the square, square root, cube, cube root and the reciprocal of all numbers up to 10,000), a copy of Creller's multiplication table (which gives at sight the product of any two numbers each less than 1000), and
tables giving the values of the trigonometric functions and logarithms
Question: Are there any available picture of this "computer desk" ?


I think that this may be the first recorded description of an workplace for numerical analysts...

REPLY [7 votes]: A little googling led to http://www-history.mcs.st-and.ac.uk/Extras/EMS_1913_Colloquium.html which quotes at length a report that appeared in The Mathematical Gazette in 1913, including the following:

...Professor Whittaker held his séances
  in the large basement hall, which has
  recently been fitted up as a
  mathematical laboratory. This, indeed,
  was the first occasion on which it had
  been used. Each student was provided
  with a specially designed desk, with a
  convenient book-rest fixed to the
  back, and with drawers and shelves for
  storing note-books, scribbling paper,
  graph paper, and books to aid
  calculation, such as Barlow's Tables
  and Crelle's Rechentafeln. At these
  desks learned professors, lecturers,
  teachers, and a few students, nearly
  eighty in all, totted up their columns
  of figures and drew their
  periodographs, and were quite elated
  when their totals came out right.

Finally, although it's about copyists rather than computers, I can't resist citing a little excerpt on ergonomics, from Herman Melville's "Bartleby the Scrivener":

Though of a very ingenious mechanical
  turn, Nippers could never get this
  table to suit him. He put chips under
  it, blocks of various sorts, bits of
  pasteboard, and at last went so far as
  to attempt an exquisite adjustment by
  final pieces of folded blotting-paper.
  But no invention would answer. If, for
  the sake of easing his back, he
  brought the table lid at a sharp angle
  well up towards his chin, and wrote
  there like a man using the steep roof
  of a Dutch house for his desk:—then he
  declared that it stopped the
  circulation in his arms. If now he
  lowered the table to his waistbands,
  and stooped over it in writing, then
  there was a sore aching in his back.
  In short, the truth of the matter was,
  Nippers knew not what he wanted. Or,
  if he wanted any thing, it was to be
  rid of a scrivener’s table altogether.<|endoftext|>
TITLE: pseudo-random algorithm allowing O(1) computation of Nth element
QUESTION [6 upvotes]: It is obvious that using seed value one can easily compute next value of some (deterministic) pseudo-random algorithm - so $N$th element can be computed in $O(N)$.
But is there such PRNG that allow to compute Nth element in $O(1)$ while still preserving good periodicity and distribution?

REPLY [2 votes]: FWIW: any reasonably-secure block cipher in CTR mode is a supposedly-perfect PRNG (if it is not - for about 20 years now, such cipher is not considered secure). 

How to calculate nth element of PRNG using this method:
-- in advance, once -- 

choose cipher C (ranging from rather poor XXTEA to rather secure Chacha20 and AES) and generate some random key K for it (it has to be done once, so K can be taken from random.org or something)

-- to calculate nth element --
1) take number n and copy it into input_block for the cipher C (filling the gaps as necessary) 
2) encrypt input_block with cipher C and key K. Output is supposed to be perfectly-random. 

Pros: (a) secure, which translates into (b) the best-quality RNG out there (any kind of correlation is a severe security weakness, and lots of effort is spent on this kind of analysis). 
Cons: performance is lower than for other PRNGs (though on a single core good implementation can easily reach gigabytes/second).<|endoftext|>
TITLE: Can one recognize this symmetric function?
QUESTION [7 upvotes]: $\newcommand{\lm}{\lambda}$ $\newcommand{\bR}{\mathbb{R}}$ Let $m$ be  an integer $>1$. Define
$$ I_m:\bR^m\to  \bR,\;\; I_m(\lm_1,\dotsc, \lm_m)=\int_{S^{m-1}}\exp\Bigl(-\sum_{j=1}^m \lm_j^2x_j^2\;\Bigr)\; |dA(x)|,  $$
where $S^{m-1}$ is the unit sphere in $\bR^m$  and $|dA(x)|$ denotes the  "area" element on $S^{m-1}$.
The function $I_m$  is real analytic  and symmetric in  the variables $\lm_1^2,\dotsc, \lm_m^2$ and in fact  it has a Taylor expansion
$$I_m (\lm_1,\dotsc, \lm_m)=2\sum_{h=0}^\infty\frac{(-1)^h}{\Gamma(\frac{m}{2}+h)}\sum_{h_1+\cdots+h_m=h}\frac{\Gamma(h_1+\frac{1}{2})\cdots \Gamma(h_m+\frac{1}{2})}{h_1!\cdots h_m!} \lm_1^{2h_1}\cdots \lm_m^{2h_m}. $$
In particular, $I_m$ can be expressed  as a function of  the  symmetric polynomials
$$s_k =\sum_{j=1}^m \lm_j^{2k},\;\; k=1,\dotsc, m. $$
Question 1. Is there  a more compact  description   of $I_m$ of the form
$$I_m(\lm_1,\dotsc, \lm_m)=F_m(s_1,\dotsc, s_m), $$
where $F_m$ is some "classical" function?
Question 2. $\DeclareMathOperator{\diag}{Diag}$ $\DeclareMathOperator{\tr}{tr}$ Consider the symmetric  matrix
$$\Lambda=\diag(\lm_1,\dotsc, \lm_m). $$
Is there  some  function $V_m:\bR\to \bR$ such that
$$ I_m(\lm_1,\dotsc, \lm_m)=e^{-\tr V_m(\Lambda)} ? $$
Can one describe such a $V_m$ explicitly? I'm vague about the term explicit, but I would be very pleased if $V_m$ were a  "special" function.
The  second question may suggest  the origin of $I_m$.     I stumbled  onto $I_m$ when I bumped into  a certain  ensemble of random real,  symmetric $m\times m$ matrices.  (The story is too bushy to include it here.)

REPLY [11 votes]: Your function is a special case of the Confluent Hypergeometric function of matrix argument, in particular it is ${}_1F_1(\frac{1}{2},\frac{n}{2},L)$, where $L$ is a diagonal matrix having $-\lambda_i^2$ ($1\le i \le n$) as its diagonal components. These functions can be more generally represented using Zonal Polynomials.
This function arises as the normalization constant of a Bingham Distribution. More details about this function and other related hypergeometric functions can be found in the textbooks:

Directional Statistics, Mardia and Jupp (2000)
Aspects of multivariate statistical theory, Muirhead (1982).<|endoftext|>
TITLE: Kostant's theorem on invariant polynomials in positive characteristic
QUESTION [7 upvotes]: Let $k$ be an algebraically closed field and let $G$ be a reductive linear algebraic group over $k$ with Lie algebra $\mathfrak g$. If the characteristic of $k$ is $0$ then, by a classical result of Kostant, $k[\mathfrak g]$ is free over the subalgebra $k[\mathfrak g]^G$. Is there an analog of this result when the characteristic of $k$ is positive?

REPLY [5 votes]: Here is an example of $G$ for which $k[\mathfrak{g}]$ is not a free
$A$-module, where $A=k[\mathfrak{g}]^G$ (I hope there is no mistake in what follows). 
Let $G=\mathrm{PGL}_p$ where $p\ge 5$, a simple algebraic group of adjoint type with $\mathfrak{g}=\mathfrak{pgl}_p$. The Chevalley Restriction Theorem holds for $G$ as follows, e.g., from the Springer-Steinberg paper (see Ch. II, 3.17). Then we have that $A\cong k[\mathfrak{t}]^W$ where $\mathfrak{t}$ is the image of the diagonal matrices in $\mathfrak g$. Of course, 
$W\cong\mathfrak{S}_p$ and
$\mathfrak{t}$ is the quotient of the natural $\mathfrak{S}_p$-module $V$ with basis $\{e_1,\ldots, e_p\}$, permuted by $\mathfrak{S}_p$, by its trivial submodule 
$k(e_1+\cdots+e_p)$. By a result of Gregor Kemper, for $p\ge 5$  the ring $S(V^*)^{\mathfrak{S}_p}\cong k[\mathfrak{t}]^W$ is not Cohen-Macaulay (see Corollary 2.8 and Example 2.9 in Kemper's paper published in J. Algebra, Vol. 215 (1999), 33--351). In particular, $k[\mathfrak{g}]^W$ is not a polynomial algebra, which resolves in the negative Problem 3.18 in the 
Springer-Steinberg paper. 
Since $\mathfrak g$ is a restricted Lie algebra and $\mathfrak t$ is its toral Cartan subalgebra of dimension $p-1$, it follows from a result I proved that
there exist homogeneous $f_0,\ldots, f_{p-2}\in A$ with ${\rm deg}\ f_i=p^{p-1}-p^i$ such that
$x^{[p]^{p-1}}=\sum_{i=0}^{p-2}f_i(x)x^{[p]^i}$ for all $x\in\mathfrak{g}$. Furthermore, the zero locus of the $f_i$'s equals the nullcone $\mathcal N$ of $\mathfrak g$. Since $\mathcal N$ is irreducible of codimension $p-1$ it follows that  {$f_0,\ldots, f_{p-2}$} a regular sequence in $k[\mathfrak{g}]$ and $k[\mathfrak{g}]$ is a free module over $A_0:=
 k[f_0,\ldots, f_{p-2}]$.
Since $A\cong k[\mathfrak{t}]^W$ has Krull dimension $p-1$, the $f_i$'s form a homogeneous sytem of parameters for $A$. If $k[\mathfrak{g}]$ is free over $A$ then $A$ is finitely generated and projective over $A_0$. But then $A$ is free over $A_0$ implying that $A$ is Cohen-Macaulay. This contradiction shows that Kostant's freeness theorem fails in our case.<|endoftext|>
TITLE: f.g. modules vs. f.g. projective modules
QUESTION [9 upvotes]: In algebraic K-theory one defines $K_0(R)$ as the result of application of the Grothendieck construction to the semigroup of isomorphism classes of left f.g. projective $R$-modules.
But we can also consider the category of left f.g. $R$-modules and apply the same construction to obtain a group (let's call it $G(R)$).
What can we say about the canonical homomorphism
$$K_0(R) \to G(R)$$
induced by the inclusion. Is it onto? or maybe is it an isomorphism?
I am mostly interested in the case when $R$ is a group ring of a finitely presented group.

REPLY [3 votes]: Typically the homomorphism here $K_0 \rightarrow G_0$ fails to be surjective.  This shows up in a wide range of examples involving group algebras of finite groups (over fields of characteristic dividing the group order), restricted enveloping algebras of modular Lie algebras, etc.   The homomorphism itself is often called the Cartan homomorphism:  in some of his early work, Elie Cartan studied the regular representation of a finite dimensional algebra ("hypercomplex system").   See for example Chapter IX.2 of the pioneering 1968 Benjamin lecture notes Algebraic K-Theory by Hyman Bass.   
The Grothendieck group formalism was popularized in modular representation theory of finite groups by Serre in his lecture notes (later translated into English as a volume in the Springer GTM series).   That's one of many sources for concrete examples showing why the Cartan map is usually not surjective; Curtis-Reiner is another source.   There are also many easy examples in the books and papers on representations of algebraic groups and their Lie algebras in prime characteristic, with analogues in characteristic 0 for quantum groups at a root of unity.  
I'd emphasize that there are many different situations involving rings and finitely generated projectives, but only in fairly simple cases is the Cartan map likely to be surjective.<|endoftext|>
TITLE: Group rings isomorphic over $\mathbf{F}_p$, but not over $\mathbf{Z}_p$?
QUESTION [18 upvotes]: Suppose given a prime $p$.
Question: Do there exist finite groups $G$ and $H$ such that ${\bf F}_p G$ is isomorphic to ${\bf F}_p H$, but such that ${\bf Z}_p G$ is not isomorphic to ${\bf Z}_p H$ ?
Variants: Suppose given $s\geqslant 2$ and replace ${\bf Z}_p$ resp. ${\bf F}_p$ by ${\bf Z}/p^s$.
Variant: Suppose $G$ and $H$ to be $p$-groups. (It is unknown whether there are nonisomorphic $p$-groups with isomorphic group rings over ${\bf F}_p$ , but still, maybe someone knows an argument in favour of 
${\bf F}_p G \simeq {\bf F}_p H$ $\Rightarrow$ ${\bf Z}_p G \simeq {\bf Z}_p H$ in this case?)

REPLY [11 votes]: In a very short paper put on the arXiv recently, García, Margolis and del Río give examples of nonisomorphic finite $2$-groups $G$ and $H$ with $\mathbb{F}_2G\cong\mathbb{F}_2H$, thus solving the modular isomorphism problem. The smallest examples have order $512$.
Combined with Roggenkamp and Scott's theorem that $\mathbb{Z}_pG\cong\mathbb{Z}_pH\Rightarrow G\cong H$ for $p$-groups, this answers the question when $p=2$.<|endoftext|>
TITLE: Distribution of Maximum of a uniform multinomial distribution
QUESTION [6 upvotes]: Hello, I'm working with a data structure which uses a uniform distribution to bucket the inputs into $k$ buckets. The efficiency of the structure is bounded by the $\frac{k_{max}}n$, where $n$ is the number of items. How many elements are in each bucket will follow a uniform multinomial distribution. What will the distribution be for the number in the largest bin. We can assume that $n$ is much larger than $k$, and an approximate answer is good. I just want to be able to say something like: the largest bucket will have at most $(1.5k)/n$ elements with probability $p$.

REPLY [6 votes]: The probability that there is at least one bin with at least $c$ items is less than or equal to the expected number of bins with at least $c$ items, which is $k$ times the probability that a particular bin has at least $c$ items. You can bound the probability that a particular bin contains at least $c$ items using the Hoeffding inequality.
$$\begin{eqnarray}Pr(\max \ge n/k + d) & \le & k Pr(\text{Binomial}(n,1/k) \ge n/k + d) \\\ & \le & k \exp(-2d^2/n).\end{eqnarray} $$
There are sharper bounds available such as the Chernoff bound, but this is simple and it sounds like it will suffice.

REPLY [4 votes]: This is addressed by Bruce Levin, 1983, "On Calculations Involving the Maximum Cell Frequency."
Also in
http://www.jstor.org/stable/2347220 .<|endoftext|>
TITLE: Does normal Riemann hypothesis follow from extended Riemann hypothesis
QUESTION [5 upvotes]: Does normal Riemann hypothesis for $\zeta_{\mathbb{Q}}$ follows from the extended Riemann hypothesis for some $K \neq \mathbb{Q}$ (i.e. the statement that all zeroes of the Dedekind zeta function $\zeta_K$ for $K$ a number field in the critical strip lie on the axis $\mathfrak{R}(s)=1/2$)?

REPLY [11 votes]: Yes, certainly: zeta functions of abelian extensions of $\mathbb Q$ factor as products of Dirichlet $L$-functions over $\mathbb Q$, including $\zeta(s)$, and those other $L$-functions have no poles (which might cancel an off-line zero of zeta), so RH for any abelian extension of $\mathbb Q$ implies that for $\mathbb Q$.
At this point in history, the fact that we do not know that all Artin $L$-functions lack poles in the critical strip severely complicates our provable understanding of an analogous assertion for not-abelian extensions.<|endoftext|>
TITLE: Non-regular Connected Hausdorff Banach Manifold
QUESTION [23 upvotes]: After reading this MO post, I am wondering: 
Is every (connected) Hausdorff Banach manifold a regular space?
Though unjustified, page 53 of this paper nonchalantly states: "Note that a Hausdorff Banach manifold X is a regular space."
But does anyone know of a proof of this statement (or a counterexample)?
Of course, the real difficulty arises in proving the statement for the infinite-dimensional version, since such a Banach manifold will not be locally compact.

Follow-up: Now that Theo Buehler has kindly pointed to a counterexample (i.e. a connected Hausdorff Banach manifold which is not regular) perhaps it will give someone an idea about how to tackle the question that provided the inspiration for this one.

REPLY [16 votes]: Apparently the answer is no, not every connected Hausdorff Banach manifold is regular, not even when it is modeled on a separable Hilbert space.
I quote (verbatim) from J. Margalef-Roig, E. Outerelo-Dominguez, Differential Topology, North Holland Mathematics Studies 173, 1992, page 44f.

It is well known the result of General Topology that every
  Hausdorff locally compact topological space satisfies the
  Tychonoff axiom [M-O-P, V.2, pg. 231]. By this and the Riesz's
  theorem every Hausdorff locally finite dimensional differentiable
  manifold satisfies the Tychonoff axiom. This last affirmation is not
  true for arbitrary Hausdorff differentiable manifolds. In
  [M.O.1] there is an example of a Hausdorff connected
  differentiable manifold $X$ of class $\infty$, such that $\partial X = \emptyset$,  $X$ is not
  regular and $X$ admits an atlas whose charts are modelled over an
  infinite dimensional separable real Hilbert space.

They continue to add the regularity hypothesis in their results whenever it is needed.
The cited references are:

[M.O.P.] MARGALEF, J.-OUTERELO, E.-PINILLA, J.L.: Topologia,
I-V. Alhambra, Madrid 1975, 79, 79, 80 and 1982.
[M.O.1] MARGALEF, J.-OUTERELO, E.: Una variedad diferenciable
de dimension infinita, separada y no regular.
Rev. Mat. Hisp.-Am, IV, V.42, 1982, 51-55. (QuickView link).


Edit: As was pointed out by Benjamin Dickman in the comments, the example also appears in English in A. Kriegl, P.W. Michor, The convenient setting of global analysis, AMS (1997), 27.6 Non-regular manifold, page 266. The book is available as a pdf file on Kriegl's homepage.<|endoftext|>
TITLE: Complex Analysis - Analytic Continuation and Residual Integration
QUESTION [5 upvotes]: At first an example which I know how to treat.
Let's say we have the following integral
$\int dx \frac{1}{x^2+a^2}$
Now we do an analytic continuation of the constant $a$ to the complex numbers: $a \rightarrow ia$. This gives
$\int dx \frac{1}{x^2-a^2}$
which now has poles at $x = \pm a$. Let's say I want to evade these poles during the integration (small half-circles into the complex plane around the singular points), then I have to take the half-circle at $x = -a$ into the lower half-plane (integration counter-clockwise), and the half-circle at $x = a$ into the upper half-plane (integration clockwise). As can be seen in this picture:
>click<
Whether the integration of a half-circle is clockwise or counter-clockwise, determines the extra sign before the residual part of the integral to be minus or plus respectively.
Now let's get to my actual question.
Imagine, instead of the simple example above we have the following integral:
$\int_{0}^{\frac{\pi}{2}} dx \frac{1}{1-f(a^2)\sin^2(x)}$
where $f(a^2) = \frac{\text{polynomial in }a^2}{\text{other polynomial in }a^2}$ and $0 < f(a^2) < 1$ for all $a$.
Since $0 < f(a^2) < 1$, there are no singularities in the integral. Now we do the analytic continuation $a \rightarrow ia$. This gives
$\int_{0}^{\frac{\pi}{2}} dx \frac{1}{1-f(-a^2)\sin^2(x)}$
and it is given that always $f(-a^2) > 1$. That causes a singularity to appear in $x = \arcsin(\frac{1}{\sqrt{f(-a^2)}})$. (Before the analytic continuation there was no singularity, because $\arcsin(y) $ is defined only for $-1\leq y\leq 1$). It is not hard to compute the residue at this singularity, but now it is less transparent, whether we should evade the pole drawing the small half-circle into the upper or into the lower half-plane. This little detail determines the sign of the residual part of the integral (I would like to separate it in the residual and principal parts). How can I determine whether I should draw the half-circle in the lower or in the upper half-plane in the general case, or in this situation at least?

REPLY [4 votes]: Notice that in the picture you linked to, the poles moved from $\pm ia$ toward the real axis along counter-clockwise arcs that did not cross the real axis itself (where the integration contour lies). It is this path of approach of the poles toward the real axis that determines how the integration contour needs to be deformed. Informally, the poles make little "dents" in the integration contour as they approach it. If the poles were moved from $\pm ia$ toward the real axis along clockwise arcs, then these dents would be in the other direction and you would get a different sign in front of the residual part of the integral and hence a different answer overall. This shows that the final answer for the integral has branch cuts in its analytic dependence on $a$.
The answer to your main question depends on the way the location of the pole at $x_z=\arcsin(f(z^2)^{-1/2})$ as a function of $z$. You need to explicitly pick a path in the complex $z$-plane that takes you from $a$ to $-ia$ such that the pole $x_z$ does not cross your integration contour. As the pole approaches the real axis, it will make a dent in your integration contour, which will determine the way you need to deform it.<|endoftext|>
TITLE: The kernel of all invariant means
QUESTION [7 upvotes]: Let $G$ be a discrete group which is amenable (i.e. it admits an left-invariant mean, i.e. a continuous positive normalised linear functional $m:\ell^\infty(G) \to \mathbb{R}$ such that $\forall g \in G, m \circ \lambda_g = m$ where $\lambda_g : \ell^\infty(G) \to \ell^\infty(G)$ is the left-regular action of $G$). Let $\mathcal{M}$ be the set of all invariant means on $G$.
Let $c_0(G)$ be the space of bounded functions decreasing to $0$ at $\infty$ (i.e. $f \in c_0(G)$ if $f$ is in the closure of the space of finitely supported functions, i.e. $f \in \ell^\infty(G)$ and $\forall \epsilon >0$ there exists a finite set $F \subset G$ with $\lVert f\rVert_{\ell^\infty(G \setminus F)} \leq \epsilon$).
It's not difficult to see that $c_0(G)$ is included in the kernel of all (left- or right- or bi-)invariant mean, provided $G$ is infinite. When $G$ is finite, there is only one such mean, and it is trivial to compute the kernel. However the cardinality of $\mathcal{M}$ is otherwise quite big (if I remember correctly, it is uncountable for $G=\mathbb{Z}$).
Question: Assuming $G$ is infinite, is there a description/characterisation of the elements of $\ell^\infty(G)$ which belong to the kernel of all the invariant means, i.e. $\bigcap_{m \in \mathcal{M}} \ker m$?

REPLY [2 votes]: (Essentially from Narutaka Ozawa's comments)
Let $K = \bigcap_{m \in \mathcal{M}} \ker m$. The trivial answer would be that $K$ is the closed linear span of
$$S =\{f-\lambda_g(f):\;g\in G,\;f\in\ell^\infty(G) \}.$$
For a non-amenable group this is all of $\ell^\infty(G)$. For an amenable group with Følner net $(F_i)$, then for any $f \in K$
$$\lim_{i\to\infty}\sup_{x\in G}\frac{\left|\sum_{y\in F_ix}f(y)\right|}{|F_i|}=0.$$
Indeed, if it's nonzero, we can find $(x_i)$ such that
$$\limsup\frac{\left|\sum_{y\in F_ix_i}f(y)\right|}{|F_i|}>0;$$
then the uniform measure on $F_ix_i$ accumulates to an invariant mean that is non-zero at $f$.
Hence, if $f \in K$, then
$$f_i(x)=\sum_{y\in F_ix}\frac{f(y)}{|F_i|}$$
has the same mean as $f$ for every invariant mean. A consequence of the above is that $\|f_i\|_{\ell^\infty} \to 0$. Thus $f - f_i \in S$ and $f-f_i \to f$ which shows that $K \subset \overline{S}$.
The inclusion $\overline{S} \subset K$ is a direct consequence of the invariance of the mean.<|endoftext|>
TITLE: Are irreducible representations subrepresentations of a symmetric power representation?
QUESTION [7 upvotes]: First of all I am far from being an expert in representation theory, so it is possible (likely) that the following question is trivial (in fact a trivial reference question):
Let $\Gamma$ be a, let's say finite (abelian) group, and $R$ be an effective representation of $\Gamma$.
Is it true, that every irreducible $\Gamma$-representation is a subrepresentation of the $k$-th symmetric power representation $S^kR$, for some $k=0,1,\dots$ ? 
(Note that it is enough to show that any regular representations of $\Gamma$ is a subrepresentation of $S^kR$.)
I strongly suspect that this is a well know fact in representation theory, but I couldn't find any reference so far and would therefore appreciate any kind of literature reference.

REPLY [10 votes]: Edit: The answer is yes for all finite groups $G$ (even, as far as this makes sense, independently of the ground field). A reference is Theorem 1 on page 45 of Alperin: "Local representation theory". While the claim of this theorem only refers to the tensor product
$V\otimes \ldots \otimes V$ for some faithful $kG$-module $V$, the proof actually constructs
a free submodule in
$$
  k[V] \cong \bigoplus_{i=0}^{\infty} S^k V
$$ 
which gives you exactly what you need.

For a finite abelian group $G$ the answer is yes. Assume $V=\langle v_1\rangle\oplus \ldots \oplus \langle v_n \rangle$ is the decomposition of a faithful $\mathbb C G$-module $V$ into irreducibles. Denote the character associated to $v_i$ by $\chi_i$. Since $V$ is faithful the $\chi_i$ must generate the character group $\textrm{Hom}(G,\mathbb C^\times)$. So we can write any irreducible character $\psi$ of $G$ as 
$$
   \psi = \chi_1^{k_1}\cdots \chi_n^{k_n}
$$
and then the vector $v_1^{k_1}\cdots v_n^{k_n} \in S^{k_1+\ldots+k_n} V$ spans a one-dimensional submodule of this symmetric power with character $\psi$.<|endoftext|>
TITLE: Characteristic subgroups of extra-special p-groups
QUESTION [5 upvotes]: Let $G$ be a extra-special $p$-group of order $p^{1+2r}$ with exponent $p$ (p odd). I want to know  if $G$ has only $3$ characteristic subgroups? 
Background: From [2],  if $G$ is extra-special $5$-group of order $5^5$ with exponent $5^2$, then $G$ has more than $3$ characteristic subgroup. How about the exponent of the extraspecial $p$-group is $p$?
There are some references about this topic.
[1]. D.R. Taunt, Finite groups having unique proper characteristic subgroups I, Proc. Cambridge Philos. Soc. 51 (1955) 25–36.
[2]. S.P. Glasby, P.P. Pálfyb, Csaba Schneider p-groups having a unique proper non-trivial characteristic subgroup Journal of Algebra 348 (2011) 85–109

REPLY [9 votes]: An extraspecial $p$-group of exponent $p$ contains exactly three characteristic subgroups, $1$, $G$ and the center of $G$.
Let $Z$ be the center of $G$ (so $Z=[G,G]=\Phi(G)$).  The elementary abelian group $G/Z$ is a vector space of dimension $2r$ over the field of order $p$.  The commutator map on $G$ induces a nondegenerate alternating bilinear form on $G/Z$.  As shown in a paper of D. L. Winter in the Rocky Mountain Journal (1972), $Aut(G)$ has a subgroup $H$ of index $p-1$ such that $H/Inn(G)$ is isomorphic to the full stabilizer of the given form (this does not hold if $G$ does not have exponent $p$).  Since this stabilizer is irreducible on $G/Z$, no characteristic subgroup of $G$ (other than $G$) strictly contains $Z$.  Now assume for contradiction that $G$ has some nontrivial proper characteristic subgroup $X$ that does not contain $Z$.  Then $XZ$ is characteristic in $G$ and strictly contains $Z$, which forces $XZ=G$.  Now $X$ is maximal in $G$.  However, this forces $Z=\Phi(G)\leq X$, a contradiction.<|endoftext|>
TITLE: Prime-like elements of rings
QUESTION [5 upvotes]: An element $p$ of a commutative ring $R$ is called "prime" if, for any $a,b\in R$, whenever $ab$ is a multiple of $p$, either $a$ or $b$ is a multiple of $p$.  
Is there a word for the "prime-like" property that, whenever $ab$ is a multiple of $p^2$, either $a$ or $b$ is divisible by $p$?  Or another, more usual concept in ring theory that this is connected to?
I ask because the "prime-likeness" of $2$ in $R$ seems to control whether the quadratic formula can be made to work for monic polynomials over $R$ (as long as $2$ is also not a zero-divisor).  This is because, if the discriminant of $x^2 + bx + c$ is a square $b^2 - 4c = d^2$, then $(-b+d)(-b-d) = 4c$, so at least one (and hence both) of $(-b+d)$ and $(-b-d)$ are multiples of $2$ in $R$.  Their halves are the two roots of $x^2 + bx + c$.
For example, $2$ is "prime-like" in $\mathbb{Z}[\sqrt{2}]$, which is easy to verify elementarily.  Hence a monic quadratic over $\mathbb{Z}[\sqrt{2}]$ factors iff its discriminant is a square.  But $2$ is not "prime-like" in $\mathbb{Z}[\sqrt{5}]$, since $(\sqrt{5}-1)(\sqrt{5}+1) = 4$.  And indeed, the discriminant of $x^2 -x-1$ is a square in $\mathbb{Z}[\sqrt{5}]$, but the polynomial doesn't factor there.

REPLY [2 votes]: I assume (based on your example) that you're primarily interested in the case where $R$ is the ring of integers in an algebraic extension of $\mathbb Q$. Then your property of being "prime-like" is equivalent to the property of generating a primary ideal. 
So assume $R$ is a Dedekind domain (in particular ideals factor uniquely into products of prime ideals), and let $p\in R$ be an arbitrary element. Then 

$p$ is "prime-like" if and only if $pR=P^k$ for some prime ideal $P$ of $R$ and some $k\in \mathbb N$

i.e. "prime-like" in Dedekind domains is the same as "primary". An element $x$ in the fraction field of $R$ lies in $R$ iff the valuations $\nu_Q(x)$ are non-negative for all prime ideals $Q$ of $R$. So let $a,b\in R$ and assume $p^2\mid ab$. Then $\nu_Q(a/p)=\nu_Q(a)\geq 0$ and $\nu_Q(b/p)=\nu_Q(b)\geq 0$ for all primes $Q\neq P$. So to see that either $a/p$ or $b/p$ lies in $R$, one just has to check that one of them has positive $P$-valuation. But

$p^2\mid ab$ implies $\nu_P(a)+\nu_P(b)=\nu_P (ab) \geq \nu_P(p^2)=2\nu_P(p)$  which implies $\nu_P(a) \geq \nu_P(p)$ or $\nu_P(b) \geq \nu_P(p)$

so either $\nu_P(a/p)\geq 0$ or $\nu_P(b/p)\geq 0$. On the other hand, if the ideal $pR$ isn't primary then $p$ is not prime-like (the construction in Julian's comment can be generalized).
Of course I'm not sure what exactly you're looking for, but at least this clears up what is going on in your last example: while $2$ isn't "prime-like" in $\mathbb Z[\sqrt{5}]$, it is prime like in its integral closure $\mathbb Z[\frac{1+\sqrt{5}}{2}]$ (which is however unspectacular because it remains a prime in that ring).<|endoftext|>
TITLE: Is there a semisimple $\mathbf{Q}_\ell$-representation of $G_F$ ramified at an infinite set of places?
QUESTION [5 upvotes]: See http://math.uni.lu/~wiese/galois/Boeckle-Luxemburg-Notes.pdf, Theorem 1.4(a): Is there an example of a semisimple $\mathbf{Q}_\ell$-representation $V$ of $G_F$ ($F$ a global field) ramified at a set $S$ of places where $S$ is not finite (for every $\dim{V} \geq 1$)?

REPLY [6 votes]: In the case $\dim V=2$, my advisor Ravi Ramakrishna has shown in his paper Infinitely ramified Galois representations (Ann. Math. 151) that there are surjective representations $\rho\colon G_\mathbf{Q}\to \mathrm{GL}_2(\mathbf{Z}_p)$ ramified at infinitely many primes.<|endoftext|>
TITLE: When k[G/H] is multiplicity free G module ? 
QUESTION [9 upvotes]: Consider finite group G and its subgroup H, and representation  of G in k[G/H] i.e. functions on G/H.
Question: What is known about the question: when k[G/H] is multiplicity free ? (Let us consider k - complex numbers for simplicity).
More generally consider $Ind^G_H V$ some induced representation of G from $H$, same question. 
(In case V= trivial representation we get previous question).
Are there some general conditions ? What is know about the cases G = S_n, A_n, GL_n(F_q) ? 
Example: $S_{n-1} \in S_n$ is  multiplicity free since G/H = C^n - standard representation.
Particular question $H=GL_{n}(F_q) \subset G= GL_{n+1}(F_q)$ is C[G/H] multiplicity free ?
(This is true from GL(C) and leads to Gelfand-Zeitlin basis theory, well , may be it is not correct). 
More generally what about induced representations in this case ? 
Morally multiplicity free means that $H$ is "big" subgroup of G, however I do not know any precise way to say what "big" means.
PS
I googled "Multiplicity-free permutation representations of the alternating groups", which promises survey on the topic, but but I have no access to this file:(
https://www.ideals.illinois.edu/handle/2142/22828
If someone can help ... would be great...

REPLY [6 votes]: The fact (mentioned in Florian Eisele's answer) that the endomorphism algebra must be commutative gives a further necessary condition for $\mathbb{C}_H\uparrow^G$ to be multiplicity free: $N_G(H)/H$ must be abelian. For $\mathbb{C} N_G(H)/H$ appears as a subalgebra of the endomorphism algebra, coming from those double cosets labelled by an element of the normalizer.
In the case when $G$ is the symmetric group $S_n$, the problem of which permutation characters are multiplicity-free is completely solved for $n \geq 66$ by work of Mark Wildon: http://arxiv.org/abs/0903.2864<|endoftext|>
TITLE: When are $9^{n} - 2$ and $9^{n} + 2$ both prime?
QUESTION [9 upvotes]: Inspired by the discussion at a since-deleted thread:

Question: For $n > 2$, are  $9^{n} - 2$ and $9^{n} + 2$ ever both prime?"

It seems that many different techniques were tried, ranging from Wilson's Theorem to a generalization of Clement's Theorem, and by looking at a couple of relevant OEIS sequences (A090649; A128455).
Note that $9^{n} + 2$ is prime for $n$ beginning with:
$$0, 1, 2, 4, 5, 7, 12, 13, 18, 49, 55, 63, 193, 247, \ldots$$
Similarly, $9^{n}-2$ is prime for $n$ beginning with:
$$1, 2, 3, 11, 45, 51, 260, 324, 390, 393, 1112, 3092, 4445, 10373 \ldots$$

REPLY [9 votes]: A solution was ultimately published in the Pi Mu Epsilon Journal, 13(7), Fall 2012, pp. 431 (JSTOR).
It was attributed to Paul S. Bruckman (separate PME interview) and reads, in part, as follows.<|endoftext|>
TITLE: When Do a Few Eigenvectors of Graph Laplacians Not Determine the Graph?
QUESTION [12 upvotes]: Essentially as the title, but I'll give a little bit more background.
I have some finite graph $G$ with $n$ vertices and adjacency matrix $A$. Let $D$ be the $n$ by $n$ matrix with the degree of vertex $i$ at the $i,i$ entry, and 0's everywhere. Finally, let $L = D - A$ be the (unnormalized) graph Laplacian of $G$. Next, fix some collection of eigenvectors and eigenvalues of $L$.
My big-picture question is: Under what conditions are there other graphs which share those eigenvectors/values? 
With a little more precision: Approximately how many eigenvectors & eigenvalues can be specified before the answer is no? About how many graphs are there when the answer is yes?
It seems likely that the answer would be a little complicated. I know a few special cases (e.g. 2 eigenvectors/values determine cycles completely; on the other hand, as long as the average degree is fairly large, there are generally very many graphs with the same bottom eigenvector). I certainly appreciate hearing about conditions which aren't tight, as long as they are at least a little broad.
I'm interested in the situation where the eigenvectors DON'T determine the graph, so I would also appreciate any literature pointers to `relaxations' of this idea. For example, one could imagine requiring that the Laplacian contracts the eigenvectors by at least a certain amount (this certainly allows many graphs, but that space is pretty big). In another direction, it seems plausible there is some analogue in the language of graphons. 
Thanks for any help!

REPLY [10 votes]: The experimental evidence for adjacency matrices  suggests that, for a random graph, its characteristic polynomial is irreducible over the rationals. I would expect that the characteristic polynomial of the Laplacian of a random graph on $n$ vertices would have one irreducible factor of degree $n-1$. However, while it easy is to convince yourself of this by testing on random graphs, absolutely nothing has been proved.
If we move away from the generic case, things become much more complicated. 
All circulant graphs can be assumed to have the same orthogonal basis of eigenvectors
(a Vandermonde matrix) and there are examples of cospectral circulants. (Note
that for regular graphs, the adjacency matrix and the Laplacian have the same eigenvectors, and provide the same spectral information.) So now we
have nonisomorphic graphs with the same eigenvalues and eigenvectors. Hence we must assume that we are given pairs (eigenvalue, eigenvector). This is not quite enough, because our eigenvalues need not be simple. It would seem that our data
will be pairs consisting of an eigenvalue and the matrix representing orthogonal
projection onto the corresponding eigenspace.
Let $M$ be  a $d\times m$ binary matrix with linearly independent rows.
Let $X(M)$ be the graph with the elements of $\mathbb{Z}_2^d$ as its vertices,
two adjacent if and only if their difference is a column of $M$. Such a graph
is a Cayley graph for $\mathbb{Z}_2^d$, and is known as a cubelike graph.
The characters of $\mathbb{Z}_2^d$ are eigenvectors, and the eigenvalues
are integers. The eigenvalues and their multiplicities are determined by the weight enumerator of the binary code generated by the rows of $M$ - a code word of weight $k$ gives an eigenvalue $m-2k$. Since there are nonisomorphic binary codes with the same weight enumerator, there are cospectral cubelike graphs. I do not know
how much eigendata two cubelike graphs may have in common, but this would
seem a good place to look.
EDIT: Let $G$ be a graph with vertex set $\{1,\ldots,n\}$. Let $H$ be a graph with
vertex set $\{\pm1,\ldots,\pm n\}$; if $i$ and $j$ are adjacent in $G$, let
$(i,j)$ and $(-i,-j)$ be edges in $H$, and if $i$ and $j$ are not adjacent 
in $G$, let $(i,-j)$ and $(-i,j)$ be edges. (Note that $H$ is regular 
with valency $n-1$.) Call $H$ the two-graph constructed from $G$.
Let $\pi$ be the partition of $V(H)$ with cells $\{i,-i\}$.
Then the subspace of $\mathbb{R}^{2n}$ formed by the vectors constant on cells
of $\pi$ is invariant under the adjacency matrix of $H$. It is the direct
sum of the span of the constant vectors (which has dimension 1) and the
vectors constant on cells that sum to 0 over $V(H)$. The second subspace is 
an eigenspace for $A(H)$, with eigenvalue $-1$. Since the space of vectors
constant on cells does not depend on the structure, distinct two-graph
on $2n$ vertices have $n$ eigenvectors and eigenvalues in common. (The spectrum of
a two-graph is the union of the spectrum of $K_n$ and the spectrum of $2A(G)+I-J$).<|endoftext|>
TITLE: Why take 'complex powers' of pseudo-differential operators?
QUESTION [10 upvotes]: Given a pseudo-differential operator $P$ of order zero, Seeley showed that the holomorphic family of operators $\lbrace P^{z} : z\in \mathbb{C} \rbrace$  of all complex powers is contained in the class of pseudo-differential operators. 
Apart from knowing that we can take powers of these operators, is there any application of this theory? I understand the utility of raising operators to fractional powers. But, irrational and complex powers are not clear.

REPLY [4 votes]: Bounded imaginary powers of differential operators may give informations about maximal regularity of evolution equations. see Buzano &Nicola : Complex powers of hypoelliptic pseudodifferential operators.<|endoftext|>
TITLE: Continuous notions with compelling discrete analogues
QUESTION [12 upvotes]: Following up on the previous MO question "Are there any important mathematical concepts without discrete analogue?", I'd like to ask the opposite: what are examples of notions in math that were not originally discrete, but have good discrete analogues?  While a few examples arose in the answers to that earlier MO question, this wasn't what that question was asking, so I'm sure there are many more examples not mentioned there or at least not really explained there.  What reminded me of this older MO question was seeing an MO question "Why is the Laplacian ubiquitous?", since that is an instance of an important notion which has a discrete analgoue. 
In an answer, it would be interesting to hear about the relationship between the continuous and discrete versions of the notion, if possible, and references could also be helpful.  Thanks!

REPLY [7 votes]: Finite graphs are a rich source of discrete analogues (I will be partially repeating the OP and some other answers here): 

The Laplacian on a finite graph is a discrete analogue of the Laplacian on a Riemannian manifold. In particular, it is possible to formulate the heat equation, the wave equation, and the Schrödinger equation on a finite graph. There are actually two Laplacians, a vertex Laplacian and an edge Laplacian, which give a discrete analogue of Hodge theory. 
The Ihara zeta function of a finite graph is a discrete analogue of the Selberg zeta function of a Riemannian manifold. A regular graph satisfies an analogue of the Riemann hypothesis if and only if it is a Ramanujan graph. There is also an analogue of the Selberg trace formula in this setting; Terras has written extensively about this kind of thing. 
The Picard group (or critical group, or sandpile group) of a finite graph is a discrete analogue of the Picard group of an algebraic curve. More generally a lot of the theory of algebraic curves can be transported to this setting, e.g. the Riemann-Roch theorem. 

(Finite graphs are also a rich source of other kinds of analogues; for example the Ihara zeta function is also analogous to the Dedekind zeta function of a number field, with coverings of graphs analogous to extensions of number fields and the Picard group analogous to the class group. There is even an analogue of the analytic class number formula in this setting although I have forgotten the reference.)

REPLY [2 votes]: Continuous-time random walks on graphs are in some sense a discrete analogue of diffusions on a Riemannian manifold (of course, the reverse can be argued, but I think that diffusions play a more central role in modern probability theory).  Of course, the most important diffusion is Brownian motion, i.e., the Markov process associated with the Laplace-Beltrami operator.  From my perspective, the natural analogue of Brownian motion is the operator $\mathcal{L}_V$ given by (we use unweighted graphs for simplicity)
\begin{equation*}
(\mathcal{L}_Vf)(x) := \sum\_{y\sim x}(f(y)-f(x)).
\end{equation*}
A more 'common' choice might be the rate-1 continuous time random walk with generator $\mathcal{L}_C$ given by
\begin{equation*}
(\mathcal{L}_Cf)(x) := \frac{1}{\deg(x)}\sum\_{y\sim x}(f(y)-f(x)).
\end{equation*}
However, this choice of generator has several 'bad' properties if you want to view it as an analogue of Brownian motion -- for example, the generator is always bounded on $L^2(\deg)$, it cannot have discrete spectrum, and the associated random walk cannot explode; in contrast, the operator $\mathcal{L}_V$ may be unbounded, and discrete spectrum and explosiveness are possible.
Once you have this discrete (space) analogue of Brownian motion on a Riemannian manifold, a natural question is to ask what the discrete analogue of the Riemannian metric should be for this process.  It is not too hard to find examples that show that the graph metric is a bad analogue, since the Riemannian metric governs heat flow (in some sense) on a Riemannian manifold (see e.g. here), but Gaussian heat kernel estimates do not hold for the random walk associated with $\mathcal{L}_V$ if you take the manifold heat kernel estimates and replace the distance function with the graph metric.  A reasonable analogue has been formulated recently, see e.g. here and here.<|endoftext|>
TITLE: Differential Calculus and the De Rham Homotopy Operator
QUESTION [8 upvotes]: Suppose $M$ is a smooth manifold, $\Omega^k(M)$ the vector space (or $C^\infty(M)$-module in case this is better suited) of $k$-differentialforms on $M$ and
$$I: \Omega^k(M\times \mathbb{R}) \to \Omega^{k-1}(M)$$ the De Rham homotopy operator defined by $I(\omega):= \int_0^1i_{\partial_t}\omega(t)dt$. (Where $t$ is assumed to be the global coordinate of $\mathbb{R}$ and $\partial_t$ its appropriate tangent coordinate)
...
I would like to know, how $I$ interacts with the operators of differential calculus:
First we have the well known equation:
$$d\circ I + I \circ d = i_1^* − i_0^*$$
where $i_j^*$ is the pullback defined by the inclusion 
$i_j : M \to M \times \mathbb{R}; x \mapsto (x,j)$. This is the 'interaction' of $I$ with the exterior differential $d$. 
1.) How does $I$ interacts with the interior product 
$i_X: \Omega^k(M \times \mathbb{R}) \to \Omega^{k-1}(M \times \mathbb{R})$ for a vector field $X$ on $M$ (or $M \times \mathbb{R}$) or (via inclusion) with 
$i_X: \Omega^k(M) \to \Omega^{k-1}(M)$ ?
2.) How does $I$ interact with the exterior product?

Edit: If someone knows a different definition of a homotopy operator, with a known behavior 
related to the interior product $i_X$, I would like to know it, too.

REPLY [2 votes]: If $X$ is a vector field on $M$ (time independent) then 
$i_X \circ I = - I \circ i_X$. If $X$ is also time dependent you can play with the cases $\omega$ exact etc.<|endoftext|>
TITLE: Fundamental group of the moduli stack of elliptic curves
QUESTION [26 upvotes]: I've heard that the étale fundamental group of the moduli stack of elliptic curves (over $\mathbb{Z}$) is trivial. Is there an easy proof of that? (Note that there are plenty of étale covers once one inverts a prime $p$, given by taking elliptic curves with some form of a level $p^n$ structure.)
More generally, I'd be interested in how it works out for the stack of cubic curves which are allowed either to be smooth or to have a nodal singularity.

REPLY [32 votes]: Yes, there is a proof which is long but one might consider to be "easy" after digesting it. 
Let's first show that the question (of triviality of connected finite etale covers) for the moduli stack $M_1$ is equivalent to its counterpart for the "Deligne-Rapoport" compactification $\overline{M}_1$ (a regular proper DM stack), since the method of the harder direction will be used in our argument for the case of $M_1$. The easier direction is that if the case of ${M}_1$ is known then we can settle the case of $\overline{M}_1$. It suffices to show that if a normal noetherian DM (or Artin) stack $X$ has a dense open substack $U$ with no nontrivial connected finite etale cover then the same holds for $X$. It suffices to show more generally that if $U$ is a dense open substack of $X$ and $X' \rightarrow X$ is a finite etale cover and $s:U \rightarrow X'|_U$ is a section over $U$ then $s$ uniquely extends over $X$. The uniqueness allows us to work over a smooth scheme chart, so we're reduced to the well-known case when $X$ is a scheme (though one can also adapt to the case of stacks the proof in the scheme case via chasing connected components).
The more interesting direction is the converse: the case of $\overline{M}_1$ implies the case of $M_1$.  For this we just have to show that for any connected finite etale cover $T$ of $M_1$ the connected finite normalization $\overline{T} \rightarrow \overline{M}_1$ (which is flat since $\overline{M}_1$ is a regular DM stack of dimension 2) is etale around the closed complement $\infty$ of $M_1$ in $\overline{M}_1$. Since $\overline{M}_1$ is regular and $\infty$ is a relative Cartier divisor that is regular with generic characteristic 0, it follows from the relative Abyhankar Lemma (Exp. XIII, SGA1 for schemes, easily adapts to DM stacks by usual etale-localization stuff) that $\overline{T}$ is "relatively tamely ramified" along $\infty$. Thus, the etale-local structure near $\infty$ is given by an $e$th-root extraction of a local generator of the ideal of the connected substack $\infty$ with $e$ a unit along $\infty$. But (as in Saito's argument mentioned by Minhyong) ${\rm{Spec}} \mathbf{Z}$ supports points of every possible prime residue characteristic, so $e$ has no prime factors and hence $e = 1$, so $\overline{T}$ is etale over $\overline{M}_1$ as desired.  
Now we directly attack the case of $M_1$ (though one can also solve the case of $\overline{M}_1$ directly in a more illuminating "topological" manner over $\mathbf{C}$ after some preliminaries with the triviality of $\pi_1({\rm{Spec}}(\mathbf{Z}))$ to handle geometric connectivity of connected components, but that gets caught up in "foundational" issues related to analytification of stacks). We shall initially work with the open substack $M$ of elliptic curves with $j \ne 0, 1728$ on fibers; i.e., the stack of elliptic curves whose automorphism scheme is the constant group $\langle \pm 1 \rangle$ (exercise: equivalent to impose this condition on automorphism groups of geometric fibers, since that constant group has no nontrivial automorphisms); one might say this is the moduli stack of elliptic curves with "no extra automorphisms". I claim that $M$ has exactly one nontrivial connected finite etale cover, a scheme cover of degree 2, and we'll use this to bootstrap to get a handle on the entire moduli stack.
Let $Y$ be the open subscheme of $\mathbf{A}^1_{\mathbf{Z}} = {\rm{Spec}}(\mathbf{Z}[j])$ defined by $j(j-1728)$ being a unit. Early in Silverman's first book on elliptic curves you'll find an elliptic curve $E$ over $Y$ with $j$-invariant $j$, and I claim that the corresponding morphism $f:Y \rightarrow M$ is a finite etale $\mathbf{Z}/(2)$-torsor. The $j$-invariant of the universal elliptic curve over $M$ defines a morphism $j:M \rightarrow Y$, so it makes sense to form the elliptic curve $j^{\ast}(E)$ over $M$. One sees that $f$ is precisely the Isom-stack between $j^{\ast}(E)$ and the universal elliptic curve over $M$, and this Isom-stack is a torsor for the automorphism functor of the universal elliptic curve over $M$, which is to say for the constant group $\mathbf{Z}/(2)$ over $M$, because any two elliptic curves with no extra automorphisms are isomorphic etale-locally on the base if they have the same $j$-invariant (exercise in deformation theory, etc.).
Let's grant that $\pi_1(Y) = 1$, and see how to conclude.  Then at the end we will prove $\pi_1(Y) = 1$. Consider a connected finite etale cover $q:M' \rightarrow M$ of degree $> 1$. I claim it is isomorphic to $f$. Consider the pullback $Y' \rightarrow Y$ of $q$ along $f:Y \rightarrow M$.  Since $Y$ has trivial fundamental group, this pullback splits as a disjoint union of copies of $Y$. Choosing such a component of the pullback defines a morphism $s:Y \rightarrow M'$ over $M$.  But $f$ and $q$ are finite etale maps, so $s$ is also a finite etale map.  But $M'$ is connected, so the open and closed image of $s$ is full; i.e., $s$ identifies $Y$ as a finite etale cover of $M'$, so we conclude that $M'$ is sandwiched inside the degree-2 finite etale cover $f$.  But $q$ has degree $> 1$, so it follows that $q = f$ as desired. 
OK, now we can solve the original problem for $M_1$ (conditional on the triviality of $\pi_1(Y)$). The moduli stack $M_1$ is regular and connected with $M$ a dense open substack that we have just seen has exactly one nontrivial connected finite etale cover. Let $M'_1 \rightarrow M_1$ be a connected finite etale cover with degree $> 1$.  We seek a contradiction.  The restriction over $M$ is a finite etale cover $M' \rightarrow M$ of degree $> 1$, and since $M'$ is open in the connected regular stack $M'_1$ it must also be connected.  Thus, $M'$ is $M$-isomorphic to $Y$ (over $M$ via $f$).  Hence, the elliptic curve $E$ over $Y$ extends to an elliptic curve $E'_1$ over $M'_1$ (namely, the pullback of the universal elliptic curve over $M_1$!). 
The integral structure has done its job, and now to get the contradiction we consider a connected etale scheme neighborhood $(S,s)$ of a point $\xi$ with $j=0$ (or $j=1728$) on the DM stack $M'_1$ considered over $\mathbf{Q}$. We extend $S$ to a smooth connected complete curve $\overline{S}$ (with constant field that might be larger than ${\mathbf{Q}}$, but that won't matter for the ramification considerations we are about to undertake). Clearly $\overline{S}$ is a finite flat cover of the projective $j$-line over $\mathbf{Q}$ and at the point $s$ over $j=0$ (or $j=1728$) it has ramification degree 4 or 6 (I can't remember which is which) because of etaleness over $M_1$ and the fact that $M_1$ over the $j$-line has ramification over $j=0$ (or $j=1728$) equal to 4 or 6 (due to deformation theory considerations).  By design, there is an elliptic curve over the open curve $S$ (namely, the pullback of the elliptic curve $E'_1$ over $M'_1$ that extends the elliptic curve $E$ over $Y$) whose discriminant in the function field of $\overline{S}$ is $(j(j-1728))^{-1}$ (well-defined up to 12th powers of nonzero elements, of course). But the "good reduction" at $s \in S$ forces the discriminant of any model over the function field of $S$ to have valuation at $s$ that is a multiple of 12, whereas for $(j(j-1728))^{-1}$ this valuation is $-4$ or $-6$ (since $S$ at $s$ has ramification over the $j$-line equal to 4 or 6). This is a contradiction, so $M_1$ has no nontrivial connected finite etale cover, assuming 
$\pi_1(Y) = 1$.
Finally, we prove $\pi_1(Y) = 1$. Note that $Y$ is the open complement in $\mathbf{P}^1_{\mathbf{Z}}$ of the union of the sections $\infty$, $j=0$, and $j=n$ with $n = 1728$. We will now work with any nonzero integer $n$. Since $\infty$ is disjoint from the others, by Saito's argument with the relative Abhyankar's Lemma as explained above, we see that any finite etale cover of $Y$ has normalization over that projective $j$-line over $\mathbf{Z}$ that is etale over $\infty$.  Hence, it suffices to show that the open complement of $j(j-n)=0$ in $\mathbf{P}^1_{\mathbf{Z}}$ has trivial $\pi_1$. Making the change of coordinates $t = 1/j$ (which moves $j = 0$ out to $\infty$), this open complement is identified with the open complement $U_n$ in the affine $t$-line $\mathbf{A}^1_{\mathbf{Z}}$ of the locus $nt=1$.  So it is enough to prove that $\pi_1(U_n) = 1$.  Equivalently, we claim that $U_n$ has no nontrivial Galois connected finite etale covers. This will rest on three  special facts about $\mathbf{Z}$: the triviality of $\pi_1({\rm{Spec}}(\mathbf{Z}))$, the triviality of ${\rm{Pic}}(\mathbf{Z})$, and the smallness of the group of roots of unity in $\mathbf{Z}$. Beware that $U_n(\mathbf{Z})$ is empty when $n \not\in \mathbf{Z}^{\times}$.
Let $h:V \rightarrow U_n$ be a Galois connected finite etale cover with degree $> 1$, so over $\mathbf{Q}$ we get a nontrivial connected finite etale cover $V'$ of the $\mathbf{Q}$-fiber $U'_n$ of $U_n$. We first claim that $V'$ must be geometrically connected over $\mathbf{Q}$. Since we're in characteristic 0, this amounts to the condition that $V'$ has constant field $\mathbf{Q}$.  If we let the number field $K$ be its constant field then by normality of $V$ it follows that $h$ factors through $(U_n)_{O_K}$, with $V \rightarrow (U_n)_{O_K}$ necessarily surjective. This forces $(U_n)_{O_K}$ to be etale over $U_n$ (since $h$ is a finite etale cover), so since $U_n$ is fpqc over ${\rm{Spec}}(\mathbf{Z})$ it follows that ${\rm{Spec}}(O_K)$ is etale over ${\rm{Spec}}(\mathbf{Z})$. This forces $K = \mathbf{Q}$, as desired. 
By the coordinate change $x = t/n$ we identify $U'_n$ with ${\rm{GL}}_1$, so $V'$ is a geometrically connected cover of ${\rm{GL}}_1$ over $\mathbf{Q}$. Thus, for $d = {\rm{deg}}(h) > 1$ we see that over an algebraically closed extension $k$ of $\mathbf{Q}$ the map $h_k$ is identified with the endomorphism $x^d$ of ${\rm{GL}}_1$.  That is, the map $h'$ induced by $h$ between $\mathbf{Q}$-fibers is a "$\mathbf{Q}$-form" of the $\mu_d$-torsor ${\rm{GL}}_1$ over ${\rm{GL}}_1 = U'_n$. The set of isomorphism classes of such forms is given by $${\rm{H}}^1({\rm{GL}}_1,\mu_d) = (\mathbf{Q}^{\times}/({\mathbf{Q}}^{\times})^d) \times x^{\mathbf{Z}/d\mathbf{Z}}$$
(since ${\rm{GL}}_1$ has trivial Pic and has unit group $\mathbf{Q}^{\times} x^{\mathbf{Z}}$).
Explicitly, for $q \in \mathbf{Q}^{\times}$ and $j \in \mathbf{Z}$ the finite etale cover of ${\rm{GL}}_1 = U'_n$ associated to the class of $(q,j \bmod d)$ is given by the covering equation $y^d = q x^j$.  As a covering of ${\rm{GL}}_1$ with coordinate $x$, this has geometric covering group $\mu_d(\overline{\mathbf{Q}})$ via scaling on $y$, so by inspection this geometric covering group action is not defined over $\mathbf{Q}$ (i.e., the automorphism group scheme for the covering is not a constant group over $\mathbf{Q}$, or in other words not all of these geometric automorphisms are defined over $\mathbf{Q}$) except when $d = 2$.  Ah, but recall that we arranged for $h$ to be a Galois covering, so in our setting with the covering $h'$ the geometric covering group must be defined entirely over $\mathbf{Q}$ (as a constant group).  In particular, $h$ must have degree $d = 2$. Also, the geometric connectedness of the covering forces ${\rm{gcd}}(j,d) = 1$.
To summarize, we have proved that $V \rightarrow U_n$ viewed over $\mathbf{Q}$ is given by $y^2 = q(t/n)$ for some $q \in \mathbf{Q}^{\times}$.  By changing $y$ by a $\mathbf{Q}^{\times}$-scaling (as we may certainly do), we can change $q$ by any square multiple we wish, so we can arrange that $q/n$ is equal to a squarefree integer $r$. Then $V$ is identified with the normalization of $\mathbf{Z}[t][1/(nt-1)]$ in the $(nt-1)$-localization of $\mathbf{Z}[y,t]/(y^2 - rt)$.  Using that $r$ is a squarefree integer, we claim that $\mathbf{Z}[y,t]/(y^2 - rt)$ is normal (in contrast with the situation for $\mathbf{Z}[y]/(y^2 - r)$ when $r$ is odd!).  This is clear after inverting 2, so by Serre's homological criterion the only issue is to check the normality at the generic points in characteristic 2, which is to say that the maximal ideal of the local ring at these points is principal. If $r$ is even (so it is twice an odd integer) then $y$ lies in such primes, so $(2,y)$ is the only such prime and $t$ isn't in this prime.  Hence, in such cases $y$ is a local generator (as the equation $rt = y^2$ with $t$ a local unit and $r$ twice an odd integer makes $2$ a local unit multiple of $y^2$).  If $r$ is odd then $(2)$ is itself prime because the reduction of $y^2 - rt$ modulo 2 is the element $y^2 - t \in \mathbf{F}_2[t,y]$ that is irreducible.
We conclude that
$$V = {\rm{Spec}}(\mathbf{Z}[y,t]/(y^2 - rt))_{nt-1}$$
over $U_n = {\rm{Spec}}(\mathbf{Z}[t])_{nt-1}$.  Ah, but this is not etale over $U_n$, since passing to characteristic 2 turns this into a dense open piece of a purely inseparable quadratic cover in characteristic 2.  Contradiction, so $\pi_1(U_n) = 1$. QED<|endoftext|>
TITLE: The image of the point-pushing group in the hyperelliptic representation of the braid group
QUESTION [22 upvotes]: Let $B_{2g+1}$ be the Artin braid group on $2g+1$ strands.  There is a symplectic representation 
$\rho:  B_{2g+1} \rightarrow Sp_{2g}(\mathbf{Z})$
called the "hyperelliptic representation," which can be described as follows.  The braid group is the fundamental group of the moduli space of configurations of 2g+1 points on the disc; each such configuration gives you genus-g surface which double covers the disk, ramified at those 2g+1 points and at the boundary; the representation is the usual monodromy action of the fundamental group on the homology of the fiber.
Alternately, we can think of $\rho$ as the specialization of the Burau representation to $t=-1$.
On the other hand, inside $B_{2g+1}$ there is a "point-pushing subgroup" H -- this can be thought of as the group of braids in which the first $2g$ strands stay fixed in place while the last strand is allowed to wind around the others.  The group is thus naturally identified with the fundamental group of a disc with 2g punctures.  It's a subgroup of the pure braid group, and it's the kernel of the Birman exact sequence.
Question:  What is the image $\rho(H)$ of the point-pushing subgroup in the hyperelliptic representation?
The image of the pure braid group under $\rho$ is the congruence subgroup $\Gamma(2)$, so $\rho(H)$ is a subgroup of that.  It is known to be Zariski dense.  Is $\rho(H)$ all of $\Gamma(2)$?  Is it at least finite index?  
Update:  OK, this is slightly embarrassing; I asked this question because I thought that a statement equivalent to it had been proved in an unpublished manuscript of J-K-Yu, but when I looked again at the ms.  I though it was only proving something weaker.  But now I see that Yu did prove this after all!  However, I am very happy to know how to do it the way Agol explained below.

REPLY [13 votes]: In fact it contains the level 4 subgroup.  Using the lantern relation, you can show that the point-pushing subgroup contains all 4th powers of transvections (use the fact that the square of a Dehn twist about an odd curve acts trivially on the homology of the double cover), and Mennicke proved that these generate level 4.  Also, level 4 mod level 2 is just sp(2g,Z_2), so it shouldn't be too hard to compute the exact image.<|endoftext|>
TITLE: $6j$-symbols for $U_q({\mathfrak{sl}}_n)$ and colored HOMFLY polynomials
QUESTION [10 upvotes]: Explicit expression of quantum $6j$-symbolos for $U_q({\mathfrak{sl}_2})$ have been known due to the work of Kirillov and Reshitikhin. 
My Question:

How much are known about quantum $6j$-symbolos for $U_q({\mathfrak{sl}_n})$? Are there partial results, say when the highest weights are given by small rank symmetric representation (a few horizontal boxes of Young Tableaux)?

I do not know how one can calculate colored HOMFLY-PT polynomials for non-torus knots without information about quantum $6j$-symbols for $U_q({\mathfrak{sl}_n})$. Physicists guessed the form of HOMFLY-PT polynomials colored by symmetric representation for the figure-eight (See the paper by the ITEP group and the recent paper). However, I want to know more examples for colored HOMFLY-PT polynomials.

Are there explicit formulae of colored HOMFLY-PT polynomials of non-torus knots? 

In addition, 

is there any way to calculate colored HOMFLY-PT polynomials of non-torus knots without using quantum $6j$-symbolos for $U_q({\mathfrak{sl}_n})$?

A similar question can be found here

REPLY [2 votes]: Recently, there has been some progress on quantum $6j$-symbols for $U_q(\mathfrak{sl}_N)$: Nawata-Ramadevi-Zodinmawia for symmetric representations, and Gu-Jockers for the representation specified by the (2,1) Young diagram. In particular, the latter one deals with quantum $6j$-symbols with multiplicity structure.<|endoftext|>
TITLE: Picard groups of abelian étale covers
QUESTION [5 upvotes]: Let $X$ be a scheme (you can assume that $X$ is proper and smooth over an algebraically closed field) and $T$ is a finite subgroup of $\text{Pic } X$ (of order prime to the characteristic). Does there exist a finite étale cover $f:Y\to X$ (possibly Galois with Galois group $T$) for which the map $f^* :\text{Pic }X\to\text{Pic }Y$ identifies $\text{Pic } Y$ with the quotient $(\text{Pic }X)/T$?
Of course (except for the claim about $f$ being Galois) it suffices to treat the case when $T$ is cyclic (of order $m$ say). Then we take a generator $L$ of $T$ and 
$$ Y = \text{Spec} {}_X \bigoplus_{i=0}^{m-1} L^i $$
(the standard construction showing that $H^1(X, \mu_m)$ classifies such covers). I
think I can show that the kernel of $\text{Pic }X\to\text{Pic }Y$ is $T$, but I don't
know how to prove that it is surjective...
It also reminds me of the Hilbert class field: I don't know much
algebraic number theory, but I think that the construction is similar:
$K$ is a number field, $X = \text{Spec }\mathcal{O}_K$, $T = \text{Pic }X$ is the class group, then $Y = \text{Spec }\mathcal{O}_H$ where $H$ is the Hilbert class field of $K$, its maximal abelian unramified extension.

REPLY [8 votes]: The answer is "no," even for curves. Say $X$ is a genus 2 curve over $\mathbf{C}$, and $f:Y \to X$ is any finite etale degree $d$ morphism of degree $> 1$. Then $Y$ is a smooth projective curve of genus $d+1 > 2$. In particular, $\mathrm{Pic}(Y)$ has dimension $>2$, so it cannot be a quotient of $\mathrm{Pic}(X)$.<|endoftext|>
TITLE: Minimizing the excursion of a sum of unit vectors
QUESTION [16 upvotes]: I have $n$ unit-length vectors $v_i$ in $\mathbb{R}^3$, whose
sum is zero:
$$ v_1 + v_2 + \cdots + v_n = 0 \; .$$
Now I form the closed polygon $P$ in space by placing them head to tail.
So the vertices of $P$ are 
$$ 0, v_1, (v_1+v_2), \ldots, (v_1+\cdots +v_{n-1}), 0 \; .$$
My question is:

What is the minimum excursion from the origin achievable by
  shuffling the vectors by a permutation of $(1,2,\ldots,n)$?

For example, the $12$ red vectors below wander $\sqrt{10}$
from the (purple) origin, but the light blue vectors—the same in  a different order—stay within $\sqrt{3}$.

          


(These particular vectors derive from the vertices of a
cuboctahedron, so some are 
negations of others.)
Is there some constant $r_{\min}$ independent of $n$ such that the sum
can always be arranged to be at most $r_{\min}$,
i.e., lie within an origin-centered ball of that radius?  Or must $r_{\min}$ depend on $n$?
Is there some natural algorithm for minimizing the excursion, or must I
(in the worst case)
try all $n!$ permutations?
Of course the same question can be asked in any dimension $\mathbb{R}^d$, but my focus is $\mathbb{R}^3$.
Thanks for ideas and/or pointers!
Update1.
The suggestion (in the comments) that $r_{\min} = \sqrt{d}$ in $\mathbb{R}^d$, based on an answer to
the previous MO question, "Bounding a signed sum of complex numbers,"
is intriguing, and may be true.  But I do not see that it is proved by that answer.
Update2.
With key phrases suggested by Nik Weaver, I found a 1981 paper by Imre Bárány, "A Vector-Sum Theorem and its Application to Improving Flow Shop Guarantees" (Math. Oper. Res. link), which shows that $r_{\min} < \frac{3}{2} d$.

REPLY [12 votes]: This is a famous open problem, $r_{\min}\le d$ is known as the Steinitz-lemma. It is conjectured that $r_{\min}= O(\sqrt d)$ but even $r_{\min}= o(d)$ is open. See also http://www.renyi.hu/~barany/cikkek/steinitz.pdf , section 3.<|endoftext|>
TITLE: Analytic elements in non-archimedean geometry
QUESTION [10 upvotes]: Let $(k,|.|)$ be a complete non-archimedean valued field. Let $D$ be the open unit disc over $k$. (Anything I write could be adapted to the case of an open annulus.) The ring $\mathcal{O}(D)$ of analytic functions on $D$ admits an explicit description: it consists of the series $\sum_{n\ge 0} a_n T^n$ such that for any $r<1$, the sequence $(|a_n|r^n)_{n\ge 0}$ tends to 0. If such a function is bounded, its uniform norm is precisely $\sup_{n\ge 0}(|a_n|)$.
For some purposes, it is convenient to consider smaller rings: the ring of bounded functions or the ring of analytic elements, for instance. Let me recall that this latter ring $\mathcal{H}(D)$ is defined as the completion of $k(T) \cap \mathcal{O}(D)$ (i.e. rational functions with no poles in $D$) for the uniform norm on the disc. Remark that this definition depends on the choice of a coordinate $T$ on $D$. Those functions have nice properties: for example, they only have a finite number of zeros on $D$. We refer to Gilles Christol's book (chapter I) for a more detailed account. 
Consider the disc $D$ inside the affine line $\mathbf{A}^1$. Let $X$ be an affinoid domain of $\mathbf{A}^1$ that contains $D$. One may check that any function on $X$ restricts to an analytic element on $D$, whatever the choice of the coordinate (because the approximation property actually holds for any function on $X$). 
The question is the following: if $D$ is embedded inside an arbitrary affinoid space $X$, do functions on $X$ still restrict to analytic elements? 
The issue is that the choice of a coordinate in the definition of $\mathcal{H}(D)$ makes it difficult to use outside the affine line. I would appreciate any idea that helps recognizing analytic elements on a disc inside an arbitrary curve.

REPLY [5 votes]: Vladimir Berkovich has indicated to me that the answer is no: on a general curve, functions need not restrict to analytic elements. Let me copy here his argument.
The algebra of analytic elements $\mathcal{H}(D)$ can be defined as the completion of the inductive limit of the $k$-affinoid algebras $\mathcal{A}_X$ with respect to the supremum norm on $D$ taken over all affinoid domains $X$ in the affine line that contain $D$. It follows that the spectrum of $\mathcal{H}(D)$ is the closure of $D$, i.e. the union of $D$ with the Gauss point $x$, and the non-Archimedean field $\mathscr{H}'(x)$ associated to $x$ with respect to the Banach algebra $\mathcal{H}(D)$ coincides with the similar field $\mathscr{H}(x)$ of the Gauss point in the affine line. In particular, the residue field of $\mathscr{H}'(x)$ is the field of rational functions on the projective line over the residue field of $k$.
Let now $Y$ be the analytification of the generic fiber of a smooth projective curve $Z$ over the ring of integers of $k$. The space $Y$ contains a unique "generic point" $y$, whose image under the reduction map is the generic point  of the closed fiber of $Z$, and the complement of $y$ in $Y$ is a disjoint union of open unit discs (assume for simplicity that $k$ is algebraically closed). Pick up one of them $D$, and consider the similar Banach algebra $\mathcal{H}'(D)$ which is the completion of the intersection of $k(Z)$ and $\mathcal{O}(D)$ with respect to the supremum norm on $D$. It coincides with the completion of the inductive limit of the $k$-affinoid algebras $A_X$ with respect to the supremum norm on $D$ taken over all affinoid domains $X$ in $Y$ that contain $D$. The spectrum of $\mathcal{H}'(D)$ coincides with the closure of $D$ in $Y$, i.e. the union of $D$ with the point $y$, and the non-Archimedean field $\mathscr{H}'(y)$ associated to $y$ with respect to the Banach algebra $\mathcal{H}'(D)$ coincides with the similar field $\mathscr{H}(y)$ of $y$ with respect to $Y$. It follows that the residue field of $\mathscr{H}'(y)$ is the field of rational functions on the closed fiber of $Z$. If the genus of $Z$ is positive, then so is the genus of its closed fiber. This implies that the functions on $X$ (in $Y$) do not necessarily restrict to analytic elements on $D$.<|endoftext|>
TITLE: Kahn-Kalai-Linial for intersecting upsets
QUESTION [5 upvotes]: Is there any known improvement on the Kahn-Kalai-Linial inequality (on the influences of boolean functions) in the special case in which $f$ is the indicator function of an intersecting monotonic set system? More concretely, is there an absolute constant $C>0$ such that the following statement holds:

If $f:\mathcal{P}([n]) \to \lbrace 0,1\rbrace$ is the indicator of an intersecting upset, then there exists $x\in [n]$ such that $I_x(f)\geq C/\sqrt{n}$.

(Note that the "tribes" example of a half-sized system in which all influences are $\ll \log n /n$ is certainly not intersecting.)
Background: The $x$th influence of a boolean function $f$ is defined as
$$I_x(f) = \mathbf{E}(f(X)\neq f(X \Delta \lbrace x\rbrace),$$
($\Delta$ is symmetric difference) where $X$ is drawn randomly and uniformly from $\mathcal{P}([n])$. In particular, if $f$ is the indicator of a monotonic set system $\mathcal{U}\subset\mathcal{P}([n])$ (monotonic meaning $X\subset Y$ and $X\in\mathcal{U}$ implies $Y\in\mathcal{U}$), $I_x(f)$ is the number of sets $X\in \mathcal{U}$ containing $x$, minus the number of such sets not containing $x$, divided by $2^{n-1}$.

REPLY [7 votes]: This is a natural question and indeed the property of being intersecting is quite interesting also in various aspect of influences. However, you cannot improve KKL's theorem if f is intersecting: An example is this: consider your variables on a circle and let f=1 if the longest run of 1's is larger than the longest 1's of 0's and in case of equality consider the second longest run (and continue lexicographically). In this case f is intersecting and the influence of every variable is logn/n. 
This example is symmetric under rotations and therefore all influences are the same. The sum of influences can be described as the integral over all configurations x of h(x) the number of pivotal variables. Here a variable is pivotal if changing its value changes the value of f. Given x with f(x)=1 typically a variable is pivotal only if it belongs to the largest run which is of expected size logn. There are cases of equality (between 1 runs and 0 euns or between two runs) that the number of pivotal variables will be larger than log n but those are rare. This explains why every influence is proportional to logn/n but for a complete proof some more work is needed.<|endoftext|>
TITLE: common dominating measure for a family of measures
QUESTION [7 upvotes]: Given a family $\{\mu \}_{i\in I}$ on a Polish space (complete, separable metric space) $X$. When does there exist a measure $\lambda$ such that 
$$\mu_i=f_i \lambda$$
where the $f_i$ are densities (Radon-Nikodym) of $\mu_i$ with respect to $\lambda$. 
EDIT: What is a verifiable condition in the case $I$ is uncountable.

REPLY [3 votes]: At least for finite measures (the tags seem to suggest this question concerns mainly probability measures), this is equivalent to $\{\mu_i:i\in I\}$ being separable under the variation norm.
The variation norm on finite measures dominated by some finite measure $\nu$ agrees with the $L_1(\nu)$-norm of the corresponding Radon-Nikodym derivatives, so separability of $\{\mu_i:i\in I\}$ follows in this case from separability of $L_1(\nu)$.
Conversely, if the sequence of nonzero finite measures $\langle\lambda_n\rangle$ is dense in $\{\mu_i:i\in I\}$ under the variation norm, then whenever $\mu_i(A)>0$ for some $i\in I$, then there is some $n$ such that $\lambda_n(A)>0$. It follows that $\{\mu_i:i\in I\}$ is dominated by the probability measure $\kappa$ given by
$$\kappa(A)=\sum_{n=1}^\infty \frac{1}{2^n \lambda_n(X)}\lambda_n(A).$$
Relating this to the comment by  Jochen Wengenroth: If $I$ is compact and the function $i\mapsto\nu_i$ is continuous when the range is endowed with the variation distance, the range will be compact and be separable as a compact metrizable space, so in this case, a finite dominating measure will exist.<|endoftext|>
TITLE: Cohomology groups of quotient by finite group
QUESTION [6 upvotes]: I know there are already lots of questions about (co)homology groups of a quotient manifold, but please let me ask one more question. 
Let $G$ be a finite group acting on a manifold $M$ without fixed point. The standard spectral sequence argument shows that 
$$
H^k(M,\mathbb{Q})^G\cong H^k(M/G,\mathbb{Q}). 
$$
This in particular shows that rank $H^k(M,\mathbb{Z})^G$=rank $H^k(M/G,\mathbb{Z})$. 
We also have a natural map $\pi^{*}:H^k(M/G,\mathbb{Z})\rightarrow H^k(M,\mathbb{Z})^G$ via the quotient map $\pi:M\rightarrow M/G$.
Is it true that $\pi^{*}$ is injection mod torsion (and hence that $H^k(M/G,\mathbb{Z})\subset H^k(M,\mathbb{Z})^G$ (mod torsions) is of finite index)? Or are there any relation between $\mathbb{Z}$-coefficient (co)homology groups?

REPLY [4 votes]: Yes. Consider the commutative diagram involving the maps $H^k(M/G,\mathbb Z) \to H^k(M,\mathbb Z)^G \to H^k(M,\mathbb Q)^G$ and $H^k(M/G,\mathbb Z) \to H^k(M/G,\mathbb Q) \to H^k(M,\mathbb Q)^G$
The kernel of the map $\pi^*$ is contained in the kernel of the second sequence. Since the second map in the second sequence is an isomorphism, the kernel of the second sequence is the kernel of its first map, which is a tensoring with $\mathbb Q$, so its kernel is torsion.
The commutative diagram comes from the naturality of the change of coefficients map.<|endoftext|>
TITLE: countable topological spaces of uncountable weight
QUESTION [5 upvotes]: I read somewhere the following stament:
There are countable normal $T_1$ spaces of uncountable weight.
Can someone give an example or a reference?

REPLY [4 votes]: Another favourite example of mine: the Hewitt-Marczewski-Pondiczery theorem says that the product of continuum or fewer separable spaces is separable. So $\mathbb{R}^\mathbb{R}$ has a countable dense subset $D$ and it is clear any countable regular space is normal (from being Lindelöf), and it's not hard to show that every point of $D$ has a local base of $\mathfrak{c}$ many open sets (and not fewer). So it's not like the other examples mentioned, in that there are no isolated points.<|endoftext|>
TITLE: Wanted: Odd-dimensional integral cohomology class whose square is nonzero
QUESTION [10 upvotes]: Does anyone know of a nice simple example of a space $X$ with an odd-dimensional integral cohomology class $a\in H^{2k+1}(X;\mathbb{Z})$ whose square is nonzero?
I once thought that the one-dimensional generator $a\in H^1(K;\mathbb{Z})\cong\mathbb{Z}$ in the cohomology of the Klein bottle  had $a^2\in H^2(K;\mathbb{Z})\cong\mathbb{Z}/2$ nonzero, but it appears this is not the case.

REPLY [7 votes]: As Ralph is being modest, I have decided to make his comment into a CW answer.
Recall that the short exact coefficient sequence $0\to \mathbb{Z}\to \mathbb{Z}\to \mathbb{Z}/2\to 0$ leads to a long exact sequence
$$
\cdots \to H^\ast(X;\mathbb{Z})\to H^\ast(X;\mathbb{Z})\stackrel{\rho}{\to} H^\ast(X;\mathbb{Z}/2)\stackrel{\beta}{\to} H^{\ast+1}(X;\mathbb{Z})\to\cdots $$
for any space $X$, where $\rho$ denotes reduction mod 2 and $\beta$ is the Bockstein homomorphism. Recall also that $\rho$ is a ring homomorphism, and that $\rho\circ\beta = Sq^1$, the first Steenrod square. 
Let $X=\mathbb{R}P^\infty\times\mathbb{R}P^\infty$, and let $x,y\in H^1(X;\mathbb{Z}/2)$ denote the generators over each axis. Note that $c=\beta(xy) \in H^3(X;\mathbb{Z})$ is nonzero, since $$\rho(c) = Sq^1(xy) = Sq^1(x)y + x Sq^1(y) = x^2y + xy^2 \neq 0.$$  
Similarly $c^2 \in H^6(X;\mathbb{Z})$ is nonzero, since we have
$$ \rho(c^2) = \rho(c)^2 = (x^2y+xy^2)^2 = x^4y^2 + x^2 y^4 \neq 0.
$$
As Will Sawin noted in his comment, we can get a finite-dimensional example by taking $Y=X^{(6)}$, the $6$-skeleton of $X$ (this is the smallest possible dimension, by Tom Goodwillie's comment). We could also take $Y=\mathbb{R}P^4\times\mathbb{R}P^4$ (but $ \mathbb{R}P^3\times\mathbb{R}P^3$ won't work, since its six-dimensional cohomology is torsion-free).<|endoftext|>
TITLE: Is primary decomposition still important?
QUESTION [53 upvotes]: On p.50 of Atiyah and Macdonald's Introduction to Commutative Algebra, in the introduction to the chapter on primary decomposition, it says

In the modern treatment, with its
  emphasis on localization, primary
  decomposition is no longer such a
  central tool in the theory.

However, my understanding is that scheme theory, which fundamentally uses localization and is probably part of the modern treatment they refer to, does not eliminate the need for primary decomposition. Rather, one can and does (e.g. as in Eisenbud and Harris) apply primary decomposition to schemes.
Therefore, is it really true that localization somehow obviates the need for primary decomposition? Are there examples in which primary decomposition can be replaced by other arguments, making primary decomposition merely a classical but obsolete notion? Otherwise, what do the authors mean by this statement?

REPLY [13 votes]: Part of your question is about the relationship between primary decomposition and localization. Here, I think the basic connection is that if $I$ is an ideal in a Noetherian ring $R$ whose minimal primary decomposition is $I = J_1 \cap \cdots \cap J_n$ and $S$ is any multiplicative set in $R$, then a minimal primary decomposition of $(S^{-1} I) \cap R$ consists of those $J_i$ whose associated primes are disjoint from $S$. For example, in Karl's answer, this statement gives two equivalent definitions of the symbolic power by taking $I$ to be a power of a prime ideal $P$ and $S = R \setminus P$.
The point is that one way to look at the usefulness of localization is that it simplifies ideals by replacing them with the intersections of only some of their primary components. From this perspective, primary decomposition seems like a more useful way of analyzing ideals since it gives you each primary component by itself. However, the limitation of primary decomposition is that it is not unique (Eisenbud's book explains this very well if I remember correctly). Thus, primary decomposition gives you a mix of useful information about the original ideal together with some information coming from some arbitrary choices you made during the computation. Which aspects of the primary decomposition are uniquely determined by the original ideal? Well, it's the set of associated primes as well as the intersections between any subset of primary components whose associated primes form an downward-closed subset of poset of the associated primes of the original ideal, i.e. exactly those ideals which show up as $(S^{-1} I) \cap R$ for some multiplicative set $S$. I think that some of the distaste for primary decomposition on the part of, for example, Bourbaki, comes from the non-uniqueness, whereas localization can be seen as a way of capturing exactly the unique part of primary decomposition.
As came up in the comments, primary decomposition is very useful computationally. Here, the non-uniqueness is not a problem since computations typically depend on many choices anyways. I suspect that in most cases, primary decomposition in and of itself is of less interest than other questions which can be answered from it: the irreducible components of the scheme, which of these are generically reduced, the irreducible components of the non-reduced locus, or whether or not the projective scheme corresponding to a homogeneous ideal is reduced (which is not necessarily the same as whether the affine scheme is reduced). Each of these questions could be answered from the set of associated primes together with the localizations at the minimal primes, but the primary decomposition provides a convenient presentation of all of them.<|endoftext|>
TITLE: $n$ lines in a general position and the number of empty triangles
QUESTION [7 upvotes]: Question. Consider $n \geq 5$ lines in a general position (i.e. no two lines are parallel and no triple intersections are allowed) in $\mathbb{R}^2$. Let $T(n)$ denote the maximal number of empty triangles (here empty triangle means that it does not contain other triangle). What would be best upper and lower bounds for $T(n)$? I know $(n-2) \leq T(n)$ holds, but I am hoping for a better lower bound. Is it true that $n \leq T(n)$? Also, is it possible to compute $T(n)$ it for small $n$ (where small means $6 \leq n \leq 10$)? I think $T(6) = 6$, but I am not able to show $6$ is an upper bound as well.

REPLY [2 votes]: You can easily show that $3k$ lines can be arranged to have at least $\frac34 k^2$ triangles, which proves that asymptotically, $T(n) \ge n^2/12$. 
The following picture shows the example for $k=4$ -- there are three groups of $k$ lines each, group 1 and group 2 forming a "squarish" $k\times k$ chessboard, which is being then intersected by the lines from group 3. They will produce $(k+1)+(k+1)+k+k+(k-1)+(k-1)+...$ triangles which comes to approximately $\frac34k^2$ triangles. And I am not even counting the ones that these groups will produce far away from the chessboard, inside their "own domain" so to speak.
picture with 3 $k$ lines:<|endoftext|>
TITLE: Measure theory treatment geared toward the Riesz representation theorem
QUESTION [14 upvotes]: I'm looking for recommendations for books (or lecture notes) that develop measure theory in sufficient detail to state and prove the Riesz representation theorem (which is the characterization of the topological duals of linear spaces of continuous functions on a completely regular topological space). In particular, I'd like to see the cases of $C(X)$ and $C_K(X)$ of all continuous functions and all compactly supported continuous functions for non-compact $X$, with the usual Fréchet and locally convex topologies, covered as well as the more common case of compact $X$.
I know that this is basically in Dunford & Schwartz, but I always find it helpful to have multiple references at hand.
(24 Aug 2012) Update: There have been a number of helpful recommendations, however they are not all explicit about which version of the theorem is covered there. I've not yet had time to check them all, so let me collect here what I know so far, by theorem strength. I'm not aiming for a complete list. But it seems useful to have some list, since the stronger versions of the theorem don't appear to be so well known.

Up to $C_K(X)$, locally convex topology. Many texts.

Halmos, Measure Theory.
Rudin, Real and Complex Analysis.
Folland, Real Analysis.
Aliprantis, Border, Infinite Dimensional Analysis.
Fremlin, Topological Riesz Spaces and Measure Theory.

Up to $C(X)$, Fréchet topology.

Dunford, Schwartz, Linear Operators, Part 1.
Berg, Christensen, Ressel, Harmonic Analysis on Semigroups.
A short and self contained treatment in the first two chapters.

Up to $C(X)$, vector lattice from cone of positive units, order dual (since not a topological vector space).
To be honest, I'm not sure what the statement of the theorem is in this case, or what the standard associated to it. However, there there appear to be some results for this case.

König, Measure and Integration. Though, hard to interpret without an in depth reading.


Please leave a comment or an answer if you know where to place other references in this list.

REPLY [6 votes]: May I add some information on this topic?
Firstly, the space $C(X)$ is not usually a Frechet space---you need some countability condition on the compact subsets of $X$, e.g.,  it being $\sigma$-compact and locally compact.  It is not even complete in the general case---for that you need the condition that it be a $k_R$-space.  The dual of $C(X)$ can be identified, with the aid of some abstract locally convex theory and the RRT for compact spaces, with the space of measures on $K$ with compact support
(i.e. those arising from measures on some compact subset in the natural way).  If $X$ is locally compact, then Bourbaki used the dual of the space of continuous functions with compact support as the {\it definition} of the space of (unbounded) measures on $X$.  One can then interpret its members as  measures in the classical sense (i.e. as  functions defined on a suitable class of sets) by the usual extension methods.  I would suggest that the most useful extension of the Riesz representation theorem is the one for bounded, Radon measures on a (completely regular) space.  For this one has to go beyond the more common classes of Banach or even locally convex spaces, something which  was done by Buck in the 50's. He introduced a locally convex topology on $C^b(X)$ (the bounded, continuous functions) using weighted seminorms for which exactly the kind of representation theorem one would expect and hope for obtains.
He did this for locally compact spaces but it was soon extended to the general case, using the methods of mixed topologies and Saks spaces of the polish school.  There are many indications that this is the correct structure---the natural versions of the Stone-Weierstrass theorem hold for it and its spectrum (regarding $C^b(X)$ as an algebra) is identifiable with $X$ so that one has a form of the Gelfand-Naimark theory.  Further indications of its suitability are that if one considers generalised spectra, i.e.,
continuous, algebraic homomorphisms into more general algebras then one obtains interesting results and concepts.  The important case is where $A$ is $L(H)$ (or, more generally, a von Neumann algebra).  One then gets spaces of observables (in the sense of quantum theory) in the case where the underlying topological space is the real line and this provides them in a natural way with a structure which opens a path to a natural and rigorous approach to analysis in the context of spaces of observables---distributions, analytic functions, ...).<|endoftext|>
TITLE: Proof of the transcendence of the Champernowne Constant with Thue-Siegel-Roth
QUESTION [6 upvotes]: It is well know that the Champernowne Constant 
0,1234567891011121314151617....
is transcendental. This was shown by Kurt Mahler in 1937. But the proof of the transcendence should also work with the famous Thue-Siegel-Roth theorem (http://en.wikipedia.org/wiki/Thue%E2%80%93Siegel%E2%80%93Roth_theorem), that was proved in 1955. 
I'm looking for a reference of the transcendence-proof where this theorem is used.

REPLY [9 votes]: Here is an extract from van der Poorten's "Obituary. Kurt Mahler (1903--1988)" (see p.353 in J. Austral. Math. Soc. Ser. A 51 (1991)):

In a more unexpected way, Mahler's
  arguments led to the following amusing
  result: Suppose $f$ is a non-constant
  polynomial taking integer values at
  the nonnegative integers. Then the
  concatenated decimal $$
> \phi=0.f(1)f(2)f(3)\dots $$ is
  transcendental. In particular
  Champernowne's normal number $$
> 0.123\dots910111213\dots $$ is transcendental. Mahler's argument
  relies on the observation that one
  readily obtains rational
  approximations to $\phi$ with
  denominators high powers of the base
  10, thus composed of the primes 2 and
  5 alone. Perhaps disappointingly,
  Roth's definitive form of the
  Thue--Siegel inequalities permits a
  more immediate argument obviating the
  need for an appeal to the $p$-adic
  results.

This is to say that Roth's argument is more superior than Mahler's but it appeared some 20 years later...<|endoftext|>
TITLE: Covering a polygon with rectangles
QUESTION [11 upvotes]: I am trying to cover a simple concave polygon with a minimum rectangles. My rectangles can be any length, but they have maximum widths, and the polygon will never have an acute angle.
I thought about trying to decompose my concave polygon into triangles that produce a set of minimally overlapping rectangles minimally bounding each triangle and then merging those rectangles into larger ones. However, I don't think this will work for small notches in the edges of the polygon. The triangles created by the reflex vertices on those notches will create the wrong rectangles. I am looking for rectangles that will span/ignore notches.
I don't really know anything about computational geometry, so I'm not really sure on how to begin asking the question.
I found other posts that were similar, but not what I need:

split polygon into minimum amount of rectangles and triangles
Covering an arbitrary polygon with minimum number of squares
Find $k$ rectangles so that they cover the maximum number of points
Algorithm for finding the fewest rectangles to cover a set of rectangles

Some examples: Black is the input. Red is the acceptable output.

Another example: The second output is preferred. However, generating both outputs and using another factor to determine preference is probably necessary and not the responsibility of this algorithm.


Polygons that mimic curves are extremely rare. In this scenario much of the area of the rectangles is wasted. However, this is acceptable because each rectangle obeys the max width constraint.

Also, I found this article to be close to what I need:

Covering with rectangular pieces by Paul Iacob, Daniela Marinescu, and Cristina Luca

Maybe a better question is "How can I identify rectangular-like portions of a concave polygon?"

Here is an image showing the desired implementation:

The green is the actual material usage. The red rectangles are the layouts. The blue is the MBR of the entire polygon. I am thinking I should try to get little MBRs and fill them in. The 2-3 green rectangles in the upper left corner that terminate into the middle of the polygon are expensive. That is what I want to minimize. The green rectangles have a min and max width and height, but I can use as many rows and columns necessary to cover a region. Again, I must minimize the number of rectangles that do not span across the input. I can also modify the shape of the green rectangle to fit in small places that is also very expensive. In other words, getting as many rectangles as possible to span as much as possible is ideal.

REPLY [7 votes]: Here is one relevant result, by Michael Hoffmann,
"Covering polygons with few rectangles," 2001. (PDF download link)
        
He shows that minimal covering by two or three congruent axis-aligned rectangles of a collection
of polygons with a total of $n$ vertices
can be found in $O(n)$ time.
He also says that the more general problem—covering a set of polygons by $p>3$ congruent rectangles—cannot
be approximated by better than a factor of $2$ unless $P=NP$
(because of the relation to the $p$-center problem).
He does not directly address covering just one polygon, or with covering by incongruent rectangles.
This latter problem (your problem) is partially addressed computationally in the paper 2011,
"Covering a polygonal region by rectangles,"
Computational Optimization and Applications
(Springer link). It appears that they
start with a set of rectangles, and extend them until they form a cover.<|endoftext|>
TITLE: Must the composition of projective morphisms be projective?
QUESTION [25 upvotes]: The notion of a projective morphism in algebraic geometry is surprisingly subtle.  It is not quite clear what the definition is!  For example, the definition in EGA differs from that in Hartshorne.  For the purposes of this question, I take the definition in EGA$^1$.
Suppose $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ are projective morphisms.  I know that $g \circ f$ is projective if $Z$ is quasicompact (see for example Exercise 18.3.B in the August 2012 version of the notes here).  Is it true even without $Z$ quasicompact, or is there a counterexample?
(I suspect this is in one of the standard sources, but I haven't stumbled upon it.)
$^1$ EGA II, 5.5.1-5.5.2: $X$ is called projective over $Y$ if there is a closed $Y$-immersion $X \hookrightarrow \mathbb{P}(\mathcal{E})$ for some quasi-coherent sheaf $\mathcal{E}$ on $Y$ of finite type; equivalently if $X=\mathrm{Proj}(A)$ for some graded quasi-coherent $\mathcal{O}_Y$-algebra $A$ which is generated by $A_1$, which is of finite type.

REPLY [4 votes]: This is now  Tag 0C4P with an assumption that the schemes in question are qcqs as predicted in the comments by Moret-Bailly.<|endoftext|>
TITLE: Power series expansions of $L$-series
QUESTION [7 upvotes]: Let $\zeta_K(s)$ be the Dedekind zeta function for a number field $K$. We can understand the first non-vanishing coefficient of its Laurent series via the class number formula. Is anything known/conjectured about the next term?
On a related note, the BSD conjecture predicts the value of the first non-vanishing Taylor coefficient of the Hasse-Weil $L$-function of (say) an elliptic curve. Are there any conjectures about the coefficients after that?

REPLY [3 votes]: There is a related conjecture for the L-function attached to certain Galois representations, due to Colmez and stated in the beautiful survey paper Periods by Kontsevich and Zagier (see section 3.6). This survey paper refers to an Annals paper of Colmez: Périodes des variétés abéliennes à multiplication complexe (Jstor link), and to some papers of Hiroyuki Yoshida.
There is also a book by Hiroyuki Yoshida called Absolute CM-periods, where some of this might be discussed. 
I have no idea whether this is related or not to the Rössler-Maillot/Kudla story mentioned above in the comments.
The conjecture as stated by Kontsevich and Zagier is for a Galois representation sending complex conjugation to minus the identity matrix. This is a rather restrictive condition, and Kontsevich-Zagier also state (without explanation) that "in general, one does not expect any interesting number-theoretic property for subleading coefficients".<|endoftext|>
TITLE: Any text book or lecture notes regarding the algebraic part of geometry?
QUESTION [8 upvotes]: I know there are text books of Algebraic topology. There are books of Differential geometry. But when I read papers, for example lots of papers talking about fundamental groups or higher homotopy groups of certain manifolds, sometimes lots of terminologies from abstract algebra pop out - nilpotent, solvable or amenable etc. I can understand those definitions, but I feel very uncomfortable that I don't have a geometric feeling of those languages. 
So I ask for a good reference, ideally written by geometer, that covers the material of this part?
Thanks in advance for any suggestions.

REPLY [3 votes]: As you mentioned higher homotopy groups, it's worth noting that these are always abelian, which is a stronger condition than each of nilpotent, solvable and amenable. So you probably won't find these adjectives used when describing higher homotopy groups.
Fundamental groups need not be abelian, but they may be almost abelian, and each of the three properties you mention give different ways of formalizing what this should mean. As Igor mentions in his answer, the study of group properties via their realization as fundamental groups of complexes is the topic of geometric group theory.
As for references, I do not know the geometry literature all that well, but Chapter 10 of G. W. Whitehead's book "Elements of Homotopy Theory" contains a nice introductions to nilpotency for topologists.<|endoftext|>
TITLE: What is the homotopy type of the space of simple closed curves isotopic to a given one?
QUESTION [7 upvotes]: For surfaces there are many statements along the lines of: if two simple closed curves are homotopic, they are isotopic. I'm interested in such questions for families of curves.
More precisely, let $\Sigma$ be a hyperbolic surface, possibly with boundary. We fix an essential simple closed curve $\gamma$ on $\Sigma$. It is true that the subspace of $Emb(S^1,\Sigma)$ consisting of those curves that are isotopic to $\gamma$ is homotopy equivalent to a circle? Here the circle would come from reparametrisation of the curves.
This statement is true if we instead look at the space of all continuous (or smooth) maps of $S^1$ into $\Sigma$ that are homotopic to $\gamma$. Also note that this seems to be false for the torus, as for any essential simple closed curve we get at least $S^1 \times S^1$.

REPLY [5 votes]: Earlier than Grayson, the determination of the homotopy-types of these spaces was done by Gramain.  
There are a few special cases, like the torus and sphere and the non-orientable analogue, the case of null curves.  But if they're not null homotopic the components of the embedding space have the homotopy type of $S^1$ -- the reparametrizations of the given curve.  
Andre Gramain, Le type d'homotopie du groupe des diffeomorphisms d'une surface compacte.  Ann. Sci. l'ENS $4^e$ serie tome 6 $n^o$ 1 (1973) 53--66<|endoftext|>
TITLE: Implications of a hypothetical blow-up of Navier-Stokes for the mathematical model 
QUESTION [6 upvotes]: Let us suppose that there exists a (initially smooth) solution of NSE that blows up in finite time. Then, in particular, the corresponding velocity field becomes unbounded as time progresses. Which assumptions in the common modelling of fluid flow (leading to NSE) will be violated and in which way will NSE need to be adjusted? 
I read that the viscous term is some kind of expansion and higher derivatives of the velocity field u, like \Delta^2 u, should, in fact, be retained. (It´s well-known that NSE with hyperdissipation admits global smooth solutions.) Unfortunately, i couldn't find any physical details on this expansion. Does someone know about it?
Thanks a lot in advance! :-)

REPLY [3 votes]: This is a comment about Michael's answer and the comment by Daniel about compressible NSE.
1- There are several models for viscous fluids, among which we may select incompressible/compressible NSE. It is widely believed that the compressible case behaves wildlier than the incompressible one. Therefore, saying that a blow up singularity in the incompressible NSE means that incompressibility is no longer a valid assumption is not the end of the story. Because then we have to use the compressible model, which likely displays the same trouble.
2- In the compressible case, the issue of cavitation is subtle and so far remains an open problem. Except in one space dimension (ha! ha!), we don't know whether vaccum may occur at positive time if the initial density is strictly positive. But the one-D case is interesting: D. Hoff and myself have a paper in which we show that a constant viscosity coefficient yields an unphysical behaviour at very low densities ; something like an ill-posedness result for the Cauchy problem. This suggests to take in account the dependence of the viscosities upon the state (density, temperature). Actually, the derivation of NSE from Boltzmann yields a dependence upon the temperature alone. In isentropic flows, this amounts to saying that the viscosities depend upon the density ; in which case the viscosity vanishes with the density. This is what is needed to avoid the ill-posedness mentionned above.<|endoftext|>
TITLE: Are there better upper bounds on the rank of the commutant of a fusion module than the global dimension?
QUESTION [5 upvotes]: Suppose I have a fusion category $\mathcal{C}$ and an indecomposable module category $\mathcal{M}$ over it. The commutant $\mathcal{C}_\mathcal{M}^*$ is the category of module endofunctors, and gives another fusion category, Morita equivalent to the original $\mathcal{C}$.

Can I bound the rank of $\mathcal{C}_\mathcal{M}^*$?

Recall that the rank is the number of isomorphism classes of irreducible objects.
Certainly $\mathcal{C}$ and $\mathcal{C}_\mathcal{M}^*$ have the same global dimension, so easily $\operatorname{rank}(\mathcal{C}_\mathcal{M}^*) \leq \operatorname{dim}(\mathcal{C})$. Are there better upper bounds available?
Update: I'm happy to consider all the 'decategorified' data of $\mathcal{C}$ and $\mathcal{M}$, that is, the Grothendieck groups of both, along with the ring and module structures thereon, when trying to come up with an estimate, not just the rank of $\mathcal{C}$.
As examples:

the Haagerup subfactor gives a Morita equivalence between two fusion categories with ranks
4 and 6, and global dimension $\approx 35.725$
$\operatorname{Rep}(G)$ and  $\text{Vec}_G$ are Morita equivalent, with global dimension $|G|$. Here $\operatorname{rank}(\text{Vec}_G) = |G|$, while when $G$ is non-commutative $\operatorname{rank}(\operatorname{Rep}(G))$ may be much smaller.

REPLY [4 votes]: If $\mathcal{C}$ and $\mathcal{D}$ are Morita equivalent by a pair $({_\mathcal{C}}\mathcal{M}_{\mathcal{D}}, {_\mathcal{D}}\mathcal{N}_{\mathcal{C}})$ of bimodules, then there is a natural map of fusion bimodules ${_D}N \otimes_{C} M_{D} \to {_D}D_D$ that preserves Frobenius-Perron dimension (see section 5.1 of Noah's latest preprint with Pinhas Grossman). So the Frobenius-Perron dimensions of elements of $N \otimes_C M$ (which do not depend on knowing $D$) will give Frobenius-Perron dimensions of elements of $D$. Then I think you may be able to use the known possible small Frobenius-Perron dimensions of objects (from your paper with Noah and Frank Calegari) to determine some nontrivial lower bounds on Frobenius-Perron dimensions of simple objects in $\mathcal{D}$, and thus improve on the global dimension bound.
(Or do arbitrarily small numbers of the form $2 \cos (\pi / n)$ already generate the full ring of real cyclotomic integers? Even if so, this method could at least reduce the bound by 1 for weakly integral categories, although that's not a great improvement.)<|endoftext|>
TITLE: quantum groups... not via presentations
QUESTION [28 upvotes]: Given a semisimple Lie algebra $\mathfrak g$ with Cartan matrix $a_{ij}$, the quantum group $U_q(\mathfrak g)$ is usually defined as the $\mathbb Q(q)$-algebra with generators $K_i$, $E_i$, $F_i$ (the $K_i$ are invertible and commute with each other) and relations
$$
\begin{split}
K_iE_j &K_i^{-1}=q^{\langle\alpha_i,\alpha_j\rangle}E_j\qquad\qquad
K_iF_j K_i^{-1}=q^{-\langle\alpha_i,\alpha_j\rangle}F_j\,
\\\
&[E_i,F_j]=\delta_{ij}\frac{K_i-K_i^{-1}}
{\quad q^{\langle\alpha_i,\alpha_i\rangle/2}
-q^{-\langle\alpha_i,\alpha_i\rangle/2}\quad}\,
\end{split}
$$
along with two more complicated relations that I won't reproduce here.
One then defines the comultiplication, counit, and antipode by some more formulas.
Is there a way of defining $U_q(\mathfrak g)$ that doesn't involve writing down all those formulas?
In other words, is there a procedure that takes $\mathfrak g$ as input, produces $U_q(\mathfrak g)$ as output, and doesn't involve the choice of a Cartan subalgebra of $\mathfrak g$?

REPLY [24 votes]: $\newcommand\g{\mathfrak{g}}$The answer to your question "is there a procedure that takes $\g$ as input, produces $U_q(\g)$ as output, and doesn't involve the choice of a Cartan subalgebra of $\g$?" is No.  Not if you want it "canonical" in any sense.  (Of course, if I wanted to cheat my way to a "yes," I could make choices that are equivalent to choosing a Cartan but not stated as such — I would construct for you a certain canonical homogeneous space, and then ask you to pick a point in it, ....)
The problem is that the automorphism group of $\g$ does not lift to $U_q(\g)$.  Recall that the inner automorphism group is precisely the simplest group $G$ integrating $\g$ (take any connected group integrating $\g$ and mod out by its center).  On the other hand, the inner automorphism "group" of $U_q(\g)$ is $U_q(\g)$ (or "$\operatorname{spec}(\operatorname{Fun}_q(G))$", depending on your point of view) itself.  This certainly deforms the automorphisms.  But there is not a procedure like you ask for, because you have to break some symmetry.
Here's a way to say this correctly: In a precise sense $U_q(\g)$ degenerates to $U(\g)$ as $q\to 1$, and this is part of the structure that I take you to mean when you write "$U_q(\g)$".  In this degeneration, you can also study $\frac{\partial}{\partial q}(\dots)$ at $q=1$.  In particular, looking at $\frac{\partial}{\partial q}\bigr|_{q=1}$ of the comultiplication on $U_q(\g)$ recovers the Lie cobracket on $\g$.  But the Lie cobracket knows the Cartan subalgebra: it is precisely the kernel of the Lie cobracket.  So $\operatorname{Aut}(\g)$ cannot lift to $U_q(\g)$ compatibly with all of this structure.
What does exist without choosing a Cartan subalgebra is the braided category of representations of $U_q(\g)$, although I would have to think a moment to recall how to write it down explicitly.  (In the asymptotic limit $q = e^\hbar$ with formal $\hbar$, I do know how to write down $\operatorname{Rep}(U_{e^\hbar}\g)$ explicitly for any choice of Drinfel'd associator.)  In particular, as a braided monoidal category, this category does have an action by $\operatorname{Aut}(\g)$.
But the category is strictly less data than the Hopf algebra.  Namely, "Tannakian reconstruction" is the statement that $U_q(\g)$ is the Hopf algebra of endomorphisms of a certain braided coalgebra in $\operatorname{Rep}(U_{q}\g)$ (or it is the Hopf dual of the Hopf algebra of co-endomorphisms of a certain braided algebra in the category of Ind-objects in $\operatorname{Rep}(U_{q}\g)$, if for you representations are finite-dimensional), and you cannot choose this coalgebra canonically.  This coalgebra is unique up to isomorphism, but certainly not up to unique isomorphisms (or else $U_q(\g)$ would be trivial).  The failure of this coalgebra to exists up to canonical isomorphism is essentially the same problem as above.
In a precise way, this is failure of there to exist a canonical isomorphisms between different choices of the coalgebra is analogous of the failure of the "fundamental group" of a topological space to be an honest group.  Recall that a pointed topological space has a fundamental group, which is a group determined up to canonical isomorphism.  For comparison, a non-pointed but path-connected topological space has a group assigned to each point, and these are non-canonically isomorphic.  Thus a non-pointed path-connected topological space has a "fundamental group up to conjugation," also known as a connected groupoid.
What this should all mean, although I'm not going to try to write down the correct definition, is that there does exist canonically associated to $\g$ a "Hopf algebroid" which is noncanonically equivalent as a Hopf algebroid to any choice of $U_q(\g)$.  Or, at least, I'm confident of everything in my answer in the $\hbar\to 0$ asymptotics, and have thought less about the finite-$q$ case, and so I'm generalizing intuition from that setting, but I think it's all correct.<|endoftext|>
TITLE: Hales work on Kepler conjecture
QUESTION [8 upvotes]: It seems that the Fulkerson prize has been attributed to Thomas Hales for this work. What is the present status of the conjecture, then?

REPLY [14 votes]: The original 1998 proof appeared in 2005/6, in an abridged version in Annals of Mathematics:

A proof of the Kepler conjecture.  Ann. of Math. (2)  162  (2005),  no. 3, 1065–1185. 

and unabridged as volume 36 of Discrete & Computational Geometry (2006).
It is my understanding that meanwhile there are no actual doubts in this proof (though, possibly, still some reservations by some, due to the complexity and the computer aid).
Thus, there is an ongoing effort to simplfy/clean up/formalize the proof. For example: 
Hales, Thomas C.; Harrison, John; McLaughlin, Sean; Nipkow, Tobias; Obua, Steven; Zumkeller, Roland "A revision of the proof of the Kepler conjecture"  Discrete Comput. Geom.  44  (2010),  no. 1, 1–34.
From the MR review of that paper (by Uwe Schnell):

Since the original proof makes use of computer programs, the so-called Flyspeck project was started to give a formal proof of the Kepler conjecture. This article summarizes the current status of this initiative and gives a list of minor errata in the original proof.

Regarding this Flyspeck project see for example a recent presentation by Hales (given March 2012, the date on the slides seems wrong), where he says that this project is 80% complete. On the web one can easily find further information related to this project.  
Update: The Flyspeck program was completed in August 2014. See this Announcement

OP asks regarding the meaning of the awarding of the Fulkerson prize: 
In some sense there is no specific meaning to it (as concerns acceptance of the community of the proof) as the Fulkerson prize is not a prize that is a priori awarded for 'The Kepler Conjecture', as opposed to the situation, say, for the Wolfskehl prize for FLT or the Clay prizes for the seven problems. By contrast 'The Delbert Ray Fulkerson Prize recognizes outstanding papers in the area of discrete mathematics.'  So that in principle it would even be possible that the prize would be awarded while people only believe that the papers make significant progress towards a proof. However, this seems not the case as the Notices AMS article on the 2009 awarding of this prize contains 'After four centuries, Ferguson and Hales have now proven Kepler’s assertion.' Thus, yes, this is further evidence that the proof is generally accepted.
However, also note that this was not the first prize they received for their work. They also received the 2007 Robbins prize. The citation contains "The Kepler conjecture asserts [...]  The proof of this
result is a landmark achievement." but also at the end "Some controversy has surrounded this proof, with its large computer component, concerning its reliable checkability by humans. Addressing this issue, Hales has an ongoing project, called the 'Flyspeck' project, whose object is to construct a 'second-generation' proof which is entirely checkable by computer in a formal logic system."
So, in summary, this prize is a recognition of their work and that it was awarded can be considered as further evidendence that the proof is accepted as correct. However, I would not consider the awarding of this prize as something like the moment when the proof in any sense 'officially' got accepted as such. 
(Indeed, as far as I know, it was already widely accepted several years before that, e.g., otherwise the Annals would not have published this paper; except for the reservations I mentioned above, but then those also did not get addressed by the prize-awarding.)<|endoftext|>
TITLE: Tensor product of sheaves separated?
QUESTION [5 upvotes]: Let $(X, O_X)$ be an arbitrary scheme and $\mathcal{F}$ a presheaf of $O_X$-modules. The presheaf $\mathcal{F}$ is said to be separated if the natural map to it's sheafification $\mathcal{F}^+$ is an injection. That is, for each open set $U$, $\mathcal{F}(U) \hookrightarrow \mathcal{F}^+(U)$. 
Let $\mathcal{F}, \mathcal{G}$ be two sheaves of $O_X$ modules. Is the presheaf tensor product $\mathcal{F} \otimes_{O_X} \mathcal{G}$ a separated presheaf? Is this true more generally if $(X,O_X)$ is just a ringed space?
This tells you for example, that if $\mathcal{F}, \mathcal{G}$ have some global sections then so does their sheaf tensor product (assuming the presheaf tensor product has global sections).
I wasn't able to find this in the stacks project. Is this proved in books on sheaf theory?

REPLY [11 votes]: If I understand correctly your question, then the answer is "no, the presheaf tensor product need not be separated."  Let $(X,\mathcal{O}_X)$ be the projective line $(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1})$ over a base ring $R$.  Let $\mathcal{F}$ and $\mathcal{G}$ both be the invertible sheaf $\mathcal{O}_{\mathbb{P}^1}(1)$.  Since these are sheaves, they are separated.  Consider the presheaf tensor product $\mathcal{E}$.  In particular, the space of global sections of this presheaf is $$ H^0(\mathbb{P}^1,\mathcal{E}) = H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(1))\otimes_R H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(1)).$$  This is a free $R$-module of rank $4$.  On the other hand, the sheafification of the presheaf tensor product is $\mathcal{O}_{\mathbb{P}^1}(2)$, whose space of global sections is a free $R$-module of rank $3$.<|endoftext|>
TITLE: Second-order term in first-order logic?
QUESTION [10 upvotes]: Could a function in FOL take functions as arguments? FOL only limits on the order of the individuals being quantified, but if an expression does not involve quantifying over second-order or higher terms, would it still be valid in FOL? Say, $f(g)$ where  $f : (A \to B) \to C$ and $g : A \to B$.
So another way to put my question is that does the rule of formation in FOL say that $f(g)$ is valid as long as $g$ is in the domain of $f$, despite the type of $f$? Could $f$ be a 4th order function and would still be valid in FOL?

REPLY [20 votes]: I think that the spirit of this question, combined with the clarifications in comments, is:

What is it that makes first-order logic "first order"? 

Unfortunately, the terms "first order" and "second order" get used to mean various things. 
A formal but unsatisfying answer would say that first-order logic is a specific logic defined in, say, Mendelson's textbook, and any other logic is not "first order logic" strictly speaking. This is unsatisfying because we know there are many inessential variations of first-order logic - really there are many first-order logics that share a certain core. The question I quoted asks for a characterization of that core. 
One common answer is that any logic in which we intend to have quantifiers over "functions" or "sets" is higher order. This is unsatisfying because, as Andrej Bauer points out, such theories can be syntactically expressed in multi-sorted first-order logic.  There are many theories of "second order arithmetic", for example, which allow us to express set and function quantification but which are treated as first-order theories. Unfortunately, the terminology "second order" is established for these theories and cannot be avoided.  
Recall that a logic consists of both a syntax and a semantics. The truly defining feature of a first-order logic is the semantics. First-order semantics begins with the notion of a structure (also called a model), as defined in every introductory textbook on first-order logic.  
Consider how we would express function quantification in (multi-sorted) first-order logic, as in Andrej's answer. Each structure must interpret two sorts. It uses a set of individuals for the quantifiers over individuals and a separate set of functions for the quantifiers over functions. This set of functions, in an arbitrary structure, might be a proper subset of the collection of all functions on the set of individuals; nothing in the definition of a structure requires otherwise. Indeed some structures will have an infinite set of individuals but a finite set of functions. 
Full second order semantics changes the class of allowable structures so that only those whose function set includes all the functions are allowed. This does not affect the syntax in any way, but it deeply changes the semantics. Because fewer structures are being considered, more formulas will be logically valid, and fewer will be satisfiable.  Thus there are more categorical theories in these semantics, such as the well known categorical second-order axiomatizations of the natural numbers.  Those same axiomatizations are syntactically fine in first-order logic, where the simple difference is that they are no longer categorical. 
Thus the key difference between function quantification in multi-sorted first order logic (or type theory) and function quantification in full second-order semantics is not the existence of syntactic quantifier symbols that allow quantification over functions. The difference is in the meaning of those quantifiers, which derives from the way the semantics are defined. In the first-order case, we have little control over the range of quantifiers. In full second-order semantics, once the set of individuals is fixed, the range of the function quantifiers is also fixed. This distinction is only visible at the meta level, when we are studying the logic from the outside and can specify which interpretations are permissible. Nothing in the syntax of the logic tells us what collection of structures will be used to interpret it.<|endoftext|>
TITLE: Are congruence subgroups of the modular group finitely presented?
QUESTION [6 upvotes]: Are the congruence subgroups of the modular group $\Gamma\equiv\mathrm{PSL}\left(2,\mathbb{Z}\right)$ (e.g. $\Gamma\left(n\right)$, $\Gamma_{0}\left(n\right)$, $\Gamma_{1}\left(n\right)$ etc.) finitely presented? If so, is there a proof of this? Assuming they are finitely presented, are the presentations of e.g. the principal congruence subgroups $\Gamma\left(n\right)$ (for small $n$) documented anywhere? I know they could be computed using the Reidemeister-Schreier process, but it would be nice to have some independent confirmation of the presentations.
Many thanks!

REPLY [18 votes]: Not only these subgroups are finitely presented, they are all finite free products of cyclic groups; most of them (for sufficiently large $n$) are actually free of finite rank (once congruence subgroup contains no elements of order 2 and 3). For instance, you can easily check that $\Gamma(n)$ is torsion-free for all $n\ge 2$ by looking at traces for $n\ge 3$ (since $1\ne 2$ mod $n\ge 2$ and $2\ne 0$ mod $n\ge 3$) and by looking at matrix coefficients for $n=2$. Rank is easily computable if you know index of the congruence subgroup in the modular group. The magic formula is multiplicativity of the Euler characteristic: For the modular group $\Gamma$, $\chi=-1+\frac{1}{2} + \frac{1}{3}=-\frac{1}{6}$. If $\Gamma'\subset \Gamma$ is a subgroup of index $i$ then $\chi(\Gamma')=i \chi(\Gamma)$. If $\Gamma$ is free of rank $r$ then $\chi(\Gamma)= 1-r$. For instance, to find index $i$ for $\Gamma(n)$, compute the order of the quotient group $SL(2, Z_n)/\pm I$. There is a closed formula for the order of this group (in terms of prime factors of $n$) which will tell you what the index is:    
If $n$ is the product of powers of primes 
$\prod_i p_i^{k_i}$ then
$$
|PSL(2,Z_n)|= \frac{n^3}{2} ~~~\prod_i (1- p_i^{-2}). 
$$<|endoftext|>
TITLE: Undergraduate Topology
QUESTION [17 upvotes]: I am developing an introductory topology course for undergraduates, and I am wondering what topics to cover.  At my institution, real analysis is not a prerequisite for the course, so it is more than likely that the intended audience has not been exposed to this material.  Does anyone have any suggestions?

REPLY [2 votes]: I think it depends on your students' backgrounds and goals. You mention that real analysis is not a prerequisite for the course, but what are the prerequisites? Can you assume any or all of linear algebra, multivariable calculus, discrete mathematics, a transition to proofs course and abstract algebra? Are your students interested in applications of topology outside of mathematics or within mathematics, or would they prefer to learn the subject for its own sake (or all three)?
For instance, you can give the course an analysis flavor by focusing on metric spaces and general topology. On the other hand, you could go straight to homology (see Peter Giblin's Graphs, Surfaces and Homology) if your students have taken / are concurrently taking a course in group theory, or you are prepared to teach the requisite group theory during the course itself.
If you'd like to focus on general topology with applications outside of mathematics, Introduction to Topology: Pure and Applied is a nice option.
Computational topology, as mentioned in another answer, could be an option for students with strong applied interests. Robert Ghrist has an intriguing draft text on applied algebraic topology available at his website, which could be a source of inspiration. There's also a question on teaching undergraduates computational topology at the Mathematics Educators Stack Exchange, although it's aimed at engineering students rather than mathematics majors.<|endoftext|>
TITLE: Measuring contact between algebraic varieties
QUESTION [5 upvotes]: I have two regular surfaces in three space, both of which are given by an equation. I would like to measure the contact between the two surfaces using only their equations. Usually, one would find a local parametrisation for one of the surfaces, and then substitute this into the other surface's equation. This would give a function in two variables, and the singularity type of this map would give the contact between the two surfaces. However, as I have mentioned: I only want to use the equations.
Is there a way to do this? For example, by looking at the dimension of some suitable ideal?

REPLY [2 votes]: While I suspect that you are looking for some kind of homological algebra answer, here's a naive algorithm to get what you want:  
Suppose that one is trying to determine the 'contact $k$-type' of a pair of algebraic surfaces at a point $p\in\mathbb{R}^3$.  One may as well assume that $p$ is the origin and let the surfaces be defined by polynomial equations $f(x,y,z)=0$ and $g(x,y,z)=0$.  
Of course, one must have $f(0,0,0)=g(0,0,0)$ or else the surfaces don't both pass through $p$.  
Also, you are assuming that the 'surfaces are regular', by which, I am guessing that you want that $\nabla f$ and $\nabla g$ don't vanish at $p$, so I'll assume that.  If $\nabla f\wedge\nabla g$ does not vanish at $p$, then the surfaces aren't tangent at $p$, so assume that $\nabla f\wedge\nabla g$ vanishes at $p$.  Under these assumptions, you can, by a linear change of coordinates, assume that $f$ has the form
$$
f(x,y,z) = z - f_2(x,y,z),
$$
where $f_2$ vanishes to order $2$ at $p=(0,0,0)$.  Then, of course, one has 
$$
g(x,y,z) = az + g_2(x,y,z)
$$
for some $a\not=0$ and some polynomial $g_2$ that vanishes to order $2$ at $p$.  
Now define a sequence of polynomials $h_i(x,y,z)$ as follows:
$$
h_2(x,y,z) = g\bigl(x,y,f_2(x,y,z)\bigr)
$$
and, for $k\ge 2$,
$$
h_{k+1}(x,y,z) = h_k\bigl(x,y,f_2(x,y,z)\bigr).
$$
One can now prove, by induction, that, when one writes, for $k\ge 1$,
$$
h_{k+1}(x,y,z) = p_k(x,y,z) + R_{k+1}(x,y,z),
$$
where $p_k$ has degree at most $k$ and $R_{k+1}$ vanishes to order $k{+}1$ at $p$, then $p_k$ is a polynomial in $x$ and $y$ only, and it defines the $k$-th order contact  type between the two surfaces at $p$.  (Of course, $p_1=0$.)
There is still the task of determining when two $p_k$'s are equivalent under change of variable in $x$ and $y$, but that's another issue.<|endoftext|>
TITLE: Are there F_un Lie algebras ?  
QUESTION [16 upvotes]: Background See WP-article on F_1 = F_{un} = Field with one element (and also this MO question). Paraphrasing someone:
we do not know what is it, but it is not a field :). For this question it is enough to keep in mind J. Tits idea (1957) that Weyl groups should be thought as semisimple groups over F_1.
E.g. symmetric group S_n = GL_n(F_1). 
Q1 What might be Lie algebras over F_1 ? In particular for GL_n ?
What numerology should correspond to gl_n(F_1) ? I.e. are there some numbers related to
gl_n(F_q) which have a limit when q->1 (may be renormalized like with GL_n(F_q)) ? 

Comments on further questions are also welcome:
Q2 To what extent representation theory of S_n can be thought as limit q->1 of representation theory of GL_n(F_q) ? (There is some paper "Translating the Irreducible Representations of S_n into GL_n(F_q)", but I do not quite understand it).
Q3 What might be "orbit method" to construct representations of S_n = GL_n(F_1) ?
Q4 What might be Langlands correspondence over F_1 ? Should it be related to bijection between
irreps of S_n and its conj. classes (keep in mind that GL^L=GL).

REPLY [7 votes]: I will only attempt to answer the first question.
$n$-dimensional vector space over $\mathbf{F}_1$ is the same as a pointed set with $n+1$ elements. It is natural to call $GL_n(\mathbf{F}_1)$ the group of automorphisms of $\mathbf{F}_1^n$ and $\mathfrak{gl}_n(\mathbf{F}_1)$ the monoid of endomorphisms. There are at least two notions of morphisms:

Plain morphisms of pointed sets. The monoid of endomorphisms has cardinality $(n+1)^n$.
Maps of pointed sets which are injective if you throw away the basepoints (see http://arxiv.org/abs/1006.0912). It is not too hard to see that the cardinality of $End(\mathbf{F}_1)$ is
$$\sum_{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)\frac{n!}{(n-k)!}=\sum_{k=0}^n\frac{(n!)^2}{k!(n-k)!^2}.$$
Here $k$ is the number of elements that don't go to the basepoint.

Note, that in both cases the group of automorphisms is the same ($S_n$).<|endoftext|>
TITLE: Non-split extension of the rationals by the integers 
QUESTION [32 upvotes]: Can someone describe explicitly an abelian group $A$ such that the extension $$0 \to \mathbb{Z} \to A \to \mathbb{Q} \to 0$$ doesn't split ? 
Background: The Stein-Serre theorem (Hilton, Stammbach: A course in homol. algebra, Theorem 6.1) states that if $A$ is abelian of countable rank (=maximal number of linear independent elements), then $Ext(A,\mathbb{Z})=0$ implies $A$ free. When applied to $A= \mathbb{Q}$, I  conclude $Ext(\mathbb{Q},\mathbb{Z})\neq 0$. Moreover, by interpreting $A \in Ext(\mathbb{Q},\mathbb{Z})$ as extension $0 \to \mathbb{Z} \to A \to \mathbb{Q} \to 0$ of abelian groups, there must be an $A$ such that the extension doesn't split. The problem is that the proof of the theorem isn't constructive and doesn't show how to construct such an $A$.

REPLY [12 votes]: Building on Ralph's answer a bit we can get uncountably many inequivalent examples as Mark Grant's comment on the original post suggested there should be.
Let $S,T$ be a partition of the primes into two nonempty sets (or if you prefer, the multiplicative sets generated by these).  Localize at these sets and form the sequence $0\to\mathbb{Z}\to S^{-1}\mathbb{Z}\oplus T^{-1}\mathbb{Z}\to\mathbb{Q}\to 0$, where the first map is $n\mapsto (n,-n)$ and the second is $(a,b)\mapsto a+b$.  (My comment on Ralph's answer was the case $T = \{p\}$.)  Then the same partial fractions argument as in Ralph's answer shows that this is an exact sequence which does not split.
Now let $U,V$ be another such partition of the primes.  Suppose there is an isomorphism $f: S^{-1}\mathbb{Z}\oplus T^{-1}\mathbb{Z}\to U^{-1}\mathbb{Z}\oplus V^{-1}\mathbb{Z}$ making the corresponding exact sequences equivalent.  Assume WLOG that $S$ contains at least two elements $p,r\in S$ and $p\in U$.
For any $k\geq 1$, equivalence of the exact sequences gives $f(1/p^k,0) = (a_k,b_k)$ where $a_k+b_k = 1/p^k$.  Since $p\in U$ and $U^{-1}\mathbb{Z}\cap V^{-1}\mathbb{Z} = \mathbb{Z}$, we get $(a_k,b_k) = (1/p^k + m_k,-m_k)$ for some $m_k\in\mathbb{Z}$.  The map $f$ is a homomorphism, so $f(1,0) = (1 + p^km_k, -p^km_k)$.  The value $k$ was arbitrary, so the second component of $f(1,0)$ is divisible by $p^k$ for all $k\geq 1$ and must be zero.  Therefore $m_k = 0$ and $f(1/p^k,0) = (1/p^k,0)$ for all $k\geq 0$.
The same argument shows that $f(1/r,0)$ is either $(1/r,0)$ or $(0,1/r)$ depending on whether $r\in U$ or $r\in V$.  The second case would make $f(1,0) = (0,1)$, contradicting the above, so $r\in U$.  In this way we obtain $S\subseteq U$.  The same arguments applied to the isomorphism $f^{-1}$ yield $U\subseteq S$, so $S=U$.
Thus the exact sequences are equivalent if and only if $\{S,T\} = \{U,V\}$.  There are uncountably many partitions of the primes into two nonempty sets, so there are uncountably many inequivalent non-split exact sequences $0\to\mathbb{Z}\to A\to\mathbb{Q}\to 0$.<|endoftext|>
TITLE: Thurston's senior thesis at New College
QUESTION [45 upvotes]: I've been collecting some of the many unpublished manuscripts of Bill Thurston over the years.  His recent passing inspired me to ask the following.  I've seen a number of references (for instance, in his wikipedia biography here) to a senior thesis that he wrote when he was a student at New College in Florida which concerns intuitionist foundations for mathematics.  Does anyone happen to have a copy of this that they could either post electronically or mail to me?
EDIT : This was originally a comment, but I thought I'd move it up here because it might be of wider interest.  Thurston made an interesting blog comment about his early philosophical interests here. An excerpt : "BTW, I was very taken by Kleene’s book on Foundations of Mathematics when I was in college, and it motivated me to write a senior thesis on intuitionist topology. I thought I might become an intuitionist logician, but when I approached Tarski to advise me, he said that Berkeley wasn’t a good place for intuitionism, so I went into topology instead."

REPLY [18 votes]: I've emailed you a pdf copy of "A Constructive Foundation for Topology" by Bill Thurston (June 14, 1967; New College Senior Thesis; submitted to Roger Renne).
In consideration of the comments above, I'm hesitant to post the file in a publicly accessible location. Perhaps the mathoverflow community can figure out whether such a posting would be appropriate.<|endoftext|>
TITLE: Minimal distance spheres in complex projective spaces
QUESTION [7 upvotes]: My question has to do with distance spheres in $\mathbb CP^{n+1}$. I am interested in knowing what is the radius $r$ of a distance sphere $S(r)$ around a point that makes it a minimal submanifold of $\mathbb CP^{n+1}$. It is known that, in this type of geometric situation, the distance spheres have constant mean curvature and some distance sphere will be minimal, and my problem is determining at what exact radius $r_0$ that happens. I thought of two possible approaches to this problem, and they seem to provide different answers; the first gives $r_0=\arctan\sqrt{2n+1}$ and the second $r_0=\pi/4$. Although I tend to believe the second is correct, I would like to understand what is going wrong here, or what I am missing...

Distance spheres in $\mathbb CP^{n+1}$:
Consider $\mathbb CP^{n+1}$ with the standard Fubini-Study metric $g_{FS}$ and distance spheres around a point $p\in\mathbb CP^{n+1}$. These are isometric embeddings $f_r\colon S^{2n+1}\to\mathbb CP^{n+1}$, parameterized by their radius $r\in ]0,\pi/2[$, that foliate $\mathbb CP^{n+1}$, and as $r\to 0$ collapse to the (round) point $p$ and as $t\to\pi/2$, these spheres collapse to $\mathbb CP^n\subset\mathbb CP^{n+1}$. They also happen to be (principal) orbits of a cohomogeneity one action of $\mathrm{SU}(n+1)$, whose singular orbits are the fixed point $p$ and $\mathbb CP^n$.
It is well-known that the metric in such distance spheres is a Berger metric, i.e., obtained by shrinking the round metric in the direction of the Hopf fibers, see e.g. Petersen's book. In fact, one can compute the metric on the distance sphere of radius $r$ to be $$g_r:=f_r^*(g_{FS})=\sin^2 r (g+(\cos^2 r) h),$$ where we decompose the round metric as $g_{S^{2n+1}}=g+h$, such that $h$ is the component in the direction of the Hopf fiber and $g$ is the component in the directions orthogonal to the Hopf fiber.

First approach:
To find a minimal $(S^{2n+1},g_r)$ inside $\mathbb CP^{n+1}$, we compute its mean curvature, by computing its second fundamental form. According to multiple sources (e.g. Maeda's paper, p. 38 or Karcher's survey p. 220), the second fundamental form $A_r$ of $(S^{2n+1},g_r)$ is given by: $$A_r(\xi)=(2\cot 2r)\xi, \quad A_r(u)=(\cot r)u,$$ where $\xi$ is the vector tangent to the Hopf fiber and $u$ is any tangent vector orthogonal to $\xi$. Therefore, since we can form an orthonormal basis of eigenvectors of $A_r$ with $1$ vector in the direction of $\xi$ and $2n$ vectors orthogonal to $\xi$, we have that the mean curvature is the sum of the corresponding eigenvalues: $$H=2\cot2r+2n\cot r=0 \Leftrightarrow r=\arctan\sqrt{2n+1}.$$
Second approach:
According to a theorem of Hsiang (Hsiang, p. 6), a $G$-orbit is minimal iff it has extremal volume among orbits of the same type. The volume of $(S^{2n+1},g_r)$ can be computed using Fubini's Theorem, as $$Vol(S^{2n+1},g_r)=Vol(\mathbb CP^n) \ Vol(\mbox{Hopf fiber}),$$ and the Hopf fiber is a circle of length $2\pi\sin r\cos r$. Differentiating the above with respect to $r$, since the only factor depending on $r$ is the volume of the Hopf fiber, we find that $(S^{2n+1},g_r)$ has extremal volume (and is hence minimal) iff $r=\pi/4$.

REPLY [5 votes]: A clearer and more correct way to do the second calculation is to look at the volume of the preimage of $S_r^{2n+1}$ in standard $S^{2n+3}$.  The preimage is a certain torus $T_r$, and one has
$$\mathop{Vol} T_r = 2\pi\mathop{Vol} S_r^{2n+1}.$$
At the same time,
$$\mathop{Vol} T_r \propto (\cos r)(\sin r)^{2n+1},$$
because $T_r$ is the Riemannian Cartesian product of a circle of radius $\cos r$ and an $S^{2n+1}$ of radius $\sin r$.  If you calculate when the derivative of this expression vanishes, you get agreement with your first calculation.
Addendum:  You can also directly check that the scale of the $\mathbb{C}P^n$ quotient of $S^{2n+1}_r$, if you want to do the calculation that way, is $\sin r$, which yields a factor of $(\sin r)^{2n}$, in addition to the factor of $\sin 2r \propto (\sin r)(\cos r)$ from the Hopf fiber.  So it all fits together.<|endoftext|>
TITLE: proofs of ergodicity of Sl(2, Z) action on R^2 without using duality
QUESTION [8 upvotes]: The group $
G=SL(2, R)$ acts linearly on $\mathbb R^2$. The Lebesgue measure of $\mathbb R^2$ is invariant and ergodic for $G$. There is a proof using duality theorem:
Let $U$ be the upper triangular unipotent subgroup of $G$ and let $\Gamma=SL(2, \mathbb Z)$. Then $U$ acts ergodically on $G/\Gamma$. Therefore duality tells us that $\Gamma$ acts ergodically on $G/U\cong \mathbb R^2\backslash 0$.
This method is good but it can not be extended to the group action on non homogeneous manifolds. Are there any other proofs? It is better that I can see different proofs to learn methods.

REPLY [3 votes]: I recommand the work of Thomas Roblin  "Ergodicité et équidistribution en courbure négative"  Mémoires de la SMF 95 (2003),  in addition to the references mentioned in the first answer. 
In particular, he does not need any compactness or finite volume assumption, he works with variable negative curvature, and even on CAT(-1) spaces $X$, not only on manifolds. 
In this framework ergodic properties of the horospherical foliation are related (by duality) to ergodic properties of the action of $\Gamma$ on the space of horospheres $\partial\tilde{X}\times \mathbb{R}$ (which identifies with $\mathbb{R}^2\setminus \{0 \}/\pm$ in the case of hyperbolic surfaces). 
These ergodic properties are strongly related to the mixing property of the geodesic flow. 
As mentioned in the beginning of the first answer, in the much earlier approach of Furstenberg, harmonic analysis is used on $\mathbb{R}^2$ to get unique ergodicity of the action of a cocompact fuchsian group $\Gamma$ on $\mathbb{R}^2\setminus\{0\}$, and the unique ergodicity of the horocyclic flow on $SL(2,\mathbb{R})/\Gamma$ is deduced by duality. 
The modern approach (as in Roblin, or Coudène, or others) through mixing is more dynamical.<|endoftext|>
TITLE: Criteria for irreducibility of polynomial
QUESTION [23 upvotes]: If $f, g\in \mathbb C[a,b]$ are  polynomials in two variables, are there easy criteria that allow to see if $f(x,y)-g(t,z)\in \mathbb C[x,y,t,z]$ is irreducible? 
Thank you very much,
best

REPLY [11 votes]: David is correct: my REU students have determined the complex polynomials $F(x)$ and $G(y)$ for which $F(x)-G(y)$ has an irreducible factor defining a curve of genus 0 or 1.  By Faltings' theorem (a.k.a. Mordell's conjecture), this lets us write down all $F(x)$ and $G(y)$ with algebraic coefficients for which there is a number field $K$ such that $F(K)$ and $G(K)$ have infinite intersection.  By one of Picard's theorems, it also means we've solved the functional equation $F\circ A = G \circ B$ in complex polynomials $F,G$ and meromorphic functions $A,B$.  This last problem has been studied in the context of finding variants of Nevanlinna's theorem that a nonconstant meromorphic function is uniquely determined by its preimages at each of five points: our equation implies that the preimage under $A$ of the multiset of zeroes of $F(x)$ comprises the same multiset as the preimage under $B$ of the multiset of zeroes of $G(x)$.
We crucially use the genus condition.  If one asks for reducibility of $F(x)-G(y)$ without imposing any hypothesis on the genus, then (as noted previously) one can find all examples when $F$ and $G$ are indecomposable: this was the goal of the paper by Cassou-Nogues and Couveignes, although it should be noted that their list of pairs $(F,G)$ is incomplete, so another thing we did this summer was to find the full list of pairs $(F,G)$ in this situation.  In the decomposable case, much remains to be discovered about reducibility of $F(x)-G(y)$, although progress has been made.  In particular, Mike Fried showed that, if $F$ is indecomposable but $G$ is decomposable, then reducibility of $F(x)-G(y)$ implies that  $G=A \circ B$  for some polynomials $A$ and $B$ such that  $A$  is indecomposable  and  $F(x)-A(y)$  is reducible.  It follows that the pair $(F,A)$ occurs on the (corrected) Cassou-Nogues--Couveignes list, and in particular, we may assume that either $F = A$ or $\deg(F)=\deg(A)\le 31$.  This is shown in [Michael Fried, The field of definition of function fields and a problem in the reducibility of polynomials in two variables], via a novel argument combining Galois theory and representation theory (if you're interested in this, please ask me, since one of my REU students found a very simple proof of Fried's result that I'd love to share).  I note that, while the Cassou-Nogues--Couveignes result depends on the classification of finite simple groups (via the classification of finite groups $G$ that have a cyclic subgroup $C$ which acts transitively in two inequivalent doubly transitive permutation representations of $G$).  However, Fried's result is elementary.
When both $F$ and $G$ are decomposable, in the same paper Fried proved the following result: if $F(x)-G(y)$ is reducible, then we can write  $F = A \circ B$  and  $G = C \circ D$  in such a way that  $A(x)-C(y)$  is reducible  and  the splitting field of  $A(x)-t$  over  $\mathbb{C}(t)$  equals the splitting field of  $C(x)-t$  over  $\mathbb{C}(t)$  (where $t$ is transcendental over $\mathbb{C}$).  The condition on equal splitting fields is extremely restrictive -- for instance, it implies that $\deg(A)=\deg(C)$ (by considering the size of the inertia groups at infinite places), and also that $A$ and $C$ have the same critical values, or more precisely that, for each complex number $\theta$, the least common multiple of the multiplicities of the roots of $A(x)-\theta$ equals the corresponding least common multiples for $C(x)-\theta$.  Fried's proof is remarkably simple; a nice exposition of it is Theorem 8.1 in [Yuri Bilu and Robert Tichy, The Diophantine equation $f(x)=g(y)$, Acta Arith. 95 (2000), 261--288].
Surprising phenomena in the decomposable case can be found in Peter M\"uller's papers [Kronecker conjugacy of polynomials, Trans. Amer. Math. Soc. 350 (1998), 1823--1850] and [An infinite series of Kronecker conjugate polynomials, Proc. Amer. Math. Soc. 125 (1997), 1933--1940].  The most recent work on the decomposable case (to my knowledge) is [Michael Fried and Ivica Gusic, Schinzel's problem: Imprimitive covers and the monodromy method, arXiv:1104.1740].  I have not digested everything in these three papers, so I would be thrilled if someone wanted to summarize their main achievements.<|endoftext|>
TITLE: Effective Serre Vanishing
QUESTION [7 upvotes]: Suppose that $X = \mathbb{P}^n_k$ and $G$ is a coherent sheaf on $X$.  
Question: Is there a way to determine some integer $n_0$ such that $H^1(X, G \otimes O_X(n)) = 0$ for all $n \geq n_0$?  (obviously $n_0$ depends on $G$)
Of course, the higher cohomologies would be very interesting as well, but perhaps the above is simpler.  One can also ask this for the relative case as well $f : Y \to X$ a projective morphism.
What I'd be particularly interested in is something that can be implemented into a computer. 
In my particular case, I know I only have to compute some other things until Serre vanishing hits, at that point, I know I can stop computing.  Unfortunately, I don't know how to tell when I've arrived.  You can assume I have a presentation of my module $G$ if it helps.
PS: Actually, what I really want is vanishing of $$R^1 f_* (G(-nE))$$ for a blow-up $f : Y \to X$ with $E$ the pullback of the blownup ideal.

REPLY [7 votes]: Take $n_0$ as (Castelnuovo–Mumford) regularity of $G$ minus $1$.
This generalizes to higher cohomology: for $H^i$ take the regularity minus $i$
Let $M$ be a module representing $G$.
Then, the (Castelnuovo–Mumford) regularity of $M$ is an upper bound of the regularity of $G$.
So for an implementation you can take the regularity of $M$, which can be computed using a free resolution of $M$.
If you need a better bound then you can compute a "better" module for $G$ by approximating its module of global sections.
However, this is rather expensive to my experience.<|endoftext|>
TITLE: Regularity of simplices
QUESTION [5 upvotes]: A triangle is regular, provided it is equilateral, or, also, equiangular. How these conditions generalize to characterizations of regularity of simplices?
In particular, it turns out that 

a simplex, all of whose facets meet with the same angle, is regular.

I think I have a proof this fact, but it's not so nice and direct as one would like it to be (I'll sketch it at request, though I hesitate to do it now, as it could be somehow misleading). I suspect there is a more elegant argument, e.g. based on duality. Can you see a proof, or do you have a reference, for this and other regularity results for simplices? (A counterexample of the above statement is also very welcome, of course, provided it is false).

REPLY [7 votes]: I am not sure what you mean by a direct proof, but here is a reasonably straightforward argument. Consider your equi-angular simplex $S$, and consider the link $L(v)$ of a vertex $v$ (the intersection of a small sphere centered at $v$ with your simplex, scaled so that the sphere has radius $1.$ The dihedral angles of $L(v)$ are equal to the corresponding dihedral angles of $S,$ so the links of all vertices are equiangular. We can show that an equiangular spherical simplex is regular by polar duality (as you had suspected):  The dual simplex $T^\ast$ of a spherical simplex $T$ is one whose vertices are the face normals of $T.$ This polar duality interchanges distances with exterior dihedral angles, and transforms the Gram matrix in a very simple way, described in this paper by Kokkendorff (2005). The regular case is particularly simple, since the gram matrix has the form $I + c J,$ (where $J$ is the matrix of all ones). $J$ is idempotent, so $(I + c J)^-1 = I + c_1 J,$ by expanding the inverse in a power series (notice that in the spherical case the Gram matrix is positive definite, so $c < 1,$ so we are allowed to do this). 
Anyway, when the smoke clears, we see that all of the links of our original simplex $S$ are regular spherical simplices, congruent amongst themselves. The facets of the links correspond to the links of the facets of $S,$ and the result follows by induction (the base case is the two dimensional case which you already know how to deal with).<|endoftext|>
TITLE: Tensor product of simple modules
QUESTION [5 upvotes]: Let $M$ a right simple module and $N$ be a left simple module over a ring $R$. I'm seeking a kind of Schur's lemma, with $\mathrm{Hom}_R (M,N)$ replaced by $M \otimes_R N$. So my questions are:
Can we describe $M \otimes_R N$ explicitly?
In particular, for a fixed $M$, is $N$ such that $M \otimes_R N \neq 0$ unique up to isomorphism? If not, can we classify such $N$'s in a reasonable way?

REPLY [6 votes]: I would guess the answer in general is hopeless, even for nice (noncommutative) algebras. For example, take $A_q := \mathbb C \langle X^{\pm 1}, Y^{\pm 1}\rangle / (XY=q^2YX)$, the quantum torus. If $q \in \mathbb C^*$ is not a root of unity, this is a simple algebra with trivial center. Let $M = P_k = \mathbb C[x^{\pm 1 }]$ as vector spaces with $1$ mapping to $m_0$ and $p_k$, respectively. Give $M$ and $P_k$ right and left $A_q$-module structures using
$f(x)m\cdot X = f(q^{-2}x) m$ and $f(x)m\cdot Y = xf(x)m$
$X\cdot f(x) p_k = xf(x)p_k$ and $Y\cdot f(x)m = q^{-k}x^k f(q^{-2}x)p_k$.
Claim: All the $P_k$ are non-isomorphic, each vector space $M \otimes_{A_q} P_k$ is 1-dimensional (spanned by $m\otimes p_k$), and if $q \in \mathbb C^*$ is not a root of unity, then $M$ and $P_k$ are simple.<|endoftext|>
TITLE: Can we calculate the inner product of a semicontinous function with the Dirac delta function?
QUESTION [5 upvotes]: Dear all,
It is clear that if $f:R\mapsto R$ is a continuous function, than $< f, \delta_x >=f(x)$. Now, if $f$ is only semicontinous, can we say that $< f, \delta_x >=f(x)$? I think this is true at all continuous points of $f$. But when $f$ has a jump at $x$, can we properly define this inner product? Does anyone know any references dealing with this matter?
By the way, I checked the wikipedia page about semicontinuous functions, from where I find Bourbaki's two volumns. But I didn't find any information about such pairing.

EDIT: 
Following the remark by Tapio Rajala, I think what I want is the following:
Suppose $f$ is a semicontinous function. Then the function 
$$
x\mapsto (f* \delta_0 )(x) = \int f(x-y) \delta_0(y)d y 
$$
is in $L^{\infty}_{loc}(R)$. 
It seems true for me. If anyone knows a reference, it would be nice, even though the proof seems not difficult. :-)

Here is another motivation of this problem:
Consider the wave equation in $R$
$$
\frac{\partial^2 }{\partial t^2} u(t,x) = \frac{\partial^2 }{\partial x^2} u(t,x) ,\quad t>0,x\in R\;,
$$ 
with vanishing initial position. Suppose that the initial velocity is a nonnegative Borel measure $\mu$. The solution is
$$
u(t,x) = \mu\left([x-t,x+t]\right) = (\mu * G_t)(x)\:,
$$
where $G_k(x)$ is the fundamental solution:
$$
G_t(x) = 1_{|x|\le t}
$$
which is a semicountinuous function. In particular, by letting $\mu=\delta_0$, we have the problem of pairing a delta function with a semicontinuous function. What I need is simply that the solution $u(t,x)$ is in $L^\infty_{loc}(R)$. I think this is true.
Thanks a lot for any hints and helps!
RIP, Bill.

REPLY [5 votes]: This is an ill posed problem. The Dirac $\delta$ is a  continuous linear map from the (locally convex) space of continuous function on $\mathbb{R}^n$ to $\mathbb{R}$.   You ask if it admits an extension to the larger  set  of semicontinuous functions.   First,  semicontinuity is not a linear condition: the sum of a lower semicontinuous function with an upper semi-continuous function may not be semi-continuous in any fashion.   (Lower semicontinuous functions do form a convex cone.)   I assume you want your extension to be linear. So you may be asking for an extension of $\delta$  as a linear function on the larger vector space  of functions which  can be represented as a sum of a lower semi-continous function with an  upper semi-continous function.  Extensions   of linear functionals from a subspace  do exist and  by no means are unique.    Even requiring some sort of continuity (?) of the extension, the result will not be unique.    Without specifying    what properties you expect your extension to have   you cannot expect  a definitive answer. You need to  phrase your question more accurately.<|endoftext|>
TITLE: Ratio of number of subgroups to the order of a finite group
QUESTION [16 upvotes]: Let $\mathcal{G}$ be the set of finite groups and for $G \in \mathcal{G}$, let $S(G)$ be the set of subgroups of $G$. I am interested in the ratio $R(G)=|S(G)|/|G|$. It is easy to show that by picking $G$ appropriately, $R(G)$ can be made arbitrarily large or arbitrarily close to zero. I am interested in some deeper properties of the set $R=(R(G) : G \in \mathcal{G})$, such as: 
(1) For which $x \in \mathbb{R}$ do there exist sequences of finite groups $G_1, G_2, ...$ such that the sequence $R(G_1), R(G_2), ... $ converges to $x$?
(2) Does $R$ contain a (non-empty) interval $(a,b) \in \mathbb{Q}$?
(3) Which integers belong to $R$?
(4) How do these properties change when $\mathcal{G}$ is replaced by the set of finite abelian groups? Finite simple groups?

REPLY [16 votes]: With regards to Q1 (and part of Q4), the numbers of the form $R(G)$ are dense in $\mathbb{R}_{\ge 0}$ even when $G$ is restricted to be abelian.
Some first results. $R(G \times H) = R(G) R(H)$ if $\gcd(|G|, |H|) = 1$. We also have $R(C_p) = \frac{2}{p}$ ($p$ is always a prime) and
$$R(C_p^4) = \frac{1}{p^4} \sum_{k=0}^4 {4 \choose k}_p = 1 + \frac{3}{p} + \frac{4}{p^2} + \frac{3}{p^3} + \frac{5}{p^4}$$
where ${n \choose k}_p$ is a Gaussian binomial coefficient. 
Lemma: Let $a_1, a_2, ... $ be a sequence of positive reals such that $\lim_{n \to \infty} a_n = 0$ but such that $\sum a_n$ diverges. Then the set of sums of finite subsequences of the $a_i$ is dense in $\mathbb{R}_{\ge 0}$.
Proof. Let $r \in \mathbb{R}_{\ge 0}$ and fix $\epsilon > 0$. Choose $N$ such that $a_n < \epsilon$ for all $n \ge N$. Then the partial sums starting from $a_N$ diverge but begin less than $\epsilon$ and increase by at most $\epsilon$ at each step, so the conclusion follows. $\Box$
Applying the lemma to the sequence $\log R(C_p^4)$ (which satisfies the hypotheses of the lemma using the fact that $\sum \frac{1}{p}$ diverges), we conclude that the numbers of the form $R(G)$ where $G$ is a product of groups of the form $C_p^4$ for distinct primes $p$ are dense in $\mathbb{R}_{\ge 1}$, and if we allow in addition the groups of the form $C_p$, then the conclusion follows.
Q2 appears to be potentially very difficult and I have not thought about Q3.<|endoftext|>
TITLE: When does a hypersurface have contact-type?
QUESTION [9 upvotes]: In a symplectic manifold $(X^{2n},\omega)$, a hypersurface $Y\subset X$ has contact-type if there is a contact form $\lambda$ such that $d\lambda=\omega|_Y$. Recall that a contact form is a 1-form with $\lambda\wedge(d\lambda)^{n-1}>0$, i.e. the opposite of a foliation.  For example, a starshaped hypersurface has contact-type in $\mathbb{R}^{2n}$, or more generally an $Y\subset\mathbb{R}^{2n}$ transverse to a Liouville vector field defined in a neighborhood of $Y$.  In particular, any contact manifold $Y$ is a contact-type hypersurface in a symplectic manifold (the symplectization $\mathbb{R}\times Y$).  
Now it's nice and useful to consider symplectic manifolds where its boundary has contact-type. And this can usually be done given a compact symplectic manifold (cobordism between contact manifolds).
This leads to the question of whether or not you can always build such a space. In other words:
Given a random hypersurface in $\mathbb{R}^{2n}$, is it of contact-type? How do you tell when it's not of contact-type?
Edit: In light of the posted responses, I think it would be appropriate to tweak one of the questions above. In particular, it was essentially pointed out twice that in the "set of hypersurfaces" there are open neighborhoods which contain no contact-type ones. But, are "most" hypersurfaces in $\mathbb{R}^{2n}$ of contact-type? i.e. Should I expect my hypersurface to be of contact-type?

REPLY [6 votes]: Here is an example of a surface which cannot be made of contact type even after isotopy. The elliptic surface $E(1)$ is obtained by blowing up $\mathbb{C}\mathbb{P}^1$ nine times. It is a symplectic four-manifold with fibre of genus 1 and 12 singular fibres. If we make a fibred connected sum between two copies of E(1) we obtain the elliptic surface $E(2)$, which is a symplectic four-manifold with fibre of genus 1 and 24 singular fibres.
Then $E(2)$ is separated by a torus $T^3$ into two pieces which are diffeomorphic to the complement of a regular fibre in $E(1)$. This shows that the separating torus cannot be made of contact type because, by a result of Chris Wendl, all strong fillings of $T^3$ are diffeomorphic to a blow up of $T^*T^2$.<|endoftext|>
TITLE: henselization and completion
QUESTION [10 upvotes]: This might not be a question appropriate for this forum, I apologize in this case...
Is it true that any element of the completion of a valued ring $R$ that is algebraic over the field of fractions of $R$ is in the henselization of $R$ ?

REPLY [11 votes]: Edit Add noetherian hypothesis because $R$ might no be a subring of $\hat{R}$ otherwise. 
The answer is no as pointed out by quasi-coherent in the comments. But suppose $R$ is a discrete valuation ring and denote $K=\mathrm{Frac}(R)$, then 

$R^h=\hat{R}\cap K^{sep}$. 

There are several equivalent properties defining henselien local rings. I will use two of them: 
let $(A, m)$ be a local ring and $k=A/m$. Then $A$ is henselian iff 
(a) for any monic polynomial $f(X)\in A[X]$, any simple root of $\bar{f}(X)\in k[X]$ lifts to root of $f(X)$ in $A$. 
or equivalently
(b) If $A\to A'$ is a local homomorphism, étale and with trivial residue extension, then $A'=A$. 
One can find a proof in Raynaud: Anneaux locaux henséliens, IV, §3. For discrete valuation rings, one can certainly find easier references. 
Now let us prove the claim above $R^h$. As $\hat{R}$ is henselian (Hensel Lemma), we have $R\subseteq \hat{R}$. 
(1) For any algebraic separable extension $F/K$ contained in $\hat{K}$, $B_F=\hat{R}\cap F$ is a DVR with ramification index $1$ over $R$ and trivial residue extension. This is easy. 
(2) Let $B=B_L$ where $L$ is the separable closure of $K$ in $\hat{K}$. Let us prove that $B$ is henselian using Property (a) above. Let $f(X)\in B[X]$ as in (a). Then any simple root of $\bar{f}(X)\in k[X]$ lifts to a root $\lambda\in \hat{K}$. It is automatically a simple root, so $\lambda$ is separable over $L$, hence $\lambda\in L\cap \hat{R}=B$. 
(3) Let $R\to R'$ be an extension to a henselian DVR. Let's us prove it factorizes throught $R\to B$, which will show that $B$ is a henselization of $R$. It is enough to prove this factorization for $B_F$ for any finite separable exension $F/K$ contained in $\hat{K}$.  As $B_F$ is an étale extension of $R$, $B_F\otimes_R R'$ is étale over $R'$, with a unique maximal ideal above the maximal ideal of $R'$ and trivial residue extension at this maximal ideal (because the quotient $B_F\otimes_R R'/(m')\simeq R'/m'$). By Property (b), this implies that $F\otimes_K K'$ has a direct factor equal to $K'$. Hence $F\subseteq K'$ (the minimal polynomial of a primitive element of $F$ has a root in $K'$) and $B_F\subseteq R'$.  
Remark Parts (1)-(2) work for any local ring $R$ such that $\hat{R}$ is a domain and such that $R\to \hat{R}$ is injective. In particular $\hat{R}\cap K^{sep}$ is henselian. (3) works for any valuation ring But in general, I don't think (3) holds because $B_F$ has no reason to be étale over $R$.<|endoftext|>
TITLE: The behavior of a certain greedy algorithm for Erdős Discrepancy Problem
QUESTION [20 upvotes]: Let $N$ be a positive integer.
We want to find a completely multiplicative functions $f(n)$ with values $\pm 1$ for $n \le N$ such that the discrepancy
$$D=\max_{n \le N} |\{\sum_{i=1}^nf(i)\}|$$
is as small as possible. This is Erdős Discrepancy problem for multiplicative functions.
Consider the following greedy algorithm:
After you assigned the values $f(2),f(3),\dots f(p_i)$  for the first $i$ primes assign the value $f(p_{i+1})$ so as to minimize the maximum discrepancy $|\{\sum_{i=1}^nf(i)\}|$ in every partial sum where unassigned entries of $f$ get the value zero.

Question:  How does this greedy algorithm perform?

Experimental or heuristic answers as well as rigorous proofs are welcomed.
For more background and related questions see this post .
Variation
Consider the same greedy algorithm when you impose the condition that $f(m)=0$ unless $m$ is square free. (If $m$ is not square free $f$ is multiplicative and has values $\pm1$.)

Question: How does our greedy algorithm performs on the square-free version?

Namely, we would like to understand the behavior of the discrepancy of the function obtained by our greedy algorithm. While for EDP there are known examples with $\log N$ discrepancy, this is not known for the square-free version.
Update:
The very nice answer by rlo suggests that the greedy algorithm gives discrepancy close to $n^{1/3}$ or so, and rlo expects it also for the square free variation. Can an upper bound of $N^{1/2-\epsilon}$ be proved? What about a lower bound of $N^{\epsilon}$. Another interesting question is if you can improve the greedy algorithm to get lower discrepancy. Our greedy ignore 0's in intervals. A greedy algorithm that ignore intervals with 0's was considered in polymath5 and to the best of my memory achieve discrepancy $n^{1/2}$. Maybe a clever interpolation between these two variants will do a better job than both?
Further meditation and a new variant
It seems that in our greedy algorithm the decisions we make for small primes are fairly irrelevant. A way to check it:

Run the algorithm for N and test what is the discrepancy for an interval [1,T] where T is,
say, $\sqrt N$. I would expect the answer to be roughly $\sqrt T$.

So now we can think about the following variation:
Let $a>1$ be a real number. We run the greedy algorithm above but our decision for $f(p)$ is based only on intervals $[1,n]$ where $n \le p^a$. (Of course we consider only $n \le N$.

Questions: Can this variant lead to lower discrepancy?
What is the optimal value of $a$?

REPLY [8 votes]: Here is a plot of rlo's data (as listed in his/her item (1)):
          

And here is the $\log D(N) / \log N$ ratio he mentioned in the comment (where the constant red line is $\frac{1}{3}$):<|endoftext|>
TITLE: There are two slightly different notions of ultraproduct.  Why is one said to be better than the other?
QUESTION [20 upvotes]: Let $I$ be a set and $\mathcal{U}$ an ultrafilter on $I$.  Let $(X_i)_{i \in I}$ be an $I$-indexed family of sets.  The ultraproduct of the family $(X_i)$ with respect to $\mathcal{U}$ is, everyone agrees, another set.  But which set is it?  There are two different definitions, and they sometimes give different results.  
For the sake of discussion, I'll call them "Type 1" and "Type 2" ultraproducts.
Type 1 $\ $ The type 1 ultraproduct of $(X_i)_{i \in I}$ with respect to $\mathcal{U}$ is
$$
\Bigl( \prod_{i \in I} X_i \Bigr) \Bigl/ \sim
$$
where 
$$
(x_i)_{i \in I} \sim (x'_i)_{i \in I} \iff \{ i \in I: x_i = x'_i \} \in \mathcal{U}.
$$
Type 2 $\ $ View the poset $(\mathcal{U}, \subseteq)$ as a category.  The type 2 ultraproduct of $(X_i)_{i \in I}$ with respect to $\mathcal{U}$ is the colimit of the functor $(\mathcal{U}, \subseteq)^{\text{op}} \to \mathbf{Set}$ defined on objects by 
$$
J \mapsto \prod_{j \in J} X_j
$$
and on maps by projection.  Explicitly, then, the Type 2 ultraproduct is
$$
\Bigl( \coprod_{J \in \mathcal{U}} \prod_{j \in J} X_j \Bigr) \Bigl/ \approx
$$
where 
$$
(x_j)_{j \in J} \approx (x'_k)_{k \in K} \iff \{ i \in J \cap K: x_i = x'_i \} \in \mathcal{U}.
$$
The difference $\ $ The two types of ultraproduct are the same if either none of the sets $X_i$ are empty or almost all of them are empty.  But in the remaining case, where at least one $X_i$ is empty but the set of such $i$ is not large enough to belong to $\mathcal{U}$, they're different: the Type 1 ultraproduct is empty but the Type 2 ultraproduct is not. 
The question $\ $ I've read in a couple of texts (both coming from the point of view of categorical logic) that the Type 2 ultraproduct is really the right one.  But why?  On what criteria is Type 2 judged to be better than Type 1?
A vague guess at an answer $\ $ I think I can guess very roughly what's going on.  There's been a tradition in logic — maybe dying out now? — of taking all structures to be nonempty by definition.  But when you move to the more general setting of categorical logic, that's no longer a satisfactory approach.  Although in the category of sets, there's just a single object with no elements, in many other categories, there are lots of interesting objects with no (global) elements: e.g. there are lots of interesting sheaves with no global sections.  
So categorical logic sometimes involves a recasting of classical, set-based logic, in order to handle empty sets/types satisfactorily.  I imagine that something of the sort is going on here.  (I only defined ultraproducts of sets, but you could of course define ultraproducts of objects of any other sufficiently complete category.)  But still, I don't see clearly why Type 2 is the right choice.  
See also This question of Joel David Hamkins, and its responses.

REPLY [6 votes]: Michael Barr pointed out one thing that goes wrong if you try to use the type 1 definition when some of the sets involved are empty.  Now that I understand this issue better, I'll point out a couple of other things that go wrong.  They're both of the form "this theorem holds cleanly for type 2 ultraproducts, but for type 1 you have to make exceptions".
I'll use the standard notation for the ultraproduct of $(X_i)_{i \in I}$ with respect to an ultrafilter $\mathcal{U}$ on $I$:
$$
\Bigl( \prod_{i \in I} X_i \Bigr) \bigl/ \mathcal{U}
$$
(for either of the two definitions).  The notation makes more sense for type 1 than type 2, but never mind.
(1) Ultraproduct with respect to a principal ultrafilter is projection.  That is, if $k \in I$ and $\mathcal{U}$ is the principal ultrafilter on $k$ then $(\prod X_i)/\mathcal{U} = X_k$.  This is true without exception for type 2 ultraproducts.  It is almost true for type 1, but fails if $X_k \neq \emptyset$ and $X_i = \emptyset$ for some $i \neq k$.
(2) Ultraproducts preserve finite coproducts.  That is, writing $+$ for the coproduct (disjoint union) of sets,
$$
\Bigl( \prod_{i \in I} (X_i + Y_i) \Bigr) \bigl/ \mathcal{U} \cong \Bigl( \prod_{i \in I} X_i \Bigr) \bigl/ \mathcal{U} + \Bigl( \prod_{i \in I} Y_i \Bigr) \bigl/ \mathcal{U}
$$
for any families of sets $(X_i)$ and $(Y_i)$.  This is true without exception for type 2 ultraproducts (using the fact that these are *ultra*$\mbox{}$filters).  But again, it isn't quite true for type 1: it can fail when some of the sets are empty.  For example, let $I$ be a set, choose any nonempty proper subset $J$ of $I$, and put
$$
X_i = \begin{cases} 1 &\text{if } i \in J\\\ \emptyset &\text{otherwise} \end{cases}
\qquad\ \qquad
Y_i = \begin{cases} \emptyset &\text{if } i \in J\\\ 1 &\text{otherwise,} \end{cases}
$$
where $1$ denotes a one-element set.  Then according to the type 1 definition, $(\prod(X_i + Y_i))/\mathcal{U} = 1$ but $(\prod X_i)/\mathcal{U} + (\prod Y_i)/\mathcal{U} = \emptyset$.
I guess all of these things that go wrong are intimately related to Łoś's theorem, which Michael alluded to.<|endoftext|>
TITLE: Question about the Yangian
QUESTION [22 upvotes]: I've a slightly technical question about the Yangian which I'm hoping an expert out there can answer.
Recall that the Yangian $Y(\mathfrak{g})$ is a Hopf algebra quantizing $U(\mathfrak{g}[z])$.    Drinfeld, in his quantum groups paper, explains that the algebra of integrals of motion of certain integrable lattice models can be described in terms of the Yangian, as follows. 
Let $C(\mathfrak{g})$ be the algebra of linear maps 
$$l :  Y(\mathfrak{g}) \to \mathbb{C}$$ with $l([a,b] ) = 0$.  In other words, $C(\mathfrak{g})$ is the linear dual of the zeroth Hochschild homology of $Y(\mathfrak{g})$. The product on $C(\mathfrak{g})$ comes from the coproduct on the Yangian, and the existence of the $R$-matrix implies that $C(\mathfrak{g})$ is commutative.
Question 1: Is there an explicit description of the algebra $C(\mathfrak{g})$?  
Let $C'(\mathfrak{g})$ be the classical analog of $C(\mathfrak{g})$, defined by replacing $Y(\mathfrak{g})$ in the above discussion with $U(\mathfrak{g}[z])$. 
One can compute that $C'(\mathfrak{g})$ is the algebra of $\mathfrak{g}[z]$-invariant formal power series on $\mathfrak{g}[z]$ (invariant under the adjoint action). 
Question 2: Is there a PBW theorem for $C(\mathfrak{g})$, stating that $C(\mathfrak{g})$ has a filtration whose associated graded is $C'(\mathfrak{g})$?  (This is equivalent to asking if part of the spectral sequence computing Hochschild homology of the Yangian from Hochschild homology of $U(\mathfrak{g}[z])$ degenerates). 
Thanks,
Kevin

REPLY [20 votes]: It seems that subalgebra in the question is the so-called "Bethe subalgebra" of the Yangian.
It is not immediate for me to recognize the connection with definition given in the question and the definition I'll give below - but I am sure that it should be simple and well-known. Can someone clarify this?  However, the sentence "One can compute that $C'(g)$ is the algebra of $g[z]$-invariant formal power series on $g[z]$ (invariant under the adjoint action)" leaves me with no doubt that this is Bethe subalgebra. (I somehow missed this point when I wrote the first version of this answer).
Some references
The subalgebra originates from the work of Faddeev's school on quantum integrable systems, and many things were known and "obvious" for Leningrad's team and it is not always easy to provide appropriate references.
In the mathematical literature it was introduced in
"Bethe Subalgebras in Twisted Yangians", by Maxim Nazarov, Grigori Olshanski
where the name "Bethe" subalgebra was proposed, Leningrad's constructions have been mathematically written up for $Y(gl_n)$ and for twisted Yangians. (Twisted Yangians for semisimple $g$, were introduced by Olshanski ~1989).
Surprisingly, a similar construction for Yangians(g) for classical $g$ (not twisted Yangian, not $gl_n/sl_n$), is quite recent (as far as I understand):
"Feigin-Frenkel center in types B, C and D", by A. I. Molev.
The moral of the papers is that these subalgebras are very similar to the center of $U(g)$.
They are free commutative algebras with generators which can be indexed by $C_{i,k}$, $k=1,\dots,\infty$ (corresponds to loop variable $z^k$) and $i=1,\dots,rank(g)$ - corresponds to generators of the center of $U(g)$. I.e. the center of $U(g)$ can be "loopified" to get commutative subalgebra in $Y(g)$.
They degenerate to appropriate subalgebras in $U(g[t])$.
Small detail: you can work with either $U(tg[t])$ or $U(g[t])$. In the first case you get maximal commutative subalgebra, in the second it is not maximal and you can add arbitrary constant matrix to make it maximal. I mean morally it is maximal, but there are details.
Similar subalgebras are in $U_q(g[t])$, elliptic algebras, many twisted versions - physics literature devoted to "Bethe ansatz" - the method to find joint eigenpairs is enourmous.
Concerning the Yangian: there are surveys:

"MNO" http://arxiv.org/abs/hep-th/9409025,
Molev http://arxiv.org/abs/math/0211288, and
Molev's boook "Yangians and Classical Lie Algebras." Mathematical Surveys and Monographs. Providence, RI: American Mathematical Society.

They do not cover the physics literature and some recent developments.
Quick facts. Everyone can understand.
Let me give some facts as abstract ones and later relate to the subalgebra in question.
Fact 1. (AKS-lemma (Adler-Kostant-Symmes)).
Consider any associative algebra $A$, and its two subalgebras $B,C$. Assume   $A$ is isomorphic to $B\otimes C$ as a vector space. Then the projection of center of $A$ to $B$ gives a commutative subalgebra in $B$. (Same for $C$). (Same for Poisson algebras).
Proof - pleasant exercise. Importance: all natural commutative subalgebras and integrable systems come in this way or around :)
Fact 2. (Poisson center of $S(g)$ vs. center of $S(g(t))$).
Consider Lie algebra $g$ and its loop algebra $g(t)$.
How are the Poisson centers of $S(g)$ and $S(g(t))$ related?
It is simple (if I am not mistaking:) - from any element $C$ of $Z(S(g))$ one can make a generating
function $C(t)= \sum t^iC_i$  for elements of the center of $Z(S(g(t)))$.
Explanation: think of the loop space for $g^*$ as infinite product of copies of $g^*$ indexed by points of the circle $S^1$. The structure of the center of the finite product is clear. We just consider the infinite limit.
Proof: Exercise or if anyone is interested I can try to write it.
The same works not for only for $g^*$, but for any Poisson manifold. (Open question - what happens when quantizing? What is general "anomaly cancellation = critical level"?  (This MO question is related).
Where do all "these" commutative subalgebras (integrable systems) come from?
Now it is pretty simple. Take Lie algebra $g$ (usually reductive).
Consider the loop algebra $g(t)$ and split it to $g[t], g[t^{-1}]$.
From fact 2 above we have a huge center in $S(g(t))$.
Let us project it to $S(g[t])$ - we get a commutative subalgebra there.
Variation on this theme - take quantum-super generalization of $g(t)$ - the same thing will work.
E.g. Yangian case.
That is all.
What is difficult? Quantization, anomalies, Langlands correspondence
Everything works well when we consider the "classical world" i.e. work with Poisson brackets,
not commutators.
We might ask what happens for $U(g(t))$, not for $S(g(t))$.
Here the story becomes interesting - the moral is that everything will work, but not in the obvious way.
In physics it is related to "anomalies" "renormalization", change of $k\rightarrow k+2$ in Chern-Simons and WZ coupling constants.
The miracle is that - if we will think properly how to construct the center of $U(g(t))$
then we will come to a form of the Langlands correspondence.
The breakthrough is due to Dmitry Talalaev: http://arxiv.org/abs/hep-th/0404153
The key point is that generators of the center of $Z(U(g(t))$ should be organized in the
differential operator of order rank(g): $\sum_i C_i(z) (\partial_z)^i $.
So we have a form of Langlands correspondence: take an irrep $V$ (automorphic side);
center acts on $V$ by scalars so we get  $\sum_i C_i(z)|_{V} (d_z)^i $ - scalar differential operator.
Differential operators should be thought as "Galois side" - their monodromy gives the "representations of the Galois group".
Well, over the complex numbers there is not a Galois group, but we should think about differential operators as a kind of representations of Galois group.
Partly thing like these are described in our paper
http://arxiv.org/abs/hep-th/0604128 .
Talking about this "Langlands related" stuff one should mention works by Feigin and E. Frenkel.
There are several surveys by E. Frenkel on the arXiv. I like the old one: http://arxiv.org/abs/q-alg/9506003 from which I learnt a lot.
Back to question. 1) Explicit description
Question 1: Is there an explicit description of the algebra $C(g)$?
Yes there is for classical $g$. For $Y(gl_n)$ I think we know it pretty well. For $Y(g)$ for classical $g$ this is recent work by Molev mentioned above, which is the starting point, hopefully there will be further works.
For exceptional I do not know anything.
Let me write something on an explicit description.
It is important to keep in mind two levels: Poisson and quantum.
In the Poisson case everything is obvious for any $g$ (even exceptional).
To describe this explicitly let me remind you of the "matrix notations" - the Yangian for $gl_n$ can be described as $RTT=TTR$ where "$T(z)$" is a matrix which contains all generators of $Y(gl_n)$. It is called the "Lax matrix" or "transfer matrix", sometimes the "monodromy matrix".

In the Poisson case:
$Trace( T^k(z)) = \sum_i T_{k,i} z^i$

or

$det(l-T(z)) = \sum_{k,i} l^kz^i C_{k,i}  $

$T_{k,i}$ and $C_{k,i}$ always (any $g$) Poisson commute among themselves (easy exercise for R-matrix calculations).
Any set $T_{k,i}$ or $C_{k,i}$ provides a set of free generators of $C(Y(g))$ (well I forget pffaffian in $so(2n)$ case).

Quantum case

It is similar to quantum groups - instead of the determinant, one should use the q-determinant.
In the Yangian case we do not have "q" but we must insert the shift in "z" e.g.

$Trace( T(z) T(z+1) T(z+2) ... T(z+k-1) ) = \sum_i T_{k,i} z^i$

This will give the generators of Bethe subalgebra - surprisingly it is recent ( http://arxiv.org/abs/0711.2236 ).
Talalaev's determinant:

$det^{column} ( 1- exp(-d/dz) T(z) ) =  \sum_{k,i} exp(-l d/dz) z^i C_{k,i}$

This will give another set of generators of the Bethe subalgebra. The generators themselves
are old (going back to the works of Faddeev's team in the late 1980's or maybe even earlier).
Talalaev's insight is to introduce $\exp(-d/dz)$ and to obtain this difference GL-oper.
This allowed him to make a degeneration to $U(g[z])$.
He proved that $\det(d/dz - L(z))$ will give generators of the Bethe subalgebra in $U(g[z])$,
they were not known before in a nice form.
The situation might be strange - in the quantum group case we've known how to describe the Bethe subalgebra for
20 years, but in the seemingly simpler case of $U(g[z])$ it was not known. However this is true - in some sense quantum groups easier to deal with than classical ones.
Let me mention that $e^{-d/dz} T(z)$ is a Manin matrix.
Back to question 2): Is there a PBW theorem for $C(g)$, stating that $C(g)$ has a filtration whose associated graded is $C'(g)$?
For $gl_n$ this should be contained in Nazarov-Olshanski and the survey MNO mentioned above.
For classical $g$ see the recent paper of Molev mentioned above. (I am not sure about $so(2n)$ and pffafian).
For exceptional I think it is not known. Maybe the Feigin-Frenkel technique works here - they do not need explicit formulas.<|endoftext|>
TITLE: Extending an assignment property from Q to R (or C)
QUESTION [10 upvotes]: Property of any odd number of nonnegative integers:
Given $x_1 \leq \ldots \leq x_{2n + 1}$ with each $x_i \in \mathbb{Z}_{\geq 0}$, suppose that for any $x_i$ we remove, the remaining numbers can be assigned to disjoint multisets $A$ and $B$ such that $|A| = |B| (= n)$ and $\sum_{x \in A} A = \sum_{x \in B} B.$ Then all the $x_i$ must be equal.
To prove this, fix any $n \geq 0$ and note that all the $x_i$ must have the same parity. Suppose there were an assignment of values such that the $x_i$ weren't all equal; using the Well-Ordering Principle, consider such a non-trivial assignment with $\sum x_i$ minimal. If $x_1 = 0,$ divide each $x_i$ by $2$ to derive a contradiction; if $x_1 > 0,$ subtract $1$ from each $x_i$ to derive a contradiction. Thus, all the $x_i$ must be equal.
This property extends easily from the nonnegative integers to the rational numbers: if we had a non-trivial assignment of rational values, we could multiply through by their lcd to obtain a non-trivial assignment of integer values, and then subtract from each the smallest integer value to obtain a non-trivial assignment of nonnegative integers, thereby contradicting our earlier conclusion.
Question: Does this property also hold for any odd number of real (or complex) numbers?
Any connections between this question and other areas would, of course, be welcomed.
Question re-stated explicitly: Given $x_1, \ldots, x_{2n + 1}$ with each $x_i \in \mathbb{R}$ (or $\mathbb{C}$), suppose that for any $x_i$ we remove that the remaining numbers can be assigned to disjoint multisets $A$ and $B$ such that $|A| = |B| (= n)$ and $\sum_{x \in A} A = \sum_{x \in B} B.$ Is it true that all the $x_i$ must be equal?

Edit 1: An equivalent problem appears as #15.23 in "Problems and Theorems in Classical Set Theory" (solution on pp. 323-324) by Péter Komjáth and Vilmos Totik (2006).

Edit 2: The integer version of this problem was Putnam Problem B1 (1973); the question about extending to the reals is posed at that solution link as well. The integer version (for 2n+1 = 23) is also included as problem #3.4.31 in "The Art and Craft of Problem Solving" (2e, p. 107) by Paul Zeitz. An earlier extension to the positive reals (easily seen as equivalent to the reals) is AMM Problem 11002 (2003), though the solver whose answer is linked to proceeds in a manner quite different from that of Pierre or Péter.

Edit 3: In a 1992 article, Liong-shin Hahn notes the connection between B1/1973 and an article on abelian groups with no nontrivial element of odd order (Respective sources: L.-S. Hahn, "Alternate Solutions to Putnam Competition Problems," Mathematics Magazine, 65(2), 1992; G.A. Martin, "A class of Abelian groups arising from an analysis of a proof," Amer. Math. Monthly, 95, 1988). 
Hahn then proceeds to solve the problem over R and C using the sort of approach used above in AMM 11002's solution. This appears to be an English version of Hahn's earlier write-up (1981), in which he claims to have solved this problem (known as Problem 4304 in Mathmedia). I couldn't find his first solution (see Edit 4); however, I did locate another citation for his having solved it, as well as the original problem. All were written in Chinese (Hahn was born in Taiwan; see his AMS Citation for Public Service), so I will type out the original problem and provide an English translation below.
Traditional Chinese:
4304 （編輯部提供） 有17個球，假定無論任取一個後，剩下的16個都可以分為重重相等的兩堆，每堆8個。試證此17個球的重量必然相同。
Simplified Chinese:
4304 （编辑部提供） 有17个球，假定无论任取一个后，剩下的16个都可以分为重重相等的两堆，每堆8个。试证此17个球的重量必然相同。
English Translation:
4304 (Proposed by the editors) Given 17 balls, suppose that no matter which one is removed, the remaining 16 can be separated into two piles of equal weight, with 8 balls in each pile. Prove that these 17 balls must all be equal in weight.

Edit 4: I managed to obtain a copy of Hahn's original solution (1981) to this problem (over $\mathbb{R}_{+}$, which easily extends to $\mathbb{C}$) by contacting an administrator at Academia Sinica, Taipei, Taiwan. His approach proceeds using only basic linear algebra with an application of Cramer's Rule as the coup de grâce. The solution in Chinese can be viewed here; I will not endeavor to translate it unless it is somehow essential to someone's research. Nonetheless, it's interesting to see a full solution that preceded the AMM solution by 22 years. Though this is nothing like Weil's popularization of the Shimura-Taniyama conjecture from a Japanese journal, it does make one wonder what other gems have been hidden away in Asian journals of mathematics.

REPLY [13 votes]: Yes, the result can be deduced from the rational case. Let $x_1, \ldots, x_{2n+1}\in \mathbb C$ be given. Fix a basis of the $\mathbb Q$-vector space spaned by the $x_i$'s and write the coordinates of $x_i$ in this basis as $(x_i^1, \ldots, x_i^r)$.
Now the same property holds for each family $x_1^k,\ldots,x_{2n+1}^k$. As those are now rationals, we conclude that they are all equal. Hence the result follows.<|endoftext|>
TITLE: What is the theoretical interest of finding closed-form solutions of infinite series?
QUESTION [9 upvotes]: I was reading this when I came across Gourevitch's conjecture.
My understanding is that solutions to these series are of practical interest. If one encounters such a series, being able to solve it exactly is more practical than having to solve it numerically.
But, not being a mathematician, I simply can't imagine what the theoretical implications of proving such conjectures are.
What are they?

REPLY [2 votes]: This is an old question, but just popped up again, so I thought I'd mention a very important situation where explicit formulas are of fundamental theoretical importance. Number theorists are very interested in understanding the ideal class group $H_p$ of the cyclotomic field $\mathbb Q(\zeta_p)$, where $\zeta_p$ is a primitive (prime) root of unity. People use explicit formulas for special values of $L$-series to construct integers that annihilate the class group, thereby obtaining information about the size of the class group. And for elliptic curves, Rubin in the CM case and Kolyvagin for curves over $\mathbb Q$ have used special values such as $L(E,1)$ and $L'(E,1)$ to prove that the Tate-Shafarevich group of $E$ is finite, which had been a long-standing conjecture. Very (very) roughly, they use the special values to construct an integer $m$ that annihilates every element of SHA, and from there the finiteness follows, since it is not hard to prove that SHA$[m]$ is finite. This is the theory of Euler Systems, which is still attracting lots of research attention.
I realize that $L$-series are Dirichlet series, and this question asks about "infinite series", which is probably meant to refer to power series. But one can move back and forth between $L$-series and power series using the Mellin transform, and indeed, for elliptic curves over $\mathbb Q$, the first step in studying $L(E,1)$ is to write it as the Mellin transform of a modular form, courtesy of Wiles et al.<|endoftext|>
TITLE: Finding the smallest eigenvalues of a large, but structured, matrix
QUESTION [8 upvotes]: I'm trying to find the eigenvector corresponding to the second smallest eigenvalue of a large $(4,000,000 \times 4,000,000)$ matrix $M$. $M$ is a Laplacian matrix, and it has the following structure: $M = D - AA'$, where $D$ is a diagonal matrix and $A$ is a large sparse matrix. $A$ has dimension $4,000,000 \times 10,000,000$, but only about $40,000,000$ non-zero entries. So I can rapidly perform matrix-vector multiplication: $Mv = Dv - A(A'v)$.
Currently I'm using Scipy (which calls ARPACK) to find the smallest eigenvalues and corresponding eigenvectors of $M$. The implementation is a variant of Lanczos method, as far as I can tell. Unfortunately the implementation fails to converge with some frequency. Even when it does converge, since I need fairly high accuracy, it takes many iterations to converge. Any suggestions?
Lanczos method is much more reliable when finding the largest eigenvalues and the corresponding eigenvectors, as I understand it. So I was thinking that I could transform the problem, perhaps instead finding the largest eigenvalues of M's left inverse. But, even if I somehow manage to find a good approximation of M's left inverse (most methods don't converge), the largest eigenvalue is infinite (the smallest eigenvalue of $M$ -- a Laplacian matrix -- is $0$), so it seems there would still be convergence issues, since I can't solve for the second largest eigenvalue without also solving for the largest, as I understand it. Is there an easier way?

REPLY [3 votes]: If it is a Laplacian then you not only know the smallest eigenvalue is zero, but you also know its corresponding eigenvector.  You can use this information by essentially adding a single step to your procedure, it's something like adding back the mean of v to each element of Mv.  Then use arnoldi to find the smallest (instead of second smallest) eigenpair.<|endoftext|>
TITLE: Square root of a positive $C^\infty$ function.
QUESTION [41 upvotes]: Suppose $f$ is a $C^\infty$ function from the reals to the reals that is never negative.  Does it have a $C^\infty$ square root?  Clearly the only problem points are those at which $f$ vanishes.

REPLY [41 votes]: The answer is "no". This is covered in great detail here:
http://www.math.polytechnique.fr/~bony/BBCP_jfa.pdf

REPLY [31 votes]: The function
$$f(x) = \begin{cases}
\sin^2 \left(\frac{1}{x} \right) e^{-1/x} + e^{-2/x} & \text{if $x > 0$,}\\
0 & \text{if $x \leq 0$,}
\end{cases}$$
is $C^\infty$ but has no $C^2$ square root. I found this example in the paper Choosing roots of polynomials smoothly by Alekseevsky, Kriegl, Losik, and Michor (available freely here). This example appears to have come from Frank Warner's (unpublished) 1963 dissertation.<|endoftext|>
TITLE: Complex evaluation of a classical (real) integral
QUESTION [14 upvotes]: There are several ways to compute the classical integral
$$
\int_{\mathbb R}e^{-x^2}dx=\sqrt{\pi}.
$$
Probably, best known are
(1) squaring the integral with subsequent change
of (now two) variables to the polar form, and
(2) the reducing to the Gamma-function at $1/2$.
I am interested though in a "complex" analysis method (namely, a use
of the residue theorem) to do the job. The reason is that several integrals like
$$
\int_0^\infty e^{-x^2}\cos ax\ dx
\qquad\text{or}\qquad
\int_0^\infty\sin x^2\ dx
$$
can be computed via the residue theorem and the above integral,
so I would like to avoid any reference to real analysis. Is there such
a complex evaluation though?!

REPLY [2 votes]: It seems it was Polya, not Mirsky, who gave the first published evaluation of the probability integral via contour integration in 1945: https://projecteuclid.org/euclid.bsmsp/1166219199 (George Polya, Remarks on Computing the Probability Integral in One and Two Dimensions - see chapter 5). Polya's paper was published only in 1949, so Mirsky was not aware of it.
This story is told in the book D. Mitrinovic and J.D. Keckic,  The Cauchy Method of Residues: Theory and Applications, Volume 1, pp. 158-168. Interestingly, the authors write: 
"Some books even state that its value cannot be found by contour methods (see, for example [1] or [2]).
In fact, this is not so. We did our best to find the earliest proof of (1) by contour methods, but we did not arrive at a decisive conclusion. In a private communication dated 3 June, 1971, Professor Copson informed us that immediately after the publication of [2] (in 1935) somebody evaluated the integral (1) by the method of residues, but that he forgot who that was".
Is it possible to find this 1935 publication, if it existed?
For one more recent variant of such a calculation, see http://ijmsa.yolasite.com/resources/60--sepp.pdf (A simplified contour integration for the Probability integral, by P.W. Gwanyama)<|endoftext|>
TITLE: Quasi-coherent sheaves on $X/G$
QUESTION [11 upvotes]: Let $X$ be a scheme and let $G$ be an abstract group acting on $X$ by scheme automorphisms. I'm happy to assume finiteness conditions on $X$ (such as locally Noetherian) and on $G$ (such as $G$ is finite) as necessary.
I understand that it is possible to enlarge the category of schemes in such a way that there is a good "quotient space" $X/G$ associated to this data. 
A. What is the correct framework in which to study $X/G$? Is this an example of a quotient stack or perhaps an algebraic space?
There should be a good category of sheaves on $X/G$. Let's agree that the "right" category of sheaves on the scheme $X$ is the category of quasi-coherent sheaves of $\mathcal{O}_X$-modules.
B. What is the "right" category of sheaves on $X/G$? 
This should be an abelian category with enough injectives. Assuming the answer to this question is "yes", let's call this category $Qcoh(X/G)$. 
C. Is it possible to realise $Qcoh(X/G)$ as a subcategory of the category of abelian sheaves $Sh(\mathcal{C})$ on some Grothendieck site $\mathcal{C}$?
One candidate for such a $\mathcal{C}$ is the set of $G$-stable open subsets of $X$, but presumably this is too naive to work in general.
D. Is there always a morphism of abelian categories $Qcoh(X) \to Qcoh(X/G)$? 
On the other hand, it is possible to form the category of $G$-equivariant quasi-coherent sheaves  $G-Qcoh(X)$ on $X$. 
E. What, if any, is the connection between $G-Qcoh(X)$ and $Qcoh(X/G)$?
I suspect that the answer to this last question is "$Qcoh(X/G)$ is equal to $G-Qcoh(X)$ by definition", in which case the question becomes "Can you shed some light on this definition"?
Finally, references to the literature where similar questions are studied would be greatly appreciated.

REPLY [5 votes]: I think this question is a good one, but don't expect an encyclopedic answer — MO is not an encyclopedia.  Here are some answers, with the disclaimer that I'm a category theorist but not an algebraic geometer.
To question A, by and large the 21st perspective will probably say that it is definitely a stack, in some notion of the word.  Certainly there are applications where you do want to consider the "space" quotient $X/G$, in which points in the same orbit are honestly identified.  This is like taking a form of "$\pi_0$" of the stack.  (Not etale $\pi_0$, certainly, but a form of $\pi_0$ that's valued in spaces rather than sets.)
To questions B, D, and E, the answer is that, as you guessed, the best definition of $\operatorname{QCoh}(X/G)$ is the category of $G$-modules in $\operatorname{QCoh}(X)$, at least when $G$ is a finite group.  The geometric intuition is that a quasicoherent sheaf on $X$ is something like a vector bundle over $X$.  In the quotient $X/G$, we add an isomorphism between any two points for each way that they are related by an element of $G$.  So a vector bundle over $X/G$ should have a fiber over each point of $X$, and an isomorphism between these fibers for each pair of $G$-related points.
There is a quotient morphism $X \to X/G$.  The $\operatorname{QCoh}$ functor is best understood as contravariant, just like $\mathcal{O}$ is contravariant.  Namely, a geometric morphisms $f: X \to Y$ correspond (modulo details) to symmetric monoidal "linear" functors $f^\ast : \operatorname{QCoh}(Y) \to \operatorname{QCoh}(X)$, which pull back a "vector bundle" along the map.  This is certainly true for $\operatorname{QCoh}(X/G) = \operatorname{QCoh}(X)^G$, with the quotient morphism corresponding to the functor "forget the $G$-action".  That said, each such functor $f^\ast$ also has a right adjoint $f_\ast : \operatorname{QCoh}(X) \to \operatorname{QCoh}(Y)$, which is not usually symmetric monoidal — it is that takes a "vector bundle" over $X$ and makes it into the "vector bundle" over $Y$ whose stock over $y\in Y$ is the space of sections over $f^{-1}(y)$ of the corresponding bundle on $X$.  In the case of the quotient map $X \to X/G$, its right adjoint is the "free" functor, assigning to a quasicoherent module $M$ the corresponding free $G$-module $G \otimes M$.  You ask for $f_\ast$ to be "a morphism of abelian categories", which is vague to me.  The best definition I know of "morphism of abelian categories" is a right-exact functor (if I have left and right correct), in which case in general pushforward maps are not morphisms — they are instead left-exact.  I think that if $G$ is finite, then in fact the pushforward along the quotient map is exact; maybe I need to include that the characteristic of the ground field does not divide the order of $G$.
As for C, as just a subcategory, I'm sure the answer is yes.  If you ask for more conditions, the answer is probably still yes, at least in the finite-group case: you should be able to take the category of cocommutative coalgebras in $\operatorname{QCoh}(X/G)$ and find this as sheaves-of-sets on something.  But I'd have to think more about details.<|endoftext|>
TITLE: Are there lower bounds on the quality of a rational approximation?
QUESTION [5 upvotes]: For which irrational numbers $\xi$ does there exist a constant $A$ such that $\left|\frac{p}{q}-\xi\right|<\frac{1}{Aq^2}$ (where $p/q$ is a rational number) has only finitely many solutions?

Background
I apologize if this is a terribly elementary question—my background is in analysis, not number theory.  If there is some well-known source that answers this question, I would appreciate a link or citation.
As far as I can tell, there are definitely some irrational $\xi$ for which such an $A$ exists.  For instance, if $\xi = \frac{1+\sqrt{5}}{2}$, then for any $A>\sqrt{5}$ the inequality $\left|\frac{p}{q}-\xi\right|<\frac{1}{Aq^2}$ has only finitely many solutions (this seems to be a very elementary result, which appears in a number of texts on elementary number theory).  Additionally, if $\xi$ is an irrational root of a quadratic polynomial, then there is some $A$ for which $\left|\frac{p}{q}-\xi\right|<\frac{A}{q^2}$ has no solutions (see the Wikipedia article on Liouville numbers).  Thus for any irrational root of a quadratic polynomial, there is some constant $A$ of the kind desired.
According to MathWorld, there are some results in this direction:  the article on Hurwitz's Irrational Number Theorem suggests that the answer to this question is be related to the existence or nonexistence of a Lagrange number associated to a particular irrational number.  However, there are only countably many Lagrange numbers, and each Lagrange number is associated to only a countable number of irrationals, so it seems that the existence/nonexistence of a Lagrange number associated to a particular irrational is only a part of the answer.
Again, I apologize if this is a really basic question, and appreciate any help or advice.

REPLY [5 votes]: Here are some standard facts that you can find in many textbooks, e.g. in Cassels: An introduction to diophantine approximation.

There exists an $A>0$ such that $\left|\frac{p}{q}-\xi\right|<\frac{1}{Aq^2}$ has only finitely many rational solutions $\frac{p}{q}$ if and only if the continued fraction expansion of $\xi$ consists of bounded digits. In particular, quadratic irrationals have this property because they have a periodic continued fraction expansion.
There exists an $A<3$ such that $\left|\frac{p}{q}-\xi\right|<\frac{1}{Aq^2}$ has only finitely many rational solutions $\frac{p}{q}$ if and only if $\xi$ lies in an $\mathrm{SL}_2(\mathbb{Z})$-orbit associated with some Markov number (see here). In particular, only countably many $\xi$'s belong to this case.
There are continuum many $\xi$'s such that $\left|\frac{p}{q}-\xi\right|\geq\frac{1}{3q^2}$ for any rational number $\frac{p}{q}$.<|endoftext|>
TITLE: Generators of associated graded algebra
QUESTION [7 upvotes]: Suppose that $A = \bigcup_{n=0}^{\infty} A_n$ is a filtered algebra over a field $k$.  The associated graded algebra is $\mathrm{gr} A = \bigoplus_{n=0}^{\infty} A_n/A_{n-1}$, where we define $A_{-1} = (0)$.  There is no canonical algebra map from $A$ to $\mathrm{gr} A$, but there is a well-defined function $\gamma : A \to \mathrm{gr} A$ given by
$$
\gamma(x) = x + A_{n-1} \in A_n/A_{n-1},
$$
where $n$ is the unique natural number (or 0) such that $x \in A_n$ but $x \notin A_{n-1}$.  (To forestall nitpicking, let's say that $\gamma(0) = 0 \in A_0$.)  Of course, this map fails to be even additive, but it does exist.
Question:
Given a set of generators $\{x_i\}$ for $A$, when is it the case that the set $\{\gamma(x_i)\}$ generates $\mathrm{gr} A$?
Here is an easy example where this fails to happen.  Let $\mathfrak{h}$ be the 3-dimensional Heisenberg Lie algebra (over $\mathbb{C}$, say), spanned by three elements $X,Y,Z$ with $[X,Y] = Z$ and $Z$ central.  Let $A = U(\mathfrak{h})$ be its universal enveloping algebra with the usual filtration.  Since $XY - YX = Z$, it follows that $A$ can be generated just by $X$ and $Y$.  But Poincare-Birkhoff-Witt tells us that
$\mathrm{gr} A \cong \mathbb{C}[X,Y,Z]$, which is certainly not generated just by $X$ and $Y$.
The problem here is with the relation $XY-YX=Z$: since $Z$ has lower degree than $XY$ and $YX$, it drops out of the relation in the associated graded.
Can anything be said about this, in general?  Are there any nice criteria on the filtration and the generating set (and the relations, obviously) that ensure things don't go wrong in this way?  Also I am amenable to making assumptions on the algebra $A$, for example that it is finitely generated or Noetherian (or ...?), if that helps.  I do not want to increase the size of the generating set.

REPLY [7 votes]: This question was posted some time ago but it's never been answered properly, so it seems worthwhile to record the answer anyway. In short, when $A$ is finitely generated, the associated graded is generated by the elements you describe if and only if the filtration is inherited from the natural grading of a free associative algebra.
Suppose that $A$ is filtered $A = \bigcup_{i\geq 0} A_i$ with $A_0 = k$ and that $A$ is generated by a finite set of elements, say $x_1,...,x_n$. We may suppose that $x_i \in A_{d_i}\setminus A_{d_i-1}.$ By the universal property of free associative algebras there is a homomorphism from $F = k\langle X_1,...,X_n\rangle$ onto $A$, say $\phi: F \twoheadrightarrow A$ which sends $X_i \mapsto x_i$. Now $F$ is also graded by placing $X_i$ in degree $d_i$ and this defines a filtration $F = \bigcup_{k \geq 0} F_k$ where $F_k$ is the span of monomials $x_{i_1} \cdots x_{i_m}$ with $\sum_{j=1}^m d_{i_j} \leq k$ and $1\leq i_1,...,i_m \leq n$. Note that $\phi$ preserves the filtration.
The image of $\operatorname{gr}\phi$ is the subalgebra of $\operatorname{gr} A$ generated by the elements $x_i + A_{d_i-1}$ and so your question can now be phrased as asking when is $\operatorname{gr} \phi$ surjective? According to Corollary 7.6.14 of McConnell and Robson "Noncommutative Noetherian rings" this is the case if and only if $A_i = \phi(F_i)$. This also tells us that, out of all the filtrations on $A$ satisfying $x_i \in A_{d_i}\setminus A_{d_i-1}$ there is a unique filtration such that $\operatorname{gr} A$ is generated by the elements $x_i + A_{d_i-1}$, and it is the minimal filtration in a precise sense.
Now consider the example of the three dimensional Heisenberg algebra $\mathfrak{h}$ which you mentioned. We can view this as a quotient of the free algebra $F = k\langle X, Y\rangle$ by the ideal generated by $[X,[X,Y]]$ and $[Y, [X, Y]]$, however if you want the element $z := \phi [X, Y]$ to lie in degree 1 of $U(\mathfrak{h})$ (in accordance with the PBW theorem) then you have $\phi(F_1)$ spanned by ${1, \phi(X), \phi(Y)}$, whilst in $U(\mathfrak{h})$ the degree 1 filtered component also contains $z$.<|endoftext|>
TITLE: What can an algebraic geometer do outside academia?
QUESTION [39 upvotes]: This question is inspired by this
and this. But it is not a duplicate, read on. Please don't close it: I choose to be anonymous just not to be identified.
About to be on the job market, disenchanted with academia, bored by the teaching load in grad school, I have to make ends meet (by, say, finding a job somewhere). I've seen math people finding non-math jobs. But they seem to know some math that can be ``applied'', e.g. stochastic process or combinatorics. Trained as an algebraic geometer, I don't have a strong background on those stuff or anything that helps job hunting ---- knowing things about the Weil conjecture doesn't seem to be a plus. I don't think I will enjoy a teaching job in college much either.
I would guess I'm not the only one in this situation. 
Question: Can anyone here give some career suggestions? 
(P.S. I'm in US now but not a US citizen, so National Security Agency or similar jobs appear in answers to similar posts are really not in my list.)

REPLY [21 votes]: Although not an algebraic geometer, I recently went through a similar experience (I actually posted one of the questions to which you linked.) One of the reasons I became disenchanted with academia was that I struggled to learn algebraic geometry for many years but failed. So I hope you appreciate that this advice is coming from someone who is not as smart as you!
After returning to my own country following an unsatisfactory year teaching at a liberal arts college, I did some part-time teaching at my old university while I researched possible careers. Attracted by the idea of earning a lot of money, my first thought was finance and I read several books about the stock market and the history of money. By the way, it's very hard to find textbooks about finance which don't "take sides". The very best book I found, coming from zero knowledge of economics, was "Introduction to Money" by Honor Croome.
We don't have quants in this country, but the career which appealed to me most was actuary, and I contacted some actuaries via the University Careers Service. The main actuarial recruiter in my city was not interested in me, saying that I was "over-qualified" (an expression you will probably soon be hearing a lot) and would find the job boring. Most of the actuaries I talked to were very friendly but were baffled that someone who had a PhD in mathematics would want to start a new career from the beginning.
Meanwhile, I realised that any job involving maths was going to require some knowledge of programming, so I began to learn Python. This was quite fun, although I did not get very far before switching to Javascript, because I wanted to be able to share my code more easily. An excellent resource is a free book called "Eloquent Javascript". I really recommend this if you have no programming experience. I didn't get very deeply into programming, but learned enough to do simple calculations and talk about things like "APIs", "dynamic typing" and other jargon.
I attempted to get a full-time teaching job at my old university but failed. However, everyone was very impressed by my lecture. I decided that I wanted to make a positive contribution to the world and so began to consider mathematical ecology, especially fisheries. I talked to some people and they recommended that I learn Bayesian Statistics, so I got a book and started to work on this. I rapidly became starstruck by the beauty and power of the Bayesian approach. It was also a source of helpful programming exercises (Gibbs samplers etc.) and motivated me to learn R, which is the industry standard among academically-oriented statisticians.
Around this time, I failed to get a job in fisheries. It turns out that people don't really care about whether you are capable of doing the job; they want you to "demonstrate an interest" in it. It's a bit like how, when applying for a liberal arts position from a research university, you have to harp on about how committed you are to the liberal arts philosophy. Otherwise your application goes directly into the trash.
I was at a loose end and a new semester was starting, so I sat in on two courses, one on Bayesian stats and the other in data mining. The data mining was helpful for learning R, because it's a very scatty language and there are all kinds of little tricks you need to know. The Bayesian stats was an opportunity to work through more Bayesian stats, with exercises.
Around this time, I contacted an ecologist in the statistics department who happened to have a problem to work on, so I started working on this. After a couple of months I was able to amke progress on it and I'm hoping we will eventually get a publication out of it, which will certainly boost my credibility with the ecology crowd.
Anyway, at this point (about a year after my search started) I suddenly got a job in the tax department. The reason why I got this was because my boss is a mathematician. It turns out that there are lots of mathematicians in industry who were once in the same position as I was, and you now are. They like to hire mathematicians, just like how people who have emigrated to a different country like to hire people from the same country as them. Another mathematician was hired at the same time as me. I don't view the new job as a permanent thing that will go on forever, but I really like my colleagues. I do have to deal with meetings and people blathering about "going forward", "taking the first cab off the rank", "passing the ball" and all manner of similar phrases. On the other hand, I can bask in the feeling that I am helping to stick it to those nasty finance people who care about nothing but money ...
Ultimately I am hoping to return to academia (as a statistician) or become a statistical consultant if I can, and I want to move to Canada one day.
I guess the most useful pieces of advice for someone in this position are probably the following:

The only way to get a job is via personal connections.
Appearances are very important.

and (more an obervation than a piece of advice)

Every real-world problem is ultimately about maximising some hideously complicated function.

I am sorry if you didn't find this too helpful, but I thought it might be useful to hear from someone who has very recently been in a similar position. It took me a year to find a job, so don't lose heart!

REPLY [10 votes]: I voted to close this question earlier, but on the principle "if you can't beat 'em, join 'em", I'll say this. I know of any number of former academic mathematicians who decided to leave academe to pursue a career as an actuary. They come from all over: algebraic geometry, number theory, logic, quantum field theory, you name it. The bit about having to have some iron in applied mathematics here just doesn't hold true: what counts to prospective employers is being able to think clearly and precisely and have good mathematical sense. Being able to pass actuarial exams (the first few of which are math-y but doable with little sweat by an ex-mathematician) goes a long way towards establishing your cred. 
The pay tends to be pretty good, and actuarial work tends to be stable, certainly compared to what struggling academics put up with. (I am not an actuary myself but I am married to one.) A lot of mathematicians I know are very happy with this career choice and have never looked back. 
Perhaps the real point is that this is merely an example. I support what Angelo said: "a PhD in mathematics is highly regarded in the 'real world', even when it is in pure mathematics." Your prospects are probably very bright.

REPLY [5 votes]: People not originally from the US (or at least not from an English speaking country) tend to have a little more trouble with this question, because they have more difficulty understanding the following:
Traditionally in the United States, higher education is not specific training for a career.  Rather, it is training to be a thoughtful contributing citizen.  Skills necessary for being a thoughtful contributing citizen tends to be useful in all kinds of intellectual and non-intellectual work.  Companies looking for intelligent employees therefore look for intelligent, hard-working people who have done well at studying something, not necessarily anything related to what they do.
As a graduate of a liberal arts college, I saw music majors and English majors get all kinds of jobs.  What were the most common?
1) Sales.  Plenty of companies need people to sell something.  One might sell large software packages to businesses.  One might sell re-insurance to insurance companies.  One might sell medical devices to doctors.  Since salespeople generally are mostly paid on commission, hiring a salesperson is low risk.  Traditionally, salespeople know almost nothing about what they are selling; they are just good at convincing people to buy whatever they are selling.
2) Technical sales support.  Someone has to understand what they are selling.  Except technical sales support people are not actually experts in what is being sold.  They've just had a little more training so they can answer the 98% of questions that aren't actually that technical.
3) Software.  If you've learned how to program, the vast majority of jobs in the software industry don't require that much software engineering knowledge, and you've probably learned enough.  We are not talking about making a database or operating system run faster; we're talking about changing the interface of MathOverflow to move the answer box 5 pixels to the right, or just making sure this webpage is displaying the answers to this question and not some other random question.
4) Consulting.  There is both software consulting (see software) and management consulting.  This involves being in a group that is hired to look at a company's operations, gather qualitative and quantitative data, and make recommendations.  The whole point is that you are mostly not experts and therefore (at least supposedly) have a fresh view on what they are doing and can recommend the obvious stupid things they are missing.
5) Investment banking.  See Sales, except you're selling financial products to companies.  Except that in this industry, sales is frequently done in teams, so your role might be closer to that of a consultant (assembling data to support the sale) than an actual salesperson.<|endoftext|>
TITLE: Which W*-algebras are the duals of C*-coalgebras?
QUESTION [10 upvotes]: A Banach algebra (assumed associative and unital) is precisely a monoid object in the monoidal category of Banach spaces, short linear maps, and the projective tensor product.  A Banach coalgebra is then a comonoid object in this monoidal category.  It's straightforward to write down what a Banach *-coalgebra is too.  It's a little less obvious what a C*-coalgebra is, and I don't know if that term appears in the literature, but I've written down my definition.
Generally, the dual space of a coalgebra is an algebra (but not conversely), and that works here too: the dual of a C*-coalgebra is a C*-algebra.  But not every C*-algebra arises in this way; obviously, since all of these C*-algebras have preduals (having been explicitly constructed so), they are W*-algebras.  But I don't know what other conditions must be satisfied.
So my question is, and I'll be grateful for incomplete answers:  Which W*-algebras arise (up to abstract isomorphism) as duals of C*-coalgebras?
Partial answers:  The sequence space $l^\infty$ is the dual of $l^1$, and $l^1$ is a C*-coalgebra.  But this doesn't work for $L^\infty(R)$; this is the dual of $L^1(R)$, but I can't make $L^1(R)$ into even a Banach coalgebra (in an appropriate way), essentially because the diagonal in $R^2$ has measure zero.  (Unless I've miscalculated something, and I'm trying to do the wrong thing here.)  Of course, these are quite limited examples: they're (co)commutative.  I'd be grateful for more.

REPLY [2 votes]: Only a partial answer, but I thought I'd stick it up to give other people something to poke holes in or build on. Hopefully I will come back to elaborate or correct.
Let $M$ be the W*-algebra whose predual $A$ is your co-algebra. Then your comultiplication is meant to land in $A\hat{\otimes A}$, I gather. So dualizing you want your multiplication on $M$ to be a well-defined map $(M_*\hat{\otimes} M_*)^*\to M$. But now I think one can invoke machinery from tensor norms of Banach spaces to say that this would imply multiplication in $M$ extends to a well-defined continuous linear map $M\check{\otimes} M \to M$ where $\check{\otimes}$ is Grothendieck's injective tensor product. Then if you want this map to be contractive I believe $M$ has to be commutative. If you want it to be bounded, I think you even need $M$ to be a finite sum of (commutative) $\otimes$ (matrices).<|endoftext|>
TITLE: Examples of finiteness of rational points for hypersurfaces in $\mathbb P^3_{\mathbb Q}$ of degree $>4$
QUESTION [8 upvotes]: Given an homogeneous polynomial $F(X,Y,Z,T)\in \mathbb Q[X,Y,Z,T]$ of degree $>4$, the surface it defines is well-known to be of general type. Suppose, moreover, that this surface doesn't contain any rational or elliptic curve (a fact that it happens "generically", according to a theorem by Geng Xu in Subvarieties of general hypersurfaces in projective space). Then, a conjecture by Bombieri and Lang (see for example the book by Marc Hindry and  Joseph H. Silverman "Diophantine Geometry: An Introduction") asserts that this surface should have a finite number of rational points (this is a special case of a strong version of the conjecture) (as pointed out below by Felipe Voloch, one needs to assume the surface is smooth) .
Some questions:

Is it known any example of such a surface which has some rational point, it is smooth, and one can prove there is only a finite number of rational points? I ask to have a rational point in order to prevent a proof by "local considerations": it has no rational points since it has no real points, or $p$-adic points, or "modulo n" points, etc...

A bit less: Is it known explicit examples of such surfaces verifying the hypothesis of the conjecture? The results I find are for "generic" cases.

For example, does there exists some $k$ such that any non-singular hypersurfaces in $\mathbb P^3_{\mathbb Q}$ of degree $>k$  does not contain any rational or elliptic curve?


I understand that Faltings' results on subvarieties of abelian varities cannot be applied here, since there is no trivial map to these varities to an abelian variety (since their albanese variety is trivial, since the dimension of their $H^1$ is zero, by standard results).

REPLY [8 votes]: For question 2, you can look for surfaces of Picard number 1. One way to find such surfaces is to reduce modulo p and show that the reduction has Picard number 1 by computing its L-function. This will only work for odd degree. For even degree, you may need to compute the reduction modulo two primes with Picard number 2 but incompatible lattices so the Picard number of the original surface is 1 (this is a trick due to R. van Luijk). These computations are not easy and there is some work, e.g. by Kedlaya, to make them feasible.
Question 3 is false. You can easily write down smooth surfaces of any degree containing a straight line.
Off the top of my head, I don't have an example for your question 1, which is unconditional. But, if believe the ABC conjecture in its multiple summands variant, then diagonal hypersurfaces give examples (e.g. $x^n+2y^n+3z^n=6w^n, n$ large).
Edit: As Jason points out in the comments, my suggestion for question 2 is not enough. A different idea would be to use the idea for question 1. Namely use the function field ABC (which is a theorem) to show that a "fewnomial" (diagonal equations contain lines) equation has no curves of genus 0 or 1. I haven't worked out the details, though.<|endoftext|>
TITLE: Algorithms to find irreducible polynomials of a given degree
QUESTION [8 upvotes]: I need to know what are the efficient algorithms to find all the irreducible polynomials of a given degree, say $d$ over a given finite field, say $\mathbb{F}_{p^n}.$
One way is to factorize the polynomial $x^{p^{dn}}-x$, which is the product of all irreducible polynomials whose degree divides $d$, using factorization algorithms and collect all the degree $d$ factors. But I guess we are doing some extra job here. Are there better algorithms to find all irreducible polynomials of degree $d$ ?
I also want to know about the algorithms to find one irreducible polynomial of a given degree over a given finite field.

REPLY [5 votes]: The last word on the second question is this paper of Couveignes and Lercier. The question is highly nontrivial.<|endoftext|>
TITLE: Finite groups such that every irrep can be induced from trivial irrep of a subgroup ? 
QUESTION [18 upvotes]: What are examples (general features) of the finite groups $G$, such that every irrep (irreducible representation) is contained (as constituent) in the representation induced from  trivial representation of some non-trivial subgroup? (I allow subgroups to vary - I mean take all subgroups, induce reps from trivial characters and require that any irrep sits in this set).
Remark: If we induce from the subgroup = {identity}, we will get a regular representation of the whole group, and every irrep lives there. So I should not allow this trivial subgroup.
Example 1: $S_n$ satisfies this property (there is some comment by David Speyer which I cannot find now, which says (as far as I remember) that induction from $S_{k1}\times S_{k2} \times\cdots\times S_{kl}$  will do the job. I have another argument but it is more complicated).
Example 2: Take $\mathbb{Z}/p\mathbb{Z}$ for prime p - it does NOT satisfy my requirement - the only subgroups are $\lbrace 1\rbrace$ and $\mathbb{Z}/p\mathbb{Z}$ itself; first one is forbidden by the rules of the game, and induction from the second will give trivial irrep.
Example 3: If we take PSL(2,7)=GL(3,2) and induce from the 3-Sylow subgroup - it will contain all irreps as constituents (see MO 104939).
In particular:  what about $GL(n,\mathbb{F}_p)$ and $A_n$?

REPLY [24 votes]: By Frobenius reciprocity, a representation is induced from the trivial representation of the subgroup $H$ if and only its restriction to $H$ includes the trivial representation. So every non-cyclic abelian group has this property, because an irreducible representation is one-dimensional, so factors through a map to a cyclic group, so has a kernel.
Moreover, this implies that if $H\subset G$ and $H$ has this property, then $G$ does as well, so a sufficient condition is that a group have an abelian subgroup that's not cyclic.
This resolves the case for $GL_n(\mathbb F_p)$, $n>1$ and $p>2$ (the diagonal subgroup), and $n>2$ and $p=2$ (an abelian subgroup of the $2$-Sylow subgroup). n=2, p=2 is just $S_3$, which manifestly has this property. (Showing that this sufficient condition is not necessary) So $GL_n$ for $n>1$ has this property. $GL_1$ does not, since it is always cyclic. (For cyclic groups, every induced representation fails to be faithful, so take a faithful 1-dimensional irrep.)
$A_n$ for $n\geq 4$ contains the Klein four subgroup of $A_4$ and so has this property. $A_3$ is cyclic and so does not.
I do not know an easy-to-check necessary condition.

REPLY [16 votes]: EDITED IN RESPONSE TO COMMENTS BY DAVID SPEYER AND F. LADISCH: Examples of finite groups which fail to have the desired property,  which are effectively exhaustive, are the finite groups which occur as Frobenius complements. These are the finite groups which admit a (necessarily faithful) representation in which every non-identity element acts without the eigenvalue $1$. Such groups have cyclic Sylow $p$-subgroups for all odd primes $p,$ and cyclic or generalized quaternion Sylow $2$-subgroups, properties which also occur in Will Sawin's answer . This is a very restricted class of groups. For example, the only perfect Frobenius complement is ${\rm SL}(2,5).$ In any case, if $G$ is a Frobenius complement, and $\chi$ is a faithful complex irreducible character such that $\langle {\rm Res}^{G}_{H}(\chi), 1 \rangle =0$ for each non-identity cyclic subgroup $H$ of $G$ (and such a $\chi$ must exist), then $\chi$ is not a constituent of any permutation character induced from the trivial character of a non-identity subgroup of $G.$
Let me justify that Frobenius complements are just those groups with an irreducible complex representation where every non-identity element acts without eigenvalue $1$ . Recall that a Frobenius group $G$ has the form $G = KH,$ where $K \lhd G$ and $H \cap K = 1,$ and, furthermore, $H \cap H^{g} = 1$ for all $g \in G \backslash H.$
Notice then that $|K| \equiv 1$ (mod $|H|$), so that ${\rm gcd}(|K|,|H|) = 1.$ Also, we certainly have $C_{G}(h) \leq H$ whenever $h$ is a non-identity element of $H$ since $h \in H \cap H^{c}$ for all $c \in C_{G}(h).$
Let $V$ be a minimal non-identity $H$-invariant subgroup of $K.$ Then using Thompson's theorem that a Frobenius kernel is nilpotent, (which is actually overkill here, since by general properties of coprime automorphism groups to be found in Gorenstein's book "Finite Groups", for example, $H$ normalizes a Sylow $q$-subgroup of $V$ for each prime divisor $q$ of $|V|$), we see that $V$ is an elementary Abelian $p$-group for some prime $p$. Then $V$ is a faithful $FH$-module, where $F = {\rm GF}(p).$ Since $p$ does not divide $|H|$, $V$ "lifts" to a complex representation, and by general (indeed the defining) properties of Brauer characters, it is still the case that each non-identity element of $H$ acts without eigenvalue $1.$ The "lifted" representation need not be irreducible as a complex representation, but its irreducible components all have the property that each non-identity element of $H$ acts without eigenvalue $1$ on them (and each is faithful).
Conversely, if $H$ is a finite group which has a complex irreducible character $\chi$ which does not contain the trivial character on restriction to any non-identity cyclic subgroup of $H,$ then a complex representation of $\chi$ may be reduced (mod $p$) for any prime $p$ which does not divide $|H|$ to afford a $kH$-module $W$ on which each non-identity element of $H$ acts without non-zero fixed points, where $k$ is algebraically closed of characteristic $p.$ Then $W$ may be realised over a finite field, and the sum of its distinct Galois conjugates may be realised over ${\rm GF}(p),$ say by  module $V$ over ${\rm GF}(p).$ It is still the case that each non-identity element of $H$acts without non-trivial fixed points on $V,$ so the semidirect product $VH$ is a Frobenius group with kernel $V$ and complement $H.$
Frobenius complements are precisely the groups which have an irreducible character $\chi$ which does not occur as a constituent of ${\rm Ind}_{H}^{G}(1)$ for any non-trivial subgroup $H$ of $G.$ For if $\chi$ is such a character, then ${\rm Res}^{G}_{H}(\chi)$ has no trivial constituent for each non-trivial subgroup $H$ of $G$, in particular for each non-trivial cyclic subgroup of $G.$ Hence each non-identity element of $G$ acts without the eigenvalue $1$ in any complex representation affording $\chi.$ Conversely, if each non-identity element of $G$ acts without eigenvalue $1$ in a representation of $G$, then there is an irreducible representation $\sigma$ with that property, and if $\sigma$ affords character $\chi,$ then $\chi$ does not occur as a constituent of ${\rm Ind}_{H}^{G}(1)$ for any non-trivial subgroup $H$ of $G.$
There are examples of non-Abelian Frobenius complements of odd order: for example, let $G = \langle x,y : x^{9} = y^{7} = 1, x^{-1}yx = y^{2} \rangle.$ Note that $G$ has an irreducible character $\chi$ of degree $3$ such that $x^{3}$ acts, as a non-identity scalar matrix, so that no non-identity $3$-element of $G$ has eigenvalue $1$ in the associated representation, while also each non-identity power of $y$ has three different primitive $7$-th roots of unity as its eigenvalues in the associated representation.
However, it is not true that if a finite group of odd order has all its Sylow subgroups cyclic, then it is a Frobenius complement: for example, a non-Abelian group of order $21$ is not a Frobenius complement (though it is a Frobenius GROUP!)

REPLY [5 votes]: Concerning the finite general linear groups (or other finite groups of Lie type), it's been known for a long time that you can't get every irreducible character as a constituent of one induced from the trivial character of a proper parabolic subgroup.   This is what makes the whole subject so challenging, going back as far as the work of Frobenius for the groups $\mathrm{SL}_2(\mathbb{F}_p)$ and extending through the work of J.A. Green on characters of finite general linears to the much more complicated work coming from Deligne-Lusztig theory.  Caveat: In this situation I'm not considering all possible proper subgroups, just those relevant to the BN-pair structure, so it's of course possible to find exceptions.  But in Lie theory, including finite general linear groups, one is really looking for uniform methods to produce character tables.
On the other hand, for groups of Lie type there is a rich theory of what can be done if you induce up from the trivial character of a parabolic subgroup and then decompose the induced character using Hecke algebra methods.  The problem is that it doesn't get everything you want.
By the way, I'd be curious to know whether there is a reasonable necessary and sufficient condition on a finite group to make the answer to your question affirmative.   (Probably not.)   In any case, your header does suggest that you want the induced representations involved to be irreducible, which misleads people at first.<|endoftext|>
TITLE: Finiteness of stable homotopy groups of spheres
QUESTION [17 upvotes]: Since the work of Serre in the early 50's on homotopy groups of spheres, it is known that the homotopy group $\pi_k(S^n)$ is finite, except when $k=n$ (in which case the group is $\mathbb{Z}$), or when $n$ is even and $k=2n-1$ (in which case the group is the direct sum of $\mathbb{Z}$ and a finite group). As a consequence, the stable homotopy groups $\pi_k^s$ are finite groups for $k>0$, and $\pi_0^s \cong \mathbb{Z}$. 
The work of Serre was done before anyone knew about stable homotopy theory and chromatic methods, and this makes me wonder about the following questions. 
Question 1: Is it possible to use methods from stable/chromatic homotopy theory to prove finiteness of stable homotopy groups of spheres directly, without having to compute any unstable homotopy groups of spheres?  
Question 2: Is there any philosophical or conceptual reason for why these groups should be finite?

REPLY [7 votes]: Let me phrase a proof of Serre's computation of the rational stable homotopy groups of spheres as stably as I can:
For every spectrum $X$, we can define its rationalization $X_{\mathbb{Q}}$ as the homotopy colimit $$\mathrm{hocolim}\, X \xrightarrow{2} X \xrightarrow{3} X \xrightarrow{4} \cdots .$$ 
As such a directed homotopy colimit translates into a usual colimit after applying homotopy groups, we obtain $\pi_* X_{\mathbb{Q}} \cong (\pi_* X) \otimes \mathbb{Q}$.
By a form of Hurewicz, we know that the negative homotopy groups of the sphere spectrum $\mathbb{S}$ are $0$ and its zeroth homotopy group is $\mathbb{Z}$. Thus, we obtain a Postnikov truncation map $\mathbb{S} \to H\mathbb{Z}$. After rationalization, this induces a map $\mathbb{S}_{\mathbb{Q}} \to H\mathbb{Q}$. We want to show that this map is an equivalence. 
As these are bounded below spectra, Hurewicz/Whitehead II imply that we just need to check on homology. As homology also translates directed homotopy colimits into colimits, we obtain that $H_*(S_{\mathbb{Q}}; \mathbb{Z}) \cong \mathbb{Q}$ (concentrated in degree $0$). It suffices thus to show that the integral homology of $H\mathbb{Q}$ is the same (as the map is then automatically an isomorphism as it is one on $\pi_0$). 
As the homology of $H\mathbb{Q}$ coincides with the homotopy of $H\mathbb{Q} \otimes H\mathbb{Z}$, we see that we can equally well compute the rational homology of $H\mathbb{Z}$. The $i$-th such homology is isomorphic to the colimit over $\widetilde{H}_{i+n}(K(\mathbb{Z}, n); \mathbb{Q})$. For $n$ odd, these groups are concentrated in degree $n$ and are $\mathbb{Q}$ there, implying the result. (This is the only place where I find it to be necessary to apply an unstable result.)
For the finite generation, I likewise need one input result, namely that the homology of $H\mathbb{Z}$ is degreewise finitely generated. Using unstable methods, this is again easy to prove as we obtain $H_i(H\mathbb{Z}; \mathbb{Z})$ as the colimit of $\widetilde{H}_{i+n}(K(\mathbb{Z}, n); \mathbb{Z})$. Each of the groups is finitely generated by an inductive argument using the Serre spectral sequence and the sequence stabilizes. 
The argument is now the following: We prove simultaneously by induction that the $n$-th homotopy group of $\mathbb{S}$ is finitely generated and that the connective cover $\tau_{\geq n+1}\mathbb{S}$ has degreewise finitely generated homology. Hurewicz implies that $\pi_n \mathbb{S} \cong \pi_n\tau_{\geq n}\mathbb{S} \cong H_n(\tau_{\geq n}\mathbb{S}; \mathbb{Z})$. The latter group is finitely generated by assumption and hence also $\pi_n\mathbb{S}$. Thus $H\pi_n\mathbb{S}$ has degreewise finitely generated homology. Thus, the same is true for the fiber of $\tau_{\geq n}\mathbb{S} \to H\pi_n\mathbb{S}$, namely $\tau_{\geq n+1}\mathbb{S}$, which finishes the induction step.<|endoftext|>
TITLE: Rational Points on $y^2=x^3-86069^5$
QUESTION [14 upvotes]: The analytic rank of the Mordell elliptic curve $y^2=x^3-86069^5$ indicates that it has rank 2. However, deriving a set of generators, and hence the regulator, is proving to be a little bit of an intractable problem.
I'm posting this question in the hope that someone may have investigated this curve already and, as such, may be able to help me out with some of the points on the curve.
Any help with this would be very much appreciated.
As further background, I have been involved in a small group searching for values of #Sha > 1 up through the ranks of Mordell curves, currently from ranks 0 to 9.
To date we have been able to progress to a rank 9 curve were #Sha would seem to be at least equal to 9. The details of which may be seen at the NMBRTHRY Archives. 
Kevin.

REPLY [4 votes]: Tom A. Fisher has kindly identified the second generator for this curve, at a height of $1715.46805605884712533431816634$, and with a regulator of $1435241.11022534407264592039437$.
Many thanks to Tom for this impressive workout of his 12 descent code.
P2:= E![6409818948420000148009253515371033320765674658334995\
509069337486116156301741657451140832523716129416946553724548\
364793698626405777843956622076380997023235142487299844326278\
568312577933328949209325079504842164930964499614652180484249\
884310193722360960206856049327696413372088438499522988880226\
096525828673972002212550418445757426791463589511241682169507\
191857404060127705996714164059762587639453970388047710151991\
644286292989545241436751637355865317439825495449344823977758\
797766025325026099986880721224779341838599953004314077314458\
400203457162779096001177667625960285991372461677348744634118\
572760595736047817251613300845834732694598202906328385669708\
044981270667425075403152129513088853355196780998723571290752\
400352784381236905647783762282233/33165803105124219179549823\
283035686736456675469805935767692444023551658331570872880192\
989613282702564119725223095907264020401187622726619827641022\
070312051693797896281152446136738277971480729817095517412821\
500028443753503375569153920181215330943019021452618523429306\
154890615692258509116181263907011892227215259838198394721409\
844161749103601296315808769202168882630133040496609244967027\
573328564357038779635203477380317914882271609699042932772428\
958728775525114597825318245131110925771677312456670934287090\
438610192382893431121092876711361026457071789640268657157836\
261872615922281539370618784278612205624356072354848973487567\
580677872395212361402429095761079520397611920420171112702594\
975559992623826579984003799616507192963910592190404, 9541721\
267275526440336656075081968921795973155189044266523956895286\
489666750185996900611973660561530367870689843326743006933458\
752307224811373451969681421106849987021781748740069623362388\
923316032840282833074591791294185661671902447470420594105932\
470690611184875968605655279379274234568020070054315280341961\
704988604471445169462057158108739193383808091705876229256447\
451432215635271166605138776098284834213663487614216411244364\
417456782497135614069447487775516105502956819954881548171764\
933506408613533330700718054166278531808088754093650937927404\
096948372079871885150489168059803000137727292308698902319300\
888120656605108547686297443217509881949442132801485412503095\
358513515604711600740777338910658971774981045705755311582771\
691405245163713860189075823123694400673272454942433819801867\
842344237200886745700934179591344744409814700762619336056693\
409131813436072830610006822421568399180878828865624130779460\
438617259828853398661927306856293129558704629256045989464076\
110208592569978237366668166654822438426674566555490387538893\
656476099175185704061248678944708728574308512767720849000403\
950411614279621166869036373101/60399838475592326331362752791\
944763132549609694041832059854653129731539771658328168942361\
906568116204357279114901731638612673017179392637587993370268\
468122751057317804766338855060441224492531078338764563745984\
361634534757917619567424588718030226164336198810659807830928\
894950054794919605549933531685874245309484554173084344795097\
778895379979816703495608964775949909264616094661840599063186\
011543131081083532339588386281198484224591916139857808728711\
337072945237794430249025950156217838942697194548707342214906\
621985660080150918372835213775235759078837614090802954692490\
955558620434772673319859076612900166605163593210918785877475\
268380688613826217264936422758722358549436515729050089664820\
517090881544523313666983596465352462319026753145000093740537\
178177489596826835806901560570709618578068241711965931639634\
809586146278457253570314819921037995164438751876277090234639\
821432381946017446571917916673348009411849176259307380401023\
993294177103548682416638846438507436013936091382661874934845\
500838177629920215935612770414110473755964022901968320874940\
92431878821676880656612635392756954758754490670429601208 ];<|endoftext|>
TITLE: Special value of $L$-function
QUESTION [5 upvotes]: Let $p$ be a prime number. Let $f$ be a newform of weight 2 on $Γ_0(p)$, and $E_f$ denote the associated newform quotient of $J_0 (N)$ over $\mathbb{Q}$. Is there a way to express the
algebraic part of the special $L$-value of $E_f$ over $K$, a totally real quadratic field, in terms of the conjugacy classes of maximal orders in the definite quaternion algebra over $\mathbb{Q}$ ramified at $p$ and $\infty$?. p.s: when $K$ is a quadratic imaginary field, there is such an expression.

REPLY [5 votes]: Elaborating David H's hilariously laconic comment... :)
Lacking any other answer: work of Shimura and Waldspurger from 20+ years ago may be the relevant stuff to see what is true. 
My not-so-well-informed perception of the situation is that whatever one is likely to prove (or is known by this date, or at least not-so-long-ago) goes by way of a theta correspondence.
But/and a theta correspondence from an orthogonal group to $SL_2(\mathbb R)$ (or a two-fold cover, if necessary) maps at archimedean places to holomorphic discrete series essentially only when the source repn of the orthogonal group is the trivial repn on a definite orthogonal group, such as $SO(2)$. In contrast, at archimedean places $SO(1,1)$ produces principal series, so such an orthogonal group's afms map to waveforms, not holomorphic things.
That is, if things go by way of Weil (-Segal-Shale) repns, then hitting holomorphic things is a significant constraint, whatever is going on at finite primes.<|endoftext|>
TITLE: Characterizing flat 2-connections by their holonomy
QUESTION [11 upvotes]: Hello,
A flat principal $G$-bundle over $X$ is determined by its holonomies, which are (after picking a trivialization) group homomorphisms $\pi_1(X)\rightarrow G$. The fiber of the bundle is not canonically identified with $G$, so these maps are only determined up conjugation by $G$. Equivalently these are gauge transformations at the basepoint where $\pi_1$ is evaluated.
Now let $H \overset{t}\rightarrow G$ present a 2-group. With respect to some trivialization, a flat 2-connection on a principal 2-bundle assigns an element of $G$ to a closed curve, and an element $h$ of $H$ to a surface bounding a curve $\gamma$ such that $\gamma$ gets $t(h)$. Flat means that $h$ only depends on the homotopy class of this surface (homotopies fixing $\gamma$). 
Thus, I expect flat 2-connections are determined by a functor from the path 2-group of $X$ into $(H\rightarrow G)$. My question is : what is the degeneracy of this presentation? In other words, what is analogous to "up to conjugation in $G$" for flat 1-bundles?
Thanks.

REPLY [12 votes]: Let $\mathcal{P}$ be a principal 2-bundle with structure 2-group a crossed module $t:H \to G$. Then, the holonomy of a connection on $\mathcal{P}$ around a surface $\Sigma$ is a well-defined element $$\mathrm{Hol}_{\mathcal{P}}(\Sigma) \in H/[G,H],$$ where $[G,H]$ is the normal subgroup of $H$ generated by all elements of the form $h^{-1}\alpha(g,h)$, where $\alpha$ denotes the action of $G$ on $H$ of the crossed module. Above quotient generalizes the concept of "conjugacy class".
This is explained in detail in my paper with Urs Schreiber

Connections on non-abelian Gerbes and their Holonomy,

see Example 5.7. 
To which extend the elements $\mathrm{Hol}_{\mathcal{P}}(\Sigma) \in H/[G,H]$ characterize the 2-bundle $\mathcal{P}$ I cannot say.
Maybe you are willing to drop "flat" upon replacing "holonomy" by "parallel transport": even in the classical world it is true that every principal $G$-bundle with connection is characterized by its parallel transport, may it be flat or not. 
The same statement remains true in the context of connections on 2-bundles: every principal 2-bundle with connection is characterized by its 2-transport. This is one of the main statements of my above-mentioned paper with Urs.<|endoftext|>
TITLE: Can the fundamental group of an intersection of a homeomorphic image of a ball with a complement of a ball in $R^3$ be perfect?
QUESTION [9 upvotes]: I have the following problem: Let $A, B\subset R^3$, $A$ is homeomorphic to a ball, while $B$ is a standard Euclidean ball. Can it happen that the fundamental group of $A\setminus B$ is a perfect group? I am interested in answers for $A$ and $B$ both closed and open, so in fact this is 4 questions. 
I am aware of disturbing examples like the infinite grope or the complement of the Alexander's horned sphere, but I still strongly believe that the answer should be no.

REPLY [10 votes]: Consider a smooth properly embedded  surface $P\subset \mathbb{R}^3$. Then $\mathbb{R}^3= X\cup Y$, where $X\cap Y=P$ and $X, Y$ are properly embedded submanifolds with $\partial X=\partial Y=P$. By Mayer-Vietoris, we have an exact sequence $0=H_2(\mathbb{R}^3)\to H_1(P)\to H_1(X)\oplus H_1(Y)\to H_1(\mathbb{R}^3)=0$, so we see that $H_1(X)\oplus H_1(Y)\cong H_1(P) \neq 0$ unless $P$ is a union of smoothly properly embedded planes and spheres. Therefore at least one component of $\mathbb{R}^3\backslash P$ does not have perfect fundamental group, or else $P$ is a union of planes and spheres (since $P$ is smoothly properly embedded, there's a nice collar neighborhood, so $H_i(X)\cong H_i(int(X))$, and same for $Y$). In the case that $P$ is a union of properly embedded planes and 2-spheres, by Seifert-Van Kampen's theorem, each $\pi_1(X,x)$ and $\pi_1(Y,y)$, for $x\in X, y\in Y$ injects into $\pi_1(\mathbb{R}^3)$, so is trivial. 
Let's apply this to your situation. I'll consider the case of $int(A)\backslash B$, since there's not issue of local connectivity for an open set. Then $P=\partial B \cap int(A)\subset int(A)\cong \mathbb{R}^3$ is a properly embedded smooth surface, so we have either there is a component of $int(A)\backslash B$ which does not have perfect fundamental group, or $P$ is a union of properly embedded planes in $int(A)$ (or a sphere), in which case each complementary region has trivial fundamental group. I think this answers at least one interpretation of your question.<|endoftext|>
TITLE: Path length of ball on tilted, perforated plane
QUESTION [10 upvotes]: Imagine that an $\epsilon$-radius hole is punched in the plane centered
on every integer-coordinate point.
Now a point "ball" is dropped on the plane at a random spot $p$.
If $p$ has not already dropped through a hole, 
the plane is tilted in a random direction,
and the ball rolls along the gradient vector $u$.

What is the expected length $L(\epsilon)$ of its roll before it falls through a hole?

If $p$ is in a hole upon generation, then $L=0$.
Otherwise $L$ is determined by the first encounter with a hole,
i.e., the ray from $p$ along direction $u$ passes within distance
$\epsilon$ of some hole center $c$.
For example, here $\epsilon=\frac{1}{4}$ and $c=(2,1)$, with $L
\approx 1.6$:
        

I am particularly interested in the growth rate of $L(\epsilon)$.
Intuition from another viewpoint suggests $\sim \frac{1}{\epsilon}$.
Shrink the holes to points, and grow the ball to an $\epsilon$-radius
disk.  This sweeps out a rectangle of area $2 \epsilon L$
for a roll of length $L$.  Very crudely, when this
area reaches $1$ (or $1-\pi \epsilon^2$), I would expect it to include a lattice point.
So perhaps $L \approx \frac{1}{2\epsilon}$.  But this reasoning is
surely not sound.
Perhaps there is an approach that employs techniques from rational approximations?
All ideas welcomed—Thanks!
Addendum.  Here are two histograms for $L(\epsilon)$ with $\epsilon=\frac{1}{4}$.
The first is of 200 random points, the second of 300 random points.
        
They begin to illustrate Doug Zare's point that very long paths occur not infrequently.
In the 300-point example, one path has $L \approx 63$, while more than half the paths
have $L \le 1$. The probability that $L \le 10$ is $>98$%, even though the expected
length is $\infty$!

REPLY [4 votes]: J. Marklof and A. Strömbergsson have pretty much complete results for the distribution you're interested in - both on the plane and in higher dimensions. Check 
J. Marklof and A. Strömbergsson, The distribution of free path lengths in the periodic Lorentz gas and related lattice point problems, Annals of Mathematics 172 (2010) 1949–2033 
J. Marklof and A. Strömbergsson, The Boltzmann-Grad limit of the periodic Lorentz gas, Annals of Mathematics 174 (2011) 225-298<|endoftext|>
TITLE: Is there a Dedekind-Frobenius group determinant for infinite groups?
QUESTION [11 upvotes]: If $G$ is a finite group and $\lbrace x_{g} \rbrace_{g\in G}$ are commuting formal variables, then one can form a matrix whose $(g,h)$ entry is $x_{gh^{-1}}$. The determinant of this matrix is a polynomial with integer coefficients and is called the group determinant. Considering the factorization of this determinant led Frobenius to discover seminal results in the representation theory of finite groups. Later on, it was shown that a group can be recovered from its group determinant.
Many people have asked me the following question over the years, and I haven't a good answer for it. One might think about looking at something like a generalized determinant over a polynomial ring in infinitely many variables indexed by the group.

Question: In the literature, does there exist a
  more or less direct attempt to generalize the Dedekind-Frobenius
  group determinant to the setting of
  infinite groups?

Moreover, if such a thing exists, can it determine the group from which it is constructed?

REPLY [7 votes]: Such a thing is defined in Steve Humphries' very interesting paper:
Humphries, Stephen P.(1-BYU)
Cogrowth of groups and the Dedekind-Frobenius group determinant. 
Math. Proc. Cambridge Philos. Soc. 121 (1997), no. 2, 193–217
I am not sure it is as general as you might like, but...<|endoftext|>
TITLE: Is there only one natural transformation from the product functor to itself?
QUESTION [8 upvotes]: Say we are working in a category which has all binary products. I guess that the identity transformation is the only natural transformation from $\times$ to $\times$. Is this really the case? If yes, how can I prove this? If not, does it help to assume that the category is cartesian closed and has all coproducts? What about natural transformations from $+$ to $+$ in CCCCs? Are they (also) unique?

REPLY [20 votes]: The key issue is how many natural transformations there are from the identity functor on $C$ to itself. Chris Schommer-Pries observed that there are many such transformations for $C = Vect$, one for each scalar. 
A product functor $\prod: C \times C \to C$ is right adjoint to the diagonal functor $\Delta: C \to C \times C$, and it is easily seen that the collection of natural transformations $[\prod, \prod]$ is in natural bijection with the collection $[\Delta, \Delta]$, by a process called "taking the mate". Under the natural equivalence 
$$(C \times C)^C \simeq C^C \times C^C$$ 
the functor $\Delta$ is taken to the pair of identity functors, and we get under this equivalence a bijection 
$$[\Delta, \Delta] \cong [1_C, 1_C] \times [1_C, 1_C].$$ 
So in situations where there is at most one natural transformation from $1_C$ to itself, we get only one natural transformation from $\prod$ to itself. 
Consider for example the case of cartesian closed categories $C$. If we compute the hom of $C$-enriched transformations, we can use the isomorphism $1_C \cong C(1, -)$ where $1$ on the right is the terminal object. By a Yoneda argument, the hom-object of enriched natural transformations is $C(1, 1) \cong 1$. (And the hom-set of enriched transformations would be $\hom_C(1, 1)$, which is a 1-element set.) More generally still, if $C$ is symmetric monoidal closed and $I$ is the monoidal unit, the set of enriched transformations from $1_C$ to itself will be in natural bijection with $\hom_C(I, I)$; this specializes to Chris's observation where we have $\hom_{Vect}(k, k) \cong k$. 
Or, if $C$ has a terminal object $1$ and $\hom(1, -): C \to Set$ is faithful, we have an injection 
$$[1_C, 1_C] \to [\hom(1, -), \hom(1, -)]$$ 
and then an ordinary $Set$-based Yoneda argument shows there is at most one transformation from $1_C$ to itself. 
Edit: Of course, I still haven't given you an example of a cartesian closed category where the identity functor has more than one self-transformation (in the unenriched sense). But these are easy to come by. Consider for example the topos $Set^G$ of permutation representations of an abelian group $G$, i.e., functors $G \to Set$ where $G$ is regarded as a one-object category. Then for any $g \in G$, the action $g \cdot - : X \to X$ on $G$-sets $X$ is a $G$-set morphism that provides a natural transformation from the identity $Set^G \to Set^G$ to itself.

REPLY [14 votes]: No, this is not the case. 
Let the category C be vector spaces (say over the real numbers). Given any real number we get a natural transformation of the identity functor on C. On components for a given vector space V, this transformation is defined to be multiplication by the given real number.
If we whisker this (on the target side) with the product functor we get an infinite family of natural transformations from $\times$ to $\times$.   
You can get a more exotic example by noting that a pair of real numbers gives an automorphism of the identity functor of $C \times C$, hence by wiskering (on the source side) we get another family. In components this is the transformation which on $V \oplus W$ scales V by the first number and W by the second.<|endoftext|>
TITLE: Continuity with values in L^2
QUESTION [5 upvotes]: Hi,
let $T>0$, $\Omega\subset\mathrm{R}^n$ be a bounded smooth domain and suppose 
$$u\in L^2(0,T;W^{1,2}(\Omega))\cap L^\infty((0,T)\times\Omega))\ \text{and } \partial_tu\in L^2(0,T;W^{-1,2}(\Omega)),$$
where $W^{-1,2}(\Omega)=(W^{1,2}_0(\Omega))^*$. Is there a result stating that from those regularities one can deduce $u\in C([0,T];L^2(\Omega))$? 
Thanks a lot,
Richard

REPLY [2 votes]: Set $V:=L^2$ and $H:=W^{-1,2}$ and define the maximal regularity space
$$MR_2(0,T;H,D):=W^{1,2}(0,T;H)\cap L^2(0,T;D),$$
where $D$ is the domain of the unbounded operator on $H$ associated with the quadratic form $Q(u):=\|u\|_V^2$. Then it is well-known (reference in some book by J. Lions, but now I cannot find an exact one) that $MR_2(0,T;H,D)$ is continuously embedded in $C([0,T];V)$. This is almost what you want. Not quite (I have not checked, but in this case it looks like $D=W^{1,2}_0$, unlike in your assumptions; and, worse, you are not assuming $u$ to be in $W^{1,2}(0,T;H)$), but also not far away, if you are able to use your assumption that $u\in L^\infty$ to define an equivalent norm on the set of those
$$\{u\in L^2(0,T;W^{1,2}) \hbox{  s.t. } u_t\in L^2(0,T;W^{−1,2})\}.$$<|endoftext|>
TITLE: Volumes of fundamental domains of maximal orders in definite quaternion algebras over Q
QUESTION [5 upvotes]: I'm looking for an explanation of the following result:
If D is a maximal order in a definite (i.e. ramified at infinity) quaternion algebra B over $\mathbb{Q}$, and $\phi : B\otimes\mathbb{R}\rightarrow\mathbb{H}$ is an isomorphism over $\mathbb{R}$ (where $\mathbb{H}$ is the Hamiltonian quaternions over $\mathbb{R}$), then the co-volume of $\phi(D)$ in $\mathbb{H}$ (with respect to the Lebesgue measure for the basis 1,i,j,k) is $(1/4) \cdot Disc(D)$ (where the discriminant is the determinant of the trace form on a basis of $D$, or, equivalently since $D$ is maximal, the product of the finite primes where $B$ ramifies). 
Right now the only way I have to see this is through an explicit calculation using the generators for maximal orders described in a paper by Ibukiyama available here: http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.nmj/1118787011 , but I feel like there must be a nice way to see this using just properties of maximal orders without having to calculate with an explicit basis. Any ideas or references?

REPLY [5 votes]: See chapters II and III of the classic book of Vigneras. Beware however that in the later parts of chapter III she assumes widely that C.E. (the "Eichler condition") is verified, and C.E. is explicitly not verified for quaternion algebras $B$ over $\mathbf{Q}$ such that $B\otimes \mathbf{R} \cong \mathbb{H}$ (so called "definite" algebras). For lots of computation with definite algebras please see Chapter V of her book.<|endoftext|>
TITLE: Square and stationary reflection
QUESTION [10 upvotes]: It is easily shown that, for any uncountable infinite cardinal $\kappa$, $\square_\kappa$ implies that for any stationary $S\subseteq \kappa^+$, there exists a stationary $T\subseteq S$ such that $T$ does not reflect at (i.e. is not stationary in) any $\alpha<\kappa$ of uncountable cofinality. The standard proof does not go through, however, when $\square_\kappa$ is replaced by the weaker notion of $\square(\kappa^+)$. Is $\square(\kappa^+)$ compatible with stationary reflection? More precisely, if $\kappa$ is an uncountable infinite cardinal, is $\square(\kappa^+)$ consistent with the statement "every stationary $S\subseteq \kappa^+$ consisting of ordinals of cofinality $<\kappa$ reflects at some $\alpha<\kappa^+$"?

REPLY [7 votes]: The answer is yes (at least for $\kappa = \omega_1$): Shelah and Harrington showed that you can force that every stationary subset of $S_{\omega}^{\omega_2}$ reflects starting with a Mahlo cardinal. See Theorem A in Some exact equiconsistency results is set theory. 
Since $\neg \square(\omega_2)$ implies that $\omega_2$ is weakly compact in $L$, if we start with a Mahlo cardinal $\kappa$ which is not weakly compact in $L$ and collapse it to $\omega_2$ using the forcing of Shelah-Harrington we'll have $\square(\omega_2)$ in the generic extension.
I don't know if this is true for $\kappa > \omega_1$ as well.<|endoftext|>
TITLE: Sarrus rules for 4 times 4
QUESTION [19 upvotes]: This question is most probably not research level, but I thought that the MO folks might like it... Feel free to close.
Here is the motivation: If you have ever teached a maths course for engineers which covered determinants and which included a written exam then you often ask the students to calculate a 4 by 4 determinant to check if they got the basic rules (e.g. using Laplace's formula to reduce to 3 by 3 if the structure is favorable or use Gauß elimination). If you did this you most probably have seen a student solving this problem by applying the "Sarrus rule for 4 by 4 matrices".  Usually, the students memorize for 3 by 3 a pattern like

(of course, the lines mean that you should multiply the numbers along the lines; green lines get a $+$, red lines get a $-$, finally add everything up). My colleagues told me, that in every exam there is at least one smart guy who happily generalizes this rule to 4 by 4 matrices with a scheme like this:

which I am going to refer to as False Sarrus Rule. Indeed, one could turn this into a working rule by assigning the right signs and repeating the procedure two times more in a different way. I wrote a small blog post here (and there is even paper on this (German description, Russian description)). Basically, I wrote this blog post to give the people who search the net for a generalized Rule of Sarrus some visual reminder that there is no easy "Sarrus Type Rule" available. Believe it or not: The post is found frequently via search terms like "sarrus rule", sarrus 4*4", "sarrus matrice 4 4" or the like. Discussing this with a colleague today, we asked ourselves the following question:

How does the set of 4 by 4 matrices for which this "False Sarrus Rule" gives the correct determinant looks like?

Basic thoughts: A matrix $A=(a_{ij})$ is in this set, if and only if the following equation is fulfilled
$$\sum_{\text{eight special permutations}\ \pi_j} \pm a_{1\pi_j(1)}\cdots a_{4\pi_j(4)} = \det(A).$$
Four out of the eight summands on the left have the right sign, the other four have the wrong sign, and hence, one could simplify a bit. However, the bottom line is: There is just this one equation which has to be fulfilled for all the sixteen entries of a 4 by 4 matrix (and this equation is a homogeneous polynomial of degree four) and hence, the set of matrices for which the False Sarrus Rule gives the right result is a 15-dimensional variety, but I have no clue how it looks like. Probably some algebraic geometers could step in and provide some insight?
Final remark: I do not plan to include this discussion in a math course for engineers (although it may help to scare some people away from the thought that "there could be an easy $4\times 4$ Rule of Sarrus").

REPLY [10 votes]: Here is one result suggested by Gerhard Paseman's comments. The False Sarrus Rule is correct on all matrices of rank $1$ and $4\times 4$ and $5\times 5$ matrices of rank $2$. It does not hold in general on matrices of rank $3$ for $n\times n$ matrices with $n\gt 3$. It also fails for some matrices of rank $2$ and dimension $6$ or greater.
To see that it fails on matrices of rank $3$, consider block diagonal matrices with a nonsingular $2\times 2$ block $A = {a~~b \choose c~~d}$ and a $J_{n-2}$ block. This is singular for $n \gt 3$ but the False Sarrus Rule produces $\det(A) \ne 0.$
$$\begin{pmatrix} a & b & 0 & \cdots & 0  \\\ c & d & 0 & \cdots & 0 \\\ 0 & 0 & 1 & \cdots & 1 \\\ \vdots & \vdots & \vdots & \ddots & \vdots \\\ 0 & 0 & 1 &\cdots &1\end{pmatrix}$$
If $M$ has rank $\le 2$, let the columns be linear combinations of $\vec{v}$ and $\vec w$, $a_i \vec{v} + b_i\vec{w}.$ Then the monomials of the False Sarrus Rule are of the form $\prod_{i\in I} a_{\pi(i)} v_i \prod_{i \in I^c} b_{\pi(i)} w_i.$ If $|I| \le 2$ or $|I^c| \le 2$ then the coefficient of the monomial is $0$ by collecting terms ($n-n$ for ranks $0$ and $1$, and $1-1$ for rank $2$). 
This fails for $|I|,|I^c| \ge 3$. For example, the False Sarrus Rule evaluates to $1$ on $P (J_3 \oplus J_{n-3}) P $, where $P$ is the permutation matrix for the $(3 ~4)$ transposition since only the main diagonal contributes. For $n=6$, this is
$$\begin{pmatrix}1 & 1 & 0 & 1 & 0 & 0 \\\ 1 & 1 & 0 & 1 & 0 & 0 \\\ 0 & 0 & 1 & 0 & 1 & 1 \\\ 1 & 1 & 0 & 1 & 0 & 0 \\\ 0 & 0 & 1 & 0 & 1 & 1 \\\ 0 & 0 & 1 & 0 & 1 & 1 \end{pmatrix}.$$<|endoftext|>
TITLE: Morse theory and homology of an algebraic surface (example)
QUESTION [18 upvotes]: Let $T_n$ denote the $n$-th Chebyshev polynomial and define $$f_n(x,y,z):=T_n(x)+T_n(y)+T_n(z)\;\;\;\text{ and}$$ $$Z_n:=\mathcal{Z}(f_n) \subseteq \mathbb{R}^3,$$
the Banchoff-Chmutov surface, where in general, $\mathcal{Z}(f_1,\ldots,f_k)$ denotes the zero set of polynomials $f_1,\ldots,f_k$, i.e. $\{(x,y,z) \in \mathbb{R}^3; f_1(x,y,z)=\ldots=f_k(x,y,z)=0\}$.
Let us prove, that this is a surface. By the implicit function theorem, it suffices to prove that the points, where $[D_x{f_n},D_y{f_n},D_z{f_n}]$ is zero, do not lie in $Z_n$ (here $D_x$ is just the partial derivative). This is quivalent to showing that the set $$\mathcal{Z}(f_n,D_xf_n,D_yf_n,D_zf_n)=\mathcal{Z}(T_n(x) + T_n(y) + T_n(z),D_xT_n(x),D_yT_n(y),D_zT_n(z))$$ is empty. This can be done by using (from wiki page) $D_xT_n(x) = nU_{n-1}(x)$ and Pell's equation $T_n(x)^2 - (x^2 - 1)U_{n-1}(x)^2 = 1$, to obtain $\mathcal{Z}(1 + 1 + 1) = \emptyset$.
Let us observe the height function $Z_n \rightarrow \mathbb{R}$, $(x,y,z) \mapsto ax + by + cz = [a,b,c][x,y,z]^t$. It is linear, so its derivative is $[a,b,c] :T_pZ_n \rightarrow T_p\mathbb{R} = \mathbb{R}$. Its critical points are therefore those, where the tangent plane $T_pZ_n$ has normal $[a,b,c]$. But the tangent plane of $\mathcal{Z}(f)$ always has normal $[D_xf,D_yf,D_zf]$. Thus the critical points of our height function are those $x,y,z$ where $[D_xf_n,D_yf_n,D_zf_n]=[a,b,c]$, i.e. the critical points are $$\mathcal{Z}(f_n,T_n(x) - a,T_n(y) - b,T_n(z) - c).$$ Now I don't know how to check if these critical points are nondegenerate. I don't even have local parametrizations to work with. 
Question: Can one calculate the homology $H_\ast(Z_n)$ by using the elementary methods from Morse theory (i.e. structural theorem, handle decomposition, Morse inequalities, Morse complex)?

REPLY [14 votes]: The  function $h(x,y,z)=z$, corresponding to $a=b=0$ will do the trick. Assume $n$ is even.   Using a bit of Morse theory I will show that
$$ \chi(Z_n)= \frac{n^2(3-n)}{2}. \tag{1} $$  
A point $(x,y,z)$ on $Z_n$ is critical for $h$ iff
$$  T'_n(x)= T_n'(y)=0, \;\; T_n(z)=-T_n(x)-T_n(y) $$
Now the critical points  of $T_n$ are all located in the interval $[-1,1]$ and can be  easily determined from the defining equality
$$ T_n( \cos t) = \cos nt, \;\;t\in [0,\pi], \tag{A} $$
so that
$$ T_n'(\cos t) = n\frac{\sin nt}{\sin t} $$
This  nails the critical points  of $T_n$ to  
$$x_k = \cos \frac{k\pi}{n},\;\; k=1,\dotsc, n-1.$$
Note that
$$ T_n(x_k)= \cos k\pi=(-1)^k $$
so that the critical points of $h$ on the surface $Z_n$  are 
$$\bigl\lbrace (x_j,x_k,z);\;\; T_n(z)+(-1)^j+(-1)^k=0,\;\;j,k=1,\dotsc, n-1 \bigr\rbrace.  $$
Now we need to count the solutions of the equations
$$T_n(x)=0,\;\pm 2. $$
The equation $T_n(x)=0$ has $n$ solutions, all situated  in $[-1,1]$.  
On the interval $[-1,1]$  we deduce from  (A) that $|T_n|\leq 1$.  The polynomial $T_n$ is even and is increasing on $[1,\infty)$. We conclude that the equation $T_n(x)=-2$ has no solutions, while the equality $T_n(x)=2$ has two solutions.   Thus the critical set of $h$  splits  into three parts
$$ C_0= \lbrace (x_j,x_k,z);\;\;j+k\in 2\mathbb{Z}+1,\;\;T_n(z)=0\rbrace, $$
$$ C_2^+= \lbrace (x_j,x_k,z);\;\;j,k\in 2\mathbb{Z}+1,\;\;T_n(z)=2, z>1\rbrace, $$
$$ C_2^-= \lbrace (x_j,x_k,z);\;\;j,k\in 2\mathbb{Z}+1,\;\;T_n(z)=2, z<-1\rbrace. $$
From the above discussion we deduce  that the points in $C_2^-$ are minima and the points in $C_2^+$ are maxima. The function $h$ is a Morse function and  the saddle points are exactly the points in $C_0$;  for a proof, click here.
Thus  the Euler characteristic of $Z_n$ is
$$ \chi(Z_n)={\rm card}\; C_2^+ +{\rm card}\; C_2^- -{\rm card}\; C_0. $$
Now  observe that  
$$ {\rm card}\; C_2^\pm = \Bigl(\;{\rm card}\; [1,n-1]\cap (2\mathbb{Z}+1) \;\Bigr)^2= \frac{n^2}{4},$$
$$ {\rm card} \; C_0 = n\times \Bigl( \frac{n(n-2)}{4}+ \frac{n(n-2)}{4}\Bigr)= \frac{n^2(n-2)}{2}. $$
(To explain the above equality note that there are $n$ independent possible choices for $z$, the zeros of $T_n$.  Then we need to choose  integers $(j,k)$ in $[1,n-1]\times [1,n-1]$ so that exactly one of them is odd.   The number of  pairs $(j,k)$ with $j$ odd, $k$ even and $1\leq j,k\leq n-1$ is $\frac{n}{2}\times \frac{n-2}{2}$. We have an equal number of pairs $(j,k)$, $1\leq j,k\leq n-1$ with $j$ even and $k$ odd.) 
We conclude  that the Euler characteristic of $Z_n$  is
$$\chi(Z_n)= \frac{n^2}{2}- \frac{n^2(n-2)}{2}=\frac{n^2(3-n)}{2}. $$ 
For $n=2$ we get that $Z_2$ is a sphere. This agrees  with  the pictures  on the site indicated by I. Rivin.
Update.  The above computations do not  explain whether $Z_n$ is connected or not.  To check that it suffices to look at the critical values  of the above  function corresponding to saddle points. These critical values are the zeros $\zeta_1<\dotsc <\zeta_n$ of $T_n$. The level zet
$$ Z_n\cap \lbrace z=\zeta_k\rbrace $$
is the algebraic curve 
$$ T_n(x)+T_n(y)=0. \tag{C} $$
This forces $|x|,|y|\leq 1$ because $T_n(x)> 1$ for $|x|> 1$ and $|T_n(x)\leq 1$ for $|x|\leq 1$.  We can use the  homeomorphism
$$[0,\pi]\ni t\mapsto x=\cos t\in [-1,1] $$
to give an alternate description to (C).  It is  the singular curve  inside  the square $[0,\pi]\times [0,\pi]$ with coordinates $(s,t)$ described by
$$\cos ns+ \cos nt =0.$$
This can be easily visualized as the intersection of the square with the  grid
$$ s\pm t\in (2\mathbb{Z}+1)\frac{\pi}{n} $$
which is  connected. Now it is not very difficult to conclude using the Morse theoretic data on $h$ that $Z_n$ is connected.
Update To better explain my answer to Leon, below is a rendition of $Z_6$ where one can see three   layers, yellow, green and blue.

The equality  (1) predicts 
$$\chi(Z_6)=\frac{6^2(3-6)}{2}=-54. $$
One can verify this directly as follows. Consider the   $1$-dimensional simplicial complex $C$ embedded in $\mathbb{R}^3$  depicted below

The surface $Z_6$ is homeomorphic to the boundary of a thin tubular neighborhood $T$ of this set in $\mathbb{R}^3$. (Think of the edges as   thin spaghetti.) For this reason
$$ \chi(Z_6)= 2\chi(T)= 2\chi(C). $$
Let me   give an alternate proof of the equality
$$\chi(C)=-27. \tag{E} $$
The complex $C$ has $8$ Green vertices of degree $3$, $12$ Red vertices of  degree $4$, $6$ Blue vertices of degree $5$ and a unique  Black vertex of degree  $6$. Thus the number $V$ of vertices of this complex is 
$$ V= 8+12+6+1=27. $$
The number $E$ of edges is half the sum of degrees of vertices. Thus
$$ E=\frac{1}{2}( 3\times 8 + 4\times 12 + 5\times 6+ 6\times 1)=\frac{1}{2} (24+48+30+6)=54. $$
Hence 
$$\chi(C)= 27=54=-27.  $$<|endoftext|>
TITLE: Explicit method to compute Macdonald/Koornwinder functions
QUESTION [8 upvotes]: I'd like to compute explicitly symmetric Macdonald functions associated to arbitrary (possibly non-reduced) root systems, using Computer Algebra System. 
Unfortunately Sage seems to only implement the A-type Macdonald polynomials http://www.sagemath.org/doc/reference/sage/combinat/sf/macdonald.html

Is there a nice paper where a combinatorial formula is provided? 
Has somebody happened to implement it in some programming language?

Of course I can perform the Gram-Schmidt orthogonalization w.r.t. the known measure, but I'll keep it as a last resort.

REPLY [4 votes]: There is a better way to compute Macdonald polynomials explicitly than through Gram-Schmidt orthogonalization: using the action of the Macdonald operators.
Details can be found here  DOI http://dx.doi.org/10.1112/S0010437X03000149 (also avaliable in a somewhat longer version at arXiv:math/0303263)
The first few Macdonald polynomials can be computed in closed form from the Pieri formula.
Details can be found here DOI http://dx.doi.org/10.1007/s00209-010-0727-0 (also available at    arXiv:1009.4482)<|endoftext|>
TITLE: Kahler manifolds with constant bisectional curvature
QUESTION [7 upvotes]: It is well known that the universal covering of a complete Kahler manifold with constant bisectional curvature is $\mathbb{C}^n$, $\mathbb{B}^n$ or $\mathbb{CP}^n$. I need original paper(s) that prove this theorem.

REPLY [7 votes]: This is theorem 7.9 in the book of Kobayashi-Nomizu "Foundations of Differential Geometry Vol.II". There the authors attribute it to Hawley and Igusa independently. These are probably the first papers where this result was proved.
Of course, as Robert Bryant points out, the correct assumption is "constant holomorphic sectional curvature".<|endoftext|>
TITLE: Divisorial contraction: when is the image an algebraic space or a stack?
QUESTION [5 upvotes]: Let $X$ be a smooth projective surface (in the category of varieties, or schemes), and let $C\subset X$ be a curve (a priori not irreducible, but the irreducible case in itself is already interesting).
There are classical notions that say when it is possible to have a morphism $X\to Y$, where $Y$ is a variety (or a scheme) which contracts $C$ (onto points) and which restricts to an isomorphism on $X\backslash C$. The matrix of intersection numbers of the components of $C$ need in particular to be negatively defined.
1) If we admit $Y$ to be an algebraic space, are the conditions weaker? (I think that the matrix has again to be negatively defined, reading Artin, "Algebraic spaces" Theorem 4.5, but are there other conditions that are weaker?)
2) If we admit $Y$ to be an algebraic stack, is the matrix again negatively defined or are there counterexamples?

REPLY [10 votes]: In order to have a contraction morphism $X\to Y$, the intersection matrix must be negative definite.
Conversely, if the intersection matrix is negative definite, the contraction morphism exists in the category of algebraic spaces. It may not exist in the category of schemes: blowing up a smooth cubic in the projective plane in $10$ appropriately chosen points, you can contract the resulting curve of genus $1$ and self-intersection number $-1$ on the blown-up surface only within the category of algebraic spaces. Quite generally, the contraction morphism exists in the category of schemes if the contraction gives rise to a rational singularity.
For details, proofs, and references, you might want to look up Badescu's book on algebraic surfaces, Chapter 3<|endoftext|>
TITLE: Hadamard-like inequalites for positive definite symmetric matrices
QUESTION [6 upvotes]: Let $S$ be any positive semi-definite symmetric matrix (Hermitian psd matrices work as well). The Hadamard inequality is that
$$\det S\le\prod_{i=1}^n s_{ii}.$$
My question is whether there are some other upper bounds of $\det S$ in terms of a partial sum 
$$\sum_{\sigma\in F}\epsilon(\sigma)\prod_{i=1}^n s_{i\sigma(i)},$$
where $F$ is some subset of ${\frak S}_n$. The Hadamard inequality corresponds to the case $F=({\rm id})$. 

More precisely, I am interested in the validity of
  $$\det S\le\sum_{\epsilon(\sigma)=1}\prod_{i=1}^n s_{i\sigma(i)}.\qquad\qquad(1)$$

This latter inequality is true at least when $n\le4$, essentially because the difference
$$\sum_{\epsilon(\sigma)=-1}\prod_{i=1}^n s_{i\sigma(i)}$$
contains enough many non-negative terms (the half if $n=4$, all of them if $n=2$ or $3$). This reason (which can be combined to the Cauchy-Schwarz inequality) fails if $n\ge5$, and therefore I suspect that the inequality (1) could be false if $n\ge5$.
Edit. Felix Goldberg's answer tells us that inequality (1) is true. More generally, Schur proved that if $G$ is a subgroup of ${\frak S}$, and $\chi$ is a character of $G$, then
$$\chi(e)\det S\le\sum_{\sigma\in G}\chi(\sigma)\prod_{i=1}^n s_{i\sigma(i)}=:d_\chi(S).$$
This gives (1) when $G={\frak A}_n$ and $\chi={\bf 1}$.
This leads me raising a second question:

If $G$ is a subgroup of ${\frak S}_n$, let $K_G$ be the set of class functions $f$ over $G$ with the property that, for every positive definite symmetric matrix $S$, one has
  $$f(e)\det S\le\sum_{\sigma\in G}\chi(\sigma)\prod_{i=1}^n s_{i\sigma(i)}=:d_\chi(S).$$
  Clearly, $K_G$ is a convex cone, which contains all the characters. Is it equal to the convex cone spanned by the ireeducible characters of $G$ ?

REPLY [9 votes]: If I interpret correctly the terms and notations, this seems to be a result of Schur. See p.4 here.<|endoftext|>
TITLE: Are the eigenvalues of Hecke operators distinct?
QUESTION [15 upvotes]: Given a Hecke operator which acts on the space of modular forms for $\mathrm{SL}_2(\mathbb{Z})$, are the eigenvalues necessarily distinct?

REPLY [2 votes]: Let $f(z)=\sum_{n\geq1}a_f(n)e^{2\pi i n z}$ be a holomorphic cuspidal Hecke eigenform of even weight $k\geq 12$ on the full modular group with $a_f(1)=1$, with $|a_f(p)|\leq 2p^{\frac{k-1}{2}}$.  A conjecture of Atkin and Serre proposes that for every prime $p$,
$|a_f(p)|\gg_{\epsilon} p^{\frac{k-3}{2}-\epsilon}$,
from which it would follow by multiplicativity that the map $n\mapsto a_f(n)$ is finite-to-1.  In this direction, Murty and Murty (http://www.numdam.org/article/BSMF_1987__115__391_0.pdf) proved that there exists an effective constant $c>0$ such that $|a_f(n)|>(\log n)^c$ when $a_f(n)$ is odd.  Their result also extends to $f$ on congruence subgroups.  Assuming the generalized Riemann hypothesis for Hecke $L$-functions, one can replace $(\log n)^c$ with $n^c$ on a subset of the integers of natural density 1.<|endoftext|>
TITLE: $W^{2,p}$ or $W^{1,q}$ regularity for the laplace on a euclidean sphere
QUESTION [5 upvotes]: Hi,
it is easy to prove the $W^{2,2}(\mathcal S^2)$ regularity for the laplace on the (2 dimensional-) standard sphere $\mathcal S^2:=\lbrace x \in\mathbb R^3: \vert x\vert=1 \rbrace\hookrightarrow\mathbb R^3$ by partial integration, getting
 $$\Vert f\Vert_{W^{2,2}(\mathcal S^2)} \le C ( \Vert \triangle f\Vert_{L^2(\mathcal S^2)} + \Vert f\Vert_{L^2(\mathcal S^2)}).$$

But: How do you prove an equivalent $W^{2,p}(\mathcal S^2)$ regularity for $p>2$ in a more or less direct matter? The only interessting part is to prove a

$L^p$-inequality for $D^2f$:
For $p>2$ there exists $1\le q<\infty$ such that
     $$\Vert \nabla(\nabla f) \Vert_{L^p(\mathcal S^2)} \le C_{p,q}( \Vert \triangle f\Vert_{L^p(\mathcal S^2)} + \Vert f\Vert_{W^{1,q}(\mathcal S^2)}).$$

As the rest follows by the sobolev-inequalites and the $W^{2,2}$-case.

It's essential for me to get a direct proof (in particular no contradiction), because I have to adapt it in an approximated sphere on an asymptotic flat Riemannian Manifold.

In fact it would be enough for me to prove
 $$\Vert \nabla(\nabla f) \Vert_{L^p(\mathcal S^2)} \le C_{p,q,r}( \Vert \triangle f\Vert_{L^r(\mathcal S^2)} + \Vert f\Vert_{W^{1,q}(\mathcal S^2)})$$
for some $1\le q<\infty$ and some $1\le r\le\infty$, because of my special situation, but I do not believe that this is true if the inequality above is not.
Thx for some hints...
Elgrimm

Edit: I mean the sphere $\mathcal S^2:=\lbrace x\in\mathbb R^3 : \vert x \vert=1\rbrace$ as riemannian submanifold of $\mathbb R^3$ and the corresponding laplace-beltrami-operator $\triangle$ on $\mathcal S^2$ not the ball as substet of $\mathbb R^3$.

Edit:
For me it would even be sufficient to prove a


$L^\infty$-inequality for $\nabla f$:
There exists $1\le p<\infty$ and $1\le r\le\infty$ such that
    $$ \Vert \nabla f\Vert_{L^\infty(\mathcal S^2)} \le C(\Vert f\Vert_{L^p(\mathcal S^2)} + \Vert\triangle f\Vert_{L^r(\mathcal S^2)}).$$


Which could be concluded from the results above with the sobolev-inequalities.

Edit (sorry for the mass of edits): By looking at $\nabla f$ instead of $f$ it's quite obvious, that it would be a even stronger result to prove the


Weak inequality:
For $p>2$ and $\frac1p+\frac1q=1$ it's true that
    $$ \Vert f\Vert_{W^{1,p}(\mathcal S^2)} \le C(\Vert f\Vert_{L^p(\mathcal S^2)} + \Vert \triangle f\Vert_{W^{-1,q}(\mathcal S^2)}), $$
    where $W^{-1,q}(\mathcal S^2)$ is the dual of $W^{1,q}(\mathcal S^2)$, in particular
    $$\Vert \triangle f\Vert_{W^{-1,q}(\mathcal S^2)}:=\sup_{\Vert\varphi\Vert_{W^{1,q}(\mathcal S^2)}=1}\left\vert\int_{\mathcal S^2}\nabla f\cdot\nabla\varphi\right\vert.$$

REPLY [3 votes]: Sobolev-Norms:
Let $(M,g)$ be a two-dimensional compact remannian-manifold without boundary $r>0$, define for $f\in\mathcal C^\infty(M)$ the sobolev-inequalites
$$ \Vert f\Vert_{W^{k,p}(M)} := \sum_{l=0}^k r^l\cdot\left\Vert \vert\nabla^l f\vert_g\right\Vert_{L^p(M)}\qquad\forall f\in\mathcal C^1(M),\ k\in\mathbb N_{\ge0}, $$
where $\nabla$ is the levi-civita connection. $W^ {k,p}(M)$ is as usual the completion of $\mathcal C^\infty(M)$ for this norm, and as usual we identifiy elements of $W^{k,p}$-with almost every-where defined functions.

$\triangle$-$W^{1,\infty}$-, -$W^{2,2}$-, -$L^\infty$-regularities:
Let $M$ be a two-dimensional compact manifold without boundary and $c,r\in\mathbb R_{>0}$, such that
$$ \vert M\vert \le cr^2, \quad \Vert \text{Ric}(M) \Vert_{L^\infty(M)} \le \frac c{r^2},\quad \Vert f\Vert_{L^2(M)} \le \frac cr\Vert f\Vert_{W^{1,1}(M)} \quad\forall f\in W^{1,1}(M) $$
and $c>\lambda_1$, where $\lambda_1$ is the smallest positiv eigenvalue of $-\triangle$.
There is a constant $C=C(c)$ such that
$$ \Vert f\Vert_{W^{2,2}(M)} \le C(r^2\Vert \triangle f\Vert_{L^2(M)}+\Vert f\Vert_{L^2(M)}) \qquad\forall f\in W^{2,2}(M), \tag{1} $$
$$ \Vert f\Vert_{L^\infty(M)} \le C(r\Vert \triangle f\Vert_{L^2(M)}+r^{-1}\Vert f\Vert_{L^2(M)}) \qquad\forall f\in W^{2,2}(M), \tag{2} $$
$$ \Vert f\Vert_{W^{1,\infty}(M)} \le Cr^{-\frac2p}(r^2\Vert \triangle f\Vert_{L^p(M)}+\Vert f\Vert_{L^p(M)}) \qquad\forall f\in W^{2,p}(M),\ p>2, \tag{3} $$
if $r>C$.

Proof [as Deane Yang suggested by a Moser-iteration]:
Be setting $g:=\vert f\vert^{\frac 12}$, we get the usual sobolev-inequalites
$$\Vert f\Vert_{L^q(M)} \le \frac cr\Vert f\Vert_{W^{1,p}(M)} \quad\forall f\in W^{1,p}(M),\ 1\le p< 2: \ q=\frac{2p}{2-p}. \tag 4$$
By an argument as as in [1] we conclude the additional sobolev-inequality
$$\Vert f\Vert_{L^\infty(M)} \le cr^{-\frac2p}\Vert f\Vert_{W^{1,p}(M)} \qquad\forall f\in W^{1,p}(M), 2< p. \tag 5 $$
Using that $div\nabla X=g^{ij}Ric(X,e_i)e_j+\nabla div X$ for any frame $\lbrace e_1,e_2\rbrace$ and any smooth vector field $X$, we conclude $(1)$ by defining $X=\nabla f$ and partial integration and therefore we can deduce $(2)$ with $(5)$.
Let $f\in W^{2,p}(M)$ be such that $\int f=0$, in particular $\Vert f\Vert_{L^2(M)} \le c \Vert\triangle f\Vert_{L^2(M)}$, and assume there exists a polnom $P$ and a constant $C$ such that for $q=\frac{2p}{p-2}$
$$ \Vert \vert\nabla f\vert^{2s}\Vert_{L^p(M)} \le P(s)\left(\left(\frac Csr^{\frac2p}\Vert\triangle f\Vert_{L^q(M)}\right)^{2r}+r^{-\frac2p}\Vert\vert\nabla f\vert^r\Vert_{L^p(M)}\right). \tag6 $$
Setting $a_i:=(r^{2^i-\frac2p}\Vert\vert\nabla f\vert^{2^i}\Vert_{L^p(M)})^{2^{-i}}$, we conclude by basic analysis
$$ r\Vert \nabla f\Vert_{L^\infty(M)} \le C\left(r^{1-\frac2p}\Vert\triangle f\Vert_{L^p(M)}+r^{1+\frac2p}\Vert\nabla f\Vert_{L^{\frac{2p}{2-p}}(M)}\right). $$
This inequality obviously also holds for $f+d$ ($d\in\mathbb R)$. So we only have to prove $(6)$ for a polnyom and a constant not depending on $f\in\mathcal C^\infty(M)$ with $\int f\ d\mu=0$. By basic analysis and some partial integrations we see
$$ \int \vert\nabla(\vert\nabla f\vert^s)\vert^2\ d\mu \le -s\int\vert\nabla f\vert^{2s-2}g(\nabla f,\triangle\nabla f). $$
We conclude by hölder-inequalities and partial integration
$$ \int \vert\nabla(\vert\nabla f\vert^s)\vert^2\ d\mu \le Csr^{\frac2{ps}}\Vert \vert\nabla f\vert^{2s}\Vert_{L^p(M)}^{\frac{s-1}s}\underbrace{\left(\begin{aligned}\Vert\text{Ric}\Vert_{L^\infty(M)}\cdot\Vert\nabla f\Vert_{L^{\frac{2p}{p-1}}(M)}^2+\Vert\triangle f\Vert_{L^{\frac{2p}{p-1}}(M)}^2 \\\\ + \Vert\triangle f\Vert_{L^2(M)}\cdot\Vert\triangle f\Vert_{L^{\frac{2p}{p-2}}(M)} \\\\ + \Vert\text{ric}\Vert_{L^\infty(M)}^{\frac12}\Vert\nabla f\Vert_{L^2(M)}\cdot\Vert\triangle f\Vert_{L^{\frac{2p}{p-2}}(M)}\end{aligned}\right)}_{=: c_f}. $$
By some Yang- and Hölderinequalities and using $\Vert f\Vert_{L^2(M)}\le c\Vert \triangle f\Vert_{L^2(M)}$ we see for $q:=\frac{2p}{2-p}$ and the inequality for $\text{ric}$
$$ c_f \le C_pr^{\frac2p}\Vert\triangle f\Vert_{L^q(M)}^2. $$
By the Sobolev- and Young-inequalities we therefore can conclude $(6)$ for $P(s)=s^6$ using the inequality for $\text{ric}$.

[1]: Stampacchia, Guido: Régularisation des solutions de problèmes aux limites
elliptiques à données discontinues. In: Proceedings of the International Sympo-
sium on Linear Spaces. Hebrew University of Jerusalem : Jerusalem Acad. Pr.,
July 1960 (A publication of the Israel Academy of Sciences and Humanities),
S. 399–408<|endoftext|>
TITLE: Real intersections of plane cubic curves 
QUESTION [5 upvotes]: Let $C$ and $D$ be plane cubic curves over the real numbers. Suppose that the real loci in the projective plane of $C$ and $D$ consist of two connected components. Denote $C_0$ and $D_0$ the bounded component (which is often referred as "oval") of $C$ and $D$, and denote $C_1$ and $D_1$ the unbounded component of $C$ and $D$ respectively. 
I have following question: 
What is the maximal number of real intersections of $C_1$ and $D_1$ in the projective plane?
Of course a trivial bound is 9 given by Bezout's theorem, but I conjecture that the bound is  rather 5. I think this should be well known, but I couldn't find any reference.

REPLY [3 votes]: Another approach, building on Noam Elkies's answer:
$$(y - 5) (x - 2 y + 10.5) (x + 2 y - 10.5) = -0.001$$
$$(x - 5) (y - 2 x + 10.5) (y + 2 x - 10.5) = -0.001$$
The images below show the same graph at different resolutions. Unfortunately, getting far enough out to see all 9 points makes it hard to see the ovals cleanly. But the basic idea is to choose two triples of lines which form triangles (not which pass through a common point as in Noam's answer) and deform a little to make the topology work.
       (source)
       (source)
       (source)<|endoftext|>
TITLE: What's the role of $H^{p}(\mathbb{R}^{n})$ in modern (harmonic) analysis?
QUESTION [11 upvotes]: The classical theory of $H^p$,due to it's heavy dependence on the complex function theory(such as Blaschke products), seemed to have an insurmountable obstacle barrying its extension to several variables.However, C.Fefferman and E.Stein's remarkable paper'$H^{p}$ spaces of several variables' showed that $H^{p}$ classes can be characterized without any recourse to analytic functions, conjugacy of harmonic functions, etc. Thus $H^{p}$ classes have an intrinsic real variable meaning of their own. Another surprising result they got was that the predual of BMO(functions of bounded mean oscillation) was exactly $H^1$.
Well,what I'm particularly interested is its applications in mordern analysis.For instance,From Ferfferman's work,I know that $H^1$ is sometimes a  proper subsitute for $L^1$,this can be seen from the CZOs(Calderon-Zygmund operators),which are bounded from $H^1$ to itself,but not on $L^{1}$. This is useful when evaluating some singular integral operators through complex interpolation.
Sometimes it's also very convenient to prove a bounded function to be $L^p$ multipliers through $H^1$,for example
$m(\xi)=\psi(\xi)e^{i|\xi|^{a}}|\xi|^{-b}$($b>0$,$a>0$,$a\neq 1$),where $\psi \in C^{\infty}$ is 0 nere 0,and 1 when $|\xi|$ large.Then m is a $L^{p}$ multiplier iff $n|\frac{1}{2}-\frac{1}{p}|\leq \frac{b}{a}$
My question is what's the role of $H^{p}$ in modern (harmonic) analysis,and how people get useful results by choosing $H^p$.
I would appreciate any good examples, as well as some historical outlines on the topics development

REPLY [2 votes]: The real Hardy spaces are equivalent to $L^p(\mathbb{R}^n)$ space when $p>1$, and they are much easier to to use than $L^p(\mathbb{R}^n)$ when $p\leq 1$. Since $H^p(\mathbb{R}^n)$ has a maximal function and singular integral generalisation, $H^p(\mathbb{R}^n)$ gives us an extension of maximal function/singular integral results to $p\leq 1$, when they were originally designed for $L^p(\mathbb{R}^n),~p>1$. For example, suppose we have a distribution $K$ with bounded Fourier transform satisfying bounds on the derivative of $K$ equivalent to the smoothness condition known as the Hörmander condition (away from the origin, ie, in $\mathbb{R}^n\setminus 0$). Then we can ascertain that the operator $T:H^p(\mathbb{R}^n)\to H^p(\mathbb{R}^n)$ defined by the convolution
\begin{equation*}
Tf=f\ast K,~~f\in H^p(\mathbb{R}^n)
\end{equation*}
is bounded for $p\leq 1$. 
Check out the following references:

"Modern Fourier Analysis" by Loukas Grafakos
"Harmonic Analysis: Real-Variable Methods, Orthogonality, and Oscilllatory Integrals" by Elias Stein
"Singular Integrals and Differentiability Properties of Functions" by Elias Stein

They cover what you are looking for in detail.<|endoftext|>
TITLE: Isomorphisms of quantum planes
QUESTION [17 upvotes]: Let $k$ be a field and $q\in k^{*}$.  The quantum plane $k_{q}[x,y]$ is the algebra $k\langle x,y\rangle/\langle xy=qyx \rangle$ (i.e. the quotient of the free non-commutative $k$-algebra on two variables $x$ and $y$ modulo the ideal given).  

Question: For $q,r\in k^{*}$ and $q\neq r$, when is $k_{q}[x,y]$ isomorphic (as an algebra) to $k_{r}[x',y']$?

I fully expect this is known but after (what I think is) fairly comprehensive literature searching, including a large proportion of the best-known quantum groups texts, I have been unable to find an answer.  A reference would be appreciated just as much as a proof.
Some comments:

I know the (algebra) automorphism group: by work of Alev-Chamarie this is $(k^{*})^2$ unless $q=-1$ (when it is a semi-direct product of the torus with the group of order two generated by the map that interchanges the two variables).  Hence I don't need to worry about $q=r$.
I want algebra isomorphisms but information on Hopf algebra maps would be nice too (NB. the Hopf automorphisms for the usual Hopf structure are also those just described)
if $q$ has finite order $N$ in $k^{*}$ and $r$ is of infinite order then the corresponding quantum planes are not isomorphic, as in the first case the centre is non-trivial (generated by $x^{N}$ and $y^{N}$) but in the second the centre is just $k$
if $q$ has order $M$ and $r$ has order $N\neq M$, then the quotients by the centres are both finite-dimensional but of different dimension, hence the quantum planes are not isomorphic
I would be happy to know the answer just for $k=\mathbb{C}$

REPLY [3 votes]: You may want to have a look at this paper: Isomorphisms of some quantum spaces. The techniques rely on the graded structure of the quantum planes and, more generally, quantum affine spaces.<|endoftext|>
TITLE: Generalizations of Oppenheim's inequality
QUESTION [5 upvotes]: The well-known Oppenheim inequality says that for two positive definite matrices $A,B$ it holds that $\det(A \circ B) \geq (\prod{a_{ii}})\det(B)$. 
There has been a lot of beautiful work done extending it to cases when $A$ or $B$ or both of them are $M$-matrices or their inverses, or totally nonnegative. 
My question is: do you know of other extensions, in which $A$ is non-symmetric in an "interesting" way?

REPLY [4 votes]: This is not a generalization to other matrix classes, but replacing determinant by permanent. Actually, it is a conjecture made by Bapat and Sunder: Under the same conditions $per(A \circ B) \leq (\prod{a_{ii}})per(B)$.
...but the following result due to Jiao [On a conjecture of H. Minc, Linear and Multilinear Algebra 32 (1992) 103–105.] couldn't surprise me more
$$per(A \circ B)+per (A) per (B) \geq (\prod{a_{ii}})per(B)+(\prod{b_{ii}})per(A).$$<|endoftext|>
TITLE: When is P(x)-Q(y) irreducible?
QUESTION [22 upvotes]: Let $k$ be an algebraically closed field (in my application, it is characteristic zero, but this probably doesn't matter so much), and let $P: k \to k$, $Q: k \to k$ be polynomials of one variable.  Then $P(x)-Q(y)$ is a polynomial of two variables $x,y$.   Generically, one expects this polynomial to be irreducible, but there are some exceptional cases where it becomes reducible.  For instance, if $P=Q$ and $P,Q$ have degree greater than $1$, then $P(x)-Q(y)$ contains $x-y$ as a nontrivial factor.  More generally, if $P = R \circ \tilde P$ and $Q = R \circ \tilde Q$ for some polynomials $\tilde P, \tilde Q, R$ with $R$ having degree greater than $1$, then $P(x)-Q(y)$ contains $\tilde P(x)-\tilde Q(y)$ as a nontrivial factor.   
My (somewhat vague) question is whether there is a way to classify all the pairs of polynomials $P, Q$ in which $P(x)-Q(y)$ becomes reducible.  Of course, this has a tautological answer - it is those $P, Q$ for which one can factorise $P(x)-Q(y) = R(x,y) S(x,y)$ for some polynomials $R,S$ - but I am looking for a criterion which is somehow simpler to verify than the general problem of determining whether an arbitrary polynomial of two variables is irreducible.  Ideally it should have the flavour of "the only obstructions to irreducibility are the obvious ones".  One could naively conjecture that the above examples are in fact the only reducible examples; indeed I could not produce any other examples, but this is likely a failure of my own imagination.
I have this vague picture of viewing the curve $\{ (x,y): P(x)=Q(y)\}$ as a relative product of the curves $\{ (x,z): z = P(x) \}$ and $\{ (y,z): z = Q(y) \}$ over the $z$-axis, so that the irreducibility of the former should somehow relate to the structure of the latter two factors (and in particular, in the location and nature of the singular points of the projection maps to the $z$-axis), but I don't know how to make this precise.  (Maybe I should have paid more attention to Riemann surfaces as a student...)
Actually, I'm ultimately interested in the multidimensional version of this question, namely to give a criterion for when the algebraic set $\{ (x,y) \in k^d \times k^d: P(x)=Q(y) \}$ is an irreducible variety, where $P, Q: k^d \to k^m$ are polynomial maps, but given that even the $d=m=1$ case seems to be non-trivial, I thought I should focus on that first.

REPLY [4 votes]: An interesting example perhaps is $$\eqalign{4 x^4 &- 4 x^2 + y^4 + 4 y^3 + 4 y^2\cr &= \left( 2{x}^{2}+2x+2xy+2y+{y}^{2} \right)  \left( 2{x}^{2}-2x-2xy+2y+{y}^{2} \right)}$$<|endoftext|>
TITLE: E an elliptic curve over Z[1/N], how many p such that E(Z/p^2) = (Z/p)^2?
QUESTION [6 upvotes]: For a fixed elliptic curve $E$ over $\mathbb{Z}[1/N]$, is there a non-trivial upper bound in terms of $x$ for the number of primes $p \leq x$ with $p \nmid N$ for which $E(\mathbb{Z}/p^2\mathbb{Z}) \cong (\mathbb{Z}/p\mathbb{Z})^2$? Anything less than $\frac{x}{(\log x)^{2}}$ would be very useful for me.
Some thoughts. The condition $E(\mathbb{Z}/p^2\mathbb{Z}) \cong (\mathbb{Z}/p\mathbb{Z})^2$ implies that $E(\mathbb{F}_p)\cong \mathbb{Z}/p\mathbb{Z}$. Conversely, given an elliptic curve $E$ over $\mathbb{Z}[1/N]$ and a prime $p$ not dividing $N$ such that $E(\mathbb{F}_p) \cong \mathbb{Z}/p\mathbb{Z}$, there are two possibilities for $E(\mathbb{Z}/p^2\mathbb{Z})$: either the extension
\[
0 \rightarrow E_1(\mathbb{Z}/p^2\mathbb{Z}) \rightarrow E(\mathbb{Z}/p^2\mathbb{Z}) \stackrel{\textrm{reduction mod $p$}}{\longrightarrow} E(\mathbb{F}_p) \rightarrow 0
\]
doesn't split, in which case we have $E(\mathbb{Z}/p^2\mathbb{Z}) \cong \mathbb{Z}/p^2\mathbb{Z}$, or it does split, so that $E(\mathbb{Z}/p^2\mathbb{Z}) \cong (\mathbb{Z}/p\mathbb{Z})^2$.
Still assuming that $E(\mathbb{F}_p)\cong \mathbb{Z}/p\mathbb{Z}$, the second case occurs if and only if $E_{\mathbb{Z}/p^2\mathbb{Z}} = E \otimes_{Z[1/N]} \mathbb{Z}/p^2 \mathbb{Z}$ is the canonical lifting of $E_{\mathbb{F}_p}$ to $\textrm{Spec}(\mathbb{Z}/p^2 \mathbb{Z})$. At least, this is what I've been told by some people working with canonical liftings, so I'm willing to go along with this. Anyway, I don't know if it helps, but somehow, the second case seems to be less "likely" to happen than the first.
Motivation. My question might seem a strange one, so I'll tell you why I am interested in this. Given an $E$ and $p$ as above, if $E(\mathbb{F}_p)$ is cyclic, we may immediately conclude that $E(\mathbb{Q}_p)$ is a procyclic topological group (i.e. topologically generated by a single element), except when  $E(\mathbb{Z}/p^2\mathbb{Z}) \cong \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$. I am interested in finding, for a given $E$, infinitely many $p$ such that $E(\mathbb{Q}_p)$ is procyclic. (I actually want this condition to hold for all quadratic twists of $E$, but I don't think I need to go into the specifics any further.) Since there are all sort of results in the literature concerning cyclicity of $E(\mathbb{F}_p)$, the only real obstacle is the occurrence of the groups $(\mathbb{Z}/p\mathbb{Z})^2$ as above.

REPLY [3 votes]: I don't know how easy it will be to prove anything along these lines, but the condition is that the Serre-Tate parameter of $E$ at $p$ satisfies $q \equiv 1 \mod p^2$. It's natural to assume that $(q-1)/p \mod p$ is random as $p$ varies, just because I know of no reason why it should be otherwise. If that's the case the number of your primes up to $x$ would be $O(\log \log x)$ kind of like Wieferich primes.<|endoftext|>
TITLE: Module in category O not generated by a finite set of HWVs.
QUESTION [5 upvotes]: For a while I've been reading J.E.Humphreys's book "Representations of semisimple Lie algebras in the BGG category $\mathcal O$" under the impression that any module in $\mathcal O$ has a finite generating set composed of highest weight vectors. Now I've realised that I'm lacking a proof. So what would serve as a counterexample? Or has my assumption been correct for some reason?
I apologize if the answer can be found further on in the book itself!

REPLY [5 votes]: This is false.  Consider the contragradient dual on a Verma module $V_\lambda$.  This is a module given by linear functions on $V_\lambda$ which kill all but finitely many weight spaces. This module has no highest weight vectors of weight other than $\lambda$; any non-zero vector $\xi$ of weight $< \lambda$ is non-zero on $F_i v$ for some $v$ (since $V_\lambda$ is generated in degree $\lambda$), and $E_i\xi\neq 0$ since it has non-zero value on $v$. 
Thus, this module is generated by weight vectors if and only if it is generated by vectors of weight $\lambda$, but this is only possible if the Verma module is irreducible (otherwise, the dual of the simple quotient of the Verma is the proper submodule generated by the vectors of degree $\lambda$).  Thus, this is a counter-example.

REPLY [5 votes]: Ben's answer is essentially correct, modulo his invented word "contragradient" (which like "indeterminant" I've been trying to stamp out, without success).   Of course, the definition shows that all modules in the category are finitely generated, so the problem involves generation by highest weight vectors.
Maybe I should comment further on the simplest counterexample, which occurs already in rank 1.   It takes a while to build up complicated examples in category $\mathcal{O}$, since I only introduce the BGG duality in Chapter 3.  While the Lie algebra acts naturally on the usual vector space dual of a module, this dual is usually too big to lie in $\mathcal{O}$, so as Ben suggests there is a more restrictive notion of duality; this stays in the caegory and preserves the formal characters.  
For the rank 1 simple Lie algebra, integral weights may be identified with ordinary integers.  In particular, you get a Verma module $M(0)$ with only two composition factors:  $L(0)$ is the one-dimensional module at the top, while $L(-2) = M(-2)$ is the unique maximal submodule.  Of course, any Verma module is generated by a highest weight vector.   But the BGG dual $M(0)^\vee$ has the same two composition factors in reverse order and in particular can't be generated by highest weight vectors since $-2$ isn't a highest weight vector here (see the discussion of Hom in my section 3.3).   
In higher ranks you start to run into more sophisticated examples where the maximal submodule of a (non-dominant) Verma module is itself not generated by its highest weight vectors.   Indeed, a Verma module might have infinitely many distinct submodules.  This was first appreciated by BGG and Conze-Duflo, but is best understood via the later Kazhdan-Lusztig theory.<|endoftext|>
TITLE: Is SL(2,C)/SL(2,Z) a quasi-projective variety?
QUESTION [22 upvotes]: Consider the complex 3-fold $SL(2,\mathbb C)/SL(2,\mathbb Z)$ (just for clarity: note that $SL(2,\mathbb Z)$ acts without stabilizers, so this is a complex manifold, not a complex orbifold).

Is $SL(2,\mathbb C)/SL(2,\mathbb Z)$ a quasi-projective variety?

The natural generalization of this question seems to be the following.  Let $G$ be a semisimple linear algebraic group over $\mathbb Q$.  Then $G(\mathbb Z)$ is well-defined up to taking a finite-index subgroup.  Thus we can ask the same question: is the complex manifold $G(\mathbb C)/G(\mathbb Z)$ a quasi-projective variety?

REPLY [26 votes]: No, the quotient is not quasi-projective. In the paper Invariant meromorphic functions on complex semisimple Lie groups by D. N. Ahiezer you can find the following result.

Theorem. Let $G$ be a connected semisimple linear algebraic group
  defined over the rationals and
  $\Gamma$ be a  subgroup of $G(\mathbb
> Q)$ which is Zariski dense in $G$.
  Then there are no invariant analytic
  hypersurfaces in $G$ invariant by the
  action of $\Gamma$. In particular, if
  the quotient $G/ \Gamma$ exists as a
  complex variety then every 
  meromorphic function on it is constant.

This implies that the quotient is not quasi-projective and even more: it cannot be holomorphically  embedded in any algebraic variety.<|endoftext|>
TITLE: Homeomorphism of the rationals
QUESTION [8 upvotes]: In working with the classification of stable vector bundles on $\mathbb{P}^2$, I've found that I need to answer a fairly basic question from analysis/point set topology.  Here it is.
Suppose $f:\mathbb{Q}\to \mathbb{Q}$ is 

strictly increasing,
not bounded above or below,
a local homeomorphism (with the topology induced from $\mathbb{R}$), and
extends to a continuous map (hence a bijection) $f':\mathbb{R}\to \mathbb{R}$.

Is $f$ a homeomorphism?  (That is, is it surjective? Or, alternately, could there exist some irrational number $c$ such that $f(c)$ is rational?)
While I am dealing with a specific function, I state things in this generality because the function itself is fairly nasty and I'd rather not have to use its explicit definition more than I have to.  My guess is that it is probably too much to hope for that this be true in this generality, so in case the general version is false here is a refined version of property (3) which incorporates a bit more about my present situation:
3'. there exists a partition $\mathbb{Q}=\bigcup_{\alpha \in A} I_\alpha$ of $\mathbb{Q}$ into countably many disjoint open intervals with irrational endpoints, such that $f$ is linear (with rational coefficients) on each interval.
The difficulty (at least for me) is that, viewing the intervals as intervals in the real numbers, their complement forms some kind of Cantor set.  
Thanks!
(EDIT: Several counterexamples have shown the first formulation, with properties 1-4 are false.  I imagine the formulation with 3' instead of 3 is also false, but it seems slightly less trivial to get a counterexample due to the condition on the irrationality of the endpoints. In particular, no two intervals can "match up" at an irrational number unless $f$ has the same slope on both intervals.)

REPLY [2 votes]: First construct the Cantor set: An uncountable closed set whose complement is a dense open set containing all the rationals. There are a bunch of ways to do this.
Pull back the regular Cantor set along a homeomorphism $\mathbb R\to \mathbb R$ that sends all the rationals to the rationals not in the Cantor set, possible by the same logic as Joseph Van Name's answer.
Take a neighborhood of the rationals with arbitrarily small measure, say finite measure. The complement has positive, indeed infinite measure, and so is uncountable.
Start with an interval and keep subdividing it. Make sure that the open interval you remove in each subdivision is the one with the smallest $f(q)$ in that interval, where $f: \mathbb Q \to \mathbb N$ is a bijection.
The complement of this set is open, therefore a countable union of open intervals. Each interval has two boundary points. Choose some element of the Cantor set that is not a boundary point, and map it to some rational number. Note that it must be irrational since it is in the Cantor set. This will be our counterexample to surjectivity.
Order the open intervals. We build a function piece by piece by choosing its values on each open interval in order. For each interval, choose some linear function with rational coefficients on it that preserves monotonicity. To preserve continuity,  choose the function to be very close to failing to preserve monotonicty. The highest $y$ value should be very close to the next $y$ value we have already defined, and similarly for the lowest. This is possible because we have two degrees of freedom and two constraints, so we can choose arbitrarily good rational approximations. The gaps between undefined $y$ values go to $0$, so we can extend continuously and monotonically as the last step, just like the standard Cantor function.
If our open intervals are all bounded, we can ensure the function is unbounded with the same trick.
It is clear our function satisfies all the properties required, except the inverse image of some rational number is an irrational number.<|endoftext|>
TITLE: Counterexamples to differentiation under integral sign?
QUESTION [25 upvotes]: I'm exploring differentiation under the integral sign (I want to be much faster and more assured in doing this common task).  So one thing I'm interested in is good counterexamples, where both expressions 
$\frac{d}{dx} \int f(x,y)dy$ 
and 
$\int \frac{\partial}{\partial x} f(x,y)dy$ 
exist at some value of x but are not equal.  Thanks in advance.

Here is some of my exploration so far.
My derivation for switching the derivative and integral is as follows:
$\frac{d}{dx} \int f(x,y)dy = \frac{d}{dx}\int f(a,y)+\int_a^x \frac{\partial}{\partial s}f(s,y)dsdy = \frac{d}{dx}\int \int_a^x \frac{\partial}{\partial s}f(s,y)dsdy$, 
provided f is absolutely continuous in the x-direction (used FTC).  Then provided $\frac{\partial}{\partial s}f(s,y)$ is integrable in some rectangle $[x,x'] \times \Omega$, we can apply Fubini and switch the order of integration,
$= \frac{d}{dx}\int_a^x \int \frac{\partial}{\partial s}f(s,y)dyds$.
Then using FTC again, the first integral and derivative cancel provided that $\int \frac{\partial}{\partial s}f(s,y)dy$ is continuous.  
Thus altogether the assumptions I need in order to switch integration order are
1) f is absolutely continuous in the x-direction
2) df/da is integrable in a rectangle where one side is a small interval containing x, and the other is the whole y-direction.
3) $\int \frac{\partial}{\partial x} f(x,y)dy$ is continuous.
4) and of course, the original expressions are defined.
This was the most general I could go.  The conditions seem messy and forgettable, so I'd like to find some nicer conditions I could use in general, if possible.
Condition 3) seemed like it might be the easiest to toss, but I found a counterexample where all other conditions are satisfied but that one:
Let y be the positive integers and dy be the counting measure, so we're just trying to flip a sum and the derivative.  Let
$f(x,y) = x$ for $\frac{1}{y+1} < x <\frac{1}{y}$ and zero elsewhere.
Then we see the integral of the derivative of $f$ is $0$ at $x=0$ and the derivative of the integral is $1$ at $x=0$, thus providing one counterexample.  Here conditions 1) and 3) were not satisfied, but the example could be easily modified to make 1) satisfied.

REPLY [9 votes]: Set 
$$
f(x,y) = \begin{cases}
\frac{x^3y}{(x^2+y^2)^2}, & {\rm if} \ x \not= 0 \ {\rm or } \ y \not= 0, \\
0, & {\rm if } \ x = 0 \ {\rm and } \ y = 0, 
\end{cases}
$$
Then the integral 
$$
F(x) = \int_0^1 f(x,y)\,{\rm d}y
$$
can be computed to equal $\frac{x}{2(1+x^2)}$ for all $x$ (check $x = 0$ separately). This is a differentiable function for all $x$, with 
$$
F'(x) = \frac{1-x^2}{2(1+x^2)^2}.
$$
In particular, $F'(0) = 1/2$. However, 
$$
\frac{\partial}{\partial x}f(x,y) = \begin{cases}
\frac{x^2y(3y^2-x^2)}{(x^2+y^2)^3}, & {\rm if } \ y \not= 0, \\
0, & {\rm if } \ y = 0, 
\end{cases}
$$
so $f_x(0,y) = 0$. Therefore the "equation"
$$
\frac{\rm d}{{\rm d}x}\int_0^1 f(x,y)\,{\rm d}y = \int_0^1 \frac{\partial}{\partial x}f(x,y)\,{\rm d}y
$$
is invalid at $x = 0$, where the left side is $1/2$ and the right side is $0$. A problem is that $f_x(x,y)$ is not a continuous function of two variables: along the line $y = x$ we have $f_x(x,y) = f_x(x,x) = 1/(4x)$ for $x \not= 0$, which does not converge as $x \rightarrow 0$ even though $f_x(0,0) = 0$ is defined.<|endoftext|>
TITLE: Is the dual of the product of infinite cyclic groups a free abelian group ?
QUESTION [10 upvotes]: By a theorem of Specker, the group $\mathrm{Hom}(\prod_{\aleph_0} \mathbb{Z},\mathbb{Z})$ is isomorphic to $\bigoplus_{\aleph_0}\mathbb{Z}$ and is in particular a free abelian group. I wonder, if this generalizes to all cardinals: 
Question: Is it true that $\mathrm{Hom}(\prod_{\kappa} \mathbb{Z},\mathbb{Z})$ is a free abelian group for each cardinal $\kappa$ ?
As the answer to such questions sometimes depends on the underlying set theory, I included the "set-theory" tag.

REPLY [15 votes]: If there are no measurable cardinals, or just if there are no measurable cardinals $\leq\kappa$, then the answer to your question is yes, and in fact all homomorphisms from $\prod_\kappa\mathbb Z$ to $\mathbb Z$ are linear combinations of the $\kappa$ projection maps.  If, on the other hand, $\kappa$ or some smaller cardinal is measurable, so $\kappa$ supports a non-trivial, countably complete ultrafilter $U$, then the "in fact" clause in the preceding sentence is false; a counterexample is given by the homomorphism sending any $f\in\prod_\kappa\mathbb Z$ to the value that $f$ takes at $U$-almost all elements of $\kappa$.  I believe that Hom($\prod_\kappa\mathbb Z,\mathbb Z)$ may nevertheless be free (though with a more complicated base than just the projections), but this probably depends on detials of the structure of the countably complete ultrafilters on $\kappa$.  (In a couple of days, I'll be back in Michigan, where I can look in my copy of the book "Almost Free Modules" by Eklof and Mekler, where the section on the Los-Eda theorem should give me a lot more information about this.  Anyone who wants to look for themselves rather than waiting for me should take into account that "Los" here is really "{\L}o\'s".)
Edit: OK, I'm back in Michigan, and I'm looking at Corollary 3.6 of the Eklof-Mekler book.  It's stated in more generality than the present question wants (using slender modules over general rings), but if I specialize the ring $R$ and all the modules $H$ and $M_i$ in this corollary to be $\mathbb Z$, it tells me that Hom($\prod_\kappa\mathbb Z,\mathbb Z)$ is freely generated by the homomorphisms that I described above, one homomorphism for each countably complete ultrafilter on $\kappa$ (including the principal ultrafilters, which give the projection homomorphisms).<|endoftext|>
TITLE: Failure of Jacobson-Morozov in positive characteristics
QUESTION [8 upvotes]: The Jacobson-Morozov theorem that any nilpotent element $e$ in the Lie algebra of a simple algebraic group $G$ can be embedded in an $\mathfrak{sl}_2$-triple, has a restriction (in terms of the Coxeter number) on the characteristic of the underlying field (assumed to be algebraically closed). This restriction is also required for the "uniqueness" of the triple, up to $C_G(e)$-action. (This result is due to Kostant.) In his 1980 paper, Pommerening had removed the restriction on the characteristic in Jacobson-Morozov's theorem, up to very small exceptions (i.e., characteristic is "bad"). Does the uniqueness as in Kostant's result also hold with this weaker restriction? If it does, then where does Jacobson-Morozov along with uniqueness result of Kostant fail in positive characteristics?

REPLY [7 votes]: Sorry to necro-post, but I have been meaning to point out for a while that I have a reasonably complete answer to the OP in some work with Adam Thomas, 'The Jacobson--Morozov Theorem and complete reducibility of Lie subalgebras' just made available in Proc. LMS, here. We were interested in $G$-complete reducibility of subalgebras of the Lie algebra $\mathfrak g$ of the reductive algebraic group $G$. A subalgebra is $G$-completely reducible if whenever it is in a parabolic subalgebra $Lie(P)$ of $\mathfrak g$ it is in a Levi subalgebra $Lie(L)$ of $Lie(P)$, a definition essentially due to Serre. It turned out that this definition matches up nicely with the Jacobson--Morozov theorem in that $G$-complete reducibility of all subalgebras guarantees you the uniqueness. We had to do a lot of case-by-case analysis, unfortunately, but we did get out the following:
Let $h(G)$ be the Coxeter number of $G$, $p$ the characteristic of the algebraically closed field $k$.

Any nilpotent element $e$ can be extended to an $\mathfrak{sl}_2$ triple $(e,h,f)\in \mathfrak{g}\times \mathfrak{g}\times \mathfrak{g}$ uniquely up to simultaneous conjugation by $C_G(e)$, if and only if $p>h(G)$.

Let $b(G)$ be the largest prime amongst those for which the Dynkin diagram contains an $A_{p-1}$ subdiagram and bad primes for $G$.

The number of nilpotent orbits and conjugacy classes of $\mathfrak{sl}_2$-subalgebras is the same if and only if $p>b(G)$. Moreover a bijection can be realised in a natural way by sending an $\mathfrak{sl}_2$-subalgebra to the nilpotent orbit of largest dimension meeting it.

The quantitative difference between the two results above essentially arises from the qualitative fact that when $p\leq h(G)$, then an $\mathfrak{sl}_2$-subalgebra can contain two non-conjugate nilpotent elements.
We also established, using the classification of nilpotent orbits, for the bad primes $p=3$ and $p=5$ in the exceptional groups, which nilpotent elements can be extended to $\mathfrak{sl}_2$-triples. All can except the exceptional orbit with label $A_2^{(3)}$ in $G_2$ when $p=3$.<|endoftext|>
TITLE: Who uses keywords (and how)?
QUESTION [30 upvotes]: Almost all journals ask for keywords and most authors comply.
Does any math researcher out there actually use them to search?  Are there mechanisms to do that?
I just wrote to someone that I assumed no one uses them because I never do.  It then occurred to me that my assumption might be wrong.
Educate me.
(BTW, the MO auto-criticism mechanism is giving me this message: "That's not a very good title. Can you add some more unique words to it?"  The implicit self references in the situation make it too difficult for me to even try.)
(Second BTW, this question either has never been asked here, or it was deemed worthy of being expunged.)

REPLY [6 votes]: A couple of remarks: 

In Zentralblatt MATH "keywords" are explicitly included. For example, see this entry for one of the OP's papers (link should work without subscription; the precise choice of the paper was random). And one can search for them (under category 'basic'); possibly 'browsing by' is the more interesting way to use this as discussed in  this question on searching related to  MSC classification. [For similar remarks for MathSciNet see Patricia Hersh's comment.] 
This is more for books than for journal articles, but (certain) electronic library catalogues contain "keywords". These then can also be used for searching. I occassioally found this useful. Say, I search for a book on some subject; I know one, I go to its entry, I click on a keyword, and get a list of books with the same 'keyword'. These keywords than can also be in some hierarchy and inter-linked. Personally, I never had a need to use this extensively as I was (essentially) always close to a math library that porvides free access to the books (and had the books ordered by rough subject), so that browsing the hard-copies seemed more efficient; however if I ever should be in a situation that the library does not provide free access but one can only order books, then I think I could find this option quite convenient.
Say to figure out which (text)books on a given subject are available in the library. (True, often the title can serve as a proxy for this, but it does not always.)
Furthermore, I think that in cases of publications in different languages 'keywords' are useful to have just some simple and short bits of information (the keywords) to translate that then already allow to link a publication into a body of literature. [This is not done by us (mathematicians) but those that need to do it depend on us to provide meaningful information to process, so that then later we can find the things]
The point that sometimes well-chosen keywords can convey something on the content of an article that title and abstract do not was already mentioned. I agree.
I think, but do not know for a fact, that meta data of articles (like keywords) is useful/used in order to make general search facilities work better. So, it might be that even if one does not search via keywords directly the fact that some general search mechanism (say general purpose search engine) will work well and turn up the article(s) one is interest will among others depend on the fact that somebody provided some good keywords and they were than incorporated in the electronic resource in a suitable way. 
Also, some electronic article repositories do suggest related ariticles; I think here also keywords might be used to figure out what is (supposedly) related. [Personally, I so far never found this feature very useful, but then I have to admit it never really paid much attention to these suggestions even. Perhaps I should.] 

So, in brief, I sometimes found keywords useful (though not very or in an essential way). Yet, I think to a considerable extent this is due to the existence of the MSC and if it would not exist they would be considerably more important [yet as was pointed out already that MSC works well depends to some extent on keywords]. Also, MSC is 'only' for math so in cases where publications of differing subjects all need to be dealt with (like a general library catalogue, on the web,...) keywords become more crucial. And, some of their use might actually happen 'behind the scenes'. Yet, likely, it is true that to some extent keywords become less relevant as fulltext search and so on becomes more feasible (but this is a reletively recent phenomenon).
Thus, for the question how to use keyword, I would answer for example for searching/browsing where this is supported like Zentralblatt MATH. And, as already mentioned, to get some quick idea what is inside the artcile.
And, for the possibly present implict question whether one should care as an author to provide meaningful keywords when asked to, I would say definitely yes; it might be indirectly of use in ways one does not fully oversee and is not that much work. (And if one does not provided them, somebody whose job it is to add them somewhere might just pick something that seems alright, with best intentions for lack of a different option, but might chose something quite wrong.)
Finally, since we are on MO, the super-brief version: 
Keywords are just tags by another name.<|endoftext|>
TITLE: Who first noticed that the Hilbert symbol is a Steinberg symbol ?
QUESTION [17 upvotes]: Hilbert reformulated the quadratic reciprocity law of Gauß as a product formula 
$$
\prod_v(a,b)_v=1
$$
for the various local Hilbert symbols.  For each place $v$ of $\bf Q$, the Hilbert symbol $(\ ,\ )_v$ is a bimultiplicative map 
$$
{\bf Q}_v^\times\times{\bf Q}_v^\times\to{\bf Z}^\times
$$
so that, by definition, $(a,b)_v=1$ if and only if $a\in {\rm Im\;} N_b$ where $N_b$ denotes the norm map ${\bf Q}_v(\sqrt b)^\times\to{\bf Q}_v^\times$.  An important property of the Hilbert symbol is that
$$
a+b=1\Longrightarrow (a,b)_v=1,
$$
which makes it a Steinberg symbol.  This property in not listed in older books such as Hasse's Number theory but it can be found in all modern treatments, such as Serre's Course in arithmetic or his Local fields, or Milnor's K-theory.
I'm curious as to who first noticed that the Hilbert symbol is a Steinberg symbol. Was it Steinberg himself ?  A precise reference will be appreciated.

REPLY [11 votes]: EDIT: After looking into the history more closely, I think it's fairly certain that the correct answer to the question is Calvin Moore.  (See my added text below.)  
The question is interesting and looks straightforward, but it may not have a simple answer.  A number of independent lines of research, differently motivated, converged miraculously in the late 1960s, so the word "noticed" in the question has to be placed in context.  For example, one needs the notion of topological Steinberg symbol in the study of simple algebraic groups over various local fields.  Here an essential contribution was made by Calvin Moore in his 1968 paper Group extensions of p-adic and adelic linear groups (IHES Publ. Math. 35).
The history of all the related developments really ought to be told by one of the living participants.   Seminars with Bass in that period got me interested for a while in the subject, leading me eventually to write an elementary introduction in the last part of my 1980 Arithmetic Groups (Springer Lect. Notes 789), where a lot of references are included; most of the key papers are now available online via numdam.org and such, while Steinberg's 1967-68 Yale lectures are posted on some webpages including his at UCLA and Bill Casselman's at UBC.   But it would also help to know who was talking to whom during that crucial period.
Briefly, Steinberg's work was purely algebraic at first and was motivated by Chevalley's 1955 Tohoku paper constructing versions of simple adjoint algebraic groups over arbitrary fields.   In his 1962 Brussels conference paper, Steinberg worked out generators and relations in order to study non-adjoint groups and in particular see how projective modular representations of the adjoint groups would lift to "universal" groups.   This led him to introduce what eventually became Steinberg symbols or cocycles.   By the time of his Yale lectures, further connections were in the air.  For instance, work on the Congruence Subgroup Problem (Bass, Lazard, Milnor, Serre) led Serre to an elegant formulation of the problem in terms of group extensions.  Matsumoto's thesis provided more evidence of the close connection with Steinberg's formalism.
Moore's work on the other hand came from his early interest in locally compact groups and their central extensions.    Here he eventually found that a topological version of Steinberg's algebraic formalism would fit well with the classical ideas in number theory (local and global class field theory) explored in modern terms by Serre and others.   There is a long and useful review of Moore's paper in Mathematical Reviews by Hyman Bass, if you have access.
However the question in the header is answered, it should be kept in mind that the motivation for arriving at such a bizarre connection came from work of all these people.   Connecting the dots required the existence of the dots.
ADDED: A few more remarks about references, influences, timing. 1) Steinberg's 1962 Brussels paper (in French!) is reprinted along with his others in a single volume published by AMS and reasonably priced; but I don't know any accessible online source.   However, most of his computations reappeared in his Yale lectures and in Matsumoto's thesis (with special cases treated in my lecture notes).  2) My best guess is that the combined work of Moore and Matsumoto filled in the connection with classical symbols and reciprocity laws.  Note that Moore''s paper was submitted in January 1968 but lists Matsumoto's thesis in the references, while Matsumoto's thesis was submitted six months later and cites Moore's published paper.   (The year of Moore's paper is either 1968 or 1969, depending on where you look.)    Matsumoto thanks Bruhat, Serre, Samuel for their advice.   On the other hand, Moore points to independent partial results by T. Kubota.  He especially credits his conversations with a number of people including Bass and Serre.<|endoftext|>
TITLE: 3-dim positively curved Alexandrov space
QUESTION [5 upvotes]: What is the classification of 3-dim positively curved Alexandrov space?
And if a 3-dim positively curved Alexandrov space has a totally (quasi)geodesic subset,then the classification?

REPLY [7 votes]: I guess you are interested in topological classification (?).
Given a 3-dimensional Alexandrov space $M$,
you can always find an other Alexandrov space $\bar M$ with isometric involution $J$
such that

$M$ is isometric to $\bar M/J$
$\bar M$ is topological manifold without boundary.

This is almost an answer to your first question.
If $M$ is compact and simply connected then $M$
has to be homeomorphic to one of the following (I might miss something):

$\mathbb S^3$
spherical suspension over $\mathbb R\mathrm P^2$
$\mathbb D^3$
ball in the cone over $\mathbb R\mathrm P^2$.

If noncompact then you get in addition $\mathbb R^3$ and cone over $\mathbb R\mathrm P^2$.
If the space has a totally quasigeodesic surface then cutting along this surface should give you an Alexandrov space with boundary.
Since the curvature is positive it should be disc or ball in cone over $\mathbb R\mathrm P^2$.
So you original space is described by an isometric involution of the boundary of one of these spaces.<|endoftext|>
TITLE: Forcing in Ackermann's Set Theory
QUESTION [5 upvotes]: How would one do forcing in Ackermann's set theory?  C. Alkor published an article entitled "Forcing in Ackermanns Mengenlehre" (Zeitshcr. f. math. Logik und Grundlagen d. Math., Bd. 25,8.265-286 (1979)) but the abstract, at least, is in German which I can't read.  I ask the question because the purely set-theoretical part of Ackermann's set theory equals ZF and wonder whether one could use the part of Ackermann's set theory dealing with classes to construct generic extensions of models of the purely set-theoretic fragment of the theory.  If one could then it would seem that Ackermann's set theory would be the proper setting for the naturalistic account of forcing.

REPLY [6 votes]: If one moves to a very similar but somewhat stronger theory than Ackerman set theory, then forcing works fine. 
Ackerman set theory is a version of set theory where one views the
set-theoretic hierarchy as continuing far past the construction of
the sets, into the construction of classes, classes of classes and
so on. In the Ackerman theory, it is as though one is building the full set-theoretic $V_\alpha$ hiearchy, but then part-way through one finds a particularly robust $V_\delta$ and declares its elements to be the real "sets", with everything above $\delta$ declared "classes". (Critics would say that Ackerman's sets are only some of the sets, since his classes behave fundamentally like sets.)
As François points out in the comments, however, the Ackerman theory seems to provide less than what one may want in the realm of classes, a weakness in the theory that is addressed by its natural strengthenings to various set theories in a more ZFC-like context. Namely, the Levy theory is
ZFC+$V_\delta\prec V+\delta$ is inaccessible, where $V_\delta\prec
V$ is the scheme asserting $\forall x\in V_\delta\
(\varphi(x)\iff\varphi(x)^{V_\delta})$, which is expressible in the
language of set theory augmented with the constant symbol
$\delta$. The set $V_\delta$ here plays exactly the role of $V$ in
Ackerman's theory, and so every model of the Levy scheme is a
model of Ackerman set theory, if one regards the elements of
$V_\delta$ as the official "sets" and the sets above $V_\delta$ as
the "classes". But the Levy theory asserts more than Ackerman,
because not only is the collection of sets existing as an object
in the theory, but also it is an elementary substructure of the
full universe. In addition, the Levy theory has a fuller treatment of classes, making them much more set-like, in that the larger universe above $\delta$, which correspond to the classes of the Ackerman theory, actually satisfy ZFC.
Meanwhile, many arguments in the literature are using forcing over
models of the Levy theory (for an example, see my paper A simple
maximality principle, and the Levy scheme is often arising in other work). If $\mathbb{P}$ is a set in $V_\delta$
and $G\subset\mathbb{P}$ is generic, then it is not difficult to
show that $V_\delta[G]\prec V[G]$ and so the theory is preserved
to the forcing extension. This argument resembles many other
arguments that forcing preserves large cardinals, and one should
look upon the Levy theory and Ackerman set theory itself
essentially as a large cardinal assertion.
Since the idea of reflection is one of the central motivations of
the Ackerman set theory, I would say that anyone tempted to use
Ackerman set theory would likely prefer the Levy theory, because
the form of reflection is more robust. Furthermore, the strength
of the Levy theory fits right into the large cardinal hierarchy,
equiconsistent with the assertion Ord is Mahlo, which is
strictly weaker than a Mahlo cardinal.
If one drops the assertion that $\delta$ is inaccessible, then one
gets ZFC+$V_\delta\prec V$, which is equiconsistent with ZFC. This
weaker theory is strengthened by the Feferman theory, which
asserts a proper class tower of such $\delta$, essentially
asserting that the entire set-theoretic universe is the union of
an endless transfinite elementary chain $V_{\delta_0}\prec
V_{\delta_1}\prec\cdots V$. This theory is also equiconsistent
with ZFC, but if one wants the $\delta$ to be inaccessible, then
one gets a large cardinal strength still below a Mahlo cardinal.
Because the Feferman theory provides such a robust universe
concept, many authors have proposed it as a natural setting for
category theory, viewing the $V_\delta$ as various universe
levels.
Lastly, let me address your proposal to use the classes of Ackerman set theory to produce generic extensions of the set part. This is impossible. If $\mathbb{P}$ is a set forcing notion, then in the Ackerman theory any subcollection of $\mathbb{P}$ that is a class is already a set. So there are no new generic filters arising as classes in the Ackerman theory that are not already in $V$. The analogous situation in the Levy theory is that if $\mathbb{P}\in V_\delta\prec V$ is a notion of forcing, then every $G\subset\mathbb{P}$ in $V$ is already in $V_\delta$, and hence is not $V_\delta$-generic, unless the forcing was trivial.<|endoftext|>
TITLE: What structure on the second order cotangent bundle ?
QUESTION [13 upvotes]: It is well-known that the total space of the cotangent bundle $T^*X$ of a given smooth manifold $X$ admits a symplectic form $\omega$. It is actually exact: $\omega=d\lambda$. The $1$-form $\lambda$ is called the Liouville form and can be defined in a quite tautological way: given an element $p=(x,\xi)\in T^*X$ (i.e. $x\in X$ and $\xi\in T^*_xX$) and a vector $v\in T_p(T^*X)$, we define $\lambda_p(v):=\xi(\pi_*(v))$, where $\pi:T^*X\to X$ (and thus $\pi_*(v)\in T_xX$). 

Does the total space of the second order cotangent bundle $T^*_2X$ also admits a "natural" geometric structure ? It seems that there is similarly a tautological "second order form" on $T^*_2X$ (in place of the Liouville tautological $1$-form $\lambda$). But then I don't see any analog for $\omega$... 

Recall that the second order cotangent bundle can be defined as the dual of the second order tangent bundle $T^2X$, which is the bundle of order $2$ differential operators on $X$ that vanish on constants. 
In case you would be an algebraic geometer, the fiber of $T^*_2X$ at a point $x$ ($X$ is now a smooth algebraic variety) is just $I_x/I_x^3$ if $I_x$ is the maximal ideal corresponding to $x$ (w.r.t. to an affine open neighbourhood of it).

REPLY [14 votes]: It's not completely clear to me what form of answer you would accept.  In one sense, the answer is 'the structure on $T^\ast_2M$ is the pseudogroup structure that is induced by prolongation of the pseudogroup $\mathrm{Diff}(M)$, but this kind of tautological answer is not useful.
If you are asking how one would characterize this kind of pseudogroup structure, it is worth going back and thinking about the cotangent bundle $T^\ast M$ for a moment.  Yes, there is a $\mathrm{Diff}(M)$-invariant symplectic structure on $T^\ast M$, but that's not the whole structure since, as you have remarked, the Liouville form $\lambda$ is also $\mathrm{Diff}(M)$-invariant and could legitimately be regarded as part of the structure on $T^\ast M$.  Note, by the way that the pseudogroup preserving $\lambda$ does not act transitively on $T^\ast M$, as it preserves the zero section (which is where $\lambda$ vanishes), so the 'natural' structure on $T^\ast M$ is not homogeneous.  There are various ways that one can 'forget' some of the natural structure and get something homogeneous.  One way is to just throw away everything except $\omega = d\lambda$ and consider only the symplectic structure, but you don't have to go that far.  For example, you could retain $\omega$ and the Lagrangian foliation whose leaves are the fibers of $T^\ast M\to M$, and this would be a homogeneous structure.  
I suspect that you might be interested in something similar for $T^\ast_2M$, that is, you would like to know whether there is some kind of weakening of the natural pseudogroup structure on $T^\ast_2M$ that is homogeneous and more 'familiar' than the full structure (which, by the way, also isn't homogeneous).  One way to do this is to recognize that, because there is a canonical submersion $\phi: T^\ast_2 M\to T^\ast M$, one can simply pull back the symplectic form $\omega$ and see that it is well-defined on $T^\ast_2M$.  (Of course, it's degenerate there, so it's no longer symplectic.)  You can then see that there is a way to interpret an element $e\in T^\ast_2M$ as an $\omega$-Lagrangian plane in the tangent space $T_{\phi(e)}(T^\ast M)$, so you can regard $T^\ast_2 M$ as a certain open subset of the $\omega$-Lagrangian Grassmannian of the symplectic manifold $(T^\ast M,\omega)$, namely the subset that is transverse to the fibers of the map $T^\ast M\to M$.  This weaker structure is homogeneous under the pseudogroup of symplectomorphisms and so might be thought of as a natural weakening that is more familiar that the full structure.
This kind of thinking might not be exactly where you want to go, but maybe it will be helpful as you try to think about what sort of weaker structure you really want to understand on $T^\ast_2 M$.